GTM211.Algebra (Serge Lang)

Graduate Texts inMathematics 211 Editorial Board S.Axler F.W. Gehring K.A. Ribet Springer New York Berlin Heidelberg Barcelona Hong Kong London Milan Paris Singapore Tokyo Serge Lang Department ofMathematics Yale University New Haven, CT96520 USA Editorial Board S.Axler Mathematics Department San Francisco State University San Francisco, CA94132 USAF.W.Gehring Mathematics Department East Hall University ofMichigan Ann Arbor, MI48109 USAK.A. Ribet Mathematics Department University ofCalifornia atBerkeley Berkeley, CA94720-3840 USA Mathematics Subject Classification (2000): 13-01, 15-01, 16-01, 20-01 Library ofCongress Cataloging-in-Publication Data Algebra ISerge Lang.-Rev. 3rd ed. p.em.-(Graduate texts inmathematics; 211) Includes bibliographical references and index. ISBN 0-387-95385-X (alk. paper) 1.Algebra. I.Title. II.Series. QA154.3.L3 2002 512--dc21 2001054916 Printed onacid-free paper. This title waspreviously published byAddison-Wesley, Reading, MA 1993. <92002 Springer-Verlag New York, Inc. Allrights reserved. This work may not betranslated orcopied inwhole orinpart without the written pennission ofthepublisher (Springer-Verlag New York, Inc., 175Fifth Avenue, New York, NY 10010, USA), except forbrief excerpts inconnection with reviews orscholarly analysis. Use inconnection with anyform ofinformation storage andretrieval, electronic adaptation, computer software, orbysimilar ordissimilar methodologynow known orhereafter developed isforbidden. The useofgeneral descriptive names, trade names, trademarks, etc., inthispublication,even ifthe former arenotespecially identified, isnot tobetaken asasign that such names, asunderstood by theTrade Marks and Merchandise Marks Act, may accordinglybeused freely byanyone. Production managed byTerry Kornak; manufactunng supervised byErica Bresler. Revisions typeset byAsco Typesetters, North Point, Hong Kong. Printed and bound byEdwards Brothers, Inc., Ann Arbor, MI. Printed intheUnited States ofAmerica. 98765432 1 ISBN 0-387 -95385-X SPIN 10855619 Springer- Verlag New York Berlin Heidelberg Amember ofBertelsmannSp ringer Science+Business Media GmbH FOREWORD The present book ismeant asabasic text for aone-yearcourse inalgebra, atthegraduate level. Aperspective onalgebra AsIseeit,thegraduatecourse inalgebra must primarily prepare students tohandle thealgebra which they will meet inallofmathematics: topology, partial differential equations, differential geometry, algebraic geometry, analysis, andrepresentation theory, not tospeak ofalgebra itself andalgebraic number theory with allitsramifications. Hence Ihave inserted throughout references to papers and books which have appeared during thelastdecades, toindicate some ofthedirections inwhich thealgebraic foundations provided bythis book are used; Ihave accompanied these references with some motivating comments, to explain how thetopics ofthepresent book fitinto themathematics that isto come subsequently invarious fields; and Ihave also mentioned some unsolved problems ofmathematics inalgebra and number theory. The abcconjecture is perhaps the most spectacular ofthese. Often when such comments and examples occur outofthelogical order, especially with examples from other branches ofmathematics, ofnecessitysome terms may not bedefined, ormay bedefined only later inthebook. Ihave tried tohelp thereader notonly bymaking cross-references within thebook, butalso byreferringtoother books orpapers which Imention explicitly. Ihave also added anumber ofexercises. Onthewhole, Ihave tried tomake the exercises complement theexamples, and togive them aesthetic appeal. I have tried tousetheexercises also todrive readers toward variations andappli- cations ofthemain text, aswell astoward working outspecial cases, and as openings toward applications beyond this book. Organization Unfortunately,abook must beprojected inatotally ordered wayonthepage axis, butthat's not theway mathematics "is", soreaders have tomake choices how toreset certain topics inparallel forthemselves, rather than insuccession. v vi FOREWORD Ihave inserted cross-references tohelp them dothis, butdifferent people will make different choices atdifferent times depending ondifferent circumstances. The book splits naturally into several parts. The first part introduces thebasic notions ofalgebra. After these basic notions, the book splits intwo major directions: thedirection ofalgebraic equations including theGalois theory in Part II;and thedirection oflinear and multilinear algebra inParts IIIand IV. There issome sporadic feedback between them, buttheir unification takes place atthe next level ofmathematics, which issuggested, forinstance, in 15of Chapter VI.Indeed, thestudy ofalgebraic extensions oftherationals can be carried outfrom twopoints ofview which arecomplementary and interrelated: representing theGalois group ofthealgebraic closure ingroups ofmatrices (the linear approach), andgivinganexplicit determination oftheirrationalities gen- erating algebraic extensions (the equations approach). Atthe moment, repre- sentations inGL2are atthecenter ofattention from various quarters, and readers will see GL2appear several times throughout thebook. For instance, Ihave found itappropriatetoadd asection describing allirreducible characters of GL2(F)when Fisafinite field. Ultimately, GL2will appearasthesimplest but typicalcase ofgroups ofLietypes, occurring both inadifferential context and over finite fields ormore general arithmetic rings forarithmetic applications. After almost adecade since thesecond edition, Ifind that thebasic topics ofalgebra have become stable, with oneexception. Ihave added two sections onelimination theory, complementing theexisting section onthe resultant. Algebraic geometry having progressed inmany ways, itisnow sometimes return- ing toolder and harder problems, such assearching fortheeffective construction ofpolynomials vanishingoncertain algebraic sets, and theolder elimination procedures oflastcenturyserve asanintroduction tothose problems. Except forthisaddition, themain topics ofthebook areunchanged from the second edition, butIhave tried toimprove thebook inseveral ways. First, some topics have been reordered. Iwas informed byreaders andreview- ersofthetension existing between havingatextbook usable forrelatively inex- perienced students, and areference book where results could easily befound in asystematic arrangement. Ihave tried toreduce this tension bymoving allthe homological algebra toafourth part, andbyintegrating thecommutative algebra with thechapter onalgebraic sets and elimination theory, thus givinganintro- duction todifferent points ofview leading toward algebraic geometry. The book asatext and areference Inteaching the course, onemight wish topush into thestudy ofalgebraic equations through Part II,orone may choose togofirst into thelinear algebra ofParts IIIand IV. One semester could bedevoted toeach, forinstance. The chapters have been sowritten astoallow maximal flexibility inthis respect, and Ihave frequently committed thecrime oflese- Bourbaki byrepeating short argu- ments ordefinitions tomake certain sections orchapters logically independent ofeach other. FOREWORD vii Granting thematerial which under nocircumstances can beomitted from a basic course, there exist several options forleading the course invarious direc- tions. Itisimpossible totreat allofthem with the same degree ofthoroughness. The precise point atwhich one iswilling tostop inanygiven direction will depend ontime, place, and mood. However, any book with the aims ofthe presentone must include achoice oftopics, pushing ahead indeeper waters, while stopping short offull involvement. There can benouniversal agreementonthese matters, not even between the author and himself. Thus the concrete decisions astowhat toinclude and what not toinclude arefinally taken ongrounds ofgeneral coherence and aesthetic balance. Anyone teaching the course will want,to impress their own personality onthematerial, and may push certain topics with more vigor than Ihave, atthe expense ofothers. Nothing inthepresent book ismeant toinhibit this. Unfortunately, thegoal topresentafairly comprehensive perspectiveon algebra requiredasubstantial increase insize from thefirst tothesecond edition, and amoderate increase inthis third edition. These increases requiresome decisions astowhat toomit inagivencourse. Many shortcuts can betaken inthepresentation ofthetopics, which admits many variations. Forinstance, one canproceed into field theory and Galois theory immediately after giving thebasic definitions forgroups, rings, fields, polynomials inonevariable, anvector spaces. Since theGalois theory gives very quicklyanimpression ofdpth,this isvery satisfactory inmany respects. Itisappropriate here torecall myoriginal indebtedness toArtin, who first taught mealgebra. The treatment ofthe basics ofGalois theory ismuch influenced bythepresentation inhis own monograph. Audience andbackground AsIalready stated intheforewords ofprevious editions, thepresent book ismeant forthegraduate level, and Iexpect most ofthose coming toittohave had suitable exposure tosome algebra inanundergraduate course, ortohave appropriate mathematical maturity. Iexpect students takingagraduatecourse tohave had some exposure tovector spaces, linear maps, matrices, andthey will nodoubt have seen polynomials atthevery least incalculus courses. My books Undergraduate Algebra and Linear Algebra providemore than enough background for agraduatecourse. Such elementary texts bring out in parallel thetwo basic aspects ofalgebra, and areorganized differently from the present book, where both aspectsaredeepened. Ofcourse, some aspects ofthe linear algebra inPart IIIofthepresent book are more "elementary" than some aspects ofPartII,which deals with Galois theory and thetheory ofpolynomial equations inseveral variables. Because Part IIhas gone deeper into thestudy ofalgebraic equations, ofnecessity theparallel linear algebraoccurs only later inthetotal ordering ofthebook. Readers should view both partsasrunning simultaneously. viii FOREWORD Unfortunately, theamount ofalgebra which one should ideally absorb during this first year inorder tohave aproper background (irrespective ofthesubject inwhich oneeventually specializes)exceeds the amount which can becovered physically byalecturer duringaone-yearcourse. Hence more material must be included than canactually behandled inclass. Ifind itessential tobring this material totheattention ofgraduatestudents. Ihope that thevarious additions andchanges make thebook easier touse as atext. By these additions, Ihave tried toexpand thegeneral mathematical perspective ofthereader, insofar asalgebra relates toother parts ofmathematics. Acknowledgements Iamindebted tomany people who have contributed comments andcriticisms fortheprevious editions, butespecially toDaniel Bump, Steven Krantz, and Diane Meuser, who provided extensive comments aseditorial reviewers for Addison- Wesley. Ifound their comments very stimulating and valuable inpre- paring this third edition. Iammuch indebted toBarbara Holland forobtaining these reviews when she was editor. Iamalso indebted toKarl Matsumoto who supervised production under very trying circumstances. Finally Ithank themany people who have made suggestions andcorrections, especially George Bergman, Chee-Whye Chin, Ki-Bong Nam, David Wasserman, and Randy Scott, who providedmewith alistoferrata. Ialso thank Thomas Shiple and Paul Vojta fortheir lists oferrata tothethird edition. These have been corrected inthe subsequent printings. Serge Lang New Haven For the2002 andbeyond Springer printings From now on,Algebra appears with Springer-Verlag, like the rest ofmy books. With thischange, Iconsidered thepossibility ofanew edition, but de- cided against it.Iview thebook asvery stable. The only addition which I would make, ifstarting from scratch, would besome ofthealgebraic properties ofSLnand GLn (over RorC),beyond theproof ofsimplicity inChapter XIII. Asthings stood, Ijust inserted some exercises concerningsome aspects which everybody should know. Readers can seethese worked outinJorgenson/Lang, Spherical Inversion onSLn(R), Springer Verlag 2001, aswell asother basic algebraic properties onwhich analysis issuperimposedsothat algebra inthis context appearsasasupporting tool. Ithank specifically Tom von Foerster, InaLindeman and Mark Spencer for their editorial support atSpringer,aswell asTerry Kornak and Brian Howe whc have taken care ofproduction. Serge Lang New Haven 2002 Logical Prerequisites We assume that thereader isfamiliar with sets, and with thesymbols n,U, ::>,C, E.IfA,Baresets, we use thesymbol ACBtomean that Aiscontained inBbut may beequaltoB.Similarly forA::>B. Iff:A Bisamapping ofone setinto another, wewrite xf(x) todenote the effect offon anelement xofA.Wedistinguish between the arrows and. Wedenote byf(A) the setofallelementsf(x), with xEA. Letf:A Bbeamapping (also called amap). We saythatfisinjective ifx=Fyimplies f(x) =Ff(y). We sayfissurjective ifgiven bEBthere exists aEAsuch thatf(a)=b.We saythatfisbijective ifitisboth surjective and injective.. Asubset AofasetBissaid tobeproper ifA=FB. Letf:A Bbeamap, and A'asubset ofA.The restriction offtoA'is amap ofA'into Bdenoted byfIA'. Iff:A Band g:B-+Care maps, then wehave acomposite map g0f such that (g0f)(x)=g(f(x)) forallxEA. Letf: A Bbeamap, and B'asubset ofB.Byf-l(B') wemean thesubset ofAconsisting ofallxEAsuch thatf(x)EB'.Wecallitthe inverse image of B'.Wecallf(A) theimage off. Adiagram Af) B\) C issaid tobecommutative if90f=h.Similarly,adiagram Af) B ] ]g C)D '" ix X LOGICAL PREREQUISITES issaid tobecommutative ifg0f=t/J0qJ. We deal sometimes with more complicated diagrams, consisting ofarrows between various objects. Such diagrams are called commutative if,whenever itispossible togofrom one object toanother bymeans oftwo sequences ofarrows, say IIA12 In-IAI)2)... )An and Al)...9m-1 )Bm=An'91B)292 then In-I0 ···0II=9m-10 ···091' inother words, thecomposite mapsareequal. Most ofourdiagrams are composed oftriangles orsquaresasabove, and toverify that adiagram con- sisting oftriangles orsquares iscommutative, itsuffices toverify that each triangle and square initiscommutative. We assume that the reader isacquainted with theintegers and rational numbers, denoted respectively byZandQ.For many ofourexamples, wealso assume that thereader knows thereal and complex numbers, denoted. byR and C. Let AandIbetwo sets. Byafamily ofelements ofA,indexed byI,one means amapf:I-.A.Thus foreach iEIwe aregivenanelement f(i)EA. Althoughafamily does notdiffer from amap, wethink ofitasdetermininga collection ofobjects from A,and write itoften as {f(i)}iel or {aJ ieI' writing aiinstead off(i). WecallItheindexing set. We assume that thereader knows what anequivalence relation is.Let A beasetwith anequivalence relation, letEbeanequivalence class ofelements ofA.We sometimes trytodefine amap oftheequivalence classes into some setB.Todefine such amapfontheclass E,wesometimes firstgive itsvalue on anelement xEE(called arepresentative ofE),and then show that itis independent ofthechoice ofrepresentative xEE.Inthat case wesaythatf iswell defined. Wehave products ofsets, sayfinite products AxB,orAtx... xAn'and prod ucts offamilies ofsets. Weshall useZorn's lemma, which wedescribe inAppendix 2. We let#(S) denote the number ofelements of asetS,also called the cardinality ofS.The notation isusually employed when Sisfinite. We also write #(S)=card(S). CONTENTS Part One The Basic Objects ofAlgebra Chapter I 1.Monoids 2.Groups7 3.Normal subgroups 4.Cyclic groups 23 5.Operations ofagroupon aset 6.Sylow subgroups 33 7.Direct sums and free abelian groups 8.Finitely generated abelian groups 9.The dual group 46 10. Inverse limit andcompletion 11.Categories and functors 53 12. Free groups 66Groups 33 13 25 36 42 49 Chapter II Rings 1.Rings andhomomorphisms 83 2.Commutative rings 92 3.Polynomials and group rings 97 4.Localization 107 5.Principal and factorial rings11183 Chapter III Modules 1.Basic definitions 117 2.The group ofhomomorphisms 122 3.Direct products and sums ofmodules 127 4.Free modules 135 5 .Vector spaces 139 6.The dual space and dual module 142 7.Modules over principal rings 146 8 .Euler-Poincare maps 155 9.The snake lemma 157 10. Direct and inverse limits 159117 xl xii CONTENTS Chapter IV Polynomials 1.Basic properties forpolynomials inone variable 2.Polynomialsover afactorial ring180 3.Criteria forirreducibility183 4.Hilbert's theorem 186 5.Partial fractions 187 6.Symmetric polynomials190 7.Mason-Stothers theorem and theabcconjecture 8.The resultant 199 9.Power series 205 Part Two Algebraic Equations Chapter V Algebraic Extensions 1.Finite andalgebraic extensions 225 2.Algebraic closure 229 3.Splitting fields and normal extensions 236 4.Separable extensions 239 5.Finite fields 244 6.Inseparable extensions 247 Chapter VI Galois Theory 1.Galois extensions 261 2.Examples andapplications 269 3.Roots ofunity 276 4.Linear independence ofcharacters 282 5.The norm and trace 284 6.Cyclic extensions 288 7.Solvable and radical extensions 291 8.Abelian Kummer theory 293 9.The equation xn -a=0 297 10. Galois cohomology 302 11. Non-abelian Kummer extensions 304 12.Algebraic independence ofhomomorphisms 13. The normal basis theorem 312 14. Infinite Galois extensions 313 15. The modular connection 315 Chapter VII Extensions ofRings 1.Integral ring extensions 333 2.Integral Galois extensions 340 3.Extension ofhomomorphisms 346308173 173 194 223 261 333 CONTENTS xiii Chapter VIII Transcendental Extensions 355 1.Transcendence bases 355 2.Noether normalization theorem 357 3.Linearly disjoint extensions 360 4.Separable andregular extensions 363 5.Derivations 368 Chapter IX Algebraic Spaces 1.Hilbert's Nullstellensatz 377 2.Algebraic sets, spaces and varieties 3.Projections and elimination 388 4.Resultant systems 401 5.Spec ofaring 405377 381 Chapter X Noetherian Rings and Modules 1.Basic criteria 413 2.Associated primes 416 3.Primary decomposition 421 4.Nakayama's lemma 424 5.Filtered andgraded modules 426 6.The Hilbert polynomial 431 7.Indecomposable modules 439.413 Chapter XI Real Fields 1.Ordered fields 449 2.Real fields 451 3.Real zeros andhomomorphisms449 457 Chapter XII Absolute Values 1.Definitions, dependence, andindependence 465 2.Completions 468 3.Finite extensions 476 4.Valuations 480 5.Completions and valuations 486 6.Discrete valuations 487 7.Zeros ofpolynomials incomplete fields 491465 Part Three Linear Algebra and Representations Chapter XIII Matrices and Linear Maps 1.Matrices 503 2.The rank ofamatrix 506503 xiv CONTENTS 3.Matrices and linear maps 507 4.Determinants 511 5.Duality 522 6.Matrices and bilinear forms 527 7.Sesquilinear duality 531 8.Thesimplicity ofSL2(F)/+1 536 9.The group SLn(F),n>3 540 Chapter XIV Representation ofOne Endomorphism 553 1.Representations 553 2.Decomposition over oneendomorphism 556 3.The characteristic polynomial561 Chapter XV Structure ofBilinear Forms 571 1.Preliminaries, orthogonalsums 571 2.Quadratic maps 574 3.Symmetric forms, orthogonal bases 575 4.Symmetric forms over ordered fields 577 5.Hermitian forms 579 6.The spectral theorem (hermitian case) 581 7.The spectral theorem (symmetric case) 584 8.Alternating forms 586 9.The Pfaffian 588 10.Witt's theorem 589 11. The Witt group 594 Chapter XVI The Tensor Product 601 1.Tensor product 601 2.Basic properties 607 3.Flat modules 612 4.Extension ofthe base 623 5.Some functorial isomorphisms 625 6.Tensor product ofalgebras 629 7.The tensor algebra ofamodule 632 8.Symmetric products 635 Chapter XVII Semisimpliclty 641 1.Matrices and linear mapsover non-commutative rings 641 2.Conditions defining semisimplicity 645 3.The density theorem 646 4.Semisimple rings 651 5.Simple rings 654 6.The Jacobson radical, base change, and tensor products 657 7.Balanced modules 660 CONTENTS XV Chapter XVIII Representations ofFinite Groups 663 1.Representations andsemisimplicity 663 2.Characters 667 3.I-dimensional representations 671 4.The space ofclass functions 673 5.Orthogonality relations 677 6.Induced characters 686 7.Induced representations 688 8.Positive decomposition oftheregular character 699 9.Supersolvable groups 702 10. Brauer's theorem 704 11. Field ofdefinition ofarepresentation 710 12.Example: GL2over afinite field 712 Chapter XIX TheAlternating Product 1.Definition and basic properties 731 2.Fitting ideals 738 3.Universal derivations and the deRham complex 4.The Clifford algebra 749731 746 Part Four Homological Al,gebra Chapter XX General Homology Theory 1.Complexes 761 2.Homology sequence 767 3.Euler characteristic and theGrothendieck group 769 4.Injective modules 782 5.Homotopies ofmorphisms ofcomplexes 787 6 .Derived functors 790 7.Delta-functors 799 8.Bifunctors 806 9 .Spectral sequences 814761 Chapter XXI Finite Free Resolutions 1.Special complexes 835 2.Finite free resolutions 839 3.Unimodular polynomial vectors 846 4.The Koszul complex 850835 Appendix 1 Appendix 2 Bibliography IndexThe Transcendence ofeand 'TT Some SetTheory867 875 895 903 Part One THE BASIC OBJECTS OF ALGEBRA This part introduces the basic notions ofalgebra, and themain difficulty forthebeginner istoabsorb areasonable vocabulary inashort time. None oftheconcepts isdifficult, butthere isanaccumulation ofnew concepts which may sometimes seem heavy. To understand the next parts ofthebook, the reader needs toknow essentially only the basic definitions ofthis first part. Ofcourse, atheorem may beused later for some specific and isolated applications, but onthe whole, wehave avoided making long logical chains ofinterdependence. CHAPTER I Groups 1. MONOIDS Let Sbeaset. Amapping SxS.:....S issometimes called alawofcomposition (ofSintoitself). Ifx,yare elements of S,theimage ofthepair (x,y)under thismapping isalso called their product under thelawofcomposition, andwill bedenoted byxy.(Sometimes, wealso write x.y,and inmany cases itisalso convenient touse anadditive notation, and thus towrite x+y.Inthat case, wecall this element the sum ofxand y. Itiscustomary touse thenotation x+yonly when therelation x+y= y+xholds.) Let Sbeasetwith alawofcomposition. Ifx,y,zareelements ofS,then we may form their product intwo ways: (xy)z andx(yz). If(xy)z=x(yz) forall x"y"zinSthen wesaythat thelawofcomposition isassociative. An element eofSsuch that ex =x=xeforallXES iscalled aunit element. (When thelawofcomposition iswritten additively, theunit element isdenoted by0,and iscalled azero element.) Aunit element isunique, forif e'isanother unit element, wehave e=ee' =e' byassumption. Inmost cases, theunit element iswritten simply1(instead ofe). For most ofthischapter, however, weshall write esoastoavoid confusion in proving the most basic properties. Amonoid isasetG,with alaw ofcomposition which isassociative, and havingaunit element (sothat inparticular, Gisnotempty). 3 4 GROUPS I,1 Let Gbeamonoid, and Xb...,Xnelements ofG(where nisaninteger> 1). Wedefine their product inductively: n nXv=Xl...Xn=(x 1...Xn-1)Xn. v= 1 Wethen have thefollowing rule: m n m+ n nXJl' nXm+v= nxv, Jl=l v=l v=l which essentially asserts that we can insert parentheses inanymanner inour product without changing itsvalue. Theproof iseasy byinduction, and weshall leave itasanexercise. One also writes m+n nXv instead of m+ln nXm+v v=l and wedefine o nXv=e. v=l As amatter ofconvention, weagree also that theempty product isequal totheunit element. Itwould bepossible todefine more general laws ofcomposition, i.e.maps S1XS2-+S3using arbitrary sets. One can then express associativity and commutativity inany setting forwhich they make sense. For instance, for commutativity weneed alawofcomposition f:SxS-+T where the two sets ofdepartureare the same. Commutativity then means f(x, y)=f(y, x),orxy=yxifweomit themappingffrom thenotation. For associativity, weleave ittothereader toformulate themost general combination ofsets under which itwill work. Weshall meet special cases later, forinstance arising from maps SxS-+Sand SxT-+T. Then aproduct (xy)z makes sense with XES, YES, and zET.The product x(yz) also makes sense forsuch elements x,Y,zand thus itmakes sense tosay that our lawofcomposition isassociative, namely tosaythat forallx,y,zas above wehave (xy)z=x(yz). Ifthelawofcomposition ofGiscommutative, wealso saythat Giscom- mutative (orabelian). I,1 MONOIDS 5 Let Gbeacommutative monoid, and Xl'...,Xnelements ofG.Let .pbea bijection ofthe setofintegers (1,...,n)onto itself. Then n n nx.;(v)=nXV' v=1 v=1 We prove thisbyinduction, itbeing obvious for n=1.We assume itfor n-1.Letkbeaninteger such that .p(k)=n.Then n k- 1 n-k nX.;(v)=nX.;(v).X.;(k)'nXt/1(k +v)111 k- 1 n-k =nX.;(v)'nX"'(k +v).X.;(k). 1 1 Define amap qJof(1,...,n-1)into itself bytherule qJ(v)=.p(v) qJ(v)=.p(v +1)if v<k, if v:>k. Then n k-l n-k nx.;(v)=nxtp(V)'n Xtp(k-1+v).Xn 1 1) 1 n-l =nXtp(v).Xn, 1 which, byinduction, isequal toXl···Xn,asdesired. Let Gbe acommutative monoid, letIbe aset, and letf:I Gbe a mapping such thatf(i)=eforalmost alliEI.(Here and thereafter, almost allwill mean allbut afinite number.) Let 10bethe subset ofIconsisting of those isuch thatf(i) =Fe.By nf(i) iel weshall mean theproduct nf(i) ielo taken inany order (the value does notdepend ontheorder, according tothe preceding remark). Itisunderstood that theempty product isequal toe. When Giswritten additively, then instead ofaproduct sign, wewrite the sum sign. There are anumber offormal rules fordealing with products which itwould betedious tolistcompletely. Wegiveoneexample. LetI,Jbetwo sets, and 6 GROUPS I,1 I:IxJ-+Gamapping into acommutative monoid which takes thevalue e foralmost allpairs (i,j).Then nrn!(i,j)]=0[O/(i,j)]. ielLeJ jeJ iel We leave theproof asanexercise. As amatter ofnotation, wesometimes write O/(i), omitting thesigns iEI,ifthereference totheindexing setisclear. Let xbeanelement ofamonoid G.For every integern>0wedefine xn tobe n Ox, 1 sothat inparticularwehave XO=e,Xl =x,x2=xx,. ...Weobviously have x(n+m) =xnxmand (xn)m=xnm .Furthermore, from our preceding rules of associativity andcommutativity, ifx,yare elements ofGsuch that xy=yx, then (xy)n=xnyn. We leave theformal proofasanexercise. IfS,S'aretwo subsets ofamonoid G,then wedefine SS' tobethesubset consisting ofallelements xy,with XES andYES'. Inductively,we can define theproduct ofafinite number ofsubsets, and wehave associativity. For in- stance, ifS,S',S"aresubsets ofG,then (SS')S"=S(S'S"). Observe that GG=G (because Ghas aunit element). IfxEG,then wedefine xStobe{x}S,where {x}isthe setconsisting ofthesingle element x.Thus xSconsists ofallelements xy,with YES. Byasubmonoid ofG, weshall mean asubset HofGcontaining theunit element e,and such that, ifx,yEHthen xyEH(we saythat Hisclosed under thelawofcomposition). Itisthen clear thatHisitself amonoid, under thelaw ofcomposition induced bythat ofG. Ifxisanelement ofamonoid G,then thesubset ofpowers xn(n=0,1,...) isasubmonoid ofG. The setofintegers>0under addition isamonoid. Later weshall define rings. IfRisacommutative ring,weshall deal with multiplicative subsets S,that issubsets containing theunit element, and such that ifx,YES then xyES.Such subsets aremonoids. Aroutine example. Let Nbethenatural numbers, Le. theintegers>o. Then Nisanadditive monoid. Insome applications, itisuseful todeal with a multiplicative version. See thedefinition ofpolynomials inChapter II,3,where ahigher-dimensional version isalso used forpolynomials inseveral variables. Aninteresting example. We assume that the reader isfamiliar with the terminology ofelementary topology. Let Mbethe setofhomeomorphism classes ofcompact (connected) surfaces. We shall define anaddition inM. Let S,S'becompact surfaces. LetDbeasmall disc inS,and D'asmall disc in S'.LetC,C'bethecircles which form theboundaries ofDand D'respectively. LetDo,Dbetheinteriors ofDand D'respectively, andglueS-Dotos'-D'o by identifying Cwith C'.Itcan beshown that theresulting surface isindependent, I,2GROUPS 7 uptohomeomorphism, ofthevarious choices made inthepreceding construc- tion. If(1,(1'denote thehomeomorphism classes ofSand S'respectively,we define (1+(1'tobethe class ofthesurface obtained bythepreceding gluing process. Itcan beshown that this addition defines amonoid structure onM, whose unit element isthe class oftheordinary 2-sphere. Furthermore, if! denotes the class ofthetorus, and 1tdenotes theclass oftheprojective plane, then every element (1ofMhas aunique expression oftheform (1=n!+m1t where nisaninteger>0and m=0,1,or2.We have 31t =!+1t. (The reasons forinserting thepreceding examplearetwofold: First to relieve the essential dullness ofthesection. Second toshow the reader that monoids exist innature. Needless tosay, theexample will not beused inany waythroughout the restofthebook.) Still other examples. AttheendofChapter III,4, weshall remark that isomorphism classes ofmodules over aring form amonoid under thedirect sum. InChapter XV, 1,weshall consider amonoid consisting ofequivalence classes ofquadratic forms. 2. GROUPS Agroup Gisamonoid, such that forevery element xEGthere exists an element YEGsuch that xy=yx=e.Such anelement yiscalled aninverse for x.Such aninverse isunique, because ify'isalso aninverse forx,then y'=y'e=y'(xy)=(y'x)y=ey=y. We denote this inverse byx-1(orby-x when the law ofcomposition is written additively). For anypositive integer n,weletx-n=(x-1)n. Then theusual rules for exponentiation hold forallintegers, notonly forintegers>0(as wepointed out formonoids in1).The trivial proofsarelefttothereader. InthedefinItions ofunit elements and inverses, wecould also define left units and leftinverses (intheobvious way). One caneasily prove that these arealso units and inverses respectively under suitable conditions. Namely: Let Gbeasetwith anassociative lawofcomposition, let ebealeftunitfor that law, and assume that every element has aleft inverse. Then eisaunit, and each leftinverse isa/so aninverse. Inparticular, Gisagroup. Toprove this, let aEGand letbEG besuch that ba=e.Then bab=eb =b. Multiplyingontheleftbyaleftinverse forbyields ab =e, orinother words, bisalso aright inverse for a.One sees also that aisaleft 8 GROUPS I,2 inverse forb.Furthermore, ae=aba=ea=a, whence eisaright unit. Example. Let Gbeagroup and SanQnempty set. The setofmaps M(S, G) isitself agroup; namely fortwo maps f,gofSinto Gwedefine fgtobethe map such that (fg)(x)=f(x)g(x), and wedefinef-1tobethemap such thatf- 1(X)=f(x)- 1.Itisthen trivial toverify thatM(S, G)isagroup. IfGiscommutative, soisM(S, G),and when thelawofcomposition inGiswritten additively,soisthelawofcomposition inM(S, G), sothat wewould writef+ginstead offg,and-finstead off- 1. Example. Let Sbeanon-empty set. Let Gbethe setofbijective mappings ofSonto itself. Then Gisagroup, thelawofcomposition being ordinarycom- position ofmappings. The unit element ofGistheidentity map ofS,and the other group propertiesaretrivially verified. The elements ofGare called permutations ofS.We also denote GbyPerm(S). For more information on Perm(S) when Sisfinite, see 5below. Example. Let usassume here thebasic notions oflinear algebra. Let kbe afield and Vavector spaceover k.LetGL(V) denote the setofinvertible k- linear maps ofVonto itself. Then GL(V) isagroup under composition of mappings. Similarly, letkbe afield and letGL(n, k)bethe setofinvertible nXnmatrices with components ink.Then GL(n, k)isagroup under the multiplication ofmatrices. For n>2,this group isnotcommutative. Example. The group ofautomorphisms. Werecommend that thereader now refer immediately to 11,where thenotion of acategory isdefined, and where several examplesaregiven. For anyobject Ain acategory, itsauto- morphisms form agroup denoted byAut(A). Permutations ofasetand thelinear automorphisms of avector space aremerely examples ofthis more general structure. Example. The setofrational numbers forms agroup under addition. The setofnon-zero rational numbers forms agroup under multiplication. Similar statements hold forthereal andcomplex numbers. Example. Cyclic groups. Theintegers Zform anadditive group. Agroup isdefined tobecyclic ifthere exists anelement aEGsuch that every element ofG(written multiplicatively) isoftheform anfor some integern.IfGiswritten additively, then every element ofacyclic group isoftheform na. One calls a acyclic generator. Thus Zisanadditive cyclic group with generator 1,and also with generator -1. There are noother generators. Given apositive integer n,then-th roots ofunity inthecomplex numbers form acyclic group oforder n.Interms oftheusual notation, e2'T1'i/n isagenerator forthis group. Soise2'T1'ir/n I,2 GROUPS 9 with rEZand rprime ton.Agenerator forthis group iscalled aprimitive n-th root ofunity. Example. The direct product. Let GI,G2begroups. Let GIxG2be the direct productassets, soGIxG2isthe setofallpairs (XI' X2) with XiEGi.We define theproduct componentwise by (XI'x2)(YI, Y2)=(xIYI,x2Y2). Then G1xG2isagroup, whose unit element is(el, e2)(where eiistheunit element ofGi).Similarly, for ngroupswedefine GIx... xGntobethe set ofn-tuples with XiEGi(i=1,...,n), andcomponentwise multiplication. Even more generally, letIbe aset, and foreach iEI,letGibe agroup. Let G=fIGibetheset-theoretic product ofthe sets Gi.Then Gisthe setofall families (Xi)iEIwith XiEGi.We can define agroup structure onGbycompo- nentwise multiplication, namely, if(Xi)iEIand(Yi)iEIaretwo elements ofG, we define their product tobe(XiYi)iE/. Wedefine theinverse of(xi)iEItobe(xi1)iE/. Itisthen obvious that Gisagroup called thedirect product ofthefamily. Let Gbe agroup. Asubgroup HofGisasubset ofGcontaining theunit element, and such that Hisclosed under thelawofcomposition and inverse (i.e. itisasubmonoid, such that ifxEHthen x-IEH). Asubgroup iscalled trivial ifitconsists oftheunit element lone. The intersection ofanarbitrary non-empty family ofsubgroups isasubgroup (trivial verification). Let Gbe agroup and Sasubset ofG.We shall saythat Sgenerates G, orthat Sisasetofgenerators forG,ifevery element ofGcan beexpressedasa product ofelements ofSorinverses ofelements ofS,i.e. asaproduct Xl...Xn where each Xior Xi-1isinS.Itisclear that the setofallsuch products isa subgroup ofG(the empty product istheunit element), and isthesmallest sub- group ofGcontaining S.Thus Sgenerates Gifandonly ifthesmallest subgroup ofGcontaining SisGitself. IfGisgenerated byS,then wewrite G=(S).By definition, acyclic group isagroup which has one generator. Given elements XI'...,xnEG,these elements generateasubgroup (X.,...,xn),namely the setofallelements ofGoftheform Xl···x:with k1,..., krE Z. Asingle element XEGgeneratesacyclic subgroup. Example. There are two non-abelian groups oforder 8.One isthegroup ofsymmetries ofthesquare, generated bytwo elements u, Tsuch that u4=T2=eand TUT-I=u3 . The other isthequaternion group, generated bytwo elements, i,jsuch that ifweput k=ijand m=i2 ,then i4=j4=k4=e, i2=j2=k2=m,ij=mji. After you know enough facts about groups, youcaneasily doExercise 35. 10 GROUPS I,2 LetG,G'bemonoids. Amonoid-homomorphism (orsimply homomorphism) ofGinto G'isamappingf: G G'such thatf(xy)=f(x)f(y) forallx,yEG, andmapping theunit element ofGinto that ofG'.IfG,G'aregroups,agroup- homomorphism ofGinto G'issimplyamonoid-homomorphism. We sometimes say:"Letf:G G'be agroup-homomorphism" tomean: "Let G,G'begroups, and letfbeahomomorphism from Ginto G'." Letf: G G'beagroup-homomorphism. Then f(x-1)=f(X)-1 because ife,e'aretheunit elements ofG,G'respectively,then e'=f(e)=f(xx- 1)=f(x)J'(x-1). Furthermore, ifG,G'aregroups andf: G-+G'isamap such that f(xy)=f(x)f(y) forallx,yinG,thenf(e)=e'because f(ee)=f(e) and also=f(e)f(e). Multiplying bytheinverse off(e) shows thatf(e)=e'. LetG,G'bemonoids. Ahomomorphismf: G-+G'iscalled anisomorphism ifthere exists ahomomorphism g:G'Gsuch thatfog and g0fare the identity mappings (inG'and Grespectively). Itistrivially verified thatfis anisomorphism ifandonly iffisbijective. The existence ofanisomorphism between two groups Gand G'issometimes denoted byG G'.IfG=G' , wesaythat isomorphism isanautomorphism. Ahomomorphism ofGinto itself isalso called anendomorphism. Example. Let Gbe amonoid and xanelement ofG.LetNdenote the (additive) monoid ofintegers>O.Then themapf: N-+Gsuch thatf(n)=xn isahomomorphism. IfGisagroup,we canextendftoahomomorphism ofZ into G(xnisdefined forallnEZ, aspointed outpreviously). The trivial proofs arelefttothereader. Let nbeafixed integer and letGbeacommutative group. Then one verifies easily that themap X1---+xn from Ginto itself isahomomorphism. Soisthe mapx1---+x- 1.The map x1---+xniscalled then-th power map. Example. LetI={i}beanindexing set, and let{Gj}beafamily ofgroups. Let G=fIGjbetheir direct product. Let Pj:G Gj betheprojection onthei-th factor. Then pjisahomomorphism. Let Gbe agroup, Sasetofgenerators for G,and G'another group. Let f:S-+G'be amap. Ifthere exists ahomomorphism IofGinto G'whose restriction toSisf,then there isonly one. I,2 GROUPS 11 Inother words, fhas atmost one extension to ahomomorphism ofG into G'.This isobvious, butwill beused many times inthesequel. Letf:G G'and g:G'-.G"betwo group-homomorphisms. Then the composite map g0fisagroup-homomorphism. Iff, gareisomorphisms then soisgof. Furthermore f-1 :G'-.Gisalso anisomorphism. Inparticular, the setofallautomorphisms ofGisitself agroup, denoted byAut(G). Letf:G-+G'beagroup-homomorphism. Let e,e'betherespective unit elements ofG,G'. We define thekernel offtobethesubset ofGconsisting ofallxsuch thatf(x)=e'.From thedefinitions, itfollows atonce that the kernel Hoffisasubgroup ofG.(Let usprove forinstance that Hisclosed under theinverse mapping. Let xEH.Then f(x-1)f(x)=f(e)=e'. Since f(x)=e',wehavef(x-1)=e',whence x-1EH. We leave the other verifications tothereader.) Letf:G-.G'beagroup-homomorphism again. LetH'betheimage off. Then H'isasubgroup ofG', because itcontains e',andiff(x),f(Y)EH', then f(xy)=f(x)f(y) lies also inH'.Furthermore,f(x-1)=f(X)-l isinH',and hence H'isasubgroup ofG'. The kernel andimage offaresometimes denoted byKerfandImf. AhOlnomorphism f:G-.G'which establishes anisomorphism between Gand itsimage inG'will also becalled anembedding. Ahomomorphism whose kernel istrivial isinjective. Toprove this, suppose that thekernel off istrivial, andletf(x)=f(y) for some x,yEG.Multiplying byf(y- 1)weobtain f(xy- 1)=f(x)f(y- 1)=e'. Hence xy-1liesinthekernel, hence xy-1=e,and x=y.Ifinparticular fis also surjective, thenfisanisomorphism. Thus asurjective homomorphism whose kernel istrivial must be anisomorphism. We note that aninjective homomorphism isanembedding. Aninjective homomorphism isoften denoted byaspecial arrow, such as f:GG'. There isauseful criterion for agroup tobeadirect product ofsubgroups: Proposition 2.1. Let Gbeagroup and letH,Kbetwosubgroups such that HnK=e,HK =G,and such that xy=yxfor allXEH andYEK. Then themap HxK-.G such that (x,y)t---+xyisanisomorphism. Proof. Itisobviouslyahomomorphism, which issurjective since HK =G. 12 GROUPS I,2 If(x,y)isinitskernel, then x=Y-I,whence xliesinboth Hand K,and x=e, sothat Y=ealso, and our map isanisomorphism. We observe thatProposition 2.1generalizes byinduction toafinite number ofsubgroups Hb...,Hnwhose elements commute with each other, such that HI...H =Gn , and such that Hi +In(H I...HJ=e. Inthat case, Gisisomorphic tothedirect product HIX... xHn. Let Gbeagroup and Hasubgroup. Aleft coset ofHinGisasubset of Goftype aH" for some element aofG.Anelement ofaH iscalled acoset representative ofaH. The map x axinduces abijection ofHonto aH. Hence any two left cosets have the same cardinality. Observe that ifa,bareelements ofGand aH, bH are cosets having one element incommon, then they areequal. Indeed, letax =bywith x,yEH. Then a=byx-I .Butyx-IEH, Hence aH =b(yx-I)H=bH, because for anyZEHwehave zH =H. Weconclude that Gisthedisjoint union oftheleft cosets ofH.Asimilar remark applies toright cosets (i.e. subsets ofGoftype Ha). The number ofleft cosets ofHinGisdenoted by(G:H), and iscalled the(left) index ofHinG. The index ofthetrivial subgroup iscalled theorder ofGand iswritten (G:1). From theabove conclusion, weget: Proposition 2.2. Let Gbeagroup andHasubgroup. Then (G:H)(H :1)=(G:1), inthe sense thatiftwooj'these indices arefinite, soisthethird andequality holds asstated. If(G:1)isfinite, theorder ofHdivides theorder ofG. More generally, letH,Kbesubgroups ofG and letH ::JK.Let{Xi} bea setof(left) coset representatives ofKinHand let{yj}beasetofcoset repre- sentatives ofHinG.Then wecontend that{YjXi}isasetofcoset representa- tives ofKinG. Proof. Note that H =UxiK i(disjoint), G=UyjH j(disjoint). Hence G=UyjxiK . i,j We must show that this union isdisjoint, i.e.that theyjXirepresent distinct cosets. Suppose I,3 NORMAL SUBGROUPS 13 Y.x.K =Y.,x.,KJI J I for apair ofindices (j,i)and(j',i').Multiplying byHontheright, andnoting that Xi'Xi'areinH,weget Y.H =Y.,HJ J' whenceYj=Yr.From this itfollows that XiK=xi,K and therefore that Xi=Xi" aswas tobeshown. The formula ofProposition 2.2 may therefore begeneralized bywriting (G:K)=(G:H)(H:K), with theunderstanding thatiftwoofthethree indices appearing inthis formula arefinite, then soisthethird and theformula holds. The above results areconcerned systematically with left cosets. For theright cosets, seeExercise 10. Example. Agroup ofprime order iscyclic. Indeed, letGhave order pand let aEG, a=1=e.LetHbethesubgroup generated bya.Then #(H) divides p and is =1=1,so#(H)=pand soH=G,which istherefore cyclic. Example. Let 1n={I,...,n}.Let Snbethegroup ofpermutations of In. We define atransposition tobe apermutationTsuch that there exist two elements r=1= SinInforwhich T(r)=S,T(S)=r,and T(k)=kforall k=1=r,s.Note that thetranspositions generate Sn.Indeed, say0"isapermutation, O"(n)=k=1=n.Let Tbethetransposition interchanging k,n.Then TO"leaves n fixed, and byinduction, we can write TO" as aproduct oftranspositions in Perm(l n-1),thus proving thattranspositions generate Sn. Next wenote that#(Sn)=n!.Indeed, letHbethesubgroup ofSnconsisting ofthose elements which leave nfixed. Then Hmay beidentified with Sn-l. If O"i(i=1,. . .,n)isanelement ofSnsuch that O"i(n)=i,then itisimmediately verified that0"1'. . .,O"nare coset representatives ofH.Hence byinduction (Sn:1)=n(H:1)=n!. Observe that forO"iwecould have taken thetransposition Ti'which interchanges iand n(except fori=n,where wecould takeO"ntobetheidentity). 3. NORMAL SUBGROUPS We have already observed that thekernel ofagroup-homomorphismisa subgroup. We now wish tocharacterize such subgroups. Letf: G-.G'beagroup-homomorphism, and letHbeitskernel. IfXisan element ofG,then xH =Hx, because both areequal tof-l(f(x)). We can also rewrite this relation asxH X-1=H. 14 GROUPS I,3 Conversely, letGbeagroup, and letHbeasubgroup. Assume that forall elements xofGwehave xH cHx(orequivalently, xHx-t cH). Ifwe write X-I instead ofx,wegetHcxHx-t ,whence xHx-t =H.Thus our condition isequivalent tothecondition XHX-l =HforallxEG.Asubgroup Hsatisfying thiscondition will becalled normal. Weshall now seethat anormal subgroup isthekernel ofahomomorphism. LetG'bethe setofcosets ofH.(Byassumption,aleftcoset isequal toaright coset, soweneed notdistinguish between them.) IfxHandyH arecosets, then their product (xH)(yH) isalso acoset, because xHyH=xyHH=xyH. Bymeans ofthisproduct,wehave therefore defined alawofcompositiononG' which isassociative. Itisclear that the coset Hitself isaunit element forthis lawofcomposition, and that x-tHis aninverse forthecoset xH. Hence G'isa group. Letf:G-.G'bethemapping such thatf(x) isthe coset xH. Thenfis clearlyahomomorphism, and(the subgroup) Hiscontained initskernel. If f(x)=H,then xH =H. Since Hcontains theunit element, itfollows that xEH.Thus Hisequal tothekernel, and wehave obtained our desired homo- morphism. The group ofcosets ofanormal subgroup Hisdenoted byG/H (which we read Gmodulo H,orGmod H). Themapfof Gonto G/H constructed above iscalled thecanonical map, andG/H iscalled thefactor group ofGbyH. Remarks 1.Let{Hi}iel beafamily ofnormal subgroups ofG.Then thesubgroup H=nH.I ieI isanormal subgroup. Indeed, ifyEH,and xEG,then xyx-tliesineach Hj, whence inH. 2.Let Sbe asubset ofGand letN =Nsbethe setofallelements xEG such that xSx-t=S.Then Nisobviouslyasubgroup ofG,called the normalizer ofS.IfSconsists ofone element Q,then Nisalso called the centralizer ofa.More generally, letZsbethe setofallelements xEGsuch that xyx-t=yforallYES. Then Zsiscalled thecentralizer ofS.The centralizer ofGitself iscalled the center ofG.Itisthesubgroup ofGconsisting ofall elements ofGcommuting with allother elements, and isobviouslyanormal subgroup ofG. Examples. We shall give more examples ofnormal subgroups later when wehave more theorems toprove thenormality. Here wegive only twoexamples. First, from linear algebra, note that thedeterminant isahomomorphism from themultiplicative group ofsquare matrices into themultiplicative group of a field. The kernel iscalled thespecial linear group, and isnormal. I,3 NORMAL SUBGROUPS 15 Second, let Gbe the set ofall maps Ta,b:R Rsuch that Ta,b(X)=ax+b,with a=t=0and barbitrary. Then Gisagroup under composition ofmappings. Let Abethemultiplicative group ofmaps oftheformTa,o (iso- morphic toR* ,the non-zero elements ofR), andletNbethegroup oftranslations Tt,bwith bER.Then thereader willverifyatonce thatTa,b.-..+ aisahomo- morphism ofGonto themultiplicative group, whose kernel isthe group of translations, which istherefore normal. Furthermore, wehave G=AN=NA, and NnA={id}. Intheterminology ofExercise 12, Gisthesemidirect product ofAand N. LetHbeasubgroup ofG.Then Hisobviouslyanormal subgroup ofits normalizer NH.We leave thefollowing statements asexercises: IfKisanysubgroup ofGcontaining Hand such that Hisnormal inK,then KcNH. IfKisasubgroup ofNH,then KH isagroup and Hisnormal inKH. The normalizer ofHisthelargest subgroup ofGinwhich Hisnormal. Let Gbeagroup and Hanormal subgroup. Let x,yEG.Weshall write x=y(mod H) ifxand ylieinthe same coset ofH,orequivalently ifxy-1 (ory-1X)lieinH. We read thisrelation" xand yarecongruent modulo H." When Gisanadditive group, then x=0(mod H) means that xliesinH,and x=y(mod H) means that x-y(ory-x)lies inH.This notation ofcongruence isused mostly foradditive groups. Let G' G!!.G" be asequence ofhomomorphisms. We shall say that this sequence isexact if 1mf=Ker g.Forexample, ifHisanormal subgroup ofGthen thesequence H..!.. G G/H isexact (where j=inclusion andqJ=canonical map). Asequence ofhomo- morphisms having more than one term, like GIIG12GIn- 1G1-+2-+3-+...--+n, iscalled exact ifitisexact ateach joint, i.e.if. 1m};=Kerh+1 foreach i=1,..., n-2.Forexample tosaythat o-+G' G!!.G" -+0 16 GROUPS I,3 isexact means thatfisinjective, that 1mf=Ker g,and that gissurjective. If H =Ker gthen this sequence isessentially the same asthe exact sequence o-.H-.G-.GIH-+O. More precisely, there exists acommutative diagram 0)G'f)G9 )G")0 jj j 0)H)G)GIH)0 inwhich thevertical mapsareisomorphisms, and the rows are exact. Next wedescribe some homomorphisms, allofwhich arecalled canonical. (i)Let G,G'begroups andf:G-.G' ahomomorphism whose kernel isH. Let cp:G-.GIH bethe canonical map. Then there exists aunique homomorphismf*:GIH-.G'suchthatf=f*0cp,andf*isinjective. Todefine f*,letxH be acoset ofH. Since f(xy)=f(x) forallyEH,we define f*(xH)tobef(x). This value isindependent ofthe choice ofcoset representative x,and itisthen trivially verified thatf*isahomomorphism, is injective, and istheunique homomorphism satisfying ourrequirements. We shall saythatf*isinduced byf Ourhomomorphismf*induces anisolI1orphism A.:GIH-.Imf ofGIH onto theimage off, andthusfcan befactored into thefollowing succes- sion ofhomomorphisms: G GIH Imf G'. Here,jistheinclusion ofImfin G'. (ii) Let Gbe agroup and Hasubgroup. LetNbetheintersection ofall normal subgroups containing H.Then Nisnormal, and hence isthesmallest normal subgroup ofGcontaining H.Let!: G-.G'beahomomorphism whose kernel contains H.Then thekernel offcontains N,and there exists aunique homomorphismf*: GIN-.G',said tobeinduced byf,making thefollowing diagram commutative: Gf) G'\1 GIN Asbefore, cpisthecanonical map. We can define f*asin(1)bytherule f*(xN)=f(x). This iswell defined, and istrivially verified tosatisfy allourrequirements. I,3NORMAL SUBGROUPS 17 (iii) LetGbegroup andH::>Ktwo normal subgroups ofG.Then Kisnormal inH,and we can define amap ofG/Konto G/Hbyassociating with each coset xKthe coset xH. Itisimmediately verified that this map isahomomorphism, and that itskernel consists ofallcosets xK such that xEH.Thus wehave a canonical isomorphism I(G/K)/(H/K) G/H. I One could also describe thisisomorphism using (i)and(ii). We leave ittothe reader toshow that wehave acommutative diagram ·G jcan )G/K)0 o)H jcan )H/K)G/H jid )G/H)0 o where the rows are exact. (iv) Let Gbe agroup and letH,Kbetwo subgroups. Assume that H iscontained inthe normalizer ofK. Then HnKisobviouslyanormal subgroup ofH,andequally obviously HK =KH isasubgroup ofG.There isasurjective homomorphism H-.HK/K associating with each xEHthe coset xKofKinthegroup HK.The reader willverify atonce that thekernel ofthishomomorphism isexactly HnK. Thus wehave acanonical isomorphism IH/(H nK) HK/K. I (v)Letf:G-.G'be agroup homomorphism, letH'be anormal sub- group ofG',and letH =f-l(H'). G·G' I I f-1(H')·H' Thenf-l(H') isnormal inG.[Proof: IfxEG,thenf(xHx- 1)=f(x)f(H)f(x)-1 iscontained inH', soXHX-l CH.] Wethen obtain ahomomorphism G-.G'-.G'/H' composing fwith thecanonical map ofG'onto G'IH',and thekernel ofthis composite isH.Hence wegetaninjective homomorphism J:GIH-.G'IB' 18 GROUPS I,3 again called canonical, giving rise tothecommutative diagram o )H ))G/H )1 )G'/H')o.)G [f)0 o)H')G' Iffissurjective, thenJisanisomorphism. Weshall now describe some applications ofourhomomorphism statements. Let Gbeagroup. Asequence ofsubgroups G=Go::JG1::JG2::J...::JGm iscalled atower ofsubgroups. The tower issaid tobenormal ifeach Gi+1is normal inGi(i=0,...,m-1).Itissaid tobeabelian (resp. cyclic) ifitis normal andifeach factor group Gi/G i+1isabelian (resp. cyclic). Letf: G-+G'beahomomorphism and let G' =Go::JG'l::J.. .::JG beanormal tower inG'.Let Gi=f-l(GD. Then the Gi(i=0,...,m)form a normal tower. IftheGform anabelian tower (resp. cyclic tower) then the Gi form anabelian tower (resp. cyclic tower), because wehave aninjective homo- morphism Gi/Gi+1-+G/G+1 foreach i,and because asubgroup ofanabelian group (resp.acyclic group) is abelian (resp. cyclic), Arefinement ofatower G=Go::JG1::J...::JGm isatower which can beobtained byinsertingafinite number ofsubgroups in thegiven tower. Agroup issaid tobesolvable ifithas anabelian tower, whose last element isthetrivial subgroup (i.e. Gm={e}intheabove notation). Proposition 3.1. Let Gbeafinite group. Anabelian tower ofGadmits a cyclic refinement. Let Gbeafinite solvable group. Then Gadmits acyclic tower, whose last element is{e}. Proof The second assertion isanimmediate consequence ofthefirst, and itclearly suffices toprove that ifGisfinite, abelian, then Gadmits acyclic tower. We useinduction ontheorder ofG.Let xbeanelement ofG.Wemay assume that x=Fe.Let Xbethecyclic group generated byx.Let G' =G/X. By induction, wecanfind acyclic tower inG',and itsinverse image isacyclic tower inGwhose lastelement isX.Ifwerefine this tower byinserting {e}attheend, weobtain thedesired cyclic tower. Example. InTheorem 6.4itwill beproved that agroup whose order isa prime power issolvable. I,3 NORMAL SUBGROUPS 19 Example. One ofthemajor results ofgroup theory istheFeit- Thompson theorem that allfinite groups ofodd order aresolvable. Cf.[Go 68]. Example. Solvable groups will occur infield theoryastheGalois groups ofsolvable extensions. SeeChapter VI, Theorem 7.2. Example. We assume thereader knows thebasic notions oflinear algebra. Let kbe afield. Let G=GL(n, k)bethegroup ofinvertible nxnmatrices in k.LetT=T(n, k)betheupper triangular group; that is,thesubgroup ofmatrices which are0below thediagonal. LetDbethediagonal group ofdiagonal matrices with non-zero componentsonthediagonal. LetNbetheadditive group ofmatrices which are0onand below thediagonal, and letV=I+N,where Iistheunit nxnmatrix. Then Visasubgroup ofG.(Note that Nconsists ofnilpotent matrices, i.e. matrices Asuch that Am=0for some positive integerm.Then (I-A)-I=I+A+A2+ . . .+Am-I iscomputed using thegeometric series.) Given amatrix AET,letdiag(A) bethediagonal matrix which has the same diagonal componentsasA.Then thereader willverify that wegetasurjective homomorphismT Dgiven by A.-+ diag(A). The kernel ofthishomomorphismisprecisely V.More generally, observe that for r>2,the setNr-I consists ofallmatrices oftheform 00 0aIr.....aIn 00 00a2,r+ 1 a2n M- 00................ an-r+l,n 00................0 00................0 LetVr=I+Nr .Then VI Uand Vr:JVr+I.Furthermore, Vr+Iisnormal inVnand thefactor group isisomorphictotheadditive group (!)kl1- r,under the themapping which sends I+Mtothe n-r-tuple (alr+l,.. .,an-r,n)Ekn-r . This n-r-tuple could becalled ther-th upper diagonal. Thus weobtain an abelian tower T:JV=VI::>V2:J . . .:JVn={I}. Theorem 3.2. LetGbeagroup andHanormal subgroup. Then Gissolvable ifandonlyifHand G/Haresolvable. Proof. We prove that Gsolvable implies that Hissolvable. Let G=Go:JGI:J . . .:JGr={e} be atower ofgroups with Gi+1normal inGi and such thatGi/G i+Iisabelian. LetHi=HnGi.Then Hi+Iisnormal inHi' and wehave anembedding Hi/Hi+l Gi/G i+l,whence Hi/Hi+l isabelian, whence proving that Hissolvable. We leave theproofs oftheother statements tothereader. 20 GROUPS I,3 LetGbeagroup. Acommutator inGisagroup element oftheform xyx-ly-l with x,yEG.Let GCbethesubgroup ofGgenerated bythe commutators. We call GCthecommutator subgroup ofG.As anexercise, prove that GCisnormal inG,and that every homomorphismf: G G'into acommutative group G' contains GCinitskernel, andconsequently factors through thefactor commutator group G/GC. Observe that G/GCitself iscommutative. Indeed, ifidenotes the image of xinG/Gc, then bydefinition we have iyi-1y-1=e, soi andycommute. Inlight ofthedefinition ofsolvability, itisclear that the commutator group isattheheart ofsolvability andnon-solvability problems. Agroup Gissaid tobesimple ifitisnon-trivial, and has nonormal subgroups other than {e} and Gitself. Examples. Anabelian group issimple ifandonly ifitiscyclic ofprime order. Indeed, suppose Aabelian andnon-trivial. Let aEA,a=t=e.Ifagenerates aninfinite cyclic group, then a2generatesaproper subgroup and soAisnot simple. Ifahasfinite period, and Aissimple, then A=(a). Let nbethperiod and supposennotprime. Write n=rswith r,s>1.Then ar=1=eand ar generatesaproper subgroup, contradicting thesimplicity ofA, soahasprime period and Aiscyclic oforder p. Examples. Using commutators, weshall give examples ofsimple groups inTheorem 5.5(the alternating group), and inTheorem 9.2ofChapter XIII (PSLn(F),agroup ofmatrices tobedefined inthatchapter). Since anon-cyclic simple group isnotsolvable, wegetthereby examples ofnon-solvable groups. Amajor program offinite group theory istheclassification ofallfinite simple groups. Essentially most ofthem (ifnotall) have natural representa- tions assubgroups oflinear maps ofsuitable vector spacesover suitable fields, inasuitably natural way. See[Go 82],[Go 86],[Sol 01]forsurveys. Gaps in purported proofs have been found. Asof200I,these arestillincomplete. Next we areconcerned with towers ofsubgroups such that thefactor groups Gi/G i+1aresimple. The next lemma isfor useintheproof oftheJordan-Holder and Schreier theorems. Lemma 3.3. (Butterfly Lemma.) (Zassenhaus) LetU,Vbesubgroups ofagroup. Let u,vbenormal subgroups ofUand V,respectively. Then u(U nv) isnormal inu(U nV), (unV)v isnormal in(U(\V)v, and thefactor groups areisomorphic, i.e. u(U nV)/u(U nv) (UnV)v/(u nV)v. Proof The combination ofgroups and factor groups becomes clear if one visualizes thefollowing diagram ofsubgroups (which gives itsname tothe lemma): I,3 NORMAL SUBGROUPS 21 u v u(unV) u v un V un v Inthisdiagram, we aregiven U,u,V, v.Alltheother points inthediagram correspond tocertain groups which can bedetermined asfollows. The inter- section oftwo line segments going downwards represents theintersection of groups. Two lines going upwards meet inapoint which represents theproduct oftwosubgroups (i.e. thesmallest subgroup containing both ofthem). We consider thetwoparallelograms representing thewings ofthebutterfly, and weshall give isomoft'hismsofthefactor groupsasfollows: u(unV)___unV u(u nv) (unV)(U nv)(UnV)v= (unV)v. Infact, thevertical side common toboth parallelograms has UnVasits top endpoint, and (unV)(U nv)asitsbottom endpoint. We have aniso- morphism (UnV)/(u nV)(U (\v) u(U nV)/u(U nv). This isobtained from theisomorphismtheorem H/(H nN) HN/N bysetting H=UnVand N=u(U nv).This givesustheisomorphismon theleft. Bysymmetryweobtain thecorresponding isomorphismontheright, which proves theButterfly lemma. Let Gbeagroup, and let G=G1::JG2::J...::JGr={e}, G=H1::JH2::J...::J Hs={e} benormal towers ofsubgroups, ending with thetrivial group. We shall say that these towers areequivalent ifr=sandifthere exists apermutation ofthe 22 GROUPS I,3 indices i=1,..., r-1,written ii',such that Gi/G i+1Hi,/H i,+1. Inother words, thesequences offactor groups inour two towers arethe same, uptoisomorphisms, and apermutation oftheindices. Theorem 3.4. (Schreier) LetGbeagroup. Two normal towers ofsubgroups ending with thetrivial group have equivalent refinements. Proof Let the two towers be asabove. For each i=1,..., r-1and j=1,..., swedefine Goo=G.+l(H,(\G.) IJ I J I. Then Gis=Gi+1,and wehave arefinement ofthefirst tower: G=G11::JG12::J... ::JG1,S-1:::>G2 =G21::JG22::J...::J Gr-l,l:::>...::J Gr-l,s- l::J{e}. Similar ly,wedefine H.. =H.+l(G.(\H,) Jl JI J' forj=1,...,s-1and i=1,...,r.This yieldsarefinement ofthesecond tower. Bythebutterfly lemma, fori=1,...,r-1andj=1,...,s-1we have isomorphisms Gij/Gi,j+ 1Hji/Hj,i+ 1. Weview each oneofourrefined towers ashaving (r-1)(s-1)+1elements, namely Gij(i=1,..., r-l;j=1,..., s-1)and{e}inthefirst case, Hjiand {e}inthe second case. The preceding isomorphism foreach pair ofindices (i,j)shows that our refined towers areequivalent,aswas tobeproved. Agroup Gissaid tobesimple ifitisnon-trivial, and has nonormal sub- groups other than {e}and Gitself. Theorem 3.5. (Jordan-Holder) Let Gbeagroup, and let G=G1:::>G2::J...:::>Gr={e} be anormal tower such that each group Gi/G i+1issimple, and Gi#=Gi+1 fori=1,..., r-1.Then anyother normal tower ofG having the same prop- erties isequivalent tothis one. Proof Given any refinement {Gij}asbefore for our tower, weobserve that foreach i,there exists precisely oneindexj such that Gi/G i+1=Gij/Gi,j+ 1. Thus the sequence ofnon-trivial factors fortheoriginal tower, ortherefined tower, isthe same. This proves our theorem. I,4 CYCLIC GROUPS 23 Bibliography [Go 68] [Go 82] [Go 83] [So01]D.GORENSTEIN, Finite groups, Harper andRow, 1968 D.GORENSTEIN, Finite simple groups, Plenum Press, 1982 D.GORENSTEIN, TheClassification ofFinite Simple Groups, Plenum Press, 1983 D.GORENSTEIN, Classifying thefinite simple groups, Bull. AMS 14No. 1 (1986), pp. 1-98 R.SOLOMON, Abrief history oftheclassification ofthefinite simple groups, Bull. AMS 38,3 (2001) pp.315-352[Go 86] 4. CYCLIC GROUPS The integers Zform anadditive group. Weshall determine itssubgroups. LetHbeasubgroup ofZ.IfHisnottrivial, letabethesmallest positive integer inH.Wecontend that Hconsists ofallelements na,with nEZ.Toprove this, letYEH.There exist integers n,rwith 0<r<asuch that Y=na+r. Since Hisasubgroup and r=y-na, wehave rEH,whence r=0,and our assertion follows. Let Gbe agroup. We shall saythat Giscyclic ifthere exists anelement aofGsuch that every element xofGcan bewritten intheform anfor some nEZ(inother words, ifthe mapf:Z-.Gsuch thatf(n)=anissurjective). Such anelement aofGisthen called agenerator ofG. Let Gbe agroup and aEG.The subset ofallelements an(nEZ)is obviouslyasubgroup ofG,which iscyclic. Ifmisaninteger such that am =e and m>0then weshall call manexponent ofa.We shall saythat m>0is anexponent ofGifxn =eforallxEG. Let Gbeagroup and aEG.Let/: Z-.Gbethehomomorphism such that f(n)=anand letHbethekernel off Two cases arise: 1.The kernel istrivial. Thenfisanisomorphism ofZonto thecyclic subgroup ofGgenerated bya,and thissubgroup isinfinite cyclic. Ifagenerates G,then Giscyclic.We also say that ahasinfinite period. 2.The kernel isnottrivial. Let dbethe smallest positive integer inthe kernel. Then discalled theperiod ofa.Ifmisaninteger such that am=ethen m=dsfor some integers.We observe that theelements e,a,. . .,ad-1are 24 GROUPS I,4 distinct. Indeed, ifar=aswith 0<:r,sc::::d-1,and say rc::::s,then as-r= e.Since 0<:s-r<dwemust have s-r=O.Thecyclic subgroup generated byahas order d.Hence byProposition 2.2: Proposition 4.1. LetGbeafinite group oforder n>1.Let abeanelement ofG,a=t=e.Then theperiod ofadivides n.Iftheorder ofGisaprime number p,then Giscyclic and theperiod ofany generator isequal top. Furthermore: Proposition 4.2. LetGbeacyclic group. Then every subgroup ofGiscyclic. Iffisahomomorphism ofG,then theimage offiscyclic. Proof. IfGisinfinite cyclic, itisisomorphictoZ,and wedetermined above allsubgroups ofZ,finding that theyareallcyclic. Iff:G G'isahomo- morphism, and aisagenerator ofG,thenf(a) isobviouslyagenerator off(G), which istherefore cyclic,sotheimage off iscyclic. Next letHbe asubgroup ofG.We want toshow Hcyclic. Let abe agenerator ofG.Then wehave a surjective homomorphism f:Z Gsuch thatf(n)=an. The inverse image f-I(H) isasubgroup ofZ,and therefore equal tomZfor some positive integer m.Sincefissurjective, wealso have asurjective homomorphism mZ H. Since mZ iscyclic (generated additively bym),itfollows that Hiscyclic, thus proving theproposition. We observe that twocyclic groups ofthe same order mareisomorphic. Indeed, ifGiscyclic oforder mwith generator a,then wehave asurjective homomorphism f:Z Gsuch thatf(n)=an, and ifkZ isthekernel, with kpositive, then we have anisomorphism Z/kZ=G, sok=m. Ifu:GIZ/mZ and v:G2Z/mZ areisomorphisms oftwocyclic groups with Z/mZ, then V-Iou: G1 G2isanisomorphism. Proposition 4.3. (i)Aninfinite cyclic group hasexactly twogenerators (ifaisagenerator, then a-1istheonly other generator). (ii)Let Gbeafinite cyclic group oforder n,and letxbeagenerator. The set ofgenerators ofG consists ofthose powers XVofx such that visrelatively prime ton. (iii) Let Gbeacyclic group, and leta,bbetwogenerators. Then there exists anautomorphism ofGmapping aonto b.Conversely, anyautomorphism ofGmaps aonsome generator ofG. (iv) LetGbeacyclic group oforder n.Letdbeapositive integer dividing n. Then there exists aunique subgroup ofGoforder d. (v)Let G1,G2becyclic oforders m, nrespectively. Ifm, narerelatively prime then G1XG2iscyclic. I,5 OPERATIONS OFAGROUP ON ASET 25 (vi) LetGbeafinite abeUan group. JfG isnotcyclic, then there exists aprime pand asubgroup ofGisomorphic toCxC, where Ciscyclic oforder p. Proof. Weleave thefirst three statements tothereader, andprove theothers. (iv) Letdln.Let m=n/d.Letf:Z Gbe asurjective homomorphism. Thenf(mZ) isasubgroup ofG,and from theisomorphism Z/mZ=G/f(mZ) weconclude thatf(mZ) hasindex minG,whencef(mZ) hasorder d.Conversely, letHbe asubgroup oforder d.Thenf-l(H)=mZ for some positive integer m, soH=f(mZ), Z/mZ=G/H, so n=md, m=n/d and Hisuniquely determined. (v)LetA=(a)and B=(b)becyclic groups oforders m,n,relatively prime. Consider thehomomorphism Z AxBsuch that k (ak ,bk).Anelement initskernel must bedivisible both bymand n,hence bytheir product since m, narerelatively prime. Conversely, itisclear that mnZ iscontained inthekernel, sothekernel ismnZ. The image ofZ AxBissurjective bytheChinese remainder theorem. This proves (v).(Areader who does notknow theChinese remainder theorem can see aproof inthe more general context ofChapter II, Theorem 2.2.) (vi) This characterization ofcyclic groups isanimmediate consequence of the structure theorem which will beproved in8, because ifGisnotcyclic, then byTheorem 8.1 and(v) we arereduced tothe case when Gisap-group, andbyTheorem 8.2there are atleast two factors inthedirect product (orsum) decomposition, and each contains acyclic subgroup oforder p,whence Gcontains their direct product (orsum). Statement (vi)is,ofcourse, easier toprove than thefull structure theorem, and itisagood exercise forthereader toformulate thesimpler arguments which yield (vi)directly. Note. For thegroup ofautomorphisms of acyclic group,see the end of Chapter II,2. 5. OPERATIONS OF AGROUP ON ASET Let Gbe agroup and letSbe aset. Anoperation oranaction ofGonS isahomomorphism 7T :G Perm(S) ofGinto thegroup ofpermutations ofS.We then call SaG-set. We denote thepermutation associated with anelement xEGby7Tx.Thus thehomomorphism isdenoted byx.-..+7Tx.Given sES,theimage ofsunder thepermutation 7Txis 7Tx(S). From such anoperationweobtain amapping GxS S, 26 GROUPS I,5 which toeach pair (x,s)with xEGand sESassociates theelement 7Tx(S). We often abbreviate the notation and write simplyxsinstead of 7Tx(S). With the simpler notation, wehave the twoproperties: For allx,yEGand SES,wehave x(ys)=(xy)s. Ifeistheunit element ofG,then es=sforall sES. Conversely, ifwe aregivenamapping GxS S,denoted by(x,s) xs, satisfying these twoproperties,then foreach xEGthemaps xsispermutation ofS,which wethen denote by7Tx(S). Then x7Txisahomomorphism ofG into Perm(S). So anoperation ofGonScould also bedefined as amapping GxS Ssatisfying theabove twoproperties. The most important examples ofrepresentations ofGasagroup ofpermutationsarethefollowing. 1.Conjugation. For each xEG,let Cx:G Gbethe map such that cx(y)=xyx-1 .Then itisimmediately verified that theassociation x 1---+ Cxisa homomorphism G Aut( G),and sothis map givesanoperation ofGonitself, called conjugation. The kernel ofthehomomorphismx 1---+ Cxisanormal sub- group ofG,which consists ofallxEGsuch thatxyx-1=yforallyEG,i.e.all xEGwhich commute with every element ofG.This kernel iscalled thecenter ofG.Automorphisms ofGoftheform Cxarecalled inner. Toavoid confusion about theoperationontheleft, wedon't write xyfor cx(y). Sometimes, one writes Cx-I(y)==x-1yx==yX, Le. one uses anexponential notation, sothat wehave therules y(xz)=(yX)Z and ye=y forallx,y,ZEG.Similarly, Xy=xyx-land Z(Xy)=zXy. We note that Galso operates byconjugation onthe setofsubsets ofG. Indeed, letSbethe setofsubsets ofG,and letAESbe asubset ofG.Then xAx-1isalso asubset ofGwhich may bedenoted bycx(A), and one verifies trivially that themap (x,A)1---+xAx-1 ofGxS-+Sisanoperation ofGonS.We note inaddition that ifAisasub- group ofGthen xAx-1isalso asubgroup,sothat Goperates onthe setof subgroups byconjugation. IfA,Bare two subsets ofG,wesaythat theyareconjugate ifthere exists xEG such that B=xAx-1 . 2.Translation. For each xEGwedefine thetranslation Tx:G Gby Tx(Y)=xy.Then themap (x,y)1---+xy=(y) defines anoperation ofGonitself. Warning: Txisnot agroup-homomorphism! Onlyapermutation ofG. I,5 OPERATIONS OFAGROUP ON ASET 27 Similarly, Goperates bytranslation onthe setofsubsets, forifAisa subset ofG,then xA =(A) isalso asubset. IfHisasubgroup ofG,then Tx(H)=xH isingeneral not asubgroup but acoset ofH,and hence we see that Goperates bytranslation onthe setofcosets ofH.We denote the setof left cosets ofHbyGIH. Thus even though Hneed not benormal, GIH isa G-set. Ithas become customary todenote the setofright cosets byH\G. The above tworepresentations ofGasagroup ofpermutations will beused frequently inthesequel. Inparticular, therepresentation byconjugation will be used throughout the next section, intheproof oftheSylow theorems. 3.Example from linear algebra. We assume the reader knows basic notions oflinear algebra. Let kbe afield and letVbe avector spaceover k.Let G=GL(V) bethe group oflinear automorphisms ofV.For AEGand vEV,the map (A,v) Avdefines anoperation ofGonV.Ofcourse, Gis asubgroup ofthegroup ofpermutations Perm(V). Similarly, letV=knbethe vector space of(vertical) n-tuples ofelements ofk,and letGbethegroup of invertible nxnmatrices with components ink.Then Goperatesonknby (A,X) AXforAEGand XEkn . LetS,S'betwo G-sets, andf: S S'amap. Wesaythatfisamorphism ofG-sets, oraG-map, if f(xs)=xf(s) forallxEGand sES.(We shall soon define categories, and seethat G-sets form acategory.) We now return tothegeneral situation, and consider agroup operatingon asetS.Let sES.The setofelements xEGsuch that xs =sisobviouslyasub- group ofG,called theisotropy group ofsinG,and denoted byGs. When Goperatesonitself byconjugation, then theisotropy group ofan element isnone other than thenormalizer ofthis element. Similarly, when G operates onthe setofsubgroups byconjugation, theisotropy group ofasub- group isagain itsnormalizer. Let Goperate on asetS.Let s,S'beelements ofS,and yanelement ofG such that ys=S'.Then Gs'=yGsY-1 Indeed, one sees atonce that yGsy-1 leaves s'fixed. Conversely, if x's'=s'then x'ys=ys,soy-Ix'yEGsandx'EyGsy-I. Thus theisotropy gro,ups ofsand s'areconjugate. LetKbethekernel oftherepresentation G Perm(S). Then directly from thedefinitions, weobtain that K=n Gs=intersection ofallisotropy groups.SES 28 GROUPS I,5 Anaction oroperation ofGissaid tobefaithful ifK={e}; that is,thekernel ofG Perm(S) istrivial. Afixed point ofGisanelement SESsuch that xs=sforallxEGorinother words, G=Gs. Let GoperateonasetS.Let sES.The subset ofSconsisting ofallelements xs(with xEG)isdenoted byGs,and iscalled theorbit ofsunder G.Ifxand y areinthe same coset ofthesubgroup H =Gs,then xs =ys,andconversely (obvious). Inthis manner, wegetamapping f:G/H-.S given byf(xH)=xs,and itisclear that this map isamorphism ofG-sets. In fact, one sees atonce that itinduces abijection ofG/H onto theorbit Gs. Consequently: Proposition 5.1. JfG isagroup operating onasetS,and sES,then theorder oftheorbit Gsisequal totheindex (G:Gs). Inparticular, when Goperates byconjugationonthe setofsubgroups, and Hisasubgroup, then: Proposition 5.2. The number ofconjugate subgroups toHisequal tothe index ofthenormalizer ofH. Example. Let Gbeagroup andHasubgroup ofindex 2.Then Hisnormal inG. Proof Note that Hiscontained initsnormalizer NH,sotheindex ofNH inGis1or2.Ifitis1,then we aredone. Suppose itis2.Let Goperate bycon- jugation onthe setofsubgroups. The orbit ofHhas 2elements, and Goperates onthis orbit. Inthis way wegetahomomorphism ofGinto the group of permutations of2elements. Since there isoneconjugate ofHunequal toH, then thekernel ofourhomomorphism isnormal, ofindex 2,hence equal toH, which isnormal, acontradiction which concludes theproof. For ageneralization and other examples,seeLemma 6.7. Ingeneral,anoperation ofGonSissaid tobetransitive ifthere isonly one orbit. Examples. Thesymmetric group Snoperates transitivelyon{I,2,. . .,n}. InProposition 2.1ofChapter VII, weshall see anon-trivial example oftransitive action ofaGalois group operatingontheprimes lying above agiven prime in theground ring. Intopology, supposewe have auniversal covering space p:X' X,where Xisconnected. Given xEX,thefundamental group 7Tl(X) operates transitivelyontheinverse image p-l(X). I,5 OPERATIONS OFAGROUP ON ASET 29 Example. LetSjbetheupper half-plane; thatis,the setofcomplex numbers z=x+iysuch that y>O.Let G=SL2(R)(2x2matrices with determinant 1).For (ab )az+b a= cdEG, weletaz= cz+d. Readers willverify bybrute force that this defines anoperation ofGonSj.The isotropy group ofiisthegroup ofmatrices (cos (Jsin (J )with (Jreal.-sin (J cos (J This group isusually denoted byK.The group Goperates transitively.You can verify allthese statements aseasy exercises. Let Goperateon asetS.Then two orbits ofGareeither disjoint orare equal. Indeed, ifGS1and GS2are two orbits with anelement sincommon, then s=XS1forsome xEG,and hence Gs =Gxs1=Gs1.Similarly, Gs =Gs2. Hence Sisthedisjoint union ofthedistinct orbits, and we canwrite S=UGSi iEI(disjoint), also denoted S=UGsi, iEI where Iissome indexing set,and the Siareelements ofdistinct orbits. IfSis finite, thisgivesadecomposition oftheorder ofSasasum oforders oforbits, which wecall theorbit decomposition formula, namely card(S)=L(G:Gs). ieI Let x,ybeelements ofagroup (ormonoid) G.Theyaresaid tocommute ifxy=yx.IfGisagroup, the setofallelements xEGwhich commute with all elements ofGisasubgroup ofGwhich wecalled the center ofG.Let Gact on itself byconjugation. Then xisinthe center ifandonly iftheorbit ofxisx itself, and thus has one element. Ingeneral, theorder oftheorbit ofxisequal totheindex ofthenormalizer ofx.Thus when Gisafinite group, theabove formula reads (G:1)=L(G:Gx) xeC where Cisasetofrepresentatives forthedistinct conjugacy classes, and the sum istaken over allxEC.This formula isalso called theclass formula. 30 GROUPS I,5 The class formula andtheorbit decomposition formula will beused systematically inthenext section onSylow groups, which may beviewed asproviding examples forthese formulas. Readers interested inSylow groups mayjump immediately tothenext section. The restofthis section deals with special properties ofthesymmetric group, which may serve asexamples ofthegeneral notions wehave developed. The symmetric group. Let Snbethe group ofpermutations of aset with nelements. This set may be taken to be the set ofintegers In={I,2,. . .,n}.Given anyUESn,and anyinteger i,I<i<n,wemay form theorbit ofiunder thecyclic group generated byu.Such anorbit iscalled acycle foru,and may bewritten [ili z·..ir], sou(i l)=iz,..., u(i r-l)=ir,u(i r)=il. Then {I ,. . .,n}may bedecomposed into adisjoint union oforbits forthecyclic group generated byu,and therefore intodisjoint cycles. Thus theeffect ofu on{I,. . .,n}isrepresented byaproduct ofdisjoint cycles. Example. The cycle [132] represents thepermutationusuch that a(l)=3, a(3)=2, and a(2)=I. We have a2(1)=2,a3(1)=1.Thus {1,3,2}istheorbit ofIunder thecyclic group generated bya. Example. InExercise 38, onewill seehow togenerate Snbyspecial types ofgenerators. Perhaps the most important part ofthat exercise isthatifnis prime,uisann-cycle and Tisatransposition, then u, Tgenerate Sn.As an application inGalois theory, ifone tries toprove that aGalois group isall ofSn(as agroup ofpermutations oftheroots), itsuffices toprove that the Galois group contains ann-cycle and atransposition. See Example 6of Chapter VI, 2. We want toassociate asign+Itoeach permutation. We dothis inthe standard way. Letfbe afunction ofnvariables, sayf:ZnZ, sowe can evaluate f(Xl,. . .,xn).Let ube apermutation ofIn. We define thefunction 7T(u)fby 7T(u)f(Xl'.. .,xn)=f(xu(l)'.. .,xu(n». Then for u, TESnwehave 7T(UT)=7T(U)7T( T).Indeed, we use thedefinition applied tothefunction g=7T(T)ftoget 1'(u)1'(T)f(x),. . .,xn)=(1'(T)f)(xU(l ),. . .,XU(I1») =f(x(TT(l)'. . .,XUT(n» =7T(UT)f(Xl'. .., xn). I,5 OPERATIONS OFAGROUP ON ASET 31 Since theidentity inSnoperatesastheidentityonfunctions, itfollows that we have obtained anoperation ofSnonthe setoffunctions. We shall write more simply atinstead of7T(u)f. Itisimmediately verified that fortwo functions f, 9wehave u(f+g)=uf+ug and u(fg)=(uf)(ug). Ifcisconstant, then u(cf)=cu(f). Proposition 5.3. There exists aunique homomorphism e:Sn {+I}such thatfor every transpositionTwehave e(T)= -1. Proof. Letdbethefunction d(x},. . .,xn)=D<.(Xj-Xi), I} theproduct being taken forallpairs ofintegers i,jsatisfying1<:i<j<:n. Let Tbe atransposition, interchanging the twointegers rand s.Sayr<s .We wish todetermine Td(xI'. . .,Xn)=D<.(xTV)-XT(i»). I} For one factor involving j=s,i=r,we see that Tchanges the factor (xs-xr)to-(x s-xr).Allother factors can beconsidered inpairsasfollows: (xk-Xs)(xk-xr)ifk>s, (xs-Xk)(Xk-xr)ifr<k<s, (xs-Xk)(X r-xk) ifk<r. Each oneofthese pairs remains unchanged when weapplyT.Hence we seethat Td= -d. Let e(u)bethesign1or-1such that ud=e(u)d for apermutationu. Since 7T(UT)=7T(U)7T( T),itfollows atonce that eisahomomorphism, and the proposition isproved. Inparticular, ifu=T}...Tmis aproduct oftranspositions, then e(u)=(-l)m. As amatter ofterminology,wecall ueven ife(u)=1,andodd ife(u)= -1.The even permutations constitute thekernel ofe,which iscalled thealternating group An. Theorem 5.4. Ifn>5then Snisnotsolvable. Proof. We shall first prove thatifH,Nare two subgroups ofSnsuch that NCHand Nisnormal inH,ifHcontains every 3-cycle, andifH/Nisabelian, then Ncontains every 3-cycle. To seethis, leti,j,k,r,sbefive distinct integers inin' and let u=[ijk] and T=[krs]. Then adirect computation gives their commutator UTU-IT-1=[rki]. 32 GROUPS I,5 Since thechoice ofi,j,k,r,swasarbitrary,we seethat thecycles [rki] alllie inNforallchoices ofdistinct r,k,i,thereby proving what wewanted. Now suppose that wehave atower ofsubgroups Sn=Ho:JHI:JH2:J·. .:JHm={e} such thatHvisnormal inHV-I for v=1,. . .,m,andHv/HV-l isabelian. Since Sncontains every 3-cycle,weconclude thatHIcontains every 3-cycle. By induction, weconclude thatHm={e}contains every 3-cycle, which isimpossible, thus proving thetheorem. Remark concerning thesign e(u). Apriori,wedefined thesign for a given n,soweshould write En(U). However, supposen<m.Then therestriction ofEmtoSn(viewed asapermutation ofJnleaving theelements ofJmnotinJn fixed) givesahomomorphism satisfying theconditions ofProposition 5.3, so this restriction isequal toEn'Thus AmnSn=An. Next weprovesome properties ofthealternating group. (a)Anisgenerated bythe3-cycles. Proof. Consider theproduct oftwo trans- positions [ij][rs]. Ifthey have anelement incommon, theproduct iseither the identity ora3-cycle. Ifthey have noelement incommon, then [ij][rs]=[ijr]Urs], sotheproduct oftwotranspositions isalso aproduct of3-cycles. Since aneven permutationisaproduct ofaneven number oftranspositions,we aredone. (b)Ifn::>5,all3-cycles areconjugate inAn.Proof: Ifl'isapermutation, then for acycle [i 1. . .im]wehave 1'[i1. . .im]Y-1=[1'<i1). . .1'<im)]. Given 3-cycles [ijk] and[i'j'k']there isapermutation l'such that ')'(i)=i', ')'(j)=j' ,andl'(k)=k'.Thus two3-cyclesareconjugate inSnbysome element y.If')'iseven, we aredone. Otherwise, byassumptionn::>5there exist r,s notequal toanyone ofthethree elements i,j,k.Then [rs] commutes with [ijk], and wereplace ybyl'[rs] toprove (b). Theorem 5.5. Ifn::>5then thealternating group Anissimple. Proof. Let Nbe anon-trivial normal subgroup ofAn. We prove that N contains some 3-cycle, whence thetheorem follows by(b). Let UEN, u=f=.id, beanelement which has themaximal number offixed points; that is,integers isuch thatu(i)=i.Itwill suffice toprove that uisa3-cycle ortheidentity. Decompose Inintodisjoint orbits of(u).Then some orbits have more than one element. Suppose allorbits have 2elements (except forthefixed points). Since Uiseven, there are atleast two such orbits. Ontheir union, Uisrepresentedas I,6SYLOW SUBGROUPS 33 aproduct oftwo transpositions [ij][rs]. Let k¥-i,j,r,s.Let T=[rsk]. Let u'=TUT-IU-J.Then u'isaproduct ofaconjugate ofUand U-J,sou'EN. Butu'leaves i,jfixed, and any element tEJmt¥-i,j,r,s,kleftfixed byu isalso fixed byu', sou'has more fixed points than u,contradictingour hypothesis. Sowe arereduced tothe case when atleast oneorbit of(u)has::>3elements, sayi,j,k,. . . .Ifuisnotthe3-cycle [ijk], then umust move atleast two other elements ofIn,otherwise uisanoddpermutation [ijkr] for some rEIn,which isimpossible. Then letumove r,sother than i,j,k,and let T=[krs]. Letu' bethe commutator asbefore. Then u'ENand u'(i)=i,and allfixed points ofuare also fixed points ofu'whence u'has more fixed points than u, a contradiction which proves thetheorem. Example. For n=4,thegroup A4isnotsimple. As anexercise, show that A4contains aunique subgroup oforder 4,which isnotcyclic, and which isnormal. This subgroup isalso normal inS4.Write down explicitly itselements asproducts oftranspositions. 6. SYLOW SUBGROUPS Letpbe aprime number. Byap-group, we mean afinite group whose order isapower ofp(i.e. pnfor some integern>0).Let Gbeafinite group and Hasubgroup. WecallHap-subgroup ofGifHisap-group. WecallH ap-Sylow subgroup iftheorder ofHispnandifpnisthehighest power ofp dividing theorder ofG.We shall prove below that such subgroups always exist. For this weneed alemma. Lemma 6.1. Let Gbe afinite abelian group oforder m,letpbe aprime number dividingm.Then Ghas asubgroup oforder p. Proof. We first prove byinduction that ifGhas exponentnthen the order ofGdivides some power ofn.LetbEG, b=F1,and letHbethecyclic subgroup generated byb.Then theorder ofHdivides nsince bn=1,and n isanexponent forG/H. Hence theorder ofG/H divides apower ofnby induction, andconsequently sodoes theorder ofGbecause (G:1)=(G:H)(H :1). Let Ghave order divisible byp.Bywhat wehave just seen, there exists an element xinGwhose period isdivisible byp.Letthisperiod bepsfor some integers.Then XS=F 1andobviouslyXShasperiod p,and generatesasubgroup oforder p,aswas tobeshown. 34 GROUPS I,6 Theorem 6.2. Let Gbe afinite group and paprime number dividing the order ofG.Then there exists ap-Sylow subgroup ofG. Proof. Byinduction ontheorder ofG.Iftheorder ofGisprime, our assertion isobvious. We now assume givenafinite group G,and assume the theorem proved forallgroups oforder smaller than that ofG.Ifthere exists a proper subgroup HofGwhose index isprime top,then ap-Sylow subgroup of Hwill also beoneofG, and ourassertion follows byinduction. Wemay therefore assume that every proper subgroup has anindex divisible byp.We now letG act onitself byconjugation. From theclass formula weobtain (G:1)=(Z:1)+L(G:Gx). Here, Zisthecenter ofG,and theterm (Z:1)corresponds totheorbits having oneelement, namely theelements ofZ.The sum ontheright istaken over the other orbits, and each index (G:Gx)isthen> 1,hence divisible byp.Since p divides theorder ofG,itfollows that pdivides theorder ofZ,hence inparticular that Ghas anon-trivial center. Let abeanelement oforder pinZ,and letHbethecyclic group generated bya. Since Hiscontained inZ,itisnormal. Letf: G-.G/H bethecanonical map. Let pnbethehighest power ofpdividing (G:1).Then pn-1divides the order ofGIH. LetK'be ap-Sylow subgroup ofG/H (byinduction) and let K =f-l(K'). Then K ::JHandfmaps Konto K'. Hence wehave aniso- morphism K/H K'. Hence Khasorder pn-1p=pn,asdesired. Fortherestofthetheorems, wesystematicallyusethenotion ofafixed point. Let Gbe agroup operatingon asetS.Recall that afixed pointsofGinSis anelement sofSsuch that xs=sforallxEG. Lemma 6.3. LetHbeap-group actingonafinite setS.Then: (a) The number offixed points ofHis==#(S) mod p. (b)IfHhasexactly onefixed point, then #(S)=1mod p. (c)IfPI#(S), then thenumber offixed points ofHis=0mod p. Proof. Werepeatedlyuse theorbit formula #(S)=L(H :Hs.).I For each fixed point Siwe haveHs;=H. ForSinot fixed, the index (H:Hs)isdivisible byp,so(a)follows atonce. Parts (b)and (c) arespecial cases of(a), thus proving thelemma. Remark. InLemma 6.3(c), ifHhas one fixed point, then Hhas atleast p fixed points. Theorem 6.4. Let Gbeafinite group. (i)IfHisap-subgroup ofG,then Hiscontained insome p-Sylow subgroup. I,6 SYLOW SUBGROUPS 35 (ii)Allp-Sylow subgroups areconjugate. (iii) The number ofp-Sylow subgroups ofGis=1mod p. Proof. LetPbe ap-Sylow subgroup ofG.Suppose first that Hiscontained inthenormalizer ofP.We prove that HCP.Indeed, HP isthen asubgroup ofthenormalizer, and Pisnormal inHP.But (HP :P)=(H:HnP), soifHP =t=P,then HP hasorder apower ofp,and theorder islarger than #(P), contradicting thehypothesis that PisaSylow group. Hence HP=Pand He P. Next, letSbethe setofallconjugates ofPinG.Then GoperatesonSby conjugation. Since thenormalizer ofPcontains P,and hastherefore index prime top,itfollows that#(S) isnotdivisible byp.Now letHbeanyp-subgroup. Then Halso acts onSbyconjugation. ByLemma 6.3(a), weknow that Hcannot have 0fixed points. LetQbe afixed point. Bydefinition this means that His contained inthenormalizer ofQ,and hence bythefirst part oftheproof, that HCQ,which proves thefirst part ofthe theorem. The second part follows immediately bytaking Htobe ap-Sylow group,so#(H)=#(Q), whence H=Q.Inparticular, when Hisap-Sylow group,we seethat Hhasonlyone fixed point,sothat(iii) follows from Lemma 6.3(b). This proves thetheorem. Theorem 6.5. Let Gbeafinite p-group. Then Gissolvable. Ifitsorder is >1,then Ghas anon-trivial center. Proof The first assertion follows from thesecond, since ifGhas center Z,and wehave anabelian tower forGIZ byinduction, we canliftthis abelian tower toGtoshow that Gissolvable. Toprove thesecond assertion, we use theclass equation (G: 1)=card(Z) +L(G:Gx)' the sum being taken over certain xforwhich (G: Gx)=F1.Then pdivides (G:1)and also divides every term inthe sum, sothat pdivides theorder ofthe center, aswas tobeshown. Corollary 6.6. Let Gbe ap-group which isnotoforder 1.Then there exists asequence ofsubgroups {e}=GocG1CG2C...cGn=G such that Giisnormal inGand Gi+I/G iiscyclic oforder p. Proof Since Ghas anon-trivial center, there exists anelement a=Fein thecenter ofG,and such that ahasorder p.LetHbethecyclic group generated bya.Byinduction, ifG=FH,we can find asequence ofsubgroupsasstated above inthefactor group GIH. Taking the inverse image ofthis tower inG givesusthedesired sequence inG. 36 GROUPS I,7 We now givesome examples toshow how toputsome ofthegroup theory together. Lemma 6.7. LetGbeafinite group and letpbethesmallest prime dividing theorder ofG.LetHbeasubgroup ofindex p.Then Hisnormal. Proof. LetN(H)=Nbethenormalizer ofH.Then N=GorN=H.If N=Gwe aredone. Suppose N=H.Then theorbit ofHunder conjugation has p=(G :H)elements, and therepresentation ofGonthis orbit givesa homomorphism ofGinto thesymmetric grouponpelements, whose order is p!.LetKbethekernel. Then Kistheintersection oftheisotropy groups, and theisotropy group ofHisHbyassumption,soKCH.IfK =f=.H,then from (G:K)=(G:H)(H:K)=p(H:K), and thefact that only thefirst power ofpdivides p!, weconclude that some prime dividing (p-I)!also divides (H:K), which contradicts theassumption that pisthesmallest prime dividing theorder ofG,and proves thelemma. Proposition 6.8. Letp,qbedistinct primes and letGbeagroup oforder pq. Then Gissolvable. Proof. Say p<q.LetQbeaSylow subgroup oforder q.Then Qhasindex p,sobythelemma, Qisnormal and thefactor group hasorder p.But agroup ofprime order iscyclic, whence theproposition follows. Example. Let Gbe agroup oforder 35. Weclaim that Giscyclic. Proof. LetH7betheSylow subgroup oforder 7.Then H7isnormal by Lemma 6.7. LetH5 be a5-Sylow subgroup,which isoforder 5.Then H5 operates byconjugationonH7,sowegetahomomorphism H5 Aut(H 7).But Aut(H 7)iscyclic oforder 6,soH5 Aut(H 7)istrivial, soevery element of H5commutes with elements ofH7.LetH5=(x)andH7=(y). Then x,ycommute with each other and with themselves, soGisabelian, and soGiscyclic by Proposition 4.3(v). Example. Thetechniques which have been developedaresufficient totreat manycases oftheabove types. Forinstance every group oforder <60issolvable, asyou will prove inExercise 27. 7. DIRECT SUMS AND FREE ABELIAN GROUPS Let{AihEI be afamily ofabelian groups. We define their direct sum A=EBAi iEI tobethesubset ofthedirect product flAiconsisting ofallfamilies (xi)iEIwith I,7 DIRECT SUMS AND FREE ABELIAN GROUPS 37 XiEAisuch that xi=0forallbut afinite number ofindices i.Then itisclear that Aisasubgroup oftheproduct. For each indexjEI,wemap Aj:AjA byletting Aj(X)betheelement whose j-th component isx,andhaving allother components equal toO.Then Aiisaninjective homomorphism. Proposition 7.1. Let{f;:Ai B}be afamily ofhomomorphisms into an abeUan group B.LetA=EBAi.There exists aunique homomorphism f:A B, such thatf0Aj=hforallj. Proof. We can define amapf:A Bbytherule f«Xi)iel)=L!i(Xi). iel The sum ontheright isactually finite since allbut afinite number ofterms are O. Itisimmediately verified that our mapfisahomomorphism. Furthermore, weclearly havef0Aj(X)=Jj(x)foreachjand each xEAj.Thusfhas the desired commutativity property. Itisalso clear that the mapfisuniquely determined, aswas tobeshown. The property expressed inProposition 7.1iscalled theuniversal property ofthedirect sum. Cf. 11 . Example. Let Abe anabelian group, and let{AihEI be afamily ofsub- groups. Then wegetahomomorphism EBAi A such that (Xi) 2:Xi. iEI Theorem 8.1willprovideanimportant specific application. Let Abe anabelian group and B,Csubgroups. IfB+C BnC={OJthen themapAand BxCA given by(x,y)1---+X+yisanisomorphism (as wealready noted inthe non- commutative case). Instead ofwriting A=BxCweshall write A=BffiC and saythat Aisthedirect sum ofBand C.We use asimilar notation forthe direct sum ofafinite number ofsubgroups Bl'.. .,Bnsuch that B1+...+Bn=A and Bi+l n(B 1+...+Bi)=O. 38 GROUPS I,7 Inthat case wewrite A=Blffi...ffiBn. LetAbeanabelian group. Let{ei}(iEI)be afamily ofelements ofA.We say that thisfamily isabasis forAifthefamily isnotempty, andifevery element ofAhas aunique expressionasalinear combination x=LXiei with XiEZand almost allXi=O.Thus the sum isactuallyafinite sum. An abelian group issaid tobefree ifithas abasis. Ifthat isthe case, itisimmediate thatifweletZi=Zforalli,then Aisisomorphictothedirect sum A=EBZi. lEI Next letSbe aset. We shall define thefree abelian group generated bySas follows. LetZ(S) bethe setofallmaps cp:S Zsuch that cp(x)=0foralmost allXES. Then Z(S) isanabelian group (addition being the usual addition of maps). Ifkisaninteger and xisanelement ofS,wedenote byk0xthemap cpsuch that cp(x)=kand cp(y)=0ify=t=x.Then itisobvious that every element cpofZ(S) can bewritten intheform qJ=k10Xl+...+kn.x n for some integers kiand elements XiES(i=1,...,n),allthe Xibeing distinct. Furthermore, qJadmits aunique such expression, because ifwehave qJ=Lkx.x=Lk.x xeS xeS then o=L(kx-k).x, xeS whence k=kxforallXES. We map Sinto Z(S) bythe map Is=Isuch thatI(x)=lox. Itis then clear thatIisinjective, and thatI(S) generates Z(S). Ifg:S-.Bisa mapping ofSinto some abelian group B,then we can define amap g.:Z(S)-.B such that g.(Lkx.X)=Lkxg(x). xeS xeS This map isahomomorphism (trivial) and wehave g*0f=9(also trivial). It istheonly homomorphism which hasthisproperty, forany such homomorphism g*must besuch that g*(10x)=g(x). I,7 DIRECT SUMS AND FREE ABELIAN GROUPS 39 Itiscustomary toidentify SinZ(S), and wesometimes omit thedot when wewrite kxx orasumLkxx. IfA:S-.S'isamapping ofsets, there isaunique homomorphism 1making the following diagram commutative: SIsZ(S) Aj jI S'Is'Z(S/) Infact, 1isnone other than (fs'0A).,with thenotation ofthepreceding para- graph. Theproof ofthis statement isleft asatrivial exercise. We shall denote Z(S) also byFab(S), and callFab(S) thefree abeUan group generated byS.Wecall elements ofSitsfree generators. As anexercise, show that every abelian group Aisafactor group ofafree abelian group F.IfAisfinitely generated, show that one can select Ftobe finitely generated also. Ifthe setSabove consists ofnelements, then wesaythat thefree abelian group Fab(S) isthe free abelian group on ngenerators. IfSisthe setofn letters Xl'...' xn,we say that Fab(S) isthe free abelian group with free generators Xl'.·.,xn. Anabelian group isfree ifandonly ifitisisomorphic toafree abelian group Fab(S) for some setS.Let Abe anabelian group, and letSbe abasis forA. Then itisclear that Aisisomorphic tothefree abelian group Fab(S). As amatter ofnotation, ifAisanabelian group and Tasubset ofelements ofA, wedenote by(T)thesubgroup generated bytheelements ofT,i.e.,the smallest subgroup ofAcontaining T. Example. The Grothendieck group. Let Mbe acommutative monoid, written additively. There exists acommutative group K(M) and amonoid- homomorphism y:M-.K(M) having thefollowing universal property. Iff: M Aisahomomorphism into anabelian group A,then there exists aunique homomorphism f.:K(M) A making thefollowing diagram commutative: M'YK(M) Af Proof LetFab(M) bethefree abelian group generated byM. Wedenote thegenerator ofFab(M) corresponding toanelement XEMby[x]. Let Bbe thesubgroup generated byallelements oftype [x+y]-[x]-[y] 40 GROUPS I,7 where x,yEM. We letK(M)=Fab(M)jB, and let y:M-.K(M) bethe map obtained bycomposing theinjection ofMinto Fab(M) given by x[x], and thecanonical map Fab(M)-.Fab(M)jB. Itisthen clear that yisahomomorphism, and satisfies thedesired universal property. The universal group K(M) iscalled theGrothendieck group. We shall saythat thecancellation law holds inMif,whenever x,y,ZEM, and x+Z=Y+z,wehave x=y. We then have animportant criterion when theuniversal map yabove is injective: Ifthecancellation law holds inM,then thecanonical map yofMinto its Grothendieck group isinjective. Proof. This isessentially the same proofaswhen one constructs thenega- tiveintegers from thenatural numbers. Weconsider pairs (x,y)with x,yEM and saythat(x,y)isequivalent to(x',y')ify+x'=x+y'.Wedefine addition ofpairs componentwise. Then theequivalence classes ofpairs form agroup, whose 0element istheclass of(0,0)[ortheclass of(x,x)forany xEM]. The negative ofanelement (x,y)is(y,x).Wehave ahomomorphism x1---+class of(0,x) which isinjective, asone sees immediately byapplying thecancellation law. Thus wehave constructed ahomomorphism ofMinto agroup, which is injective. Itfollows that theuniversal homomorphismmust also beinjective. Examples. See theexample ofprojective modules inChapter III,4.For arelatively fancy context, see: K.KATO,Logarithmic structures ofFontaine- Illusie, Algebraic Geometry, Analysis and Number Theory, Proc. JAMl Confer- ence, J.Igusa (Ed.), Johns Hopkins Press (1989) pp. 195-224. Given anabelian group Aand asubgroup B,itissometimes desirable to find asubgroup Csuch that A=BEt>C.The next lemma givesusacondition under which this istrue. Lemma 7.2. Let A1.A'be asurjective homomorphism ofabelian groups, and assume that A'isfree. Let Bbethekernel off.Then there exists a subgroup CofAsuch that therestriction offtoCinduces anisomorphism ofC with A',and such that A=BEt>c. Proof Let{Xaiel beabasis ofA',and foreach iEI,letXibeanelement of Asuch thatf(xi)=x.Let Cbethesubgroup ofAgenerated byallelements Xi'iEI.Ifwehave arelation "n.x.=0i.J I I ieI I,7 DIRECT SUMS AND FREE ABELIAN GROUPS 41 with integers ni,almost allofwhich areequal to0,then applyingfyields o=Lnif(xi)=Lnix, iel iel whence allni=O.Hence ourfamily {Xi}iel isabasis ofC.Similarly,one sees that ifZECandf(z)=0then z=O.Hence BnC=O.Let xEA.Since f(x)EA'there exist integers nbiEI,such that f(x)=Lnix. ie1 Applying ftox-LniXi' wefind that this element lies inthekernel off, iEI say x-Lnix i=bEB. iel From this we seethat xEB+C,and hence finally that A=BEt>Cisadirect sum, ascontended. Theorem 7.3. Let Abeafree abelian group, and letBbeasubgroup. Then Bisalso afree abelian group, and thecardinality ofabasis ofBis<the cardinality ofabasisforA.Any two bases ofBhave the same cardinality. Proof Weshall give theproof only when Aisfinitely generated, saybya basis {xb...,xn}(n>1),andgive theproof byinduction on n.We have an expression ofAasdirect sum: A=ZxlEt>...Et>Zxn. Letf: A-.ZXlbetheprojection, i.e.thehomomorphism such that f(mlx l+...+mnx n)=mlx l whenever miEZ.LetBlbethekernel offl B.Then Bliscontained inthefree subgroup (X2,...,xn).Byinduction, B1isfree and has abasis with <n-1 elements. Bythelemma, there exists asubgroup C1isomorphic toasubgroup ofZXl(namely theimage offlB) such that B=Bl Et>C1. Since f(B) iseither 0orinfinite cyclic, i.e.free on one generator, this proves that Bisfree. (When Aisnotfinitely generated, one can use asimilar transfinite argument. SeeAppendix 2,2, theexample after Zorn's Lemma.) We also observe that ourproof shows that there exists atleast one basis ofBwhose cardinality is<n.We shall therefore befinished when weprove the last statement, that any two bases ofBhave the same cardinality. Let S beone basis, with afinite number ofelements m.LetTbeanother basis, and suppose that Thas atleast relements. Itwill suffice toprove that r<m(one 42 GROUPS I,8 can then usesymmetry). Let pbe aprime number. Then B/pB isadirect sum ofcyclic groups oforder p,with mterms inthe sum. Hence itsorder ispm.Using thebasis Tinstead ofS,weconclude that B/pB contains anr-fold product ofcyclic groups oforder p,whence pr<pm,and r<m, aswas to beshown. (Note that wedid not assume apriori that Twasfinite.) The number ofelements inabasis ofafree abelian group Awill becalled therank ofA. 8. FINITELY GENERATED ABELIAN GROUPS The groups referred tointhetitle ofthis section occur sofrequently that itis worth while tostate atheorem which describes their structure completely. Throughout this section wewrite our abelian groups additively. Let Abeanabelian group. Anelement aEAissaid tobeatorsion element ifithasfinite period. The subset ofalltorsion elements ofAisasubgroup ofA called the torsion subgroup ofA.(If ahasperiodmand bhasperiodnthen, writing thegroup lawadditively, we seethat a+bhas aperiod dividing mn.) The torsion subgroup ofAisdenoted byAton orsimply At. Anabelian group iscalled atorsion group ifA=Ator, that isallelements ofAareoffinite order. Afinitely generated torsion abelian group isobviously finite. Weshall begin bystudying torsion abelian groups. IfAisanabelian group andpaprime number, wedenote byA(p) thesubgroup ofallelements xEAwhose period isapower ofp.Then A(p) isatorsion group, and isap-group ifitisfinite. Theorem 8.1 LetAbeatorsion abeUan group. Then Aisthedirect sumof itssubgroups A(p)forallprimes psuch thatA(p)=t=o. Proof. There isahomomorphism EBA(p) A p which toeach element (xp)inthedirect sum associates theelement LxpinA. We prove that thishomomorphism isboth surjective andinjective. Suppose x isinthekernel, soLxp=O.Let qbe aprime. Then xq=2::(-xp). p=l=q Let mbetheleast common multiple oftheperiods ofelements xpontheright- hand side, with xq=t=0and p=t=q.Then mXq=O.But alsoqrXq=0for some positive integerr.Ifdisthegreatestcommon divisor ofm,qrthen dXq=0, but d=1,soxq=O.Hence thekernel istrivial, and thehomomorphismis injective. I,8FINITELY GENERATED ABELIAN GROUPS 43 Asforthesurjectivity, foreach positive integer m,denote byAmthekernel ofmultiplication bym,Le. thesubgroup ofxEAsuch that mx=O.Weprove: Ifm=rswith r,spositive relative prime integers, then Am=Ar+AS" Indeed, there exist integers u,vsuch that ur+vs=I.Then x=urx +vsx, and urx EAswhile vsx EAn and ourassertion isproved. Repeating this process inductively,weconclude: Ifm=ITpe(p) then Am=LApe(p).plm plm Hence themapEBA(p) Aissurjective, and thetheorem isproved. Example. LetA=Q/Z. Then Q/Z isatorsion abelian group, isomorphic tothedirect sum ofitssubgroups (Q/Z)(p). Each (Q/Z)(p) consists ofthose elements which can berepresented byarational number a/pk with aEZand k some positive integer, i.e. arational number having onlyap-power inthe denominator. See also Chapter IV, Theorem 5.I. Inwhat follows weshall deal with finite abelian groups,soonlyafinite number ofprimes (dividing theorder ofthegroup) will come intoplay. Inthis case, thedirect sum is"the same as" thedirect product. Our next task istodescribe the structure offinite abelian p-groups. Let r1,...,rsbeintegers>1.Afinite p-group Aissaid tobeoftype (prt,...,prs) ifAisisomorphic totheproduct ofcyclic groups oforders pri(i=1,...,s). We shall need thefollowing remark. Remark. Let Abe afinite abelian p-group. Let bbe anelement of A,b=t=O.Let kbeaninteger>0such thatpkb=t=0,and letpmbetheperiod ofpkb. Then bhasperiod pk+m. [Proof. Wecertainly have pk+mb=0,andif pnb=0then first n>k,and second n>k+m, otherwise theperiod ofpkb would besmaller than pm.] Theorem 8.2. Every finite abelian p-group isisomorphic toaproduct of cyclic p-groups. Ifitisoftype (prl,.. .,prs) with r>r>...>r>1 1=2= =s=' then thesequence ofintegers (rI'. . .,rs)isuniquely determined. Proof. We shall prove theexistence ofthedesired product byinduction. LetalEAbe anelement ofmaximal period. We mayassume without loss of generality that Aisnotcyclic. Let A1bethecyclic subgroup generated byaI' sayofperiod prl. We need alemma. Lemma 8.3. Letbbeanelement ofA/AI' ofperiod proThen there exists a representativeaofbinAwhich also hasperiod pro 44 GROUPS I,8 Proof. Let bbeany representative of5inA.Then prb lies inAbsay prb=na1with some integern>o.We note that theperiod of5is<theperiod ofb.Ifn=0we aredone. Otherwise write n=pkJ.L whereJ.Lisprime top. ThenJ.Ul1 isalso agenerator ofAI'and hence hasperiod prl. We mayassume k<rl.Then pkJ.Ul} hasperiod prl-k. By ourprevious remarks, theelement b hasperiod pr+r1-k whence byhypothesis,r+r1-k<r1and r<k.This proves that there exists anelement CEA1such that prb=prc.Let a=b-c.Then aisarepresentative for5inAand pra=o.Since period (a)<prweconclude that ahasperiod equal topro We return tothemain proof. Byinduction, thefactor group AIA1 has a product expression AIA1=A2X... xAs into cyclic subgroups oforders pr2 ,...,prsrespectively, and wemay assume r2> ...>rs. Let aibe agenerator forAi(i=2,...,s)and letaibe a representative inAofthe same periodasai.LetAibethecyclic subgroup generated bya i.Wecontend that Aisthedirect sum ofAI'. . .,AS" Given xEA,let .xdenote itsresidue class inAIA1.There exist integers mi>0(i=2,...,s)such that .x=m2Q2+..·+msQs. Hence x-m2a2- ... -msa slies inAb and there exists aninteger m1>0 such that x=m1a1 +m2a2 +...+msa s. Hence A1+...+As=A. Conversely, suppose that mb...,msareintegers>0such that o=m1a1+...+msas. Since aihasperiod pri(i=1,...,s),wemay suppose that mi<pri.Putting abar onthisequation yields o=m2a2+...+mslis. Since AIA1isadirect product ofA2,. ..,Asweconclude that each mi=0for i=2,...,s.But then m1=0also, and hence allmi=0(i=1,...,s).From this itfollows atonce that (A 1+...+Ai)nAi+1=0 foreach i>1,and hence that Aisthedirect product ofA1,...,As, asdesired. We prove uniqueness, byinduction. Suppose that Aiswritten intwo ways asadirect sum ofcyclic groups, sayoftype (prl rs)and (ml mk) ,...,p p,...,p I,8 FINITELY GENERATED ABELIAN GROUPS 45 with rl> ...>rs>1and ml> ...>mk>1.Then pA isalso ap-group, oforder strictly lessthan theorder ofA,and isoftype (prl- 1,...,prs- 1)and (pml- 1,...,pmk- 1), itbeing understood that ifsome exponent riormjisequal to1,then thefactor corresponding to pri-lor pmj-1 inpAissimply thetrivial group o.Byinduction, thesubsequence of (r 1-1,..., rs-1) consisting ofthose integers> 1isuniquely determined, and isthe same as thecorresponding subsequence of (m 1-1,..., mk-1). Inother words, wehave ri-1=mi-1forallthose integers isuch that ri-1ormi-1>1.Hence ri=miforallthese integers i,and the two se- quences (rl rs) d(ml mk) p,..., p an p,...,p can differ only intheir lastcomponents which can beequal top.These cor- respond tofactors oftype (p,. ..,p)occurring sayvtimes inthefirst sequences and /1times inthesecond sequence. Thus for some integer n,Aisoftype (prl,,..,prn ,p,...,p) and'-v-' "times(prI,.. .,prn,p,...,p).'-v--I J1times Thus theorder ofAisequal to prl+,+rnp"=prl+. +r npJ1 , whence v=/1,and ourtheorem isproved. Agroup Gissaid tobetorsion free, orwithout torsion, ifwhenever an element xofGhasfinite period, then xistheunit element. Theorem 8.4. LetAbeafinitely generated torsion-free abeUan group. Then Aisfree. Proof Assume A=FO.Let Sbeafinite setofgenerators, and letXI'...,Xn be amaximal subset ofShaving theproperty that whenever v.,...,Vnare integers such that VIXI+...+VnX n=0, thenVj=0forallj.(Note that n> 1since A=F0).Let Bbethesubgroup generated byXI' ..., Xn. Then Bisfree. Given YEA there exist integers ml,...,mn,mnotallzero such that my+m1x1+...+mnX n=0, 46 GROUPS I,9 bytheassumption ofmaximality onXl'...,Xn.Furthermore, m=F0;other- wise all mj=o.Hence myliesinB.This istrue foreveryone ofafinite setof generators yofA,whence there exists anintegerm=F0such that mA cB. The map X1---+mx ofAinto itself isahomomorphism, having trivial kernel since AistorsIon free. Hence itisanisomorphism ofAonto asubgroup ofB.ByTheorem 7.3ofthe preceding section, weconclude that mA isfree, whence Aisfree. Theorem 8.5. LetAbe afinitely generated abelian group, and letAtor be thesubgroup consisting ofallelements ofAhaving finite period. Then Ator is finite, and AIAtorisfree. There exists afree subgroup BofAsuch that Aisthe direct sumofAtor and B. Proof Werecall that afinitely generated torsion abelian group isobviously finite. Let Abefinitely generated bynelements, and letFbethefree abelian group on ngenerators. Bytheuniversal property, there exists asurjective homomorphismFA ofFonto A.The subgroup cp-I(A tor)ofFisfinitely generated byTheorem 7.3. Hence Ator itself isfinitely generated, hence finite. Next, weprove thatAIAtor has notorsion. Letibe anelement ofAIAtor such that mi=0for some integerm =t=o.Then foranyrepresentative ofxof iinA, wehave mx EAtop whence qmx=0for some integer q=t=o.Then xEAtopsoi=0,andAIAtor istorsion free. ByTheorem 8.4,AIAtor isfree. We now use thelemma ofTheorem 7.3 toconclude theproof. The rank ofAIAtor isalso called therank ofA. For other contexts concerning Theorem 8.5, see the structure theorem for modules over principal rings inChapter III,7, and Exercises 5,6,and 7of Chapter III. 9. THE DUAL GROUP Let Abe anabelian group ofexponentm>1.This means that for each element xEAwehave mx=o.LetZmbe acyclic group oforder m.We denote byAA,orHom(A, Zm) thegroup ofhomomorphisms ofAinto Zm, and call it thedual ofA. Letf: A Bbe ahomomorphism ofabelian groups, and assume both have exponentm.Thenfinduces ahomomorphism fA:BAAA. I,9 THE DUAL GROUP 47 Namely, foreach t/1EB" wedefine f"( t/1)=t/10f.Itistrivially verified thatf" isahomomorphism. The properties id"=id and (f0g)"=g"0f" aretrivially verified. Theorem 9.1. IfAisafinite abeUan group, expressedas aproduct A=BxC,then A"isisomorphic toB" XC"(under themapping described below). Afinite abeUan group isisomorphic toitsown dual. Proof Consider thetwoprojections BxC/"'\ B C ofBxConitstwo components. Wegethomomorphisms (BxC)"7 B" C" and wecontend that these homomorphisms induce anisomorphism ofB" xC" onto (BxC)". Infact, lett/1I' t/12beinHom(B, Zm) andHom(C, Zm)respectively. Then (t/1I' t/12)EB"XC", and wehave acorresponding element of(BXC)" by defining (t/1l't/12)(x,Y)=t/11(x)+t/12(Y) , for(x,y)EBxC.Inthis waywegetahomomorphism B" xC" (BXC)". Conversely, let t/1E(8XC)". Then t/1(x, y)=t/1(x,O)+t/1(0, y). The functiont/11onBsuch thatt/11(x)=t/1(x, 0)isinB", andsimilarly the functiont/12onCsuch that t/12(Y)=t/1(0, y)isinC". Thus wegetahomomorphism (BxC)" B" XC", which isobviously inverse tothe one wedefined previously. Hence weobtain anisomorphism, thereby proving thefirst assertion inour theorem. We canwrite any finite abelian groupasaproduct ofcyclic groups. Thus toprove thesecond assertion, itwill suffice todeal with acyclic group. Let Abecyclic, generated byone element xofperiodn.Then nIm,andZm haspreciselyonesubgroup oforder n,Zn,which iscyclic (Proposition 4.3(iv)). 48 GROUPS I,9 Ift/I:A Zmisahomomorphism,and xisagenerator forA,then theperiod ofxisanexponent fort/I(x), sothat t/I(x), and hence t/I(A), iscontained inZn. Let ybe agenerator forZn. We have anisomorphism t/11:A-+Zn such that t/11(X)=y.For each integer kwith 0<k<nwehave thehomo- morphism kt/1 1such that (kt/11XX)=k.t/11(X)=t/11(kx). Inthis waywegetacyclic subgroup ofA"consisting ofthe nelements kt/ll (0<k<n).Conversely, any element t/IofA" isuniquely determined byits effect on the generator x,and must mapxon one ofthe nelements kx(0<k<n)ofZn' Hence t/Iisequal toone ofthe maps kt/ll' These maps constitute thefull group A", which istherefore cyclic oforder n,generated by t/ll.This proves our theorem. Inconsidering thedual group, wetake various cyclic groups Zm' There are many applications where such groups occur, forinstance thegroup ofm-th roots ofunity inthecomplex numbers, thesubgroup oforder mofQ/Z, etc. LetA,A'betwo abelian groups. Abilinear map ofAxA'into anabelian group Cisamap AxA' -+C denoted by (x,x')1---+(x,x') having thefollowing property. For each xEAthe function x' 1---+(x,x') isahomomorphism, andsimilarly foreach x'EA'thefunction x1---+(x,x')isa homomorphism. As aspecial case ofabilinear map, wehave the onegiven by AxHom(A, C)-+C which toeach pair(x,f) with xEAandfEHom(A, C)associates theelement f(x) inC. Abilinear map isalso called apairing. Anelement xEAissaid tobeorthogonal (orperpendicular) toasubset S' ofA'if(x,x')=0forallx'ES'.Itisclear that the setofxEAorthogonal toS' isasubgroup ofA.Wemake similar definitions forelements ofA',orthogonal tosubsets ofA. The kernel ofour bilinear mapontheleft isthesubgroup ofAwhich is orthogonal toallofA'.Wedefine itskernel ontheright similarly. Given abilinear map AxA' -+C,letB,B'betherespective kernels ofour bilinear mapontheleftandright. Anelement x'ofA'gives rise toanelement of Hom(A, C)given by x1---+(x,x'), which weshall denote by t/1x'. Since t/1x' vanishes onBwe seethat t/1x' isinfact ahomomorphism ofAIB into C. I,10INVERSE LIMIT AND COMPLETION 49 Furthermore, t/1x'=t/1y'ifx',y'areelements ofA'such that x'=y'(mod B' ). Hence t/1isinfact ahomomorphism o-+A'iB'-+Hom(AIB, C), which isinjective since wedefined B'tobethe group orthogonal toA. Similarly,wegetaninjective homomorphism o-+AIB-+Hom(A'IB' ,C). Assume that Ciscyclic oforder m.Then foranyx'EA'wehave mt/1 x'=t/1mx'=0, whence A'iB' hasexponentm.Similarly, AIB hasexponentm. Theorem 9.2. LetAxA' Cbeabilinear mapoftwo abeUan groups into acyclic group Coforder m.LetB,B'beitsrespective kernels ontheleftand right. Assume thatA'/B' isfinite. Then A/B isfinite, andA'/B' isisomorphic tothedual group ofA/B (under our map t/J). Proof The injection ofAIB into Hom(A'IB' ,C)shows that AIB isfinite. Furthermore, wegettheinequalities ordA/B<ord(A' /B')A=ordA'/B' and ordA'/B'<ord(A/B)A=ordA/B. From this itfollows that our map t/Jisbijective, hence anisomorphism. Corollary 9.3. LetAbeafinite abelian group, Basubgroup, AAthedual group, and B.lthe setofcpEAAsuch that cp(B)=O.Then wehave anatural isomorphism ofAA/B.lwith BA . Proof. This isaspecialcase ofTheorem 9.2. 10. INVERSE LIMIT AND COMPLETION Consider asequence ofgroups {G n}(n=0,1,2,. ..),and suppose given forall n::>1homomorphisms fn:GnGn-l. Suppose first that these homomorphismsaresurjective. We form infinite sequences x=(xo, XI'X2'...)such that Xn-I=fn(x n). 50 GROUPS I,10 Bytheassumption ofsurjectivity, given xnEGnwe canalways lift xntoGn+ 1 viaIn+I'sosuch infinite sequences exist, projecting toanygiven xo.We can define multiplication ofsuch sequences componentwise, and itisthen imme- diately verified that the setofsequences isagroup, called theinverse limit ofthefamily {(G n,In)}' We denote theinverse limit bylim(G n,In), orsimply lim Gnifthereference toInisclear. Example. Let Abe anadditive abelian group. Letpbe aprime number. LetpA:A Adenote multiplication byp.We saythat Aisp-divisible ifPAis surjective.We may then form theinverse limit bytaking An=Aforalln,and In=PAforall n.The inverse limit isdenoted byVp(A).We letTp(A)bethe subset ofVp(A) consisting ofthose infinite sequencesasabove such that Xo=O.LetA[pn] bethekernel ofp. Then Tp(A)=limA[pn+ 1]. The group Tp(A)iscalled theTate group associated with thep-divisible group A.Itarose infairly sophisticated contexts ofalgebraic geometry due toDeuring andWeil, inthetheory ofellipticcurves and abelian varieties developed inthe 1940s, which arefarafield from this book. Interested readers can consult books onthose subjects. The most common p-divisible groups areobtained asfollows. First, letAbe thesubgroup ofQ/Z consisting ofthose rational numbers (mod Z)which can beexpressed intheform a/pk with some positive integer k,and aEZ.Then A isp-divisible. Second, letfJ.[pn] bethegroup ofpn-th roots ofunity inthecomplex numbers. Let fJ.[pOC] betheunion ofallJL[pn] forall n.Then J.1[pOC] isp-divisible, and isomorphic tothegroup Aofthepreceding paragraph. Thus Tp(fJ.)=limfJ.[pn]. These groupsarequite important innumber theory andalgebraic geometry. We shall make further comments about them inChapter III, 10,inabroader context. Example. Suppose givenagroup G.Let{Hn} be asequence ofnormal subgroups such that Hn:JHn+l forall n.Let In:G/Hn G/Hn-l bethecanonical homomorphisms. Then wemay form theinverse limit limG/Hn. Observe that Ghas anatural homomorphism g:G limG/Hn, which sends anelement xtothesequence (.. .,Xn,. ..),where Xn=image of xinG/Hn. Example. Let Gn=Z/pn+lZ foreach n::>O.Let In:Z/pn+lz Z/pnz bethecanonical homomorphism. Then Inissurjective, and thelimit iscalled I,10 INVERSE LIMIT AND COMPLETION 51 thegroup ofp-adic integers, denoted byZp.We return tothis inChapter III, 10,where weshall seethatZpisalso aring. After these examples,wewant toconsider the more general situation when one deals notwith asequence butwith amore general type offamily ofgroups, which may not becommutative. We therefore define inverse limits ofgroupsin general. LetIbeasetofindices. Suppose givenarelation ofpartial ordering inI, namely for some pairs (i,j)wehave arelation i<jsatisfying theconditions: Foralli,j,kinI,wehave i<i;ifi<jandj<kthen i<k;ifi<jandj<i then i=j.We saythat Iisdirected ifgiven i,jEI,there exists ksuch that i-<kandj-<k.Assume thatIisdirected. By an(inversely) directed family ofgroups,we mean afamily {GihEl and foreach pair i-<jahomomorphism fl..G. G. I.JI such that, whenever k-<i-<jwehave f0f{=f{andf=ide Let G=f1Gibetheproduct ofthefamily. Letrbethesubset ofGconsisting ofallelements (Xi) withXiEGisuch that foralliandj:>iwehave f1(xj)=Xi. Then rcontains theunit element, and isimmediately verified tobe asubgroup ofG .We callrtheinverse limit ofthefamily, and write r=lim Gi. Example. Let Gbe agroup. Let bethefamily ofnormal subgroups of finite index. IfH,Karenormal offinite index, then soisHnK, so isa directed family. Wemay then form theinverse limit lim.G/H with HE.There isavariation onthis theme. Instead of,letpbe aprime number, and letp bethefamily ofnormal subgroups offinite index equaltoapower ofp.Then theinverse limit with respect tosubgroups HEpcan also betaken. (Verify thatifH,Karenormal offinite p-power index, soistheir intersection.) Agroup which isaninverse limit offinite groups iscalled profinite. Example from applications. Such inverse limits arise inGalois theory. Let kbe afield and letAbe aninfinite Galois extension. Forexample, k=Q and Aisanalgebraic closure ofQ.Let GbetheGalois group; that is,thegroup ofautomorphisms ofAover k.Then Gistheinverse limit ofthefactor groups G/H, where Hrangesover theGalois groups ofAover K,with Kranging over allfinite extensions ofkcontained inA.See theShafarevich conjecture inthe chapteronGalois theory, Conjecture 14.2 ofChapter VI. Similarly, consider acompact Riemann surface Xofgenus:>2.Let p:X' -+Xbetheuniversal covering space. LetC(X)==FandC(X')==F'be thefunction fields. Then there isanembedding n}(X) Gal(F' /F). Itis shown incomplex analysis that nl(X) isafree group with one commutator 52 GROUPS I,10 relation. The fullGalois group ofF'/Fistheinverse limit with respect tothe subgroups offinite index, asintheabove general situation. Completion ofagroup Supposenow that we aregivenagroup G,andfirst, forsimplicity, suppose givenasequence ofnormal subgroups {Hr} with Hr:JHr+ 1foralln,and such that these subgroups have finite index. Asequence {xn} inGwill becalled a Cauchy sequence ifgiven Hrthere exists Nsuch that forallm, n>Nwehave xnx;;;IEHr- We say that{xn} isanull sequence ifgivenrthere exists Nsuch that forall n>Nwehave XnEHr- As anexercise, prove that theCauchy sequences form agroup under termwise product, and that thenull sequences form anormal subgroup. The factor group iscalled thecompletion ofG(with respect tothe sequence ofnormal subgroups). Observe that there isanatural homomorphism ofGinto itscompletion; namely,anelement xEGmaps tothe sequence (x, X,X,. ..)modulo null sequences. The kernel ofthishomomorphism istheintersection nHnsoifthis intersection istheunit element ofG,then the map ofGinto itscompletionis anembedding. Theorem 10.1. Thecompletion and theinverse limitlimG/Hrareisomorphic under natural mappings. Proof. Wegive the maps. Let x={xn} be aCauchy sequence. Given r, forall nsufficiently large, bythedefinition ofCauchy sequence, theclass ofXn mod Hr isindependent of n.Let this class bex(r). Then the sequence (x(l), x(2),. ..)defines anelement oftheinverse limit. Conversely, givenan element (ibi2,. ..)intheinverse limit, withinEG/Hn, let Xnbe arepresenta- tive inG.Then the sequence {xn} isCauchy.We leave tothereader toverify that theCauchy sequence {xn} iswell-defined modulo null sequences, and that themapswehave defined areinverse isomorphisms between thecompletion and thedirect limit. We used sequences anddenumerability tomake thean':llogywith the con- struction ofthereal numbers clearer. Ingeneral, given thefamily ff=,oneconsiders families {XH}HEofelementsXHEG.Then thecondition for aCauchy family reads: given HoEff=there exists HIE ftsuch thatifK,K' arecontained inHI' then XKXK,lEHo. Inpractice,one canwork with sequences, because groups that arise naturally aresuch that the setofsubgroups offinite index isdenumerable. This occurs when thegroup Giscountably generated. More generally,afamily {Hi} ofnormal subgroups offinite index iscalled cofinal ifgiven HEft there exists isuch thatHiCH.Suppose that there exists such afamily which isdenumerable; that is,i=1,2,. . .ranges over thepositive integers. Then itisanexercise toshow that there isanisomorphism !illG/Hi=!illG/H, i HE I,11 CATEGORIES AND FUNCTORS 53 orequivalently, that thecompletion ofGwith respect tothe sequence {Hi} is uthe same" asthecompletion with respecttothefullfamily.We leave this verification tothereader. The process ofcompletion isfrequent inmathematics. Forinstance, weshall mention completions ofrings inChapter III, 10;and inChapter XII weshall deal with completions offields. 11. CATEGORIES AND FUNCTORS Before proceeding further, itwill now beconvenient tointroduce some new terminology. We have met already several kinds ofobjects: sets, monoids, groups. Weshall meet many more, and foreach such kind ofobjectswedefine special kinds ofmaps between them (e.g. homomorphisms). Some formal behavior will becommon toallofthese, namely theexistence ofidentity maps ofanobject onto itself, and theassociativity ofmaps when such mapsoccur in succession. We introduce thenotion ofcategory togiveageneral setting forall ofthese. AcategoryCIconsists ofacollection ofobjects Ob(Ci); and fortwoobjects A,BEOb(CI)asetMor(A, B)called the setofmorphismsofAinto B;and for three objects A,B,CEOb(Ci)alawofcomposition (i.e.amap) Mor(B, C)xMor(A, B)-+Mor(A, C) satisfying thefollowing axioms: CAT 1.Two setsMor(A, B)andMor(A', B') aredisjoint unless A=A' and B=B',inwhich case theyareequal. CAT 2. For each object AofCIthere isamorphism idAEMor(A, A) which acts asleft andright identity forthe elements ofMor(A, B)and Mor(B, A)respectively, forallobjects BEOb(Ci). CAT 3. The law ofcomposition isassociative (when defined), i.e.given IEMor(A, B),gEMor(B, C)and hEMor(C, D)then (h0g)0I=h0(gof), forallobjects A,B,C,DofCI. Here wewrite thecomposition ofanelement ginMor(B, C)and anelement finMor(A, B)asg0f,tosuggest composition ofmappings. Inpractice, inthis book weshall seethat most ofourmorphismsareactually mappings,orclosely related tomappings. The collection ofallmorphisms inacategory Ciwill bedenoted byAr(CI) ("arrows ofCi"). We shall sometimes use thesymbols "IEAr(CI)" tomean 54 GROUPS I,11 thatfisamorphism of(1,i.e. anelement ofsome setMor(A, B)for some A,BEOb(Ci). Byabuse oflanguage, wesometimes refer tothecollection ofobjectsasthe category itself, ifitisclear what themorphismsare meant tobe. AnelementfE Mor(A, B)isalsowrittenf:A-.Bor fA-.B. Amorphism fiscalled anisomorphism ifthere exists amorphism g:B-+A such that g0fandfogaretheidentities inMor(A, A)andMor(B, B)respec- tively. IfA=B,then wealso saythat theisomorphism isanautomorphism. Amorphism ofanobject Ainto itself iscalled anendomorphism. The setof endomorph isms ofAisdenoted byEnd(A). Itfollows atonce from ouraxioms thatEnd(A) isamonoid. Let Abeanobject ofacategoryCi.Wedenote byAut(A) the setofauto- morphisms ofA.This setisinfact agroup, because allofourdefinitions are soadjusted soastoseeimmediately that thegroup axioms aresatisfied (associa- tivity, unit element, and existence ofinverse). Thus we now begin tosee some feedback between abstract categories and more concrete ones. Examples. Let Sbethe category whose objectsare sets, and whose morphisms aremaps between sets. We saysimply that Sisthecategory ofsets. The three axioms CAT 1,2,3aretrivially satisfied. LetGrp bethecategory ofgroups, i.e.thecategory whose objects aregroups and whose morphisms aregroup-homomorphisms. Here again thethree axioms aretrivially satisfied. Similarly,wehave acategory ofmonoids, denoted by Mon. Later, when wedefine rings and modules, itwill beclear thatrings form a category, and sodomodules over aring. Itisimportant toemphasize here that there arecategories forwhich the set ofmorphisms isnot anabelian group. Some ofthe most important examples are: The category eO, whose objectsareopen sets inRnand whose morphisms arecontinuous maps. The category exwith the same objects, but whose morphisms are the Coo maps. The category 801, whose objectsareopensets inen, and whose morphisms areholomorphic maps. Ineach case theaxioms ofacategoryareverified, because forinstance for801, thecomposite ofholomorphic maps isholomorphic, and similarly fortheother types ofmaps. Thus aCD-isomorphism isacontinuous map!: u Vwhich has acontinuous inverse g:V U.Note that amap may be aCD-isomorphism but not aCoo-isomorphism. Forinstance, x x3isaCo- automorphism ofR,butitsinverse isnotdifferentiable. Inmathematics one studies manifolds inanyone ofthe above categories. The determination ofthegroup ofautomorphisms ineach category isoneofthe basic problems ofthe area ofmathematics concerned with that category. In I,11 CATEGORIES AND FUNCTORS 55 complex analysis, onedetermines early thegroup ofholomorphic automorphisms oftheunit disc asthegroup ofallmaps .c-z z e'B _ 1-cz with ()real and cEC, IcI<1. Next weconsider thenotion ofoperation incategories. First, observe that ifGisagroup, then theG-sets form acategory, whose morphismsarethemaps f:S S'such thatf(xs)=xf(s) for xEGand sES. More generally, we can now define thenotion ofanoperation ofagroup G on anobject inany category. Indeed, let CIbe acategory and AEOb(CI). Byanoperation ofGonAweshall mean ahomomorphism ofGinto thegroup Aut(A). Inpractice, anobject Aisasetwith elements, and anautomorphism inAut(A) operates onAasaset, i.e.induces apermutation ofA.Thus, ifwe have ahomomorphism p:G Aut(A), then foreach xEGwehave anautomorphism p(x) ofAwhich isapermutation ofA. Anoperation ofagroup Gon anobject Aisalso called arepresentation of GonA,and one then says that Gisrepresentedasagroup ofautomorphisms ofA. Examples. One meets representations inmany contexts. Inthisbook, we shall encounter representations ofagrouponfinite-dimensional vector spaces, with thetheory pushedtosome depth inChapter XVIII. Weshall also deal with representations ofagrouponmodules over aring. Intopology anddifferential geometry,one represents groupsasactingonvarious topological spaces, for instance spheres. Thus ifXisadifferential manifold, oratopological manifold, and Gisagroup,one considers allpossible homomorphims ofGintoAut(X), where Aut refers towhatever category isbeing dealt with. Thus Gmay be represented inthegroup ofCO-automorphims,orCoo-automorphisms,oranalytic automorphisms. Such topological theories are notindependent ofthealgebraic theories, because byfunctoriality,anaction ofGonthemanifold induces an action onvarious algebraic functors (homology, K-functor, whatever), sothat topologicalordifferential problems aretosome extent analyzable bythefunctorial action ontheassociated groups, vector spaces,ormodules. LetA,Bbeobjects ofacategory Cl.LetIso(A, B)bethe setofisomorphisms of'A with B.Then the group Aut(B) operatesonIso(A, B)bycomposition; namely, ifuEIso(A, B)and vEAut(B), then (v,u) v0ugives theoperation. IfUoisone element ofIso(A, B), then theorbit ofUoisallofIso(A, B), so v v0Uoisabijection Aut(B) Iso(A, B).The inverse mapping isgiven by u.-+UoUo1.This trivial formalism isvery basic, and isapplied constantly to each oneoftheclassical categories mentioned above. Ofcourse, wealso have 56 GROUPS I,11 asimilar bijectionontheother side, butthegroup Aut(A) operatesontheright ofIso(A, B)bycomposition. Furthermore, ifu:A Bisanisomorphism, then Aut(A) andAut(B) areisomorphic under conjugation, namely w uwu-lisanisomorphism Aut(A) Aut(B). Two such isomorphisms differ byaninner automorphism. One may visualize this system viathefollowing commutative diagram. uB !uwu-I B UA w! A Let p:G Aut(A) andp': G Aut(A') berepresentations ofagroup G ontwo objects AandA'inthe same category. Amorphism ofpinto p'isa morphism h:A A'such that thefollowing diagram iscommutative forall xEG: hA' !P'(x) A'A P(x)! A h Itisthen clear thatrepresentations ofagroup Gintheobjects ofacategoryC1 themselves form acategory. Anisomorphism ofrepresentations isthen an isomorphism h:A--+A'making theabove diagram commutative. Anisomor- phism ofrepresentations isoften called anequivalence, butIdon't like totamper with thegeneral system ofcategorical terminology. Note thatifhisanisomor- phism ofrepresentations, then instead oftheabove commutative diagram,we let[h]beconjugation byh,and wemayuse theequivalent diagram YAut(A) GP ![hJ Aut(A ') Consider next the case where C1isthecategory ofabelian groups, which we may denote byAb. LetAbeanabelian group and Gagroup. Given anoperation ofGontheabelian group A,Le. ahomomorphism p:G Aut(A), let usdenote byx·atheelement Px(a).Then we seethat forallx,yEG,a, bEA, wehave: I, 11 CATEGORIES AND FUNCTORS 57 e.a=a,x.(a +b)=x.a +x.b, x.0=O.x.(y.a)=(xy).a, Weobserve that when agroup Goperatesonitself byconjugation, then not only does Goperateonitself asasetbutalso operatesonitself asanobject inthe category ofgroups, i.e.thepermutations induced bytheoperationareactually group-automorphisms. Similarly, weshall introduce later other categories (rings, modules, fields) and wehave givenageneral definition ofwhat itmeans for agroup tooperate on anobject inanyone ofthese categories. Let CIbe acategory. We may take asobjects ofanew category ethe morphisms ofCI.Iff:A-.Band f':A'-.B'are twomorphisms inCI(and thus objects ofe),then wedefine amorphism f-.f'(ine)tobe apair of morphisms (qJ, 1/1)inCImaking thefollowing diagram commutative: AfB jj AIB' f' Inthat way, itisclear that eisacategory. Strictly speaking,aswith maps of sets, weshould index (qJ, 1/1)byfandf'(otherwise CAT 1isnotnecessarily satisfied), but such indexing isomitted inpractice. There aremany variations onthisexample. For instance, wecould restrict ourattention tomorphisms inCIwhich have afixed object ofdeparture,orthose which have afixed object ofarrival. Thus letAbeanobject ofCI,and letCIAbethecategory whose objectsare morphisms f:X-.A inCI,having Aasobject ofarrival. Amorphism inCIAfromf:X-.Ato g:Y-.Aissimplyamorphism h:X-.Y inCIsuch that thediagram iscommutative: Xh) Y\}A Universal objects Letebeacategory. An 0bject Pofeiscalled universally attracting ifthere exists aunique morphism ofeach object ofeinto P,and iscalled universally repelling ifforevery object ofethere exists aunique morphism ofPinto this object. 58 GROUPS I,11 When thecontext makes ourmeaning clear, weshall callobjects Pasabove universal. Since auniversal object Padmits theidentity morphism into itself, itisclear thatifP,P'aretwo universal objects ine,then there exists aunique isomorphism between them. Examples. Note that thetrivial group consisting only ofone element is universal (repelling and attracting) inthecategory ofgroups. Similarly, in Chapter lIon rings, youwill seethat theintegers Zareuniversal inthecategory ofrings (universally repelling). Next letSbeaset. Letebethecategory whose objects aremapsf:S A ofSinto abelian groups, and whose morphismsare the obvious ones: If f:S Aandf':S A'aretwo maps into abelian groups, then amorphism offintof'isa(group) homomorphism g:A A'such that theusual dia- gram iscommutative, namely 90f=f'.Then thefree abelian group generated bySisuniversal inthis category. This isareformulation ofthepropertieswe have proved about this group. LetMbe acommutative monoid and lety:M K(M) bethecanonical homomorphism ofMinto itsGrothendieck group. Then yisuniversal inthe category ofhomomorphisms ofMinto abelian groups. Throughout this book innumerous situtaions, wedefine universal objects. Aside from products andcoproducts which come immediately after these exam- ples,wehave direct and inverse limits; thetensor. product inChapter XVI, 1; thealternating product inChapter XIX, 1;Clifford algebras inChapter XIX, 4;adlib. We now turn tothenotion ofproduct inanarbitrary category. Products and coproducts Let C1beacategory and letA,Bbeobjects ofC1.Byaproduct ofA,BinC1 one means atriple (P,f, g)consisting ofanobject PinC1and twomorphisms P/ A B satisfying thefollowing condition: Given twomorphisms qJ:C Aand t/J:C-+B inCi,there exists aunique morphism h:C Pwhich makes thefollowing diagram commutative: c qJ/h\", Ik"fp A B Inother words, qJ=f0hand t/J=g0h. I,11 CATEGORIES AND FUNCTORS 59 More generally, givenafamily ofobjects {AJiel inC1, aproduct forthis family consists of(P,{};}iel)' where Pisanobject in C1and{};}iel isa family ofmorphisms Ii:P-+Ah satisfying thefollowing condition: Given afamily ofmorphisms gi:C-+Ah there exists aunique morphism h:C-+Psuch thatIi0h=giforalli. Example. Let Cibethecategory ofsets, and let{AihEI be afamily ofsets. Let A=OA ibetheir cartesian product, and letPi:A Aibetheprojection;EI onthei-th factor. Then (A,{Pi}) clearly satisfies therequirements ofaproduct inthecategory ofsets. As amatter ofnotation, weshall usually write AxBfortheproduct oftwo objects inacategory, andnAifortheproduct ofanarbitrary family ina ieI category, following the same notation asinthecategory ofsets. Example. Let{GihEI beafamily ofgroups, and letG=0Gibetheir direct product. LetPi:G Gibetheprojection homomorphism. Then these constitute aproduct ofthefamily inthecategory of-groups. Indeed, if{gi:G' -+Gi}iel isafamily' ofhomomorphisms, there isaunique homomorphism g:G' -+nGiwhich makes therequired diagram commutative. Itisthehomomorphism such that g(X')i=gi(X') forx'EG'and each iEI. LetA,Bbeobjects of acategory Ci.We note that theproduct ofA,Bis universal inthe category whose objects consist ofpairs ofmorphisms f:C Aand g:C BinCi,and whose morphisms aredescribed asfollows. Letf':C' Aandg':C' Bbeanother pair. Then amorphism from the first pairtothe second isamorphism h:C C'inC1,making thefollowing diagram commutative: C/lAB The situation issimilar fortheproduct ofafamily {AihE/. Weshall also meet thedual notion: Let{Adiel beafamily ofobjects ina category (t.Bytheir coproduct one means apair (S,{hLeI) consisting ofan object Sand afamily ofmorphisms {Ii:Ai-+S}, satisfying thefollowing property. Given afamily ofmorphisms {gi:Ai-+C}, there exists aunique morphism h:S-+Csuch that ho/;=giforalli. 60 GROUPS I,11 Intheproduct and coproduct, themorphism hwill besaid tobethe morphism induced bythefamily {gi}. Examples. Let Sbethecategory ofsets. Then coproducts exist. For instance, letS,S'besets. Let Tbeasethaving the same cardinalityasS'and disjoint from S.Letfl:S-.Sbetheidentity, andf2:S'-.Tbe abijection. Let Ubetheunion ofSand T.Then (U,fl,f2)isacoproduct forS,S',viewing fbf2asmaps into U. Let Sobethecategory ofpointed sets. Itsobjects consist ofpairs (S,x) where Sisasetand xisanelement ofS.Amorphism of(S,x)into (S',x')inthis category isamap g:S-.S'such that g(x)=x'.Then thecoproduct of(S,x) and(S',x')exists inthis category, and can beconstructed asfollows. LetTbe asetwhose cardinality isthe same asthat ofS', and such that TnS={x}. Let V=SUT,and let il:(S,x)-.(U,x) bethemap which induces theidentity onS.Let f2:(S',x')-.(U,x) be amap sending x'toxandinducingabijection ofS'-{x'} onT-{x}. Then thetriple «V,x),f},f2) isacoproduct for(S,x)and(S',x')inthecategory ofpointed sets. Similar constructions can bemade forthecoproduct ofarbitrary families ofsets orpointed sets. The category ofpointed sets isespecially important in homotopy theory. Coproductsareuniversal objects. Indeed, let C1be acategory, and let{Ai} be afamily ofobjects ind.We now define e .Weletobjects ofebethefamilies ofmorphisms {/;:Ai BhEI andgiven two such families, {h:Ai-.B} and{f:Ai-.B'}, wedefine amorphism from thefirst into thesecond tobeamorphism qJ:B-.B' inC1such thatqJ0h=fforalli.Then acoproduct of{Ai} issimplyauniversal object ine. Thecoproduct of{Ai} will bedenoted by UAi. iel The coproduct oftwoobjects A,Bwill also bedenoted byAIIB. Bythegeneral uniqueness statement, we seethat itisuniquely determined, up toaunique isomorphism. Example. Let Rbethecategory ofcommutative rings. Given two such rings A,Bone may form the tensor product, and there arenatural ring-homo- morphisms A A0Band B A0Bsuch that a a0 1and b 10bfor aEAand bEB. Then the tensor product isacoproduct inthecategory ofcommutative rings. I,11 CATEGORIES AND FUNCTORS 61 Fiber products and coproducts Pull-backs and push-outs Letebeacategory. LetZbeanobject ofe.Then wehave anew category, that ofobjects over Z,denoted byez.Theobjects ofezaremorphisms: f:X Zine Amorphism fromftog:Y Zinezismerely amorphism h:X Yine which makes thefollowing diagram commutative. Xh) Y\1 Z Aproduct ineziscalled thefiber product offand gineand isdenoted byXxzY,together with itsnatural morphisms onX,Yover Z,which are sometimes notdenoted byanything, butwhich wedenote byPI'P2' XXzYyX Y Z fibered products andcoproducts exist in'thecategory ofabelian groups The fibered product oftwohomomorphisms f:X Zand g:Y Zisthe subgroup ofXxYconsisting ofallpairs (x,y)such that f{x)=g{y). The coproduct oftwo homomorphisms f:Z Xand g:Z Yisthe factor group (XffiY)/W where Wisthesubgroup ofX(f)Yconsisting ofall elements (f{z), -g{z)) with zEZ. We leave thesimple verification tothereader (see Exercises 50-56). Inthefiber product diagram, one also calls PIthepull-back ofgbyf,and P2thepull-back offbyg.The fiber product satisfies thefollowing universal mapping property: Given any object Tineand morphisms making thefollowing diagram commutative: /TX Y Z 62 GROUPS I,11 there exists aunique morphism T XxzYmaking thefollowing diagram commutative: )1\ X+--T------. Y Dually, wehave thenotion ofcoproduct inthecategory ofmorphism sf:Z-+X with afixed object Zastheobject ofdeparture ofthemorphisms. This category could bedenoted byez .We reverse the arrows inthepreceding discussion. Given twoobjects fand g:Z-+Yinthiscategory,wehave thenotion oftheir coproduct. Itisdenoted byXUzY,with morphisms ql'Q2,asinthefollowing commutative diagram: X11YyX Y/z satisfying thedual universal property ofthefiber product. Wecall itthefibered coproduct. We call qlthepush-out ofgbyf,andq2thepush-out offbyg. Example. Let Sbethecategory ofsets. Given two maps f,gasabove, their product isthe setofallpairs (x,y)EXXYsuch thatf(x)=g(y). Functors Let C1,CBbecategories. Acovariant functor FofC1into CBisarule which toeach object AinC1associates anobject F{A) inCB,and toeach morphism f:A-+Bassociates amorphism F{f): F{A)-+F{B) such that: FUN 1.For allAinC1wehave F{id A)=idF(A). FUN 2.Iff: A-+Band g:B-+Caretwomorphisms ofC1then F{g0f)=F{g)0F{f). Example. Iftoeach group Gweassociate itsset(stripped ofthegroup structure) weobtain afunctor from thecategory ofgroups into thecategory of sets, provided that weassociate with each group-homomorphism itself, viewed onlyas aset-theoretic map. Such afunctor iscalled astripping functor or forgetful functor. We observe that afunctor transforms isomorphisms into isomorphisms, because fog=idimplies F{f)0F(g)=idalso. We can define thenotion ofacontravariant functor from C1into CBbyusing essentially the same definition, butreversing allarrows F(f), i.e. toeach morph- ismf:A-+Bthecontravariant functor associates amorphism I,11 CATEGORIES AND FUNCTORS 63 F{f):F{B)-+F{A) (going intheopposite direction), such that, if f:A-+Band g:B-+C aremorphisms in(1,then F{g0f)=F{f)0F{g). Sometimes afunctor isdenoted bywriting f*instead ofF{f) inthe case of acovariant functor, and bywriting f* inthe case of acontravariant functor. Example. The association S Fab(S) isacovariant functor from the category ofsets tothecategory ofabelian groups. Example. The association which toeach group associates itscompletion with respect tothefamily ofsubgroups offinite index isafunctor from the category ofgroups tothecategory ofgroups. Example. Letpbe aprime number. Letebethecategory ofp-divisible abelian groups. The association ATp(A)isacovariant functor ofeinto abelian groups (actually Zp-modules). Example. Exercise 49will show youanexample ofthe group ofauto- morphisms ofaforgetful functor. Example. LetMan bethecategory ofcompact manifolds. Then thehomol- ogy isacovariant functor from Man intograded abelian groups. Thecohomology isacontravariant functor into thecategory ofgraded algebras (over thering of coefficients). The product isthecupproduct. Ifthecohomology istaken with coefficients inafield ofcharacteristic 0(forsimplicity), then thecohomology commutes with products. Since cohomology iscontravariant, this means that the cohomology ofaproduct isthecoproduct ofthecohomology ofthefactors. It turns outthat thecoproduct isthetensor product, with thegraded product, which also givesanexample ofthe useoftensor products. See M. GREENBERG and J.HARPER, Algebraic Topology (Benjamin-Addison- Wesley), 1981, Chapter 29. Example. Letebethecategory ofpointed topological spaces (satisfying some mild conditions), Le.pairs (X,xo)consisting ofaspace Xand apoint Xo. Intopologyone defines theconnected sum ofsuch spaces (X,xo) and (Y,Yo), glueing X,Ytogether attheselected point. This connected sum isacoproduct inthecategory ofsuch pairs, where themorphismsare thecontinuous maps f:X Ysuch thatf(xo)=Yo.Let7Tldenote thefundamental group. Then (X,xo) 7Tl(X,xo)isacovariant functor from einto thecategory ofgroups, commuting with coproducts. (The existence ofcoproducts inthecategory of groups will beproved in12.) 64 GROUPS I,11 Example. Supposewehave amorphism I:X Yinacategory e.Bya section ofI,one means amorphism g:Y Xsuch that 90f=ideSuppose there exists acovariant functor Hfrom this category togroups such that H(Y)={e} andH(X)=f=.{e}. Then there isnosection ofI.This isimmediate from theformula H(g0I)=id,andH(/)=trivial homomorphism. Intopology one uses thehomology functor toshow, forinstance, that theunit circle Xis not aretract oftheclosed unit disc with respect totheinclusion mapping I. (Topologistsuse theword "retract" instead of"section".) Example. Let C1beacategory and Aafixed object inC1.Then weobtain a covariant functor MA:C1-+S byletting MA(X)=Mor(A, X)foranyobject XofC1.IflfJ:X-+X'isamor- phism, welet MA(lfJ): Mor(A, X)-+Mor(A, X') bethemap given bytherule glfJog forany gEMor(A, X), A X X'. The axioms FUN 1and FUN 2aretrivially verified. Similarly, foreach object BofC1,wehave acontravariant functor MB :C1-+S such that MB(Y)=Mor( Y,B).Ift/J:Y' -+Yisamorphism, then MB(t/J): Mor(Y, B)-+Mor(Y', B) isthemap given bytherule ff°t/J foranyIEMor(Y, B), Y' y1.B. Thepreceding two functors arecalled therepresentation functors. Example. Let C1bethecategory ofabelian groups. Fix anabelian group A.The association X Hom(A, X)isacovariant functor from ainto itself. The association X Hom(X, A)isacontravariant functor ofC1into itself. Example. We assume you know about the tensor product. Let Abe a commutative ring. LetMbe anA-module. The association X M0Xisa covariant functor from thecategory ofA-modules into itself. Observe thatproducts andcoproductswere defined inawaycompatible with therepresentation functor into thecategory ofsets. Indeed, givenaproductP I,11 CATEGORIES AND FUNCTORS 65 oftwoobjects AandB,then forevery object Xthe setMor(X, P)isaproduct ofthe sets Mor(X, A)andMor(X, B)inthecategory ofsets. This ismerelya reformulation ofthedefining property ofproducts inarbitrary categories.The system really works. Let ct,<Bbetwocategories. The functors of C1into <B(say covariant, and inonevariable) can beviewed astheobjects ofacategory, whose morphisms aredefined asfollows. LetL,Mbetwo such functors. Amorphism H:L M (also called anatural transformation) isarule which toeach object Xofct associates amorphism Hx:L(X) M(X) such that foranymorphism f:X Ythefollowing diagram iscommutative: L(X) L(f) ! L(Y)Hx)M(X) !M(f) )M(Y) Hy We can therefore speak ofisomorphisms offunctors. Afunctor isrepresentable ifitisisomorphic toarepresentation functor asabove. AsGrothendieck pointed out, one can use therepresentation functor to transport thenotions ofcertain structures onsets toarbitrary categories. For instance, let C1be acategory and Ganobject ofC1.We saythat Gisagroup object inC1ifforeach object XofC1we aregivenagroup structure onthe set Mor(X, G)insuch away that theassociation X Mor(X, G) isfunctorial (i.e. isafunctor from C1into thecategory ofgroups). One some- times denotes the setMor(X, G)byG(X), and thinks ofitasthe setofpoits of GinX.Tojustify thisterminology, thereader isreferred toChapter IX,2. Example. LetVar bethecategory ofprojective non-singular varieties over thecomplex numbers. Toeach object XinVar one can associate various groups, e.g. Pic(X) (the group ofdivisor classes forrational equivalence), which isa contravariant functor into thecategory ofabelian groups. LetPico(X)bethe subgroup ofclasses algebraically equivalent toO.Then Pico isrepresentable. Inthefifties and sixties Grothendieck was the one who emphasized the importance oftherepresentation functors, and thepossibility oftransposing to anycategory notions from more standard categories bymeans oftherepresentation functors. Hehimself proved that anumber ofimportant functors inalgebraic geometryarerepresentable. 66 GROUPS I,12 12. FREE GROUPS We now turn tothecoproduct inthecategory ofgroups. First aremark. Let G=nGibe adirect product ofgroups. We observe that each Gjadmits aninjective homomorphism into the product, onthej-th component, namely the map Aj:Gj-.nGisuch that i for xinGj,thei-thcomponent ofAj(X)istheunit element ofGiifi=Fj,and isequal toxitself ifi=j.This embedding will becalled thecanonical one. But westill don't have acoproduct ofthefamily, because thefactors commute with each other. Togetacoproductone has towork somewhat harder. Let Gbe agroup and Sasubset ofG.We recall that Gisgenerated byS ifevery element ofGcan bewritten asafinite product ofelements ofSandtheir inverses (the empty product being always taken astheunit element ofG). Elements ofSarethen called generators. Ifthere exists afinite setofgenerators for Gwecall Gfinitely generated. IfSisasetand qJ:S-.Gisamap, wesay that qJgenerates Gifitsimage generates G. Let Sbeaset,and/: S-.Famap into agroup. Letg:S-.Gbeanother map. Iff(S) (or aswealsosay,f) generates F,then itisobvious that there exists atmost onehomomorphism t/JofFinto Gwhich makes thefollowing diagram commutative: Sf)F\) G We now consider thecategory ewhose objectsare the maps ofSinto groups. Iff: S-.Gandf':S-.G'are two objects inthis category, wedefine amorphism fromfto f'tobeahomomorphism qJ:G-.G'such thatqJ0f=f', i.e.thediagram iscommutative: G js'" G' Byafree group determined byS,weshall mean auniversal element inthis category. Proposition 12.1. Let Sbe aset. Then there exists afree group (F,f) determined byS.Furthermore, fisinjective, and Fisgenerated bytheimage off. Proof (Iowe thisproof toJ.Tits.) Webegin with alemma. I,12 FREE GROUPS 67 Lemma 12.2. There exists asetIand afamily ofgroups {G;hEI such that, ifg:S Gisamap ofSinto agroup G,and 9generates G,then Gis isomorphic tosome G;. Proof This isasimple exercise incardinalities, which wecarry out. IfS isfinite, then Gisfinite ordenumerable. IfSisinfinite, then thecardinality ofG is<thecardinality ofSbecause Gconsists offinite products ofelements ofg(S). LetTbeasetwhich isinfinite denumerable ifSisfinite, and hasthe same cardin- alityasSifSisinfinite. For each non-empty subset HofT,letrHbethe setof group structures onH.For each yErH,letHybethe setH,together with the group structure y.Then thefamily {Hy}foryErHand Hrangingover subsets ofTisthedesired family. We return totheproof oftheproposition. For each iEIweletMibethe setofmappings ofSinto Gi.For each map ({JEMi,weletGi,qJbethe set- theoretic product ofG;and the setwith one element {qJ},sothatG;, qJisthe "same" groupasG;indexed by qJ.We let Fo=nnG;,qJ iel qJeMi betheCartesian product ofthegroups Gi,qJ.Wedefine amap 10:S-+F0 bysending Sonthefactor Gi,qJbymeans ofqJitself. Wecontend that givena map g:S GofSinto agroup G,there exists ahomomorphism t/!*: F0 G making theusual diagram commutative: Fo j* G That is,t/!.0fo=g.Toprove this, wemayassume that 9generates G,simply byrestricting ourattention tothesubgroup ofGgenerated bytheimage ofg. Bythelemma, there exists anisomorphism A.:G-+G;for some i,and A.0g isanelement t/JofMi. We letni,'"betheprojectiononthe(i,t/J)factor, and we lett/J.=A.-10n;,",.Then the map t/1.makes thefollowing diagramcom- mutative. gI)1:;. GA)G;,,,, We letFbethesubgroup ofFogenerated bytheimage of/ o,and weletI simply beequal to10'viewed asamap ofSinto F.We letg.betherestriction oft/J.toF.Inthis way, we seeatonce that themap g.istheunique onemaking 68 GROUPS I,12 ourdiagram commutative, and thus that(F,f) istherequired free group. Furthermore, itisclear thatfisinjective. For each set Sweselect one free group determined byS,and denote it by(F(s),ls)orbriefly byF(S). Itisgenerated bytheimage offs. One may view Sascontained inF(S), and theelements ofSarecalled free generators ofF(S). Ifg:S Gisamap,wedenote byg.:F(S) Gthehomomorphism realizing theuniversality ofour free group F(S). IfA:S S'isamap ofone setinto another, weletF(A):F(S) F(S') be themap (fs'0A).. SIs)F(S) Al IA.=F(A) S')F'(S')Is' Then wemay regard Fasafunctor from thecategory ofsets tothecategory of groups (the functorial propertiesaretrivially verified, and will beleft tothe reader). IfA.issurjective, then F(A.) isalso surjective. Weagain leave theproof tothereader. Iftwo sets S,S'have the same cardinality, then they areisomorphic inthe category ofsets (anisomorphism being inthis case abijection !),and hence F(S) isisomorphic toF(S'). IfShas nelements, wecallF(S) thefree group onngenerators. Let Gbeagroup, and letSbethe same set asG(i.e. Gviewed asaset,without group structure). We have theidentity map g:S G,and hence asurjective homomorphism g.:F(S) G which will becalled canonical. Thus every group isafactor group ofafree group. One canalso construct groups bywhat iscalled generators andrelations. Let Sbe aset, and F(S) the free group. We assume thatf:S F(S) isanin- clusion. Let Rbeasetofelements ofF(S). Each element ofRcan bewritten asafinite product n UXv v= 1 where each Xvisanelement ofSoraninverse ofanelement ofS.LetNbethe smallest normal subgroup ofF(S)containing R,i.e.theintersection ofallnormal subgroups ofF(S) containing R.Then F(S)/N will becalled thegroup deter- mined bythegenerators Sand therelations R. I,12 FREE GROUPS 69 Example. One shows easily that thegroup determined byone generator a,and therelation {a2},hasorder 2. The canonical homomorphism cp:F(S) F(S)/Nsatisfies theuniversal map- ping property forhomomorphisms t/JofF(S) into groups Gsuch that t/J(x)=e forallxER.Inview ofthis, one sometimes calls thegroup F(S)/N thegroup determined bythegenerators S,and therelations x=e(for allxER). For instance, thegroup inthepreceding example would becalled thegroup determined bythegenerator a,and therelation a2=e. Let Gbe agroup generated byafinite number ofelements, andsatisfying therelation x2=eforallxEG.What does Glook like? Itiseasy toshow that Giscommutative. Then one can view Gasavector spaceoverZ/2Z, soGis determined byitscardinality, uptoisomorphism. InExercises 34and35,youwill prove that there exist certain groups satisfying certain relations and with agiven order, sothat thegroup presented with these generators and relations can becompletely determined. Apriori, itisnot even clear ifagroup given bygenerators and relations isfinite. Even ifitisfinite, one does not know itsorder apriori. Toshow that agroup ofcertain order exists, one has touse various means, acommon means being torepresent the groupasagroup ofautomorphisms ofsome object, forinstance thesymmetries ofageometric object. This will bethemethod suggested forthegroups inExercises 34and 35, mentioned above. Example. Let Gbe agroup. For x,yEGdefine [x,y]=xyx-1y-1 (the commutator) and Xy=xyx-l(theconjugate). Then one has thecocycle relation [x,yz]=[x,y]Y[x, z]. Furthermore, suppose x,y,ZEGand [x,y]=y,[y,z]=Z, [z,x]=x. Then x=y=z=e.Itisanexercise toprove these assertions, but one sees that certain relations imply that agroup generated byx,y,zsubject tothose relations isnecessarily trivial. Next wegiveasomewhat more sophisticated example. We assume that the reader knows thebasic terminology offields and matrices asinChapter XIII, butapplied only to2x2matrices. Thus SL2(F) denotes thegroup of2x2 matrices with components inafield Fand determinant equal to1. Example. SL2(F). LetFbe afield. For bEFand aEF,a=t=0,welet u(b)=(),s(a)=(_}and w=(_). 70 GROUPS I,12 Then itisimmediately verified that: SL O.s(a)=wu(a-l)wu(a)wu(a-l). SL 1. uisanadditive homomorphism. SL2. sisamultiplicative homomorphism. SL3. w2=S(-1). SL4.s(a)u(b)s(a-l)=u(ba2). Now, conversely, suppose that Gisanarbitrary group with generators u(b) (bEF)and w,such thatifwedefine s(a) for a=t=0bySL0,then therelations SL 1through SL4aresatisfied. Then SL3and SL4show that s(-1)isinthe center, and w4=e.Inaddition, one verifies that: SL5.ws(a)=s(a-l)w. Furthermore, one has thetheorem: Let Gbethefree group with generators u(b), wand relations SL 1through SL4,defining s(a) asinSL O.Then thenatural homomorphism G SL2(F) isanisomorphism. Proofs ofalltheabove statements will befound inmySL2(R), Springer Verlag, reprint ofAddison-Wesley, 1975, Chapter XI,2.Ittakes about apage tocarry out theproof. IfF=Qpisthefield ofp-adic numbers, then Ihara [Ih66]proved that every discrete torsion free subgroup ofSL2(Qp)isfree. Serre putthis theorem inthe context of ageneral theory concerning groups actingontrees [Se80]. [lh66] Y,IHARA, Ondiscrete subgroups ofthetwobytwoprojective linear groupover p-adic fields, J.Math. Soc. Japan 18(1966) pp.219-235 [Se80] J.-P. SERRE, Trees, Springer Verlag 1980 Further examples. For further examples offree group constructions, see Exercises 54and 56. Forexamples offree groups occurring (possibly conjec- turally) inGalois theory,seeChapter VI,2,Example 9,and the end of Chapter VI, 14. Proposition 12.3. Coproducts exist inthecategory ofgroups. Proof Let{Gi}iel beafamily ofgroups. Weletebethecategory whose objects arefamilies ofgroup-homomorphisms {gi: GiG}iel I,12 FREE GROUPS 71 and whose morphismsaretheobvious ones. We must find auniversal element inthiscategory. For each index i,weletSibethe same set asGiifGiisinfinite, and weletSibedenumerable ifGiisfinite. We letSbeasethaving the same cardinalityastheset-theoretic disjoint union ofthe sets Si(i.e.their coproduct inthecategory ofsets). We letrbethe setofgroup structures onS,and for each YEr,welet<l>ybethe setofallfamilies ofhomomorphisms qJ={qJi: Gi-.Sy}. Each pair (Sy, qJ),whereqJE<l>y,isthen agroup, using qJmerelyasanindex. We let Fo=nn(Sy, qJ), yErqJEcDy and for each i,wedefine ahomomorphism /;:Gi-.F0byprescribing the component of/;oneach factor (Sy, qJ)tobethe same asthat ofqJi. Let now g={gi: Gi-.G}be afamily ofhomomorphisms. Replacing G ifnecessary bythesubgroup generated bytheimages ofthe gb we seethat card(G)<card(S), because each element ofGisafinite product ofelements inthese images. Embedding Gasafactor inaproduct GxSyfor some ')',we mayassume that card(G)=card(S). There exists ahomomorphism g.:Fo-+G such that g.0h=gi foralli.Indeed, wemayassume without lossofgenerality that G=Syfor some Yand that g=t/Jfor some t/JE<l>y.We letg.betheprojection ofF0onthe factor(SY' t/J). Let Fbethesubgroup ofF0generated bytheunion oftheimages of themaps /;foralli.The restriction ofg.toFistheunique homomorphism satisfying /;0g.=giforalli,and wehave thus constructed our universal object. Example. Let G2be acyclic group oforder 2and letG3be acyclic group oforder 3.What isthecoproduct? The answer isneat. Itcan beshown that G2UG3isthegroup generated bytwo elements S,Twith relations S2=1, (ST)3=1.The groups G2and G3areembedded inG2UG3bysending G2on thecyclic group generated bySandsending G3onthecyclic group generated byST. This isdone byrepresenting thegroupasfollows. Let G=SL2(Z)/+I. 72 GROUPS I,12 As wehave seen inanexample of5, thegroup Goperatesontheupper half- plane Sj.LetS,Tbethemaps given by S(z)=-1/z and T(z)=z+ 1. Thus Sand Tarerepresented bythematrices s=(-)and T=(), andsatisfy therelations SZ=1,(ST)3=1.Readers will find aproof ofseveral properties ofS,TinSerre' sCourse inArithmetic (Springer Verlag, 1973, Chapter VII, 1),including thefact that S,Tgenerate G.Itisanexercise from there to show that Gisthecoproduct ofGzand G3asasserted. Observe that these procedures godirectly from theuniversal definition and construction intheproofs ofProposition12. 1andProposition 12.3 tothe more explicit representation ofthefree grouporthecoproductasthe case may be. One relies onthefollowing proposition. Proposition 12.4. Let Gbe agroup and{GihEIafamily ofsubgroups. Assume: (a) Thefamily generates G. (b)If x=Xi,··.XinwithXi£lEGi£l'xi£l=t=eand iv=t=iv+ 1forall v, then x=1=e. Then thenatural homomorphism ofthecoproduct ofthefamily into Gsending Gionitself bytheidentity mapping isanisomorphism. Inother words, simply put, Gisthecoproduct ofthefamily ofsubgroups. Proof. Thehomomorphism from thecoproduct into Gissurjective bythe assumption that thefamily generates G.Supposeanelement isinthekernel. Then such anelement has arepresentation X....X.'I 'n asin(b),mapping totheidentity inG, soallXi£l=eand theelement itself is equal toe,whence thehomomorphism from thecoproduct into Gisinjective, thereby proving theproposition. Exercises 54and 56mentioned above giveone illustration oftheway Prop- osition 12.4 can beusd. We now show another way, which wecarry outfor twosubgroups. Iamindebted toEilenberg forthe neat arrangement oftheproof ofthe next proposition. I,12 FREE GROUPS 73 Proposition 12.5. LetA,Bbetwo groups whose set-theoretic intersection is {1}.There exists agroup A0Bcontaining A,Bassubgroups, such that AnB={I}, andhaving thefollowing property. Every element =t= 1ofA0B has aunique expressionasaproduct a1...an (n>1,ai=F 1alli) with aiEAoraiEB,and such thatifaiEAthen ai+1EBandifaiEBthen ai+l EA . Proof Let A0Bbethe setofsequences a=(al'. ..,an) (n>0) such that either n=0,and the sequence isemptyorn>1,and then elements inthesequence belong toAorB,are =F1,and two consecutive elements ofthe sequence donotbelong both toAorboth toB.Ifb=(b 1,...,bm),wedefine theproduct abtobethesequence (ab...,an,bb...,bm) if anEA,b1EB or anEB,b1EA, (ab...,anbl'...,bm) If an,b1EA or an,b1EB, and anb1=F1, (a1,...,an-1)(b 2,...,bm) byinduction, ifan,b1EA or an,b1EBand anb1=1. The case when n=0orm=0isincluded inthefirst case, and theempty sequence istheunit element ofA0B.Clearly, (a1'..., an)(a; 1,..., all)=unit element, soonly associativity need beproved. Let c=(cb...,cr). First consider the case m=0,i.e.bisempty. Then clearly (ab)c=a(bc) andsimilarly ifn=0orr=O.Next consider the case m=1.Let b=(x) with xEA,x=F1.We then verify ineach possiblecase that (ab)c=a(bc). These cases are asfollows: (ai'...,an,x,c1,...,cr) ifanEBand c1EB, ifanEA,anx=F I,C 1EB, if anEB,c1EA,xc1=F1, ifan=X-I and c1EB,(al'...,anX, Cb...,cr) (ai'...,an,XCI'...,Cr) (ai'...,an-1)(cl'...,Cr) 74 GROUPS I,912 (al'...,an){c2,...,Cr) ifanEBand-1 Cl=X, (al'...,an-hanXCbC2'...,cr) if an,c 1EA,a nxcl=F1, if an,ClEA and anxc1=1. (ab...,an-1){c2,...,Cr) Ifm>1,then weproceed byinduction. Write b=b'b"with b'and b" shorter. Then {ab)c={a{b'b"))c=({ab')b")c=(ab'){b"c), a{bc)=a{{b'b")c)=a{b'{b"c))=(ab'){b"c) aswas tobeshown. We have obvious injections ofAand Binto A0B,and identifying A,B with their images inA0Bweobtain aproof ofourproposition. We can prove thesimilar result forseveral factors. Inparticular, wegetthe following corollary forthefree group. Corollary 12.6. LetF(S) bethefree grouponasetS,and letx.,. . .,Xnbe distinct elements ofS.LetvI'..., Vrbeintegers=t=0and letiI'. . .,irbe integers, 1<. .<='b...,lr=n suchthat ij=Fij+1forj=1,..., r-1.Then VI Vr -J..1 Xi 1...Xir-r-. Proof Let Gb...,Gnbethecyclic groups generated byXb...,Xn. Let G=G10...0Gn.Let F{S)-+G bethehomomorphism sending each XionXi'and allother elements ofSonthe unit element ofG.Our assertion follows atonce. Corollary 12.7. Let Sbeasetwith nelements x.,. . .,Xn,n>1.Let GI, . . .,Gnbetheinfinite cyclic groups generated bythese elements. Then themap F{S)-+G10...0 Gn sending each Xionitself isanisomorphism. Proof Itisobviously surjective andinjective. Corollary 12.8. LetG.,...,Gnbegroups with G;nGj={I}ifi=t=j. Thehomomorphism G111...11Gn-+G1o...oGn oftheir coproduct into G10...0Gninduced by the natural inclusion Gi-+G10 ...0Gnisanisomorphism. Proof Again, itisobviously injective andsurjective. I,Ex EXERCISES 75 EXERCISES 1,Show that every group oforder <5isabehan. 2.Show that there are twonon-isomorphic groups oforder 4,namely thecyclic one, and theproduct oftwocyclic groups oforder 2. 3.Let Gbe agroup. Acommutator inGisanelement oftheform aba-1b-1with a, bEG. Let GCbethesubgroup generated bythecommutators. Then GCiscalled the commutator subgroup. Show that GCisnormal. Show that anyhomomorphism of Ginto anabelian group factors through G/GC . 4.LetH,Kbesubgroups ofafinite group Gwith KeNH.Show that #(H)# (K)#(HK)= #(H nK). 5.Goursat's Lemma. LetG,G'begroups, and letHbeasubgroup ofGxG'such that the twoprojections Pt:H Gand P2:H G'aresurjective. LetNbethekernel ofP2 and N'bethekernel ofPt.One canidentify Nasanormal subgroup ofG,and N'asa normal subgroup ofG'. Show that theimage ofHinGINxG'IN' isthegraph ofan isomorphism GIN G'IN', 6.Prove that thegroup ofinner automorphisms ofagroup Gisnormal inAut(G). 7,Let Gbe agroup such thatAut(G) iscyclic. Prove that Gisabelian. 8,Let Gbeagroup and letH,H'besubgroups. Byadouble coset ofH,H' one means asubset ofGoftheform HxH' . (a) Show that Gisadisjoint union ofdouble cosets. (b)Let{c} be afamily ofrepresentatives for the double cosets .For each aEGdenote by[a]H' theconjugate aH'a-IofH'. For each cwehave a decomposition intoordinarycosets H=Uxc(H n[c]H'), C where {xc} isafamily ofelements ofH,dependingon c.Show that the elements {xcc} form afamily ofleft coset representatives forH'inG;that IS, G=UUxccH', Xc Xc and theunion isdisjoint, (Double cosets will notemerge further until Chapter XVIII. ) 9.(a)Let Gbe agroup and Hasubgroup offinite index. Show that there exists a normal subgroup NofGcontained inHand also offinite index. [Hint: If (G:H)=n,find ahomomorphism ofGinto Snwhose kernel iscontained in H.] (b)Let Gbe agroup and letHI' H2besubgroups offinite index. Prove that HInH2has finite index. 10. Let Gbe agroup and letHbe asubgroup offinite index. Prove that there isonlya finite number ofrightcosetsofH,and that thenumber ofright cosets isequal tothe number ofleft cosets. 76 GROUPS I,Ex 11, Let Gbe agroup, and Aanormal abelian subgroup, Show that GIAoperatesonA byconjugation,and inthismanner.getahomomorphism ofGIA intoAut(A). Semidirect product 12. Let Gbe agroup and letH,Nbesubgroups with Nnormal. Let'Yxbeconjugation byanelement xEG. (a) Show that x'Yxinduces ahomomorphismf: H Aut(N). (b)IfHnN={e}, show that themap HxN HNgiven by(x,y) xyis abijection, and that this map isanisomorphism ifandonly iffistrivial, Le.f(x)=idNforallxEH. We define Gtobethesemidirect product ofHand NifG=NH andHnN={e}. (c)Conversely, letN,Hbegroups, and let.p:H Aut(N) be agiven homo- morphism. Construct asemidirect productasfollows. Let Gbethe setof pairs (x,h)with xENand hEH.Define thecomposition law (xI'hi)(X2,)=(x 1cf1(h.)x2,hih2). Show that this isagroup law, andyieldsasemidirect product ofNand H, identifying Nwith the setofelements (x,1)andHwith the setofelements (1,h), 13.(a) LetH,Nbenormal subgroups ofafinite group G.Assume that theorders ofH, Narerelatively prime. Prove that xy=yxforallxEHand yEN, and that HxN=HN, (b) LetHI', . .,Hrbenormal subgroups ofGsuch that theorder ofHiisrelatively primetotheorder ofHjfori=1=j.Prove that HIx ,., xHr=HI.. .Hr' Example. IftheSylow subgroups ofafinite grouparenormal, then Gisthe direct product ofitsSylow subgroups. 14, Let Gbe afinite group and letNbe anormal subgroup such that Nand GIN have relatively prime orders. (a)Let Hbe asubgroup ofGhaving the same order asGIN, Prove that G=HN, (b)Let 9beanautomorphism ofG,Prove thatg(N)=N, Some operations 15. Let Gbe afinite group operatingon afinite setSwith #(S)>2,Assume that there isonlyone orbit. Prove that there exists anelement xEGwhich has nofixed point, i.e. xs =1=sforall sES. 16. LetHbe aproper subgroup ofafinite group G,Show that Gisnottheunion ofall theconjugates ofH,(But seeExercise 23ofChapter XIII.) 17, LetX,Ybefinite sets and letCbe asubset ofXxY.For xEXletcp(x)=number ofelements yEYsuch that (x,y)EC.Verify that #(C)=Lcp(x). ../XEX I,Ex EXERCISES 77 Remark. Asubset Casinthe above exercise isoften called acorrespondence, and cp(x) isthenumber ofelements inYwhich correspond toagiven element xEX. 18. LetS,Tbefinite sets. Show that#Map(S, T)=(#T)#(S). 19. Let Gbe afinite group operatingon afinite setS. (a)For each sESshow that 2:I =1 lEGs#(Gt). (b)For each xEGdefine f(x)=number ofelements sESsuch that xs=s. Prove that thenumber oforbits ofGinSisequal to #(IG)x/(X). Throughout, pisaprime number. .. 20, LetPbe ap-group, LetAbe anormal subgroup oforder p.Prove that Aiscontained inthe center ofP, 21, Let Gbe afinite group andHasubgroup. LetPHbeap-Sylow subgroup ofH.Prove that there exists ap-Sylow subgroup PofGsuch that PH=PnH. 22. LetHbe anormal subgroup ofafinite group Gand assume that#(H)=p.Prove that Hiscontained inevery p-Sylow subgroup ofG. 23, LetP,P'bep-Sylow subgroups ofafinite group G, (a)IfP'CN(P) (normalizer ofP), then P'=P. (b)IfN(P')=N(P), then P'=P. (c) We have N(N(P»=N(P). Explicit determination ofgroups 24, Let pbeaprime number. Show that agroup oforder p2isabelian, and that there are only two such groups uptoisomorphism. 25, Let Gbeagroup oforder p3,where pISprime, and Gisnotabelian. LetZbeItScenter. Let Cbeacychc group oforder p. (a)Show that Z CandG/Z Cxc. (b)Every subgroup ofGoforder p2contaJns Zand isnormal. (c)Suppose xP=1forall xEG,Show that Gcontains anormal subgroup H Cxc. 26. (a)Let Gbe agroup oforder pq, where p,qareprimes and p<q.Assume that q=1= 1mod p.Prove that Giscyclic, (b) Show that every group oforder 15iscyclic. 27. Show that every group oforder <60issolvable. 28. Letp,qbedistinct primes, Prove that agroup oforder p2q issolvable, and that one ofitsSylow subgroups isnormal. 29, Letp,qbeoddprimes. Prove that agroup oforder 2pq issolvable. 78 GROUPS I,Ex 30. (a) Prove that oneoftheSylow subgroups ofagroup oforder 40isnormal. (b) Prove that oneoftheSylow subgroups ofagroup oforder 12isnormal. 31, Determine allgroups oforder-<1°uptoisomorphism. Inparticular, show that a non-abelian group oforder 6isisomorphictoS3' 32, Let5nbethepermutation group on nelements, Determine thep-Sylow subgroups of 53'54'55forp=2and p=3. 33. Let 0'be apermutation ofafinite setIhavingnelements, Define e(0')tobe(-I)m where m=n-number oforbits of(J. If11'...,Iraretheorbits of(J,then misalso equal tothe sum r m=L[card(l J-1]. \'=1 IfTisatransposition, show that e(O'T)= -e(0')beconsidering thetwo cases when i,jlieinthe same orbit of0',orlieindifferent orbits. Inthefirst case, O'Thas one more orbit and inthesecond case one less orbit than 0'.Inparticular, thesign ofa transposition is-1.Prove that e(0')=E(0') isthesign ofthepermutation, 34. (a) Let nbe aneven positive integer, Show that there exists agroup oforder 2n, generated bytwo elements 0', Tsuch that O'n=e=T2 ,and O'T==TO'n-l. (Draw apicture of aregular n-gon, number thevertices, and use thepictureas an inspiration toget 0',T,)This group iscalled thedihedral group. (b) Let nbeanoddpositive integer. LetD4nbethegroup generated bythematrices (° 1_ 01 ) (y and(I) where Cisaprimitive n-th root ofunity. Show that D4nhas order 4n, andgive thecommutation relations between theabove generators. 35. Show that there areexactly twonon-isomorphic non-abelian groups oforder 8.(One ofthem isgiven bygenerators (J,!with therelations (J4 = 1, !2 = 1, !(J! =(J3, The other isthequaternion group.) 36. Let 0'==[123.. .n]inSn.Show that theconjugacy class of0'has(n-I)!elements. Show that thecentralizer of0'isthecyclic group generated by0', 37. (a) Let 0'==[iI...im]be acycle, Letl'ESn. Show that 1'0'1'-1 isthecycle [1'(iI)...1'(im)]' (b)Suppose that apermutation(Jin5ncan bewritten as aproduct ofrdisjoint cycles, and letdh.,.,drbethenumber ofelements ineach cycle, inincreasing order. Let !beanother permutation which can bewritten as aproduct of disjoint cycles, whose cardinalities ared'l'...' d;inincreasing order. Prove that (Jisconjugate to!inSnifandonly ifr=sand di=dforalli=1,...,r. 38. (a) Show that Snisgenerated bythetranspositions [12], [13],., ,,[In]. (b) Show thatSnisgenerated bythetranspositions [12], [23], [34],..., [n-1,n], I,Ex EXERCISES 79 (c) Show that Snisgenerated bythecycles [12] and [123...n]. (d) Assume that nisprime. Let u=[123. ..n]and let T=[rs] beanytransposition. Show that u, Tgenerate Sn. Let Gbe afinite group operatingon asetS.Then Goperates inanatural wayon theCartesian product s(n)foreach positive integern .We define theoperation onS toben-transitive ifgivenndistinct elements (Sb' . .,sn)and ndistinct elements (s;". .,s)ofS,there exists uEGsuch that us,=sjforalli=1,...,n. 39. Show that theaction ofthealternating group Anon{I,. , .,n}is(n-2)-transitive. 40. LetAnbethealternating group ofeven permutations of{I ,. , .,n},Forj=1,. . .,n letHjbethesubgroup ofAnfixing j,soHj=An-I' and(An: Hj)=nfor n>3, Let n>3and letHbe asubgroup ofindex ninAn. (a) Show that theaction ofAnoncosets ofHbyleft translation givesaniso- morphism Anwith thealternating group ofpermutations ofAniH. (b) Show that there exists anautomorphism ofAnmapping HIonH,and that such anautomorphism isinduced byaninner automorphism ofSnifandonly ifH=Hifor some i. 41. LetHbe asimple group oforder 60. (a) Show that theaction ofHbyconjugationonthe setofitsSylow subgroups givesanimbedding H A6. (b)Using thepreceding exercise, show that H=As. (c) Show that A6has anautomorphism which isnotinduced byaninner auto- morphism ofS6. Abelian groups 42.Viewing Z,Qasadditive groups, show thatQ/ZISatorsion group, which has one and only onesubgroup oforder nforeach integern>1,and that thissubgroup iscyclic. 43. LetHbeasubgroup ofafinite abelian group G.Show that Ghas asubgroup that is isomorphic toG/H, 44. Letf:A A'be ahomomorphism ofabelian groups. Let Bbe asubgroup ofA. Denote byAIand AItheimage and kernel offinArespectively, andsimilarly forBI and BI.Show that (A:B)=(AI: BI)(A I:BI)'inthe sense that iftwo ofthese three indices arefinite, soisthethird, and thestated equality holds. 45, Let Gbe afinite cyclic group oforder n,generated byanelement (1.Assume that G operatesonanabelian group A,andlet,h g:A Abetheendomorphisms ofAgiven by f(x)=(1X-xand g(x)=x+(1X+...+(1n-1X. Define the Herbrand quotient bytheexpression q(A)=(AI:Ag)/(A g:AI), provided both indices arefinite. Assume now that Bisasubgroup ofAsuch that GB cB, (a)Define inanatural wayanoperation ofGonA/B. (b)Prove that q(A)=q(B)q(A/B) inthe sense that iftwoofthese quotientsarefinite, soisthethird, and thestated equality holds, (c)IfAisfinite, show that q(A)=1. 80 GROUPS I,Ex (This exercise isaspecialcase ofthegeneral theory ofEuler characteristics discussed inChapter XX, Theorem 3.1.After reading this, thepresent exercise becomes trivial. Why?) Primitive groups 46. Let GoperateonasetS,Let S=USibeapartition ofSintodisjoint subsets. Wesay that thepartition isstable under GifGmaps each S;onto Sjfor some j,and hence G induces apermutation ofthe sets ofthepartition among themselves, There are two partitions ofSwhich areobviously stable: thepartition consisting ofSitself, and the partition consisting ofthesubsets with oneelement. Assume that Goperates transitively, and that Shas more than one element. Prove that thefollowing two conditions are equivalent: PRIM 1.Theonly partitions ofSwhich arestable arethetwopartitions mentioned above. PRIM 2.IfHistheisotropy group ofanelement ofS,then Hisamaximal subgroup ofG. These two conditions define what isknown asaprimitive group, ormore accurately,a primitive operation ofGonS. Instead ofsaying that theoperation ofagroup Gis2-transitive, one also says that itis doubly transitive, 47. Let afinite group Goperate transitively andfaithfullyon aset Swith atleast 2 elements and letHbetheisotropy group ofsome element sofS.(All the other isotropy groupsareconjugates ofH.) Prove thefollowing: (a) Gisdoubly transitive ifandonly ifHactstransitivelyonthecomplement ofsinS. (b) Gisdoubly transitive ifandonly ifG=HTH, where Tisasubgroup ofG oforder 2notcontained inH. (c)IfGisdoubly transitive, and(G:H)=n,then #(G)=den-l)n, where distheorder ofthesubgroup fixing two elements. Furthermore, H isamaximal subgroup ofG,Le. Gisprimitive. 48. Let Gbe agroup acting transitivelyon asetSwith atleast 2elements. For each xEGletI(x)=number ofelements ofSfixed byx.Prove: (a)LI(x)=#(G). XEG (b) Gisdoubly transitive ifandonly if Lf(X)2=2#(G). XEG 49. Agroupasanautomorphism group. LetGbeagroup andletSet(G)bethecategory ofG-sets (Le. sets with aG-operation), LetF:Set(G) Set betheforgetful functor, which toeach G-set assigns the setitself. Show thatAut(F) isnaturally isomorphic toG. I,Ex EXERCISES 81 Fiber products and coproducts Pull-backs and push-outs 50.(a)Show that fiber products exist inthecategory ofabelian groups. Infact, IfX,Y are abelian groups with homomorphisms f:X-+Zand g:Y-+Zshow that XxzYisthe setofallpairs (x,y)with xEXand yEYsuch thatf(x)=g(y). The maps Pt,P2aretheprojections onthefirst and second factor respectIvely. (b)Show that thepull-back ofasurjectIve homomorphism issurjective. 51.(a)Show that fiber products exist inthecategory ofsets. (b) Inany category e,consider thecategory e7.ofobjects over Z.Leth:T-+Z beafixed object inthiscategory, LetFbethefunctor such that F(X)=Morz(T, X), where Xisanobject over Z,and Morzdenotes morphisms over Z.Show that Ftransforms fiber products over Zinto fiber products inthecategory ofsets. (Actually,once you have understood thedefinitions, this istautological.) 52,(a)Show that push-outs (i.e. fiber coproducts) exist inthecategory ofabelian groups. Inthis case thefiber coproduct oftwohomomorphisms f,gasabove isdenoted byX(f)zY.Show that itisthefactor group Xzy=(X Y)/W, where Wisthesubgroup consisting ofallelements (f(z),-g(z» with zEZ. (b) Show that thepush-out ofaninjective homomorphism isinjective. Remark. After you have read about modules over rings, you should note that the above two exercises apply tomodules aswell astoabelian groups, 53. LetH,G,G'begroups, and let f:H-+G, g:H-+G' betwo homomorphisms. Define thenotion ofcoproduct ofthese two homomor- phisms over H,and show that itexists. 54.(Tits). Let Gbe agroup and let{GJiElbe afamily ofsubgroups generating G. Suppose Goperates on asetS.For each iEI,suppose givenasubset SiofS,and let sbe apoint ofS-l)Si.Assume that foreach 9EG;-{e}, wehave , gSjCS;forallj=1=i, and g(s) ES;foralli. Prove that Gisthecoproduct ofthefamily {GJ;El' (Hint: Supposeaproduct g....gm=idonS,Apply thisproduct tos,and useProposition 12.4.) 55. LetMEGL2(C) (2x2complex matrices with non-zero determinant). We let (ab )az+bM= ,andforzECweletM(z)=d' cd cz+ Ifz=-d/ c(c=1=0)then weputM(z)=00,Then youcanverify (and you should have seen something like this inacourse incomplex analysis) that GL2(C) thus operatesonCU{oo}. Let A,A'betheeigenvalues ofMviewed asalinear mapon C2. LetW,W'bethecorresponding eigenvectors, W=f(W., w2) and W'=f(W;, w;), 82 GROUPS I,Ex Byafixed point ofMonCwe mean acomplex number zsuch thatM(z)=z.Assume that Mhas twodistinct fixed points=1=00. (a) Show that there cannot bemore than two fixed points and that these fixed pointsare w=wllw2 and w'=wi/w2. Infact one may take W=t(w, 1),W'=t(w', 1). (b) Assume that1AI<1A'I,Given z=1=w,show that limMk(z)=w'. k-oo [Hint: Let S=(W,W') and consider S-IMkS(Z)=exkz where ex=AIA'.] 56.(Tits) LetM.,. . .,MrEGL2(C) be afinite number ofmatrices. Let A;,A;bethe eigenvalues ofM;. Assume that each M;has two distinct complex fixed points, and thatIA;I<1A;I.Also assume that thefixed points forMI'. , .,Mrarealldistinct from each other, Prove that there exists apositive integer ksuch thatM,. ,,,M arethefree generators ofafree subgroup ofGL2(C), [Hint: Let wi'w;bethefixed points ofM;. LetV;be asmall disc centered atWiandV;asmall disc centered at w;.LetS;=V;UV;.Let sbe acomplex number which does notlieinany S;.Let G;=(M). Show that theconditions ofExercise 54 aresatisfied forksufficiently large.]. s. 57. Let Gbe agroup actingon asetX.Let Ybe asubset ofX.Let Gybethesubset of Gconsisting ofthose elements gsuch that gYnYisnotempty. Let Gybethe subgroup ofGgenerated byGy.Then GyYand (G-Gy)Y aredisjoint. [Hint: Suppose that there exist glEGyand g2EGbut g2$Gy,and elements YI,Y2,EY such that g2Yl=g2Y2. Then g:;lglYI=Y2,sog:;lg) EGywhence g2EGy,contrary toassumption.] Application. Suppose thatX=GY, butthatXcannot beexpressedasadisjoint union asabove unless oneofthetwo sets isempty. Then weconclude that G-Gy isempty, and therefore Gygenerates G. Example 1.Suppose Xisaconnected topological space, Yisopen, and Gacts continuously. Then alltranslates ofYare open,soGisgenerated byGy. Example 2.Suppose Gisadiscrete group acting continuously anddiscretely onX.Again suppose Xconnected and Yclosed. Then any union oftranslates ofY byelements ofGisclosed, soagain G-Gyisempty, and Gygenerates G. CHAPTER II Rings 1. RINGS AND HOMOMORPHISMS Aring Aisaset,together with two laws ofcomposition called multiplica- tion and addition respectively, and written asaproduct and asasum respec- tively, satisfying thefollowing conditions: RI1.With respect toaddition, Aisacommutative group. RI2.Themultiplication isassociative, and has aunit element. RI3.For allx,y,ZEAwehave (x+y)z=xz+yz (This iscalled distributivity.) Asusual, wedenote the unit element foraddition by0,and the unit element formultiplication by 1.We donot assume that 1=Fo.We observe that Ox =0for all xEA.Proof: We have Ox+x=(0+l)x=Ix =x. Hence Ox =o.Inparticular, if1=0,then Aconsists of0alone. For any x,YEA wehave(-x)y=-(xy). Proof: We have xy+(-x)y=(x+(-x))y=Oy=0,and z(x+y)=zx+zy. so(-x)y istheadditive inverse ofxy. Other standard laws relating addition andmultiplicationareeasily proved, forinstance (-x)(-y)=xy. We leave these asexercises. Let Abe aring, and letUbethe setofelements ofAwhich have both a right and left inverse. Then Uisamultiplicative group. Indeed, ifahas a 83 84 RINGS II,1 right inverse b,sothat ab =1,and aleft inverse c,sothat ca =1,then cab =b,whence c=b,and we seethat c(orb)isatwo-sided inverse, and that citself has atwo-sided inverse, namely a.Therefore Usatisfies allthe axioms ofamultiplicative group, and iscalled thegroup ofunits ofA.Itis sometimes denoted byA*, and isalso called thegroup ofinvertible elements ofA.Aring Asuch that 1=F0,and such that every non-zero element is invertible iscalled adivision ring. Note. The elements ofaring which areleftinvertible donotnecessarily form agroup. Example. (The Shift Operator). Let Ebethe setofallsequences a=(at, a2,a3,...) ofintegers. One can define addition componentwise. Let Rbethe setofall mappings f:E-+EofEinto itself such thatf(a +b)=f(a) +f(b). The law ofcomposition isdefined tobecomposition ofmappings. Then Risaring. (Proof?) Let T(a t,a2,a3,...)=(0,at,a2,a3,...). Verify that Tisleftinvertible but notright invertible. Aring Aissaid tobecommutative ifxy=yxforallx,YEA. Acommu- tative division ring iscalled afield. We observe that bydefinition, afield contains atleast two elements, namely 0and 1. Asubset Bofaring Aiscalled asubring ifitisanadditive subgroup, if itcontains themultiplicative unit, and ifx,YEBimplies xyEB.Ifthat is the case, then Bitself isaring, thelaws ofoperation inBbeing the same as thelaws ofoperation inA. For example, the center ofaring Aisthe subset ofAconsisting ofall elements aEAsuch that ax =xaforall xEA.One sees immediately that the center ofAisasubring. Just as weproved general associativity from theassociativity forthree factors, one can prove general distributivity. Ifx,Yt, ..., Ynareelements ofa ring A,then byinduction one sees that X(YI +...+Yn)=XYt +...+XYn' IfXi(i=1,...,n)andYj(j=1,...,m)areelements ofA,then itisalso easily proved that Ctx)CYj)=itjXiXj' Furthermore, distributivity holds forsubtraction, e.g. x(Yt-Y2)=XYt-XY2' We leave alltheproofs tothereader. II,1 RINGS AND HOMOMORPHISMS 85 Examples. Let Sbeasetand Aaring. LetMap(S, A)bethe setofmap- pings ofSinto A.Then Map(S, A)isaringifforf,gEMap(S, A)wedefine (fg)(x)=f(x)g(x) and (f+g)(x)=f(x) +g(x) forallXES. Themultiplicative unit isthe constant map whose value isthe multiplicative unit ofA.The additive unit isthe constant map whose value istheadditive unit ofA,namely O.The verification that Map(S, A)isaring under theabove laws ofcomposition istrivial and left tothereader. Let Mbe anadditive abelian group, and letAbethe setEnd(M) of group-homomorphisms ofMinto itself. We define addition inAtobethe addition ofmappings, and we define multiplication tobecomposition of mappings. Then itistrivially verified that Aisaring. Itsunit element isof course theidentity mapping. Ingeneral, Aisnotcommutative. Readers have nodoubt metpolynomials over afield previously. These pro- vide abasic example ofaring, and will bedefined officially forthis book in3. Let Kbe afield. The setofnxnmatrices with components inKisa ring. Itsunits consist ofthose matrices which areinvertible, orequivalently have anon-zero determinant. Let Sbe asetand Rthe setofreal-valued functions onS.Then Risa commutative ring. Itsunits consist ofthose functions which arenowhere O. This isaspecialcase ofthering Map(S, A)considered above. The convolution product. We shall,now give examples ofrings whose product isgiven bywhat iscalled convolution. Let Gbe agroup and letK be afield. Denote byK[G] the setofallformal linear combinations rx=Laxx with xEGand axEK,such that allbut afinite number ofaxare equal toO.(See3,and also Chapter III,4.) IfP=LbxxEK[G], then one can define theproduct rxp=LLaxbyxy=L(Laxb y)z. xeG yeG zeG xy=z With this product, the group ring K[G] isaring, which will bestudied extensively inChapter XVIII when Gisafinite group. Note thatK[G] is commutative ifand only ifGiscommutative. The second sum ontheright above defines what iscalled aconvolution product. Iff,gare two functions on agroup G,wedefine their convolution f*gby (f*g)(z)=Lf(x)g(y). xy=z Of course this must make sense. IfGisinfinite, one may restrict this definition tofunctions which are 0except at afinite number ofelements. Exercise 12will give anexample (actually on amonoid) when another type ofrestriction allows for afinite sum ontheright. Example from analysis. Inanalysis one considers asituation asfollows. Let L1=L1(R) bethe space offunctions which areabsolutely integrable. 86 RINGS II,1 Given functions f,gEL t,one defines their convolution product f*gby (f*g)(x)=tf(x-y)g(y) dy. Then thisproduct satisfies allthe axioms ofaring, except that there isno unit element. Inthe case ofthegroup ring ortheconvolution ofExercise 12, there isaunit element. (What isit?) Note that theconvolution product in the case ofLt(R) iscommutative, basic,\lly because Risacommutative additive group. More generally, let Gbe alocally compact group with a Haar measure JLThen theconvolution product isdefined bythe similar formula (f*g)(x)=Lf(xy-l )g(y) dJ1.(Y). After these examples, wereturn tothegeneral theory ofrings. Aleft ideal Qinaring Aisasubset ofAwhich isasubgroup ofthe additive group ofA,such that AQ CQ(and hence AQ =Qsince Acontains 1).Todefine aright ideal, werequire QA =Q,and atwo-sided ideal isa subset which isboth aleft and aright ideal. Atwo-sided ideal iscalled simplyanideal inthis section. Note that (0)and Aitself areideals. IfAisaring and aEA,then Aaisaleftideal, called principal. We say that aisagenerator of Q(over A).Similarly, AaA isaprincipal two-sided ideal ifwedefine AaA tobethe setofall sumsLXiaYi with Xi'YiEA.Cf. below thedefinition oftheproduct ofideals. More generally, letat, ..., an beelements ofA.We denote by(at, ...,an)the setofelements ofAwhich can bewritten intheform Xtat+...+xna n with XiEA. Then this setofelements isimmediately verified tobe aleft ideal, and at, ..., anarecalled generators oftheleftideal. If{Qi} ieIisafamily ofideals, then their intersection nQi ieI isalso anideal. Similarly forleft ideals. Readers will easily verify that if Q=(at, ...,an)' then Qisthe intersection ofallleft ideals containing the elements at,...,an. Aring Aissaid tobecommutative ifxy=yxforall X,YEA. Inthat case, every left orright ideal istwo-sided. Acommutative ring such that every ideal isprincipal and such that 1=F0 iscalled aprincipal ring. Examples. The integers Zform aring, which iscommutative. Let Qbe anideal =FZand =FO.IfnEQ,then-nEQ.Let dbethe smallest integer >0lying in Q.IfnEQthen there exist integers q,rwith 0<r<dsuch that n=dq+r. II,1 RINGS AND HOMOMORPHISMS 87 Since aisanideal, itfollows that rliesina,hence r=o.Hence aconsists of allmultiples qdofd,with qEZ,and Zisaprincipal ring. Asimilar example isthering ofpolynomials inone variable over afield, aswill beproved inChapter IV,also using theEuclidean algorithm. Let Rbethering ofalgebraic integers in anumber field K.(For definitions, seeChapter VII.) Then Risnotnecessarily principal, but letp be aprime ideal, and letRpbethering ofallelements a/bwith a,bERand brtp.Then inalgebraic number theory, itisshown that R"isprincipal, with oneprime idealmpconsisting ofallelements a/basabove but with aEp. See Exercises 15,16,and 17. Anexample from analysis. Let Abethe setofentire functions onthe complex plane. Then Aisacommutative ring, and every finitely generated ideal isprincipal. Given adiscrete setofcomplex numbers {Zi} and integers mi>0,there exists anentire function Ihaving zeros atZiofmultiplicity mi and noother zeros. Every principal ideal isoftheform AIfor some such I. The group ofunits A*inAconsists ofthefunctions which have no zeros. It isanice exercise inanalysis toprove the above statements (using the Weierstrass factorization theorem). We now return togeneral notions. Let a,bbeideals ofA.We define ab tobethe setofallsums X1Yl +...+XnYn with XiEaand YiEb.Then one verifies immediately that abisanideal, and that the setofideals forms amultiplicative monoid, theunit element being thering itself. This unit element iscalled theunitidea and isoften written (1). Ifa,bareleftideals, wedefine their product ab asabove. Itisalso aleftideal, andifa,b,careleftideals, then weagain have associativity: (ab)c=a(bc). Ifa,bare left ideals ofA,then a+b(the sum being taken asadditive subgroup ofA)isobviouslyaleft ideal. Similarly forright and two-sided ideals. Thus ideals also form amonoid under addition. We also have distributivity: Ifa1,..., an'bareideals ofA,then clearly b(a 1+...+an)=ba1+...+ban' andsimilarly ontheother side. (However, the setofideals does not form a ring! ) Let abealeftideal. Define aAtobethe setofallsums a1X1+...+anX n with aiEaand XiEA.Then aAisanideal (two-sided). Suppose that Aiscommutative. Let a,bbeideals. Then trivially abcan b, butequality does notnecessarily hold. However, asanexercise, prove that if a+b=Athen ab =anb. Asshould beknown tothereader, theintegers Zsatisfy another property besides every ideal being principal, namely unique factorization into primes. 88 RINGS II,1 We shall discuss thegeneral phenomenon in. Beitnoted here only that if aring Ahas theproperty ofunique factorization into prime elements, and p isaprime element, then theideal (p)isprime, and thering R(p) (definedas above) isprincipal. See Exercise 6.Thus principal rings may beobtained in anatural way from rings which arenotprincipal. AsDedekind found out, some form ofunique factorization can be re- covered insome cases, replacing unique factorization into prime elements by unique factorization of(non-zero) ideals into prime ideals. Example. There are cases when the non-zero ideals give rise toagroup. Let 0be asubring ofafield Ksuch that every element ofKisaquotient of elements of0;that is,oftheform a/bwith a,bE0and b=FO.Byafractional ideal Qwe mean anon-zero additive subgroup ofKsuch that OQCQ(and therefore OQ =Qsince 0contains theunit element); and such th(!t there exists anelement CEO,C=F0,such that CQCo.Wemight say that afractional ideal has bounded denominator. ADedekind ring isaring0asabove such that thefractional ideals form agroup under multiplication. Asproved in books onalgebraic number theory, thering ofalgebraic integers inanumber field isaDedekind ring. Do Exercise 14toget the property ofunique factorization into prime ideals. See Exercise 7ofChapter VII for asketch of thisproof. IfaEK, a=F0,then oaisafractional ideal, and such ideals are called principal. The principal fractional ideals form asubgroup. The factor group iscalled theideal class group, orPicard group of0,and isdenoted byPic(0). See Exercises 13-19 for some elementary facts about Dedekind rings. Itis abasic problem todetermine Pic(o) forvarious Dedekind rings arising in algebraic number theory and function theory. See mybook Algebraic Num- berTheory forthebeginnings ofthetheory innumber fields. Inthe case of function theory, one isled toquestions inalgebraic geometry, notably the study ofgroups ofdivisor classes onalgebraic varieties and allthat this entails. The property that thefractional ideals form agroup isessentially associated with thering having "dimension 1"(which wedo not define here). Ingeneral one isledinto thestudy ofmodules under various equiva- lence relations; seeforinstance the comments attheend ofChapter III,4. We return tothegeneral theory ofrings. Byaring-homomorphism one means amapping f:A-+Bwhere A,Bare rings, and such thatfisamonoid-homomorphism for themultiplicative structures onAand B,and also amonoid-homomorphism fortheadditive structure. Inother words, fmust satisfy: f(a +a')=f(a) +f(a'), f(l)=1,f(aa')=f(a)f(a'), f(O)=0, foralla,a'EA.Itskernel isdefined tobethekernel offviewed asadditive homomorphism. II,1 RINGS AND HOMOMORPHISMS 89 The kernel ofaring-homomorphism f:A-.Bisanideal ofA, as one verifies atonce. Conversely, let abeanideal ofthering A.We can construct thefactor ring A/aasfollows. Viewing Aand aasadditive groups, letA/a bethe factor group. We define amultiplicative law ofcomposition onA/a: If x+aand y+aare two cosets ofa,wedefine (x+a)(y +a)tobethe coset (xy+a).This coset iswell defined, forifXl'Ylareinthe same coset asx,y respectively, then one verifies atonce that XlYlisinthe same coset asxy. Our multiplicative law ofcomposition isthen obviously associative, has a unit element, namely the coset 1+a,and the distributive law issatisfied since itissatisfied forcoset representatives. We have therefore defined aring structure onA/a, and thecanonical map f:A-.A/a isthen clearlyaring-homomorphism. Ifg:A-.A'isaring-homomorphism whose kernel contains a,then there exists aunique ring-homomorphism g.:A/a-.A'making thefollowing dia- gram commutative: Ag) A'f\I. A/a Indeed, viewing f,gasgroup-homomorphisms (for the additive struc- tures), there isaunique group-homomorphism g.making our diagram commutative. We contend that g.isinfact aring-homomorphism. We could leave thetrivial proof tothereader, but wecarry itout infull. If xEA,then g(x)=g.f(x).Hence forx,YEA, g.(f(x)f(y))=g.(f(xy))=g(xy)=g(x)g(y) =g.f(x)g.f(y). Given ",.,EA/a, there exist x,YEA such that,=f(x) and,.,=f(y). Since f(1)=1,wegetg.f(1)=g(1)=1,and hence the two conditions thatg.bea multiplicative monoid-homomorphismaresatisfied, aswas tobeshown. The statement we have just proved isequivalent tosaying that the canonical mapf:A-.A/a isuniversal inthecategory ofhomomorphisms whose kernel contains a. Let Abe aring, and denote itsunit element byeforthe moment. The map A:Z-.A such that A(n)=neisaring-homomorphism (obvious), and itskernel isan ideal (n),generated byanintegern>O.We have acanonical injective homo- morphism Z/nZ-.A,which isa(ring) isomorphism between ZjnZ and a 90 RINGS II,1 subring ofA.IfnZisaprime ideal, then n=0orn=pfor some prime number p.Inthefirst case, Acontains asasubringaring which isisomorphic toZ,and which isoften identified with Z.Inthat case, wesaythat Ahascharacteristic o.Ifontheother hand n=p,then wesay that Ahascharacteristic p,and A contains (anisomorphic image of)Z/pZasasubring. We abbreviate Z/pZ by Fp./ IfKisafield, then Khas characteristic 0orp>O.Inthefirst case, K contains asasubfield anisomorphic image oftherational numbers, and in the second case, itcontains anisomorphic image ofFp.Ineither case, this subfield will becalled theprime field (contained inK). Since thisprime field isthe smallest subfield ofKcontaining1and has noautomorphism except theidentity, itiscustomary toidentify itwith QorFpasthe case may be. By theprime ring (inK) we shall mean either theintegers ZifKhas characteristic 0,orFpifKhascharacteristic p. Let Abe asubring ofaring B.Let Sbe asubset ofBcommuting with A;inother words wehave as =saforall aEAand sES.We denote by A[S] the setofallelements a Sit. ..Sin ii'.,in1 n, the sum ranging over afinite number ofn-tuples (i1, ...,in)ofintegers>0, andait."inEA, S1,,,.,SnES. IfB=A[S], we say that 8is aset of generators (or more precisely, ring generators) for Bover A,orthat Bis generated by Sover A.IfSisfinite, we say that Bisfinitely generated as aring over A.One might say that A[8] consists ofallnot-necessarily- commutative polynomials inelements of8with coefficients inA.Note that elements of8may not commute with each other. Example. The ring ofmatrices over afield isfinitely generated over that field, but matrices don't necessarily commute. Aswith groups,weobserve that ahomomorphism isuniquely determined by itseffect ongenerators. Inother words, letf:A-.A' be aring- homomorphism, and letB=A[S] asabove. Then there exists atmost one extension offtoaring-homomorphism ofBhaving prescribed values onS. Let Abearing,aanideal, and 8asubset ofA.Wewrite S=0(mod a) ifSea. Ifx,YEA, wewrite x=Y(mod a) ifx-YEa. Ifaisprincipal, equal to(a),then wealso write x=Y(mod a). Iff:A-.A/a isthe canonical homomorphism, then x -Y(mod a)means thatf(x)=f(y). The congruence notation issometimes convenient when we want toavoid writing explicitly thecanonical mapf II,1 RINGS AND HOMOMORPHISMS 91 The factor ring A/a isalso called aresidue class ring. Cosets of ainA are called residue classes modulo a,and ifxEA,then the coset x+ais called theresidue class ofxmodulo a. We have defined thenotion ofanisomorphism inany category, and soa ring-isomorphism isaring-homomorphism which has atwo-sided inverse. Asusual wehave thecriterion: Aring-homomorphism f:A-.Bwhich isbijective isanisomorphism. Indeed, there exists aset-theoretic inverse g:B-.A,and itistrivial toverify that gisaring-homomorphism. Instead ofsaying "ring-homomorphism" we sometimes say simply "homomorphism" ifthe reference torings isclear. We note that rings form acategory (themorphisms being thehomomorphisms). Letf:A-.Bbe aring-homomorphism. Then theimage f(A) offisa subring ofB.Proof obvious. Itisclear that aninjective ring-homomorphism f:A-.Bestablishes a ring-isomorphism between Aand itsimage. Such ahomomorphism will be called anembedding (ofrings). Letf:A-.A'be aring-homomorphism, and leta'be anideal ofA'. Then f-l(a') isanideal ainA,and wehave aninduced injective homo- morphism A/a-.A'/a'. The trivial proof isleft tothereader. Proposition 1.1. Products exist inthecategory ofrings. Infact, let{Ai}ielbe afamily ofrings, and letA=nAibetheir product asadditive abelian groups. We define amultiplication inAinthe obvious way: If(Xi)iel and (Yi)iel are two elements ofA,wedefine their product to be(XiYi)i eI'i.e. wedefine multiplication componentwise, justaswedid for addition. Themultiplicative unit issimply theelement oftheproduct whose i-th component istheunit element ofAi.Itisthen clear that weobtain a ring structure onA,and that theprojectiononthe i-th factor isaring- homomorphism. Furthermore, Atogether with these projections clearly satisfies therequired universal property. Note that the usual inclusion ofAi onthe i-th factor isnot aring- homomorphism because itdoes not map theunit element eiofAiontheunit element ofA.Indeed, itmaps eionthe element ofAhaving eiasi-th component, and 0(=0i)asallother components. Let Abe aring. Elements x,YofAaresaid tobezero divisors ifx=F0, Y=F0,and xy=O.Most oftherings without zero divisors which we con- sider will becommutative. Inview ofthis, wedefine aring Atobeentire if 1=F0,ifAiscommutative, and ifthere are no zero divisors inthering. (Entire rings are also called integral domains. However, linguistically, Ifeel 92 RINGS II,2 the need for anadjective. "Integral" would do, except that inEnglish, "integral" has been used for"integral over aring"asinChapter VII, 1.In French, asinEnglish, two words exist with similar roots: "integral" and "entire". The French have used both words. Why not dothe same in English? There isaslight psychological impediment, inthat itwould have been better ifthe use of"integral" and "entire" were reversed tofitthe long-standing French use. Idon't know what todoabout this.) Examples. The ring ofintegers Ziswithout zero divisors, and isthere- fore entire. IfSisasetwith more than 2elements, and Aisaring with 1=F0,then thering ofmappings Map(S, A)has zero divisors. (Proof?) Let mbe apositive integer =F1.The ring Z/mZ has zero divisors ifand only ifmisnot aprime number. (Proof left asanexercise.) The ring of nxnmatrices over afield has zero divisors ifn>2.(Proof?) The next criterion isused very frequently. Let Abeanentire ring, and leta,bbenon-zero elements ofA.Then a,b generate the same ideal ifandonlyifthere exists aunit uofAsuch that b=au. Proof Ifsuch aunit exists we have Ab =Aua=Aa. Conversely, assume Aa =Ab. Then we can write a=beand b=adwith some elements c,dEA.Hence a=adc, whence a(1-dc)=0,and therefore de =1.Hence cisaunit. 2. COMMUTATIVE RINGS Throughout this section, weletAdenote acommutative ring. Aprime ideal inAisanideal p=FAsuch that A/p isentire. Equiva- lently, wecould say that itisanideal p=FAsuch that, whenever x,YEA and xyEp,then xEporYEp. Aprime ideal isoften called simplyaprime. Let mbe anideal. We say that misamaximal ideal ifm=FAand if there isnoideal Q=FAcontaining mand =Fm. Every maximal ideal isprime. Proof Let mbemaximal and letx,YEA besuch that xy Em.Suppose xftm. Then m+Ax isanideal properly containing m,hence equal toA. Hence we can write l=u+ax with UEmand aEA.Multiplying byYwefind II,2 COMMUTATIVE RINGS 93 y=yu+axy, whence yEmand mistherefore prime. Let Qbeanideal =FA.Then Qiscontained insome maximal ideal m. Proof. The setofideals containingQand =FAisinductively ordered by ascending inclusion. Indeed, if{bi}isatotally ordered setofsuch ideals, then 1ftbiforany i,and hence 1does not lieintheideal b=Ubi'which dominates allbi.Ifmisamaximal element inour set,then m=FAand mis amaximal ideal, asdesired. The ideal {O}isaprime ideal ofAifandonlyifAisentire. (Proof obvious.) We defined afield Ktobe acommutative ring such that 1=t=0,and such that themultiplicative monoid ofnon-zero elements ofKisagroup (i.e. such that whenever xEKand x=t=0then there exists aninverse forx).We note that theonly ideals ofafield KareKand the zero ideal. Ifmisamaximal ideal ofA,then Aim isafield. Proof IfxEA,wedenote byxitsresidue class mod m. Since m=FA wenote that Aim has aunit element =Fo.Any non-zero element ofAim can bewritten asxfor some xEA,xftm.Tofind itsinverse, note that m+Ax isanideal ofA=Fmand hence equal toA.Hence we can write 1=u+yx with uEmandYEA. This means that yx=1(i.e.=1)and hence that xhas aninverse, asdesired. Conversely, weleave itasanexercise tothereader toprove that: Ifmisanideal ofAsuch thatAim isafield, then mismaximal. Letf:A-+A'beahomomorphism ofcommutative rings. Letp'beaprime ideal ofA',and letp=f-1(p'). Then pisprime. To prove this, letx,YEA, and xyEp.Supposexftp.Then f(x) ftp'. Butf(x)f(y)=f(xy)Ep'.Hence f(y)Ep',asdesired. As anexercise, prove that iffissurjective, and ifm'ismaximal inA', thenf-1(m/)ismaximal inA. Example. LetZbethering ofintegers. Since anideal isalso anadditive subgroup ofZ,every ideal =t={O} isprincipal, oftheform nZfor some integer n>0(uniquely determined bytheideal). Let pbe aprime ideal =t={O}, p=nZ. Then nmust be aprime number, asfollows essentially directly from thedefinition ofaprime ideal. Conversely, ifpisaprime number, then pZ is aprime ideal (trivial exercise). Furthermore, pZ isamaximal ideal. Indeed, suppose pZcontained insome ideal nZ. Then p=nmfor some integer m,whence n=porn=1,thereby proving pZmaximal. 94 RINGS II,2 Ifnisaninteger, the factor ring Z/nZIScalled the nng ofintegers modulo n.We also denote Z/nZ=Z(n). Ifnisaprime number p,then thering ofintegers modulo pisinfact afield, denoted byFp.Inparticular, themultiplicative group ofFpiscalled the group ofnon-zero integers modulo p.From theelementary properties of groups, wegetastandard fact ofelementary number theory: Ifxisan integer =1=0(mod p),then xp-t=1(mod p).(For simplicity, itiscustomary towrite mod pinstead ofmod pZ, and similarly towrite mod ninstead of mod nZforanyinteger n.)Similarly, given anintegern>1,theunits inthe ring Z/nZ consist ofthose residue classes mod nZwhich arerepresented by integersm=F0andprime to n.The order ofthegroup ofunits inZ/nZ is called bydefinition qJ(n) (where qJisknown asthe Euler phi-function). Consequently, ifxisaninteger prime ton,then xqJ(n) =1(mod n). Theorem 2.1. (Chinese Remainder Theorem). Let at, ..., anbeideals of Asuch that ai+aj=Aforalli=Fj.Given elements xt,..., XnEA,there exists xEAsuch that x=Xi(modai)foralli. Proof Ifn=2,wehave anexpression 1=al+a2 for some elements aiEaj,and weletx=X2at +Xta2' For each i>2we can find elements aiEatand biEaisuch that ai+bi=1, i>2. n The product n(ai+bi)isequal to1,and liesin i=2 n al+nai' i=2 i.e.inat+a2...an'Hence n at+nai=A. i=2 Bythetheorem for n=2,we can find anelement YlEAsuch that Yl=1(mod Qt), Yt=0(mod.Ii11;). 1=2 Wefindsimilarly elements Y2, ..., Ynsuch that Yj=1(mod aj) andYj=0(mod ai) fori=Fj. Then x=XlYl+...+XnYn satisfies ourrequirements. II,2 COMMUTATIVE RINGS 95 Inthe same vein asabove, weobserve that ifa1,..., anare ideals ofa ring Asuch that a+...+a=A1 n' andifV1,..., Vnarepositive integers, then al+...+an=A. Theproof istrivial, and isleft asanexercise. Corollary 2.2. Let a1,..., anbeideals ofA.Assume that ai+aj=Afor ii=j.Let n f:A-.nAlai=(Ala 1)x...X(Alan) i=1 bethe map ofAinto theproduct induced bythecanonical map ofAonto n Alai for each factor. Then the kernel offisnai'andfissurjective, thus giving anisomorphismi=1 Aln ainAlai. Proof That the kernel offiswhat we said itIS, ISobvious. The surjectivity follows from thetheorem. The theorem and itscorollary arefrequently applied tothering of integers Zand todistinct prime ideals (P1)' ..., (Pn). These satisfy the hypothesis ofthetheorem since they aremaximal. Similarly, one could take integers m1, ..., mnwhich arerelatively prime inpairs, andapply thetheorem totheprincipal ideals (m1)=m1Z,...,(m n)=mnZ. This istheultraclassical case oftheChinese remainder theorem. Inparticular, let mbeaninteger> 1,and let nr.m=PiI i be afactorization ofminto primes, with exponents ri>1.Then wehave a ring-isomorphism: Z/mZ nZlp?Z. i IfAisaring,wedenote asusual byA*themultiplicative group ofinvertible elements ofA.We leave thefollowing assertions asexercises: Thepreceding ring-isomorphism ofZlmZ onto theproduct induces agroup- isomorphism (ZlmZ)* n(ZlpiZ)*. i Inview ofourisomorphism,wehave qJ(m)=nqJ(pi). i 96 RINGS II,2 Ifpisaprime number and raninteger>1,then cp(pr)=(p_l)pr-l. One proves this last formula byinduction. Ifr=1,then Z/pZ isafield, and themultiplicative group ofthat field has order p-1.Let rbe >1,and consider thecanonical ring-homomorphism Z/pr+l Z-.Z/prz, arising from theinclusion ofideals (pr+l)C(pr). We getaninduced group- homomorphism A:(Z/pr+l Z)*-.(Z/prz)*, which issurjective because any integerawhich representsanelement of Z/prz and isprime topwill representanelement of(Z/pr+l Z)*. Let abean integer representing anelement of(Z/pr+l Z)*, such that A.(a)=1.Then a=1(mod prz), and hence we can write a=1+xpr (mod pr+lZ) for some xEZ.Letting x=0,1,..., p-1gives rise topdistinct elements of (Z/pr+l Z)*, allofwhich areinthekernel ofA.Furthermore, the element x above can beselected tobe one ofthese pintegers because every integer is congruent toone ofthese pintegers modulo (p). Hence thekernel ofAhas order p,and ourformula isproved. Note that thekernel ofA.isisomorphic toZ/pZ. (Proof?) Application: The ring ofendomorphisms of acyclic group. One ofthe first examples ofaring isthering ofendomorphisms ofanabelian group. In the case ofacyclic group,wehave thefollowing complete determination. Theorem 2.3. Let Abe acyclic group oforder n.For each kEZlet fk:A Abetheendomorphism x kx(writing Aadditively). Then k fk induces aring isomorphism Z/nZ=End(A), and agroup isomorphism (Z/nZ)*=Aut(A). Proof Recall that the additive group structure onEnd(A) issimply addition ofmappings, and themultiplication iscomposition ofmappings. The fact that k1---+hisaring-homomorphism isthen arestatement ofthe formulas 1a=a, (k+k')a=ka+k'a, and (kk')a=k(k'a) fork,k'EZand aEA.Ifaisagenerator ofA,then ka =0ifand only if k=0mod n, soZ/nZ isembedded inEnd(A). On the other hand, let f:A-.Abeanendomorphism. Again for agenerator a,wehave f(a)=ka II,3 POLYNOMIALS AND GROUP RINGS 97 for some k,whence f=hsince every xEAisofthe form mafor some mEZ,and f(x)=f(ma)=mf(a)=mka =kma =kx. This proves theisomorphism ZjnZ End(A). Furthermore, ifkE(ZjnZ)* then there exists k'such that kk' =1mod n,soAhas theinverse h,andhis anautomorphism. Conversely, given anyautomorphism fwith inverse g,we know from thefirst part oftheproof thatf=fk,g=gk'for some k,k',and fog=idmeans that kk'==1mod n,sok,k'E(Z/nZ)*. This proves the isomorphism (Z/nZ)*=Aut(A). Note that ifAiswritten asamultiplicative group C,then the map fkis given byx1---+xk .For instance, letJlnbethegroup ofn-th roots ofunity inC. Then allautomorphisms offinaregiven by '1---+,k with kE(ZjnZ)*. 3. POLYNOMIALS AND GROUP RINGS Although allreaders will have metpolynomial functions, this section lays theground work forpolynomials ingeneral. One needs polynomials over arbitrary rings inmany contexts. For onething, there arepolynomialsover afinite field which cannot beidentified with polynomial functions inthat field. One needs polynomials with integer coefficients, and one needs to reduce these polynomials mod pforprimes p.One needs polynomialsover arbitrary commutative rings, both inalgebraic geometry and inanalysis, for instance thering ofpolynomial differential operators. We also have seen the example ofaring B=A[S]generated byasetofelements over aring A. We now giveasystematic account ofthe basic definitions ofpolynomials over acommutative ring A. We want togiveameaning toanexpression such as ao+atX+...+anXn , where aiEAand Xisa"variable". There are several devices fordoing so, and wepick one ofthem. (Ipicked another inmyUndergraduate Algebra.) Consider aninfinite cyclic group generated byanelement X.We letSbethe subset consisting ofpowers xrwith r>O.Then Sisamonoid. We define the setofpolynomials A[X] tobethe setoffunctions S-.Awhich areequal to0except for afinite number ofelements ofS.For each element aEAwe denote byaXnthefunction which has thevalue aonxnand thevalue 0for allother elements ofS.Then itisimmediate that apolynomialcan be written uniquelyasafinite sum 98 RINGS II,3 aoXo +...+anXn for some integernENand aiEA.Such apolynomial isdenoted byf(X). The elements aiEAare called the coefficients offWe define theproduct according totheconvolution rule. Thus, given polynomials n f(X)=LaiXi i=Oandm g(X)=LbjXj j=O wedefine theproduct tobe f(X)g(X)=tC+ka;bj)Xk. Itisimmediately verified that this product isassociative and distributive. We shall give the details ofassociativity inthe more general context ofa monoid ring below. Observe that there isaunit element, namely lXo. There isalso anembedding A-.A[X] givenbya1---+aXo. One usually does notdistinguishafrom itsimage inA[X], and one writes a instead ofaXo. Note that for CEAwehave then cf(x)=LcaiXi. Observe that byourdefinition, wehave anequality ofpolynomials LaiXi=LbiXi ifandonly ifai=biforalli. Let Abe asubring ofacommutative ring B.Let xEB.IffEA[X] isa polynomial,wemay then define theassociated polynomial function fB:B-.B byletting fB(X)=f(x)=ao+atx+...+anxn . Given anelement bEB,directly from the definition ofmultiplication of polynomials,wefind: The association evb:ft-+ f(b) isaring homomorphism ofA[X] into B. This homomorphism iscalled theevaluation homomorphism, and isalso said tobeobtained bysubstituting bforXinthepolynomial. (Cf.Proposition 3.1below.) Let xEB.We now seethat thesubring A[x]ofBgenerated byxover A isthering ofallpolynomial values f(x), forfEA[X]. Iftheevaluation map fl---+f(x) gives anisomorphism ofA[X] with A[x], then we say that xis II,3 POLYNOMIALS AND GROUP RINGS 99 transcendental over A,orthat xisavariable over A.Inparticular, Xisa variable over A. Example. Let rx=fieThen the setofallreal numbers ofthe form a+brx,with a,bEZ,isasubring ofthe real numbers, generated byfie Note that rxisnot transcendental over Z,because thepolynomial X2-2lies inthekernel oftheevaluation mapff(fi).Ontheother hand, itcan be shown that e=2.718... and 1taretranscendental over Q. SeeAppendix1. Example. Let pbe aprime number and letK =Z/pZ. Then Kisa field. Letf(X)=XP-XEK[X]. Thenfisnot the zero polynomial. But fKisthe zero function. Indeed, fK(O)=O.IfxEK, x=F0,then since the multiplicative group ofKhas order p-1,itfollows that xp-1=1,whence xP=x, sof(x)=o.Thus anon-zero polynomial gives rise tothe zero function onK. There isanother homomorphism ofthepolynomial ring having todo withthecoefficients. Let cp:A-.B be ahomomorphism ofcommutative rings. Then there isanassociated homomorphism ofthepolynomial rings A[X]-.B[X], such that f(X)=LaiXi Lcp(ai)Xi=(cpf)(X). The verification that this mapping isahomomorphism isimmediate, and further details will begiven below inProposition 3.2, in amore general context. We callf cpfthereduction map. Examples. In some applications the map cpmay be anisomorphism. Forinstance, iff(X) has complex coefficients, then itscomplex conju- gatef(X)=LaiXiisobtained byapplying complex conjugation toits coefficients. Let pbe aprime ideal ofA. Letcp:A-.A'bethe canonical homo- morphism ofAonto A/p. Iff(X) isapolynomial inA[X], then cpfwill sometimes becalled thereduction offmodulo p. For example, taking A=Zand p=(p)where pisaprime number, we can speak ofthepolynomial 3X4-X+2asapolynomial mod 5,viewing thecoefficients 3,-1,2asintegers mod 5,i.e.elements ofZ/5Z. We may now combine the evaluation map and the reduction map to generalize theevaluation map. Letq>:A Bbeahomomorphism ofcommutative rings. Let xEB.There isaunique homomorphism extending 'P A[X]-.B such that Xx, andforthishomomorphism, LaiXiLcp(ai)xi . 100 RINGS II,3 Thehomomorphism oftheabove statement may beviewed asthecomposite A[X]----+B[X] B where thefirst map applies lfJtothe coefficients ofapolynomial, and the second map istheevaluation atxpreviously discussed. Example. InChapter IX,2and3, weshall discuss such asituation in several variables, when (((Jf)(x)=0,inwhich case xiscalled azero ofthe polynomial f n When writingapolynomial f(X)=LaiXi, ifan=F0then wedefine n i=l tobethedegree offThus thedegree offisthe smallest integernsuch that ar=0for r>n.Iff=0(i.e.fisthe zero polynomial), then by con- vention wedefine thedegree offtobe-00. We agree tothe convention that -00 +-00 =-00, -00 +n=-00,-00<n, forallnEZ, and noother operation with -00 isdefined. Apolynomial of degree1isalso called alinear polynomial. Iff=F0anddegf=n,then we call antheleading coefficient offWe call aoitsconstant term. Let g(X)=bo+·..+bmxm beapolynomial inA[X], ofdegree m,and assume g=FO.Then f(X)g(X)=aob o+·..+anbmXm+n . Therefore: Ifwe assume that atleast oneoftheleading coefficients anorbmisnot a divisor of0inA,then deg(fg)=degf+deg g and theleading coefficient offgisanb m.This holds inparticular when anor bmisaunit inA,orwhen Aisentire. Consequently, when Aisentire, A[X] isalso entire. Ifforg=0,then westill have deg(fg)=degf+deg g ifweagree that -00 +m=-00 foranyinteger m. One verifies trivially that foranypolynomial f,gEA[X] wehave deg(f +g)<max(deg f,degg), again agreeing that -00 <mforevery integerm. II,3 POLYNOMIALS AND GROUP RINGS 101 Polynomials inseveral variables We now gotopolynomials inseveral variables. Let Abe asubring of acommutative ring B. LetXl'''.'XnEB. For each n-tuple ofintegers (Vl, ..., vn)=(v) ENn ,we use vector notation, letting (x)=(Xl' ..., xn),and M(v)(x)=X;1...x;n. The set ofsuch elements forms amonoid under multiplication. Let A[x]=A[Xl' ...,xn] bethesubring ofBgenerated byXl'.'"Xnover A. Then every element ofA[x] can bewritten asafinite sum La(v)M(v)(x)witha(v)EA. Using theconstruction ofpolynomials inone variable repeatedly,wemay form thering A[X 1,...,Xn]=A[X 1][X 2]...[X n], selecting Xntobe avariable over A[Xl'...,Xn-1].Then every element fof A[Xl'.. .,Xn]=A[X] has aunique expressionasafinite sum dn f=Ljj(X 1,...,Xn-1)xj j=OwithjjEA[Xl'.. ·,Xn-1]. Therefore byinduction we can writefuniquelyasasum f=(LaV1".vnX;1...X;11)x;n vn-o Vl'.",V n-l =La(v)M(v)(X)=La(v)X;1...x;n with elementsa(v)EA,which arecalled thecoefficients off.The products M(v)(X)=X;1...x;n will becalled primitive monomials. Elements ofA[X] arecalled polynomials (in nvariables). Wecalla(v)itscoefficients. Just asintheone-variable case, wehave anevaluation map. Given (x)= (xl'...,xn)andfasabove, wedefine f(x)=La(v)M(v)(x)=La(V)x;1...x;n. Then theevaluation map ev(x):A[X]-+B such thatff(x) isaring-homomorphism. Itmay beviewed asthecomposite ofthe suc- cessive evaluation maps in one variable Xi Xifor i=n,...,1,because A[X]cB[X]. Just asfor one variable, iff(X)EA[X] isapolynomial in nvariables, then weobtain afunction 102 RINGS II,3 fB:Bn B by (x) f(x). We saythatf(x) isobtained bysubstituting (x)for(X)inf,orbyspecializing (X) to(x). Asfor one variable, ifKisafinite field, andfEK[X] one may havef=F0butfK=O.Cf.Chapter IV,Theorem 1.4and itscorollaries. Next let cp:A Bbe ahomomorphism ofcommutative rings. Then we have thereduction map (generalized inProposition 3.2below) f(X)=La(v)M(v)(X) LlfJ(a(v»M(v)(X)=(lfJf)(X). We can also compose theevaluation and reduction. Anelement (x)EBnis called azero offif(lfJf)(x)=O.Such zeros will bestudied inChapter IX. Go back toAas asubring ofB. Elements xl'...,XnEBare called algebraically independent over Aiftheevaluation mapff(x) isinjective. Equivalently,wecould say thatiffEA[X] isapolynomial and f(x)=0,thenf=0;inother words, there are nonon-trivial polynomial relations among Xl'...,Xnover A. Example. Itisnot known ifeand 1tarealgebraically independent over therationals. Itisnot even known ife+1tisrational. We now come tothenotion ofdegree forseveral variables. Bythedegree ofaprimitive monomial M (X)=XVI...XVn(v) 1 n weshall mean theinteger Ivl=V1+...+Vn(which is>0). Apolynomial aXVI...XVn1 n (aEA) will becalled amonomial (not necessarily primitive). Iff(X) isapolynomial inA[X] written as f(X)=La(V)X;1...X;n, then eitherf=0,inwhich case wesaythat itsdegree is-00, orf=F0,and then wedefine thedegree offtobethe maximum ofthedegrees ofthe monomialsM(v)(X)such thata(v)=Fo.(Such monomials are said tooccur in thepolynomial.) We note that thedegree offis0ifandonly if f(X)=aoXp...Xno for some aoEA,ao=FO.We also write thispolynomial simply f(X)=ao, i.e. writing1instead of XO...XO 1 n, inother words, weidentify thepolynomial with the constant ao. II,3 POLYNOMIALS AND GROUP RINGS 103 Note that apolynomial f(X 1,...,Xn)innvariables can beviewed asa polynomial inXnwith coefficients inA[Xl'...,Xn-1](ifn>2).Indeed, we can write dn f(X)=Ljj(X l'...,Xn-1)X1, j=O where jjisanelement ofA[Xl'...,Xn-1].Bythedegree offin Xnweshall mean itsdegree when viewed as apolynomial inXnwith coefficients in A[Xl'...,Xn-1].One sees easily that ifthis degree isd,then disthelargest integer occurring asanexponent ofXninamonomial aXVI... XVn(v) 1 n witha(v)=Fo.Similarly, we define the degree offineach variable Xi (i=1,.. .,n). The degree offineach variable isofcourse usually different from its degree (which issometimes called thetotal degree ifthere isneed toprevent ambiguity). For instance, XfX 2+xi hastotal degree 4,and hasdegree 3inXland 2inX2. As amatter ofnotation, weshall often abbreviate "degree" by"deg." For each integer d>0,given apolynomial f,letfed) bethe sum ofall monomials occurring infandhaving degree d.Then f=Lf(d). d Suppose f=FO.We saythatfishomogeneous ofdegree diff=fed); thusf can bewritten intheform f(X)= aXVI... XVn(v) 1 n with V1+...+Vn=d ifa(v)=FO. We shall leave itasanexercise toprove that anon-zero polynomial finn variables over Aishomogeneous ofdegree difandonly if,for every setof n+1algebraically independent elements u,tl'..., tnover Awehave f(ut l'·..,utn)=u4j(t l'.. .,tn). We note that iff,garehomogeneous ofdegree d,erespectively, and fg=F0,then fgishomogeneous ofdegree d+e.Ifd=eandf+g=F0,then f+gishomogeneous ofdegree d. Remark. Inview oftheisomorphism A[Xl'...,Xn] A[tl'...,tn] between thepolynomial ring innvariables and aring generated over Abyn 104 RINGS II,3 algebraically independent elements, we canapply alltheterminologywehave defined forpolynomials, toelements ofA[tl'.. .,tn].Thus we can speak of thedegree ofanelement inA[t],and therules forthedegree ofaproductor sum hold. Infact, weshall also call elements ofA[t]polynomials in(t). Algebraically independent elements will also becalled variables (orindepen- dent variables), and any distinction which wemake between A[X] and A[t] ismore psychological than mathematical. Suppose next that Aisentire. Bywhat weknow ofpolynomials inone variable and induction, itfollows that A[Xl'...,Xn] isentire. Inparticular, suppose fhasdegree dand ghasdegreee.Write f=fed) +terms oflower degree, g=gee)+terms oflower degree. Then fg=f(d)g(e) +terms oflower degree, and iffg=F0then f(d)g(e) =Fo. Thus wefind: deg(fg)=degf+deg g, deg(f +g)<max(deg f,degg). We are now finished with the basic terminology ofpolynomials. We end this section byindicating how theconstruction ofpolynomials isactuallya special case ofanother construction which isused inother contexts. Inter- ested readers canskip immediately toChapter IV,giving further important properties ofpolynomials. See also Exercise 33ofChapter XIII for har- monic polynomials. The group ring ormonoid ring Let Abe acommutative ring. Let Gbe amonoid, written multiplica- tively. Let A[G] bethe setofallmaps :G-.Asuch that(x)=0foralmost all xEG. We define addition inA[G] tobetheordinary addition of mappings into anabelian (additive) group. If, pEA [G], wedefine their product pbytherule (P)(z)=L(x)P(y). xy=z The sum istaken over allpairs (x,y)with x,yEGsuch that xy=z.This sum isactually finite, because there isonlyafinite number ofpairs of elements (x,y)EGxGsuch that(x)P(y) =Fo.We also seethat(P)(t)=0 foralmost allt,and thus belongs toour setA[G]. The axioms for aringaretrivially verified. We shall carry out theproof ofassociativityasanexample. Let, p,YEA [G]. Then II,3 POLYNOMIALS AND GROUP RINGS 105 «(p)y) (z)=L(P)(x)y(y) xy=z =xzLxo((U)p(v)]y(y) -xzLxO((U)P(V)y(y)] =L(u)P(v)y(y), (U,v,y) uvy=z this last sum being taken over alltriples (uv,y)whose product isz.This last sum isnow symmetric, and ifwehadcomputed (a(f3y»(z),wewould have found this sum also. This proves associativity. The unit element ofA[G] isthe function bsuch that b(e)=1and b(x)=0forall xEG,x=Fe.Itistrivial toverify that =b =bforall EA[G]. We shall now adoptanotation which will make the structure ofA[G] clearer. Let aEAand xEG.We denote bya.x(and sometimes also byax) thefunction whose value atxisa,and whose value atyis0ify=Fx.Then anelement EA[G] can bewritten asasum =L(x).x. xeG Indeed, if{ax}xeGisasetofelements ofAalmost allofwhich are0,and we set p=Lax.x, xeG then forany yEGwehave P(y)=ay(directly from thedefinitions). This also shows that agiven element admits aunique expressionasasumLax.x. With our present notation, multiplicationcan bewritten (Lax.X)(Lby.Y)=Laxby.xy xeG yeG x,y and addition can bewritten Lax.x+Lbx.x=L(ax+bx).x, xeG xeG xeG which looks theway wewant ittolook. Note that theunit element ofA[G] issimply1.e. We shall now seethat we can embed both Aand Ginanatural way in A[G]. LetCPo: G-.A[G] bethe map given byCPo(x)=1.x.Itisimmediately verified thatCPoisamultiplicative monoid-homomorphism, and isinfact injective, i.e. anembedding. Let10:A-.A[G] bethemap given by 10(a)=a.e. 106 RINGS II,3 Itisimmediately verified that foisaring-homomorphism, and isalso an embedding. Thus weview Aas asubring ofA[G]. One calls A[G] the monoid ring ormonoid algebra ofGover A,orthe group algebra ifGisa group. Examples. When Gisafinite group and A=kisafield, then thegroup ring kEG] will bestudied inChapter XVIII. Polynomial rings arespecialcases ofthe above construction. In nvari- ables, consider amultiplicative free abelian group ofrank n.LetXl' ...,Xn begenerators. Let Gbethemultiplicative subset consisting ofelements X;l...X;" with Vi>0forall i.Then Gisamonoid, and the reader can verify atonce that A[G] isjust A[Xl'...,Xn]. As amatter ofnotation weusually omit thedot inwritinganelement of theringA[G], sowewrite simply Laxxforsuch anelement. More generally, letI={i}be aninfinite family ofindices, and letSbe thefree abelian group with freegenerators Xi'written multiplicatively. Then we canform thepolynomial ring A[X]bytaking themonoid toconsist ofproducts M(v)(X)=nXiVi , ieI where ofcourse allbut afinite number ofexponents Viareequal too.IfAis asubring ofthecommutative ring B,and Sisasubset ofB,then weshall also usethefollowing notation. Let v:S-.Nbeamapping which is0except for afinite number ofelements ofS.Wewrite M(v)(S)=nxvex). xeS Thus wegetpolynomials ininfinitely many variables. One interesting exam- pleofthe useofsuch polynomials will occur inArtin's proof oftheexistence ofthealgebraic closure ofafield, cf.Chapter V,Theorem 2.5. We now consider the evaluation and reduction homomorphisms inthe present context ofmonoids. Proposition 3.1. Let cp:G-.G'be ahomomorphism ofmonoids. Then there exists aunique homomorphism h:A[G]-.A[G'] such that h(x)= cp(x)forallxEGandh(a)=aforall aEA. Proof Infact, let rx=LaxxEA[G]. Define h(rx)=Laxcp(x). Then hisimmediately verified tobeahomomorphism ofabelian groups, and h(x)=cp(x). Letp=LbyY.Then h(rxP)=C%axby)qJ(z). We get h(rxP)=h(rx)h(P) immediately from the hypothesis that cp(xy)= II,4 LOCALIZATION 107 qJ(X)qJ(Y). If eisthe unit element ofG,then bydefinition qJ(e)=e', so Proposition 3.1follows. Proposition 3.2. Let Gbeamonoid and letf:A-+Bbeahomomorphism ofcommutative rings. Then there isaunique homomorphism such thath:A[G]-+B[G] h(Laxx)=Lf(ax)x. xeG xeG Proof Since every element ofA[G] has aunique expressionas asum Laxx,theformula giving hgivesawell-defined map from A[G] into B[G]. This map isobviouslyahomomorphism ofabelian groups. Asformultipli- cation, let Thenil=Laxx and p=LbyY. heap)=ZGfCZ axby)Z =LLf(ax)f(by)z zeGxy=z =f(il)f(P). We have trivially h(l)=1,sohisaring-homomorphism,as was tobe shown. Observe that viewing Aasasubring ofA[G], therestriction ofhtoAis thehomomorphism fitself. Inother words, ifeistheunit element ofG, then h(ae)=f(a)e. 4. LOCALIZATION Wecontinue toletAbeacommutative ring. Byamultiplicative subset ofAweshall mean asubmonoid ofA(viewed asamultiplicative monoid according toRI2).Inother words, itisasubset Scontaining 1,and such that, ifx,YES, then xYES. We shall now construct thequotient ring ofAbyS,also known asthe ring offractions ofAbyS. Weconsider pairs (a,s)with aEAand sES.We define arelation (a,s) (a',s') between such pairs, bythecondition that there exists anelement SlESsuch 108 RINGS II,4 that S1(s'a-sa')=O. Itisthen trivially verified that this isanequivalence relation, and the equivalence class containingapair (a,s)isdenoted bya/so The setof equivalence classes isdenoted byS-1A. Note that if0ES,then S-1Ahaspreciselyone element, namely 0/1. Wedefine amultiplication inS-1Abytherule (a/s)(a'/s')=aa'/ss'. Itistrivially verified that this iswell defined. This multiplication has aunit element, namely 1/1, and isclearly associative. We define anaddition inS-1Abytherule aa's' a+sa'-+-= s s' ss' Itistrivially verified that this iswell defined. As anexample, wegive the proof indetail. Leta1/s1=a/s, and leta;/s=a'/s'. We must show that (s;a 1+S1a;)/s1s;=(s'a+sa')/ss'. There existS2, S3ESsuch that s2(sa 1-S1a)=0, (" , ')0S3Sa1-S1a =. Wemultiply thefirst equation byS3S'S; and the second by S2SS1. We then add, and obtain S2S3[s's(sa1-s1a)+ss1(s'a -sa')] =0. Bydefinition, this amounts towhat we want toshow, namely that there exists anelement ofS(e.g. S2S3) which when multiplied with ss'(sa1+S1a;)-S1s(s'a+sa') yields O. We observe that givenaEAand s,s'ESwehave a/s=s'a/s's. Thus this aspect oftheelementary properties offractions still remains true in ourpresent general context. Finally, itisalso trivially verified that our two laws ofcomposition on S-1Adefine aring structure. We let ({Js:A S-1A bethe map such thatCPs<a)=a/I. Then one sees atonce thatCPsIS a II,4 LOCALIZATION 109 ring-homomorphism. Furthermore, every element ofCfJs(S) isinvertible In S-1A(the inverse ofs/lisl/s). Letebethecategory whose objectsarering-homomorphisms f:A-+B such that foreverysES,theelement f(s) isinvertible inB.Iff:A-+Band f':A-+B'are two objects ofe, amorphism goffintof'isahomo- morphism g:B-+B' making thediagram commutative: Af) Bf\f B' Wecontend thatCfJsisauniversal object inthis category e. Proof. Suppose that a/s=a'/s', orinother words that thepairs (a,s) and (a',s')areequivalent. There exists S1ESsuch that s1(s'a-sa')=O. Letf:A-+Bbeanobject ofe.Then f(S1) (f(s')f(a)-f(s)f(a')J=O. Multiplying byf(s 1)-1, and then byf(S')-1 andf(S)-1, weobtain f(a)f(s)-1=f(a')f(s')-1. Consequently,we can define amap h:S-1A-+B such that h(a/s)=f(a)f(s)-1, foralla/sES-1A.Itistrivially verified that h isahomomorphism, and makes the usual diagram commutative. Itisalso trivially verified that such amap hisunique, and hence thatCfJsisthe required universal 0bject. Let Abeanentire ring, and letSbeamultiplicative subset which does not contain O.Then CfJs:A-+S-1A isinjective. Indeed, bydefinition, ifa/I=0then there exists sESsuch that sa =0, and hence a=o. The most importantcases ofamultiplicative setSarethefollowing: 1.Let Abe acommutative ring, and let Sbethe setofinvertible elements ofA(i.e. the setofunits). Then Sisobviously multiplicative, and is 110 RINGS 1I,4 denoted frequently byA*.IfAisafield, then A*isthemultiplicative group ofnon-zero elements ofA.Inthat case, S-1Aissimply Aitself. 2.Let Abeanentire ring, and letSbethe setofnon-zero elements ofA. Then Sisamultiplicative set, and S-1Aisthen afield, called thequotient field orthefield offractions, ofA.Itisthen customary toidentify Aasa subset ofS-1A,and we can write als=S-la for aEAand sES. We have seen in3that when Aisanentire ring, then A[X 1,...,X nJis also entire. IfKisthequotient field ofA,thequotient field ofA[X 1,...,XnJ isdenoted byK(X l'...,Xn).Anelement ofK(X l'...,Xn)iscalled arational function. Arational function can bewritten as aquotient f(X)/g(X) where f,garepolynomials. If(b1,...,bn)isinK(n), and arational function admits anexpressionas aquotient fig such that g(b) =F0,then we say that the rational function isdefined at(b). From general localization properties,we seethat when this isthe case, we can substitute (b)intherational function to getavalue f(b)/g(b). 3.Aring Aiscalled alocal ring ifitiscommutative and has aunique maximal ideal. IfAisalocal ring and misitsmaximal ideal, and xEA, x m,then xisaunit(otherwise xgeneratesaproper ideal, notcontained inm, which isimpossible). Let Abe aring and paprime ideal. Let Sbethe com- plement ofpinA.Then Sisamultiplicative subset ofA,andS-IAisdenoted byAp.Itisalocal ring (cf.Exercise 3)and iscalled thelocal ring ofAatp.Cf. theexamples ofprincipal rings, and Exercises 15, 16. Let Sbe amultiplicative subset ofA.Denote byJ(A) the setofideals of A.Then we can define amap t/ls:J(A)-+J(S-1 A); namely welett/ls(a)=S-1 abethe subset ofS-1Aconsisting ofallfractions als with aEaand sES.The reader will easily verify that S-1 aisan S-1A-ideal, and that t/lsisahomomorphism for both the additive and multiplicative monoid structures onthe setofideals J(A). Furthermore, t/ls also preserves intersections and inclusions; inother words, forideals a,bof Awehave: S-I(a +b)=S-1 a+S-1b, S-I(ab)=(S-1a)(S-1 b), S-I(a nb)=S-1anS-1b. As anexample, weprove this last relation. Let xEanb.Then xis isin S-1 aand also inS-1b,sothe inclusion istrivial. Conversely, suppose we have anelement ofS-1Awhich can bewritten asals=bls' with aEa,bEb, and s,s'ES.Then there exists s1ESsuch that sls'a=slsb, II,5 PRINCIPAL AND FACTORIAL RINGS 111 and this element liesinboth aand b.Hence a/s=sls'a/s1s's liesinS-l(anb), aswas tobeshown. 5. PRINCIPAL AND FACTORIAL RINGS Let Abeanentire ring. Anelement a=F0iscalled irreducible ifitisnot a unit, and ifwhenever one can write a=bewith bEAand eEAthen bor e isaunit. Let a=F0be anelement ofAand assume that theprincipal ideal (a) is prime. Then aisirreducible. Indeed, ifwewrite a=be,then bor elies in (a), say b.Then we can write b=adwith some dEA,and hence a=acd. Since Aisentire, itfollows that cd =1,inother words, that eisaunit. The converse ofthepreceding assertion isnot always true. We shall discuss under which conditions itistrue. Anelement aEA,a=F0,issaid to have aunique factorization into irreducible elements ifthere exists aunit u and there exist irreducible elements Pi(i=1,...,r)inAsuch that r a=unPi' i=l andifgiven two factorizations into irreducible elements, r s a=unPi=u'nqj,i=l j=l wehave r=s,and after apermutation oftheindices i,wehave Pi=uiqi for some unit UiinA,i=1,..., r. We note that ifPisirreducible and uisaunit, then upisalso irreducible, so we must allow multiplication byunits in afactorization. Inthering ofintegers Z,theordering allows ustoselect arepresentative irreducible element (aprime number) out oftwo possibleones differing byaunit, namely +P,byselecting thepositiveone. This is,ofcourse, impossible in more general rings. Takingr=0above, weadopt the convention that aunit ofAhas a factorization into irreducible elements. Aring iscalled factorial (oraunique factorization ring) ifitisentire andif every element =F0has aunique factorization into irreducible elements. We shall prove below that aprincipal entire ring isfactorial. Let Abeanentire ring and a,bEA,ab =FO.We saythat a-divides band write aIbifthere exists eEAsuch that ac =b.We saythat dEA,d=F0,isa greatest common divisor (g.c.d.) ofaand bifdla,dlb, andifany element e ofA, e=f=.0,which divides both aand balso divides d. 112 RINGS II,5 Proposition 5.1. Let Abe aprincipal entire ring and a,bEA,a,b=FO. Let(a,b)=(c). Then cisagreatest common divisor ofaand b. Proof Since blies intheideal (c), we can write b=xcfor some xEA, sothat clb. Similarly, cia. Let ddivide both aand b,and write a=dy, b=dzwith y,zEA.Since cliesin(a,b)we can write c=wa+tb with some w,tEA. Then c=wdy+tdz =d(wy +tz),whence die, and our proposition isproved. Theorem 5.2. Let Abeaprincipal entire ring. Then Aisfactorial. Proof We first prove that every non-zero element ofAhas afactoriza- tion into irreducible elements. Let Sbethe setofprincipal ideals =F0whose generators donot have afactorization into irreducible elements, and suppose Sisnotempty. Let(at) beinS.Consider anascending chain (at) (a2)...(an).. . ofideals inS.We contend that such achain cannot beinfinite. Indeed, the union ofsuch achain isanideal ofA,which isprincipal, sayequal to(a). The generator amust already lieinsome element ofthechain, say(an)' and then we seethat (an)c(a)c(all)' whence thechain stopsat(an). Hence Sis inductively ordered, and has amaximal element (a). Therefore any ideal ofA containing (a)and -:1=(a)has agenerator admittingafactorization. We note that ancannot beirreducible (otherwise ithas afactorization), and hence we can write a=bewith neither bnor cequaltoaunit. But then (b)-:1=(a)and(c) -:1=(a)and hence both b,cadmit factorizations into irreducible elements. The product ofthese factorizations isafactorization for a,contra- dicting theassumption that Sisnotempty. Toprove uniqueness, wefirst remark that ifpisanirreducible element of Aand a,bEA,plab, then pia orplb. Proof: Ifpa,then theg.c.d. ofp,a is1and hence we can write 1=xp+ya with some x,YEA. Then b=bxp +yab, and since plab weconclude that plb. Suppose that ahas two factorizations a=Pt...Pr=qt...qs into irreducible elements. Since Ptdivides theproduct farthest totheright, Ptdivides one ofthefactors, which wemay assume tobeqtafter renum- bering these factors. Then there exists aunit Utsuch that qt=UtPt. We can now cancel Ptfrom both factorizations and get II,5 PRINCIPAL AND FACTORIAL RINGS 113 P2...Pr=U1q2...qs. The argument iscompleted byinduction. We could call two elements a,bEAequivalent ifthere exists aunit u such that a=bu. Let usselect one irreducible element pout ofeach equivalence class belonging tosuch anirreducible element, and let usdenote byPthe setofsuch representatives. Let aEA, a=FO.Then there exists a unit uand integers v(p)>0,equal to0foralmost allpEP such that a=unpV(P). peP Furthermore, theunit uand theintegers v(p)areuniquely determined by a. We callv(p) theorder ofaatp,also written ordpa. IfAisafactorial ring, then anirreducible element pgeneratesaprime ideal (p). Thus inafactorial ring, anirreducible element will also becalled a prime element, orsimplyaprime. We observe that one can define the notion ofleast common multiple (l.c.m.) ofafinite number ofnon-zero elements ofAintheusual manner: If aI' ..., anEA aresuch elements, wedefine al.c.m. forthese elements tobeanyCEAsuch that forallprimes pofAwehave ordpc=max ordpai. i This element Ciswell defined uptoaunit. Ifa,bEAare non-zero elements, wesay that a,bare relaively prime if theg.c.d. ofaand bisaunit. Example. The ring ofintegers Zisfactorial. Itsgroup ofunits consists of 1and-1.Itisnatural totake asrepresentative prime element the positive prime element (what iscalled aprime number) pfrom the two possible choices pand-p.Similarly, weshall show later that thering of polynomials inone variable over afield isfactorial, and one selects represen- tatives fortheprime elements tobetheirreducible polynon1ials with leading coefficient 1. Examples. Itwill beproved inChapter IVthat ifRisafactorial ring, then thepolynomial ring R[Xl'...,Xn] innvariables isfactorial. Inpartic- ular, ifkisafield, then thepolynomial ringk[X 1,...,Xn] isfactorial. Note thatk[X1]isaprincipal ring, but for n>2,theringk[X l'...,Xn] isnot princi pal. InExercise 5you will prove that thelocalization ofafactorial ring is factorial. InChapter IV,9 we shall prove that the power series ring k[[X l'...,Xn]] isfactorial. This result isaspecialcase ofthe more general statement that aregular local ring isfactorial, but wedonot define regular local rings inthis book. You can look them upinbooks oncommutative 114 RINGS II,Ex algebra. Irecommend: H.MATSUMURA, Commutative Algebra, second edition, Benjamin-Cummings, New York, 1980 H.MATSUMURA, Commutative Rings, Cambridge University Press, Cambridge, UK, 1986 Examples from algebraic andcomplex geometry. Roughly speaking, reg- ular local rings arise inthefollowing context ofalgebraic orcomplex geom- etry. Consider thering ofregular functions intheneighborhood ofsome point on acomplex oralgebraic manifold. This ring isregular. Atypical example isthering ofconvergent power series inaneighborhood of0inen. InChapter IV, weshall provesome results onpower series which give some algebraic background for those analytic theories, and which are used in proving thefactoriality ofrings ofpower series, convergentornot. Conversely totheabove examples, singularities ingeometric theories may give rise toexamples ofnon-factoriality. Wegive examples using notions which aresufficiently basic sothat readers should have encountered them in more elementary courses. Examples ofnon-factorial rings. Let kbe afield, and letxbe avariable over k.Let R=k[x2 ,x3].Then Risnotfactorial (proof?). The ring Rmay beviewed asthering ofregular functions onthe curve y2=x3 ,which has a singularity attheorigin,asyoucan seebydrawing itsrealgraph. Let Rbethe setofallnumbers oftheform a+b,where a,bEZ. Then theonly units ofRare +1,and the elements 3,2+ ,2- areirreducible elements, giving rise toanon-unique factorization 32=(2+ )(2-). (Do Exercise 10.) Here thenon-factoriality isnot due tosingularities but due toanon-trivial ideal class group ofR,which isaDedekind ring. For a definition see the exercises ofChapter III, orgostraight tomy book Algebraic Number Theory, forinstance. AsTrotter once pointed out(Math. Monthly, April 1988), therelation sin2x=(1+cosx)(l-cosx) may beviewed as anon-unique factorization inthering oftrigonometric polynomials R[sin x,cosx],generated over Rbythe functions sinxand cos x.This ring isasubring ofthering ofallfunctions, orofalldifferenti- able functions. See Exercise 11. EXERC1SES WeletAdenote acommutative ring. 1.Suppose that 1:F0inA.Let Sbeamultiplicative subset ofAnotcontaining o. Let pbeamaximal element inthe setofideals ofAwhose intersection with Sis empty. Show that pisprime. II,Ex EXERCISES 115 2.Letf:A A'be asurjective homomorphism ofrings, and assume that Aislocal, A':Fo.Show that A'islocal. 3.Let pbeaprime ideal ofA.Show thatA"has aunique maximal ideal, consisting ofallelements a/swith aEpand sp. 4,Let Abeaprincipal ring and Samultiplicative subset with 0 S,Show thatS-lAis principal. 5.Let Abeafactorial ring and Samultiplicative subset with 0S.Show thatS-lAis factorial, and that theprime elements ofS-lAare those primes pofAsuch that (p)nSisempty. 6.Le!Abe afactorial ring and paprime element. Show that thelocal ring A(p)is principal. 7.Let Abe aprincipal nng and a1,..., a" non-zero elements ofA. Let (a1, ...,a,,)=(d). Show that dis agreatest common divisor for the ai (i=1,.. .,n). 8.Let pbeaprime number, and letAbetheringZ/prz (r=integer>1).Let Gbe the group ofunits inA,i.e.the group ofintegers prime top,modulo pro Show that Giscyclic, except inthe case when p=2, r>3, inwhich case itisoftype (2,2r-2).[Hint: Inthegeneral case, show that Gis theproduct ofacyclic group generated by1+p,and acyclic group oforder p-1.Intheexceptional case, show that Gistheproduct ofthe group {+1} with thecyclic group generated bytheresidue class of5mod 2r.] 9.Let ibethecomplex numberR.Show that theringZ[i] isprincipal, and hence factorial. What aretheunits? 10.Let Dbe aninteger>1,and letRbethe setofallelement a+b with a,bEZ. (a)Show that Risaring. (b)Using the fact that complex conjugation isanautomorphism ofC,show that complex conjugation induces anautomorphism ofR. (c)Show that ifD>2then theonly units inRare +1. (d)Show that 3,2+R,2-Rareirreducible elements inZ[R ]. 11.Let Rbethering oftrigonometric polynomialsasdefined inthe text. Show that Rconsists ofallfunctions fonRwhich have anexpression oftheform " f(x)=ao+L(am cos mx+bmsinmx), m=1 where ao, am, bmare real numbers. Define thetrigonometric degree degtr(f) tobe themaximum oftheintegers r,ssuch that ar,bs:Fo.Prove that degtr(fg)=degtr(f) +degtr(g). Deduce from this that Rhas nodivisors of0,and also deduce that thefunctions sin xand 1-cos xareirreducible elements inthat ring. 116 RINGS II,Ex 12.Let Pbethe setofpositive integers and Rthe setoffunctions defined onPwith values inacommutative ring K.Define the sum inRtobetheordinary addition offunctions, and define theconvolution product bytheformula (f*g)(m)=Lf(x)g(y), xy=m where the sum istaken over allpairs (x,y)ofpositive integers such that xy=m. (a)Show that Risacommutative ring, whose unit element isthefunction such that(1)=1and(x)=0ifx:F1. (b)Afunction fissaid tobemultiplicative iff(mn)=f(m)f(n) whenever m, nare relatively prime. Iff,garemultiplicative, show thatf*gismultiplicative. (c)LetJ1.betheMobius function such that J1.(I)=1,J1.(Pt...Pr)=(_I)r ifPt, ..., Pr are distinct primes, and J1.(m)=0ifmisdivisible byp2for some prime p. Show thatJ1.*tpt=,where tptdenotes the constant function having value 1.[Hint: Show first thatJ1.ismultiplicative, and then prove the assertion forprime powers.] The Mobius inversion formula ofelementary number theory isthen nothing else but therelationJ1.*tpt*f=f. Oedekind rings Prove thefollowing statements about aDedekind ringo.Tosimplify terminology, by anideal we shall mean non-zero ideal unless otherwise specified. We letK denote thequotient field ofo. 13.Every ideal isfinitely generated. [Hint: Given anideal a,letbbethefractional ideal such that ab =o.Write 1=Laibi with aiEaand biEb.Show that a=(at, ...,all).] 14.Every ideal has afactorization asaproduct ofprime ideals, uniquely determined uptopermutation. 15.Suppose0hasonly one prime ideal p.Let tEPand tp2. Then p=(t)is princi pal. 16.Let 0beany Dedekind ring. Let pbeaprime ideal. Let0"bethelocal ring at p.Then0"isDedekind and hasonly oneprime ideal. 17.Asfortheintegers, wesaythat alb(adivides b)ifthere exists anideal csuch that b=ac.Prove: (a)albifandonly ifbca. (b)Let a,bbeideals. Then a+bistheir greatestcommon divisor. Inparticular, a,barerelatively prime ifandonly ifa+b=o. 18.Every prime ideal pismaximal. (Remember, p:F0byconvention.) Inparticular, ifPI' ..., PIIaredistinct primes, then the Chinese remainder theorem applies to h.'1 rU h.telrpowers PI' ..., PII".setIStoprove: 19.Let a,bbeideals. Show that there exists anelement CEK(the quotient field of 0)such that caisanideal relatively prime tob.Inparticular, every ideal class in Pic(o) contains representative ideals prime toagiven ideal. Foracontinuation, seeExercise 7ofChapter VII. CHAPTER III Modules Although thischapter islogically self-contained andprepares forfuture topics, inpractice readers will have had some acquaintance with vector spaces over a field .Wegeneralize this notion here tomodules over rings. Itisastandard fact (tobereproved) that avector space has abasis, butformodules this isnotalways the case. Sometimes they do; most often they donot. We shall look into cases where they do. Forexamples ofmodules andtheir relations tothose which have abasis, the reader should look atthe comments made attheend of4. 1. BASIC DEFINITIONS Let Abearing. Aleftmodule over A,oraleftA-module Misanabelian group, usually written additively, together with anoperation ofAonM(viewing Aasamultiplicative monoid byRI2),such that, forall Q,bEAand x,yEM wehave (a+b)x=ax+bx and a(x+y)=ax+aYe We leave itasanexercise toprove that a(-x)=-(ax) and that Ox =O.By definition ofanoperation, wehave 1x=x. Inasimilar way, one defines aright A-module. Weshall deal only with left A-modules, unless otherwise specified, and hence call these simply A-modules, oreven modules ifthereference isclear. 117 118 MODULES III,1 LetMbeanA-module. Byasubmodule NofMwe mean anadditive sub- group such that AN cN.Then Nisamodule (with theoperation induced by that ofAonM). Examples We note that Aisamod ule over itself. Any commutative group isaZ-module. Anadditive group consisting of0alone isamodule over anyring. Any leftideal ofAisamodule over A. Let} beatwo-sided ideal ofA.Then thefactor ringAI} isactuallyamodule over A.IfaEAand a+}isacoset of}inA,then one defines theoperation tobea(x+})=ax+}.The reader canverify atonce that this defines amodule structure onAI}. More general, ifMisamodule and Nasubmodule, weshall define thefactor module below. Thus ifLisaleft ideal ofA,thenAlLis also amodule. For more examples inthisvein, see4. Amodule over afield iscalled avector space. Even starting with vector spaces,one isledtoconsider modules over rings. Indeed, letVbeavector space over thefield K.The reader nodoubt already knows about linear maps (which will berecalled below systematically). Let Rbethering ofalllinear maps ofV into itself. Then Visamodule over R.Similarly, ifV=Kndenotes the vector space of(vertical) n-tuples ofelements ofK,and Risthering ofnxnmatrices with components inK,then Visamodule over R.For more comments along these lines, see theexamples attheendof2. Let Sbe anon-emptyset and M anA-module. Then the setofmaps Map(S, M) isanA-module. We have already noted previously that itisacom- mutative group, and forfEMap(S, M), aEAwedefine aftobethe map such that(aj)(s)=af(s). The axioms for amodule arethen trivially verified. For further examples,seetheend ofthis section. For the restofthissection, wedeal with afixed ring A,and hence may omit theprefix A-. Let Abe anentire ring and letMbe anA-module. We define thetorsion submodule Mtortobethe subset ofelements xEMsuch that there exists aEA,a=f=.0such that ax=o.Itisimmediately verified thatMtorisasubmodule. Itsstructure inanimportantcase will bedetermined in 7 . Let abe aleftideal, and Mamodule. We define aM tobethe setofall elements atXt +...+anx n with aiEaand XiEM.Itisobviouslyasubmodule ofM.Ifa,bareleftideals, then wehave associativity, namely a{bM)={ab)M. III,1 BASIC DEFIN ITIONS 119 We also have some obvious distributivities, like (a+b)M=aM +bM. If N,N'aresubmodules ofM,then a(N +N')=aN+aN'. LetMbeanA-module, and Nasubmodule. We shall define amodule structure onthefactor group M/N (for theadditive group structure). Let x+Nbe acoset ofNinM,and let aEA. We define a(x +N)tobethe coset ax+N.Itistrivial toverify that this iswell defined (i.e.ifyisinthe same coset asx,then ayisinthe same coset asax), and that this isanopera- tion ofAonM/Nsatisfying therequired condition, making M/Ninto a module, called thefactor module ofMbyN. Byamodule-homomorphism one means amap f:M-.M' ofone module into another (over the same ring A),which isanadditive group- homomorphism, and such that f(ax)=af(x) forall aEAand xEM.Itisthen clear that thecollection ofA-modules isa category, whose morphisms are themodule-homomorphisms usually also called homomorphisms forsimplicity, ifnoconfusion ispossible. Ifwewish torefer tothering A,wealso saythatfisanA-homomorphism, oralso that itisanA-linear map. IfMisamodule, then theidentity map isahomomorphism. For any module M', themap (:M-.M'such that(x)=0forallxEM isahomo- morphism, called zero. Inthe next section, weshall discuss thehomomorphisms of amodule into itself, and asaresult weshall give further examples ofmodules which arise in practice. Here wecontinue totabulate thetranslation ofbasic properties ofgroups tomodules. LetMbe amodule and Nasubmodule. We have thecanonical additive group-homomorphism f:M-.M/N and one verifies trivially that itisamodule-homomorphism. Equally trivially, one verifies thatfisuniversal inthecategory ofhomo- morphisms ofMwhose kernel contains N. Iff:M-.M' isamodule-homomorphism, then itskernel and imageare submodules ofMandM'respectively (trivial verification). Let!: M M'beahomomorphism. Bythecokernel of!wemean thefactor module M'/Im!=M'/!(M). One may also mean thecanonical homomorphism 120 MODULES III,1 M' M'/!(M) rather than themodule itself. The context should make clear which ismeant. Thus thecokernel isafactor module ofM' . Canonical homomorphismsdiscussed inChapter I,3applytomodules mutatis mutandis. For the convenience ofthe reader, we summarise these homomorphisms: LetN,N'betwo submodules ofamodule M. Then N+N'isalso asub- module, and wehave anisomorphism Nj(N nN') (N+N')jN'. IfM ::JM' ::JMil aremodules, then (MjM")j(M'jM") MjM'. Iff:M-.M'isamodule-homomorphism, and N'isasubmodule ofM', then f-l(N') isasubmodule ofMand wehave acanonical injective homomorphism J:Mjf-l(N')-.M'jN'. Iff issurjective, thenJisamodule-isomorphism. The proofs areobtained byverifying that allhomomorphisms which ap- peared when dealing with abelian groups are now A-homomorphisms of modules. We leave theverification tothereader. Aswith groups,weobserve that amodule-homomorphism which isbijective isamodule-isomorphism. Here again, theproof isthe same asforgroups, adding only theobservation that theinverse map, which weknow isagroup- isomorphism, actually isamodule-isomorphism. Again,weleave theverifica- tion tothereader. Aswith abelian groups,wedefine asequence ofmodule-homomorphisms M'1.M Mil tobeexact ifImf=Ker g.We have anexact sequence associated with a submodule Nofamodule M,namely o-.N-.M-.MjN-.0, themap ofNinto Mbeing theinclusion, and thesubsequent map being the canonical map. The notion ofexactness isdue toEilenberg-Steenrod. Ifahomomorphismu:N-.Missuch that O-.NM isexact, then wealso saythat uisamonomorphism oranembedding. Dually, if uN-.M-.O isexact, wesaythat uisanepimorphism. III,1 BASIC DEFINITIONS 121 Algebras There are some things inmathematics which satisfy alltheaxioms ofaring except forthe existence of aunit element. We gave theexample ofL}(R) in Chapter II, 1.There arealso some things which donotsatisfy associativity, butsatisfy distributivity. For instance letRbe aring, and forx,yERdefine thebracket product [x,y]=xy-yx . Then this bracket product isnotassociative inmost cases when Risnot com- mutative, butitsatisfies thedistributive law. Examples. Atypical example isthering of-differential operators with Coo coefficients, operatingonthering ofCoofunctions on anopen setinRn.The bracket product [D}, D2]=D}0D2-D20D} oftwo differential operators isagainadifferential operator. Inthetheory ofLie groups, thetangent spaceattheorigin also has such abracket product. Such considerations lead ustodefine amore general notion than aring. Let Abe acommutative ring. LetE,Fbemodules. Byabilinear map g:ExE F we mean amap such that given xEE,the map y.-..+g(x, y)isA-linear, and given yEE,the mapx g(x, y)isA-linear. ByanA-algebrawe mean a module together with abilinear map g:ExE E.We view such amapasa lawofcompositiononE.But inthisbook, unless otherwise specified,weshall assume that ouralgebrasareassociative and have aunit element. Aside from theexamples already mentioned, we note that thegroup ring A[G] (ormonoid ring when Gisamonoid) isanA-algebra, also called thegroup (ormonoid) algebra. Actually thegroup algebra can beviewed as aspecial case ofthefollowing situation. Letf:A Bbe aring-homomorphism such thatf(A) iscontained inthe center ofB,i.e.,f(a) commutes with every element ofBforeveryaEA. Then wemay view BasanA-module, defining theoperation ofAonBbythemap (a,b) f(a)b forall aEAand bEB.The axioms for amodule aretrivially satisfied, and the multiplicative lawofcomposition BxB Bisclearly bilinear (i.e., A-bilinear). Inthisbook, unless otherwise specified, byanalgebra over A,weshall always mean aring-homomorphismasabove. We say that thealgebra isfinitely gen- erated ifBisfinitely generatedasaring overf(A). Several examples ofmodules over apolynomial algebra oragroup algebra will begiven inthe next section, where we also establish thelanguage of representations. 122 MODULES III,2 2. THE GROUP OF HOMOMORPHISMS Let Abearing, and letX,X'beA-modules. Wedenote byHomA(X', X) the setofA-homomorphisms ofX'into X.Then HomA(X', X)isanabelian group, thelawofaddition being that ofaddition formappings into anabelian group. IfAiscommutative then we can make HomA(X', X)into anA-module, by defining affor aEAandfE HomA(X', X)tobethemap such that (af)(x)=af(x). Theverification that theaxioms foranA-module aresatisfied istrivial. However, ifAisnotcommutative, then weview HomA(X', X)simplyasanabelian group: We also view HomAasafunctor. Itisactuallyafunctor oftwo variables, contravariant inthefirst and covariant inthe second. Indeed, letYbe an A-module, and let X'!.X beanA-homomorphism. Then wegetaninduced homomorphism HomACt: Y):HomA(X, Y)-.HomA(X', Y) (reversing thearrow!) given by ggof This isillustrated bythefollowing sequence ofmaps: X'!.X Y. The fact that HomACt: Y)isahomomorphism issimplyarephrasing ofthe property (g 1+g2)0f=g10f+g20f,which istrivially verified. Iff=id, then composition withf acts asanidentity mappingong,i.e.g0id =g. Ifwehave asequence ofA-homomorphisms X'-.X-.X", then wegetaninduced sequence HomA(X', Y) HomA(X, Y) HomA(X", Y). Proposition 2.1. Asequence X' X-.X" -.0 isexact ifandonlyifthesequence HomA(X', Y) HomA(X, Y) HomA(X", Y) 0 isexact forallY. III,2 THE GROUP OFHOMOMORPHISMS 123 Proof This isanimportant fact, whose proof iseasy. For instance, suppose thefirst sequence isexact. Ifg:X" -+YisanA-homomorphism, its image inHomA(X, Y)isobtained bycomposing gwith thesurjective map of XonX". Ifthiscomposition is0,itfollows that g=0because X-+X"is surjective. Asanother example, consider ahomomorphism g:X-+Ysuch that thecomposition X'!.X!!.Y isO.Then gvanishes ontheimage ofA..Hence we can factor gthrough the factor module, XIImA. I X9Y Since X-+X"issurjective, wehave anisomorphism X/lmA.+-+X". Hence we can factor gthrough X",thereby showing that thekernel of HomA(X', Y) HomA(X, Y) iscontained intheimage of HomA(X, Y) HomA(X", Y). The other conditions needed toverify exactness arelefttothereader. Soisthe con verse. We have asimilar situation with respect tothe second variable, but then thefunctor iscovariant. Thus ifXisfixed, and wehave asequence ofA- homomorphisms Y'-+Y-+Y", then wegetaninduced sequence HomA(X, Y')-+HomA(X, Y)-+HomA(X, Y"). Proposition 2.2. Asequence o-+Y'-+Y-+Y", isexact ifandonlyif o-+HomA(X, Y')-+HomA(X, Y)-+HomA(X, Y") isexact forallX. 124 MODULES III,2 The verification will belefttothereader. Itfollows atonce from thedefini- tions. We note that tosaythat o Y' Y isexact means that Y'isembedded inY,i.e.isisomorphic toasubmodule of Y.Ahomomorphism into Y'can beviewed asahomomorphism into Yifwe haveY'c:Y.This correspondstotheinjection o HomA(X, Y') HomA(X, Y). LetMod(A) andMod(B) bethecategories ofmodules over rings Aand B, and letF:Mod(A) Mod(B) be afunctor. One says that Fisexact ifF transforms exact sequences into exact sequences. We see that the Horn functor ineither variable need not beexact iftheother variable iskept fixed. Inalater section, wedefine conditions under which exactness ispreserved. Endomorphisms. LetMbe anA-module. From therelations (g1+g2)0I=g10I+g20I and itsanalogueontheright, namely g0(/1+12)=g011+g0/2, and thefact that there isanidentity forcomposition, namely idM ,weconclude that HomA(M, M) isaring, themultiplication being defined ascomposition ofmappings. Ifnisaninteger>1,we can write Intomean theiteration ofIwith itself ntimes, and define 1° tobeideAccording tothegeneral definition ofendomorphisms inacategory, wealso write EndA(M) instead of HomA(M, M), and wecallEndA(M) thering ofendomorphisms. Since anA-module Misanabelian group,we seethat Homz(M, M)(= set ofgroup-homomorphisms ofMinto itself) isaring, and that wecould have defined anoperation ofAonMtobearing-homomorphism A Homz(M, M). LetAbecommutative. Then Misamodule over EndA(M). IfRisasubring ofEndA(M) then Misafortioriamodule over R.More generally, letRbe a ring and letp:R EndA(M) be aring homomorphism. Then piscalled a representation ofRonM.This occurs especially ifA=Kisafield. The linear algebra ofrepresentations of aring will bediscussed inPart III, inseveral contexts, mostly finite-dimensional. Infinite-dimensional examplesoccur inanal- ysis, but then therepresentation theory mixes algebra with analysis, and thus goes beyond thelevel ofthis course. Example. Let Kbe afield and letVbe avector spaceover K.Let D: V Vbe anendomorphism (K-linear map). For every polynomial P(X) EK[X], P(X)=La;Xiwith a;EK, we can define III,2 THE GROUP OFHOMOMORPHISMS 125 P(D)=La;D;: v V asaner:tdomorphism ofV.The association P(X) P(D) givesarepresentation p:K[X] EndK(V), which makes Vinto aK[X]-module. Itwill beshown inChapter IVthatK[X] isaprincipal ring. In7weshall giveageneral structure theorem formodules over principal rings, which will beapplied totheabove example inthe context oflinear algebra forfinite-dimensional vector spaces inChapter XIV, 3.Readers acquainted with basic linear algebra from anundergraduate course may wish to read Chapter XIV already atthispoint. Examples for infinite-dimensional vector spaces occur inanalysis. For instance, letVbethe vector space ofcomplex-valued Coofunctions onR.Let D=d/dtbethederivative (iftisthevariable). Then D:V Visalinear map, andC[X] has therepresentation p:C[X] Endc(V) given byP P(D). A similar situation exists inseveral variables, when weletVbethevector space ofCoofunctions innvariables on anopen setofRn.Then weletD;=a/atibe thepartial derivati vewith respect tothei-thvariable (i=1,. . .,n).Weobtain arepresentation p:C[X},. . .,Xn] Endc(V) such that p(X i)=Di. Example. LetHbeaHilbert space and letAbe abounded hermitian oper- ator onA.Then one considers thehomomorphism R[X] R[A] CEnd(H), from thepolynomial ring into thealgebra ofendomorphisms ofH,and one extends thishomomorphism tothealgebra ofcontinuous functions onthe spec- trum ofA.ct.myReal and Functional Analysis, Springer Verlag, 1993. Representations form acategoryasfollows. We define amorphism of a representation p:R EndA(M) into arepresentation p':R EndA(M'), orin other words ahomomorphism ofonerepresentation ofRtoanother, tobe anA-module homomorphism h:M M' such that thefollowing diagram is commutative foreveryaER: M p(a)j Mh )M' jp'(a) )M'h Inthe case when hisanisomorphism, then wemay replace theabove diagram bythecommutative diagram EndA(M) RYj[hI EndA(M') 126 MODULES III,2 where thesymbol [h]denotes conjugation byh,i.e.forfEEndA(M )wehave [h]f=h0f0h-l . Representations: from amonoid tothe monoid algebra. Let Gbe a monoid. Byarepresentation ofGon anA-module M, we mean ahomomor- phism p:G EndA(M) ofGinto themultiplicative monoid ofEndA(M). Then wemay extend ptoahomomorphism ofthemonoid algebra A[G] EndA(M), byletting P(Laxx)=Laxp(x). XEG XEG Itisimmediately verified that thisextension ofptoA[G] isaringhomomorphism, coinciding with thegiven ponelements ofG. Examples: modules over agroup ring. The next examples will follow a certain pattern associated with groups ofautomorphisms. Quite generally, sup- posewehave some category ofobjects, and toeach object Kthere isassociated anabelian group F(K), functorially with respecttoisomorphisms. This means thatifu:K K'isanisomorphism, then there isanassociated isomorphism F(u): F(K') F(K') such thatF(id)=idandF(UT)=F(u)0F(T). Then the group ofautomorphisms Aut(K) ofanobject operatesonF(K); that is,wehave anatural homomorphism Aut(K) Aut(F(K)) given byu F(u). Let G=Aut(K). Then F(K) (written additively)can bemade into amodule over thegroup ring Z[G] asabove. Given anelement a=LauuEZ[G], with auEZ,and anelement xEF(K), wedefine ax=LauF(u)x. The conditions definingamodule aretrivially satisfied. Welistseveral concrete cases from mathematics atlarge,sothere are noholds barred ontheterminology. LetKbe anumber field (i.e. afinite extension oftherational numbers). Let Gbeitsgroup ofautomorphisms. Associated with Kwehave thefollowing objects: thering ofalgebraic integers OK; thegroup ofunits Ok; thegroup ofideal classes C(K); thegroup ofroots ofunity (K). Then Goperatesoneach ofthose objects, and oneproblem istodetermine the structure ofthese objectsasZ[G]-modules. Already forcyclotomic fields this III,3 DIRECT PRODUCTS AND SUMS OFMODULES 127 determination gives rise tosubstantial theories and to anumber ofunsolved problems. Suppose that KisaGalois extension ofkwith Galois group G(see Chapter VI). Then wemay view Kitself asamodule over thegroup ring k[G].InChapter VI, 13weshall prove that Kisisomorphic tok[G] asmodule over k[G] itself. Intopology,oneconsiders aspace Xoand afinite covering X.Then Aut(X/ Xo) operatesonthehomology ofX, sothishomology isamodule over thegroup rIng. With more structure, suppose that Xisaprojective non-singular variety, say over thecomplex numbers. Then toXwe can associate: thegroup ofdivisor classes (Picard group) Pic(X); inagiven dimension, thegroup ofcycle classes orChow group CHP(X); theordinary homology ofX; thesheaf cohomology ingeneral. IfXisdefined over afield Kfinitely generatedover therationals, we can associate afancier cohomology defined algebraically byGrothendieck, andfunc- torial with respect totheoperation ofGalois groups. Then again allthese objectscan beviewed asmodules over thegroup ring ofautomorphism groups, andmajor problems ofmathematics consist indeter- mining their structure. Idirect thereader here totwo surveys, which contain extensive bibliographies. [CCFT 91] P.CAssou-NoGUES, T. CHINBURG, A. FROHLICH, M. J.TAYLOR, L-functions and Galois modules, inL1unctions and Arithmetic J.Coates andM,J,Taylor (eds,), Proceedings oftheDurham Symposium July 1989, London Math, Soc. Lecture Note Series 153, Cambridge University Press (1991), pp, 75-139 [La82] S,LANG, Units and class groups innumber theory andalgebraic geometry, Bull. AMS Vol. 6No.3 (1982), pp. 253-316 3. DIRECT PRODUCTS AND SUMS OF MODULES LetAbe aring. Let{M;hEI be afamily ofmodules. Wedefined their direct productasabelian groups inChapter I,9.Given anelement (X;);EI ofthedirect product, and aEA,wedefine a(x;)=(ax;). Inother words, wemultiply byan element ac0mponentwise. Then thedirect product 11M; isanA-module. The reader will verify atonce that itisalso adirect product inthecategory of A-modules. 128 MODULES III,3 Similarly, let M=EBMi ieI betheir direct sum asabelian groups. Wedefine onMastructure ofA-module: If(XJiel isanelement ofM,i.e. afamily ofelements XiEMisuch that Xi=0 foralmost alli,andifaEA,then wedefine a(xJiel=(axJiel, that iswedefine multiplication byacomponentwise. Itistrivially verified that this isanoperation ofAonMwhich makes Minto anA-module. Ifone refers back totheproof given fortheexistence ofdirect sums inthecategory ofabelian groups, one seesimmediately that thisproof now extends inthe same way to show that Misadirect sum ofthefamily {Mi}ielasA-modules. (For instance, themap Aj:Mj-.M such thatA.J{x)hasj-th component equal toXand i-thcomponent equal to0 for ii=jisnow seen tobe anA-homomorphism.) This direct sum isacoproduct inthecategory ofA-modules. Indeed, the reader can verify at once that givenafamily ofA-homomorphisms {Ii:MiN}, themapIdefined asintheproof forabelian groups isalso anA- isomorphism and has therequired properties. SeeProposition 7.1ofChapter I. When Iisafinite set,there isauseful criterion for amodule tobeadirect product. Proposition 3.1. LetMbe anA-module and naninteger>1.f'or each i=1,..., nletC{Ji:M-.MbeanA-homomorphism such that n LC{Ji=id andC{Ji0C{Jj=0 i=1ifii=j. Then C{Jf=C{Jifor alli.LetMi=C{Ji(M), and let C{J:M-.nMibesuch that C{J(x)=(C{Jt(x),...,C{Jn(x)). ThenC{JisanA-isomorphism ofMonto thedirect product nMi. Proof Foreachj, wehave n C{Jj=C{Jj0id =C{Jj0LC{Ji=C{Jj0C{Jj=C{Jf, i=1 thereby proving thefirst assertion. Itisclear thatC{JisanA-homomorphism. Let Xbeinitskernel. Since n X=id(x)=LC{Ji(X) i=.t III,3 DIRECT PRODUCTS AND SUMS OFMODULES 129 weconclude that x=0,so({Jisinjective. Given elements YiEMiforeach i=1,..., n,let x=Yt+.. .+Yn. Weobviously have({JJ{Yi)=0ifi=Fj. Hence ({Jj(x)=Yj foreachj=1,..., n.This proves that({Jissurjective, and concludes theproof ofourproposition. Weobserve that when Iisafinite set,thedirect sum and thedirect product areequal. Just aswith abelian groups,we usethesymbol Et>todenote direct sum. LetMbeamodule over aring Aand letSbe asubset ofM.Byalinear combination ofelements ofS(with coefficients inA)one means asum Laxx xeS where {ax} isasetofelements ofA,almost allofwhich areequal toO.These elements axare called the coefficients ofthelinear combination. Let Nbe the setofalllinear combinations ofelements ofS.Then Nisasubmodule of M,forif Laxx and xeSLbxx xeS aretwo linear combinations, then their sum isequal to L(ax+bx)x, xeS andifCEA,then C(Laxx)=Lcaxx, xeS xeS and these elements areagain linear combinations ofelements ofS.Weshall call Nthesubmodule generated byS,and wecall Sasetofgenerators forN.We sometimes write N =A(S). IfSconsists ofoneelement x,themodule generated byxisalso written Ax, orsimply (x),and sometimes wesaythat(x)isaprincipal module. Amodule Missaid tobefinitely generated, oroffinite type, orfinite over A,ifithas afinite number ofgenerators. Asubset Sofamodule Missaid tobelinearly independent (over A)ifwhen- ever wehave alinear combination Laxx xeS 130 MODULES III,3 which isequal to0,then ax=0forallXES. IfSislinearly independent andif two linear combinations Laxx andLbxx areequal, then ax=bxforallXES. Indeed, subtracting one from theother yields L(ax-bx)x=0,whence ax-bx=0forallx.IfSislinearly indepen- dent weshall also saythat itselements arelinearly independent. Similarly,a family {Xi}iel ofelements ofMissaid tobelinearly independent ifwhenever we have alinear combination a.x. =0I I , ieI then ai=0foralli.Asubset S(resp.afamily {Xi}) iscalled linearly dependent ifitisnotlinearly independent, i.e.ifthere exists arelation Laxx=0resp. xeS"a.x.=0I I ieI with notallax(resp. ai)=O.Warning. Let xbeasingle element ofMwhich islinearly independent. Then thefamily {Xi} i=1, ..., nsuch that Xi=Xforalli islinearly dependent ifn>1,butthe setconsisting ofXitself islinearly inde- pendent. LetMbeanA-module, and let{Mi}iel beafamily ofsubmodules. Since wehave inclusion-homomorphisms Ai:Mi-+M wehave aninduced homomorphism A.:EBMi-+M which issuch that foranyfamily ofelements (Xi)i eI'allbut afinite number of which are0,wehave A.«Xi))=LXi. ieI IfA.isanisomorphism, then wesaythat thefamily {MJieIisadirect sum decomposition ofM. This isobviously equivalent tosaying that every element ofMhas aunique expressionasasum LXi with XiEMb and almost allXi=O.Byabuse ofnotation, wealso write M =EBMi inthis case. III,3DIRECT PRODUCTS AND SUMS OFMODULES 131 Ifthefamily {M i}issuch that every element ofMhas some expressionasa sumLXi(not necessarily unique), then wewrite M =LMi. Inany case, if {M;} isanarbitrary family ofsubmodules, theimage ofthehomomorphism A.. above isasubmodule ofM,which will bedenoted byLMi. IfMisamodule and N,N' are two submodules such that N+N' =M and N(\N' =0,then wehave amodule-isomorphism M NffiN', just aswith abelian groups, andsimilarly with afinite number ofsubmodules. We note, ofcourse, that our discussion ofabelian groups isaspecialcase ofour discussion ofmodules, simply byviewing abelian groups asmodules over Z.However, itseems usually desirable (albeit inefficient) todevelop first some statements forabelian groups, and then point out that theyarevalid (obviously) formodules ingeneral. LetM,M',Nbemodules. Then wehave anisomorphism ofabelian groups HomA(M ffiM',N)AHomA(M, N)xHomA(M', N), andsimilarly HomA(N, M xM')AHomA(N, M)xHomA(N, M'). The first one isobtained asfollows. Iff:MffiM' -.Nisahomomorphism, thenfinducesahomomorphismf1:M-.Nand ahomomorphismf2:M' -.N bycomposing fwith theinjections ofMand M'into their direct sum re- spectively: M-.Mffi{O}cMffiM'!.N, M' -.{O}ffiM' cMffiM'!.N. We leave ittothereader toverify that theassociation f(fbf2) givesanisomorphismasinthefirst box. Theisomorphism inthesecond box isobtained inasimilar way. Given homomorphisms f1:N M and f2:N-.M' 132 MODULES III,3 wehave ahomomorphism f:N-+M xM'defined by f(x)=(fl(X),f2(X)). Itistrivial toverify that theassociation (fbf2) f givesanisomorphism asinthesecond box. Ofcourse, thedirect sum anddirect product oftwo modules areisomorphic, but wedistinguished them inthenotation forthe sake offunctoriality, and to fittheinfinite case, seeExercise 22. Propoition 3.2. Let 0-+M'1.M!!..Mil -+0be an exact sequence of modules. Thefollowing conditions areequivalent: 1.There exists ahomomorphism cp:Mil -+Msuch that g0cp=ide 2.There exists ahomomorphism 1/1:M -+M'such that1/10f=ide Ifthese conditions aresatisfied, then wehave isomorphisms: M =Imfffi Ker 1/1, M =Ker gffi1mcp, M M'ffiMil. Proof. Let uswrite thehomomorphismsontheright: M#.Mil -+O. qJ Let xEM.Then x-qJ(g(x)) isinthekernel ofg,and hence M =Ker g+1mqJ. This sum isdirect, forif x=y+z with yEKer gand zE1mqJ,Z=qJ(w)with wEMil, and applying gyields g(x)=w.Thus wisuniquely determined byx,and therefore zisuniquely determined byx.Hence soisy,thereby proving the sum isdirect. The arguments concerning theother side ofthe sequence aresimilar and will beleft asexercises, aswell astheequivalence between ourconditions. When these conditions aresatisfied, the exact sequence ofProposition 3.2 issaid to split. One also says that fjJsplitsfandcpsplits g. III,3 DIRECT PRODUCTS AND SUMS OFMODULES 133 Abelian categories Much inthetheory ofmodules over aring isarrow-theoretic. Infact, one needs only thenotion ofkernel and cokernel (factor modules). One can axi- omatize thespecial notion ofacategory inwhich many oftheargumentsare valid, especially thearguments used inthischapter. Thus wegive this axi- omatization now, although forconcreteness, atthebeginning ofthechapter, wecontinue tousethelanguage ofmodules. Readers should strike their own balance when they want toslide into the more general framework. Consider first acategoryC1such that Mor(E, F)isanabelian group for each pair ofobjects E,FofC1,satisfying thefollowing two conditions: AD 1.The law ofcomposition ofmorphisms isbilinear, and there exists azero object 0,i.e.such that Mor(O, E)andMor(E, 0)have precisely one element foreach object E. AD2.Finite products and finite coproducts exist inthecategory. Then wesaythat C1isanadditive category. Given amorphism E1.FinC1,wedefine akernel offtobe amorphism E' -+Esuch that forallobjects Xinthecategory, thefollowing sequence is exact: oMor(X, E')--+Mor(X, E)-+Mor(X, F). Wedefine acokernel forfto beamorphism F--+F"such that forallobjects X inthecategory, thefollowing sequence isexact: o Mor(F", X) Mor(F, X) Mor(E, X). Itisimmediately verified that kernels and cokernels areuniversal inasuitable category, and hence uniquely determined uptoaunique isomorphism ifthey exist. AD3. Kernels and cokernels exist. AD 4.Ifj:E-+Fisamorphism whose kernel is0,thenjisthekernel ofitscokernel. Ifr:E Fisamorphism whose cokernel is0, thenfisthe cokernel ofitskernel. Amorphism whose kernel and cokernel are0isanisomorphism. Acategory asatisfying theabove four axioms is.called anabeUan category. Inanabelian caegory, thegroup ofmorphisms isusually denoted byHorn, sofortwoobjects E,Fwewrite Mor(E, F)=Hom(E, F). Themorphismsareusually called homomorphisms. Given anexact sequence o-+M' --+M, 134 MODULES III,3 wesaythat M'isasubobject ofM,orthat thehomomorphism ofM'into Misa monomorphism. Dually, inanexact sequence M -+Mil -+0, wesaythat Mil isaquotient object ofM, orthat thehomomorphism ofMto Mil isanepimorphism,instead ofsaying that itissurjectiveasinthecategory of modules. Although itisconvenient tothink ofmodules and abelian groups to construct proofs, usually such proofs will involve only arrow-theoretic argu- ments, andwill therefore applytoanyabelian category. However, alltheabelian categoriesweshall meet inthis book will have elements, and thekernels and cokernels will bedefined inanatural fashion, close tothose formodules, so readers may restrict their attention tothese concrete cases. Examples ofabeUan categories. Ofcourse, modules over aring form an abelian category, the most common one. Finitely generated modules over a Noetherian ring form anabelian category,tobestudied inChapter X. Let kbe afield. We consider pairs (V,A)consisting ofafinite-dimensional vector space Vover k,and anendomorphism A:V V.Byahomomorphism (morphism) ofsuch pairsf:(V,A) (W, B) we mean ak-homomorphism f:V Wsuch that thefollowing diagram iscommutative: Vf )W Aj jB Vf)W Itisroutinely verified that such pairs and theabove defined morphisms form an abelian category. Itselements will bestudied inChapter XIV. Let kbe afield and letGbe agroup. LetModk(G) bethecategory offinite- dimensional vector spaces Vover k,with anoperation ofGonV,i.e. ahomo- morphism G Autk(V). Ahomomorphism (morphism) inthat category isak- homomorphism f:V Wsuch thatf(ax)=af(x) forallxEVand aEG.It isimmediate that Modk(G)isanabelian category. This category will bestudied especially inChapter XVIII. InChapter XX, 1weshall consider thecategory ofcomplexes ofmodules over aring. This category ofcomplexes isanabelian category. Intopology and differential geometry, thecategory ofvector bundles over atopological space isanabelian category. Sheaves ofabelian groupsover atopological space form anabelian category, which will bedefined inChapter XX,6. III,4 FREE MODULES 135 4. FREE MODULES LetMbeamodule over aring Aand letSbeasubset ofM.Weshall saythat Sisabasis ofMifSisnotempty, ifSgenerates M,andifSislinearly independent. IfSisabasis ofM,then inparticular M =F{O}ifA=F{O}and every element of Mhas aunique expressionasalinear combination ofelements ofS.Similarly, let{X;}iel beanon-empty family ofelements ofM. We saythat itisabasis of Mifitislinearly independent and generates M. IfAisaring, then asamodule over itself, Aadmits abasis, consisting ofthe unit element 1. LetIbe anon-empty set, and foreach iEI,letAi=A,viewed asanA- module. Let p'=ffiA i. ie1 Then Fadmits abasis, which consists oftheelements eiofFwhose i-th com- ponent istheunit element ofAi'andhaving allother components equal toO. Byafree module weshall mean amodule which admits abasis, orthe zero module. Theorem 4.1. Let Abearing andMamodule over A.LetIbeanon-empty set, and let{Xi}iel beabasis ofM. Let NbeanA-module, and let{Yi}iel be afamily ofelements ofN. Then there exists aunique homomorphism f:M-.Nsuch thatf(Xi)=Yiforalli. Proof Let xbeanelement ofM. There exists aunique family {ai}iel of elements ofAsuch that X=Laixi. ie1 Wedefine f(x)=LaiYi. Itisthen clear thatfisahomomorphism satisfyingourrequirements, and that itistheunique such, because wemust have f(x)=Laif(xi). Corollary 4.2. Let thenotation beasinthetheorem, and assume that {Yi} ie1 isabasis ofN. Then thehomomorphism fisanisomorphism, i.e. amodule- isomorphism. Proof Bysymmetry, there exists aunique homomorphism g:N-.M 136 MODULES III,4 such that g(Yi)=Xiforalli,andfog and g0faretherespective identity map- pIngs. Corollary 4.3. Two modules having bases whose cardinalities areequalare isomorphic. Proof. Clear. Weshall leave theproofs ofthefollowing statements asexercises. LetMbeafree module over A,with basis {Xi}iel,sothat M =EBAXi. ie1 Let abe atwo sided ideal ofA.Then aM isasubmodule ofM.Each ax;isa submodule ofAx;. Wehave anisomorphism (ofA-modules) M/aM EBAXi/axi. ie1 Furthermore, each Ax;/axi isisomorphic toA/a,asA-module. Suppose inaddition that Aiscommutative. Then A/a isaring. p'urthermore M/aM isafree module over A/a, and each Ax;/axi isfreeover A/a.IfXiisthe image ofXiunder thecanonical homomorphism AXi-+Ax;/ax;, then thesingle element Xiisabasis ofAx;/axi over A/a. Allofthese statements should beeasily verified bythereader. Now letAbe anarbitrary commutative ring. Amodule Miscalled principal ifthere exists anelement xEMsuch that M=Ax. The map a ax(for aEA) isanA-module homomorphism ofAonto M,whose kernel isaleft ideal a,and inducinganisomorphism ofA-modules A/a M. LetMbe afinitely generated module, with generators {VI'. . .,vn}.Let F be afree module with basis {eI'. . .,en}. Then there isaunique surjective homomorphismf: F Msuch thatf(ei)=Vi.The kernel offisasubmodule MI. Under certain conditions, M1isfinitely generated (cf.Chapter X, 1on Noetherian rings), and the process can becontinued. The systematic study of this process will becarried out inthechaptersonresolutions ofmodules and homology. III,4 FREE MODULES 137 Ofcourse, even ifMisnotfinitely generated, one can carry out asimilar construction, byusinganarbitrary indexing set.Indeed, let{Vi}(iEI)beafamily ofgenerators. For each i,letFibefree with basis consisting ofasingle element ei,soFi:::::::A.LetFbethedirect sum ofthemodules Fi(iEI), asinProposi- tion 3.1. Then weobtain asurjective homomorphism f:F Msuch that f(ei)=Vi.Thus every module isafactor module ofafree module. Just aswedidforabelian groups inChapter I,7, we can also define the free module over aringAgenerated byanon-empty setS .WeletA(S)bethe setoffunctionscp:S Asuch that cp(x)=0foralmost allXES. IfaEAand XES, wedenote byaxthemap cpsuch that cp(x)=aand cp(y)=0fory=f=.x. Then asforabelian groups, given cpEA(S)there exist elements aiEAand XiESsuch that cp=alxl+... +anxn. Itisimmediately verified that thefamily offunctions {8x} (xES)such that 8x(x)=1and 8x(Y)=0fory=f=.xform abasis forA(S). Inother words, the ex- pression ofcpas2:aixi above isunique. This construction can beapplied when Sisagroupor amonoid G,andgives rise tothegroup algebraasin Chapter II, 5. Projective modules There exists another important type ofmodule closely related tofreemodules, which we now discuss. Let Abearing and Pamodule. Thefollowing properties areequivalent, and define what itmeans forPtobeaprojective module. P1.Given ahomomorphism f:P-.Mil and surjective homomorphism g:M-.Mil, there exists ahomomorphism h:P-.Mmaking the following diagram commutative. /l M Mil 0 9 P2.Every exact sequence 0-.M' -.Mil -.P-.0splits. P3. There exists amodule Msuch that PEt>Misfree, orinwords, Pisa direct summand ofafree module. P4. The functor M 1---+HomA(P, M)isexact. We prove theequivalence ofthefour conditions. 138 MODULES III,4 Assume P1.Given the exact sequence ofP2,weconsider themapf=id inthediagram P;/ jid Mil)p)0 Then hgives thedesired splitting ofthesequence. Assume P2.Then represent Pasaquotient ofafree mod ule(cf.Exercise 1) F-.P-.0,andapplyP2tothis sequence togetthedesired splitting, which represents Fasadirect sum ofPand some module. Assume P3.Since HomA(X ffiY,M)=HomA(X, M)ffiHomA(Y, M), and since M 1---+HomA(F, M) isanexact functor ifFisfree, itfollows that HomA(P, M)isexact when Pisadirect summand ofafreemodule, which proves P4. Assume P4.Theproof ofP1will beleft asanexercise. Examples. Itwill beproved inthe next section that avector spaceover a field isalways free, i.e. has abasis. Under certain circumstances, itisatheorem thatprojective modules arefree. In7weshall prove that afinitely generated projective module over aprincipal ring isfree. InChapter X,Theorem 4.4 we shall prove that such amodule over alocal ring isfree; inChapter XVI, Theo- rem 3.8 weshall prove that afinite flatmodule over alocal ring isfree; and in Chapter XXI, Theorem 3.7, weshall prove theQuillen-Suslin theorem that ifA=k[XI'. . .,Xn]isthepolynomial ring over afield k,then every finite pro- jective module over Aisfree. Projective modules give rise totheGrothendieck group.Let Abe aring. Isomorphism classes offinite projective modules form amonoid. Indeed, ifP isfinite projective, let[P]denote itsisomorphism class .We define [P] +[Q]=[PffiQ]. This sum isindependent ofthechoice ofrepresentatives P,Qintheir class. The conditions definingamonoid areimmediately verified. Thecorresponding Groth- endieck group isdenoted byK(A). We canimposeafurther equivalence relation that Pisequivalent toP'if there exist finite free modules Fand F'such that PEBFisisomorphic to P'EBF'.Under thisequivalence relation weobtain another group denoted by Ko(A). IfAisaDedekind ring (Chapter II,1and Exercises 13-19) itcan be shown that this group isisomorphic inanatural way with thegroup ofideal classes Pic( A)(defined inChapter II, 1).See Exercises 11,12, 13.Itisalso a III,5 VECTOR SPACES 139 problemtodetermine Ko(A) for asmany ringsaspossible,asexplicitlyaspos- sible. Algebraic number theory isconcerned with Ko(A)when Aisthering of algebraic integers ofanumber field. TheQuillen-Suslin theorem shows ifAis thepolynomial ringasabove, then Ko(A) istrivial. Of course one can carry out asimilar construction with allfinite modules. Let[M] denote theisomorphism class ofafinite module M.We define the sum tobethedirect sum. Then theisomorphism classes ofmodules over thering form amonoid, and we can associate tothis monoid itsGrothendieck group. This construction isapplied especially when thering iscommutative. There are many variations onthis theme. See forinstance thebook byBass, Algebraic K-theory, Benjamin, 1968. There isavariation ofthedefinition ofGrothendieck groupasfollows. Let Fbethefree abelian group generated byisomorphism classes offinite modules over aring R,orofmodules ofbounded cardinalitysothat wedeal with sets. Inthis free abelian groupweletfbethesubgroup generated byallelements [M]-[M']-[M"] forwhich there exists anexact sequence 0 M' M M" o.The factor group FIfiscalled theGrothendieck group K(R). We shall meet this group again in8,andinChapter XX,3.Note that wemay form asimilar Grothendieck group with anyfamily ofmodules such that Misinthefamily ifandonly ifM' andM" areinthefamily. Taking forthefamily finite projective modules, one sees easily that the twopossible definitions oftheGrothendieck group coincide inthat case. 5. VECTOR SPACES Amodule over afield iscalled avector space. Theorem 5.1. Let Vbe avector space over afield K,and assume that V=F{O}. Letrbeasetofgenerators ofVover Kand letSbeasubset ofr which islinearly independent. Then there exists abasis CBofVsuch that ScCBcr. Proof Let bethe setwhose elements aresubsets Tofrwhich contain S and arelinearly independent. Then isnotempty (itcontains S),and we contend that isinductively ordered. Indeed, if{} isatotally ordered subset 140 MODULES III,5 of (byascending inclusion), thenUisagain linearly independent and con- tains S.ByZorn's lemma, letCBbeamaximal element of. Then CBislinearly independent. Let Wbethesubspace ofVgenerated byCB.IfW =FV,there exists some element xErsuch that xrtW. Then CBu{x} islinearly inde- pendent, forgivenalinear combination LayY +bx=0, ye<Bay,bEK, wemust have b=0,otherwise weget x= -Lb-IayYEW. ye<B Byconstruction, we now seethat ay=0forall YECB,thereby proving that CBu{x} islinearly independent, andcontradicting themaximality ofCB.It follows that W =V,and furthermore that CBisnotempty since V=F{o}. This proves our theorem. IfVisavector space =F{O}, then inparticular,we seethat every setof linearly independent elements ofVcan beextended toabasis, and that abasis may beselected from agiven setofgenerators. Theorem 5.2. Let Vbeavector space overafield K. Then two bases ofV over Khave the same cardinality. Proof. Let usfirst assume that there exists abasis ofVwith afinite number ofelements, say{VI'...'Vrn},m>1.We shall prove that any other basis must also have melements. For thisitwill suffice toprove: IfWI'..., Wn are elements ofVwhich arelinearly independent over K,then n<m(for we can then usesymmetry). Weproceed byinduction. There exist elements CI'...,CrnofKsuch that (1) WI=CIV I+...+CrnV rn' and some Ci,sayCI'isnotequal toO.Then VIlies inthe space generated byWI'V2,...,Vrnover K,and this space must therefore beequal toVitself. Furthermore, WI'V2,...,Vrnarelinearly independent, forsuppose bI'...,brn areelements ofKsuch that blWI+b2V2+...+brnVrn=O. IfbI=F0,divide bybIand expressWIasalinear combination ofV2,...,Vrn. Subtracting from (1)would yieldarelation oflinear dependence among the v;,which isimpossible. Hence bl=0,and again we must have allbi=0 because the Viarelinearly independent. III,5 VECTOR SPACES 141 Suppose inductively that after asuitable renumbering ofthe Vi'wehave found Wb.. .,Wr(r<n)such that {WI'...,Wr,Vr+b...,vm} isabasis ofV.We expressWr+1asalinear combination (2)Wr+1=CtWl+...+CrW r+ Cr+1Vr+1+...+CmV m withCiEK.The coefficients ofthe Viinthis relation cannot allbe0;otherwise there would be alinear dependence among the wj.SayCr+1=Fo.Usingan argument similar tothat used above, we canreplacevr+1bywr+1and still have abasis ofV.This means that we can repeat theprocedure until r=n,and therefore that n<m,thereby provingour theorem. We shall leave thegeneralcase ofaninfinite basis asanexercise tothe reader. [Hint: Use thefact that afinite number ofelements inone basis is contained inthespace generated byafinite number ofelements inanother basis.] Ifavector space Vadmits one basis with afinite number ofelements, say m, then weshall saythat Visfinite dimensional and that misitsdimension. In view ofTheorem 5.2, we seethat misthenumber ofelements inany basis ofV.IfV={O}, then wedefine itsdimension tobe0,and say that Vis O-dimensional. We abbreviate dimension" by"dim" or dimK"ifthe reference toKisneeded forclarity. When dealing with vector spaces over afield, we use thewords subspace and factor space instead ofsubmodule and factor module. Theorem 5.3. Let Vbeavector space overafield K,and letWbeasubspace. Then dimKV=dimKW+dimKV/W. Iff: V-+Uisahomomorphism ofvector spaces over K,then dim V=dim Kerf+dim1m! Proof. The first statement isaspecialcase ofthesecond, taking forfthe canonical map. Let{uiLel be abasis ofImf, and let{wj}jeJbe abasis of Kerf.Let{v;} ieIbe afamily ofelements ofVsuch thatj'(v;)=Uiforeach iEI.Wecontend that {Vi'Wj}iel,jeJ isabasis for V.This willobviously proveour assertion. 142 MODULES III,6 Let xbeanelement ofV.Then there exist elements {aJiel ofKalmost allofwhich are0such that f(x)=Laiui. ieI Hencef(x-LaiV;)=f(x)-Laif(vi)=O.Thus X-LaiVi isinthekernel off, and there exist elements {bj}jeJofKalmost allofwhich are osuch that x-"a.v.= b.w.i..J II Jr From this we seethat x=Laivi +Lbjwj,and that {Vi'Wj}generates V. Itremains tobeshown that thefamily {Vi'Wj}islinearly independent. Suppose that there exist elements Ci,djsuch that o= c.v. +"d.w. II i..J Jr Applyingfyields o=Lcif(Vi)=LCiUb whence allCi=O.From this weconclude atonce that alldj=0,and hence that ourfamily {VbWj}isabasis forVover K, aswas tobeshown. Corollary 5.4. Let Vbeavector space and Wasubspace. Then dim W<dim V. IfVisfinite dimensional anddim W =dim Vthen W =v. Proof Clear. 6. THE DUAL SPACE AND DUAL MODULE Let Ebe afree module over acommutative ring A.We view Aas afree module ofrank lover itself. Bythedual module EVofEweshall mean the module Hom(E, A). Itselements will becalled functionals. Thus afunctional onEisanA-linear mapf:E A.IfxEEandfEEV, wesometimes denote f(x) by(x,f).Keepingxfixed, we seethat thesymbol (x,f) asafunction of fEEVisA-linear initssecond argument, and hence that xinduces alinear map onEV ,which is0ifandonly ifx=O.Hence wegetaninjection E EVV which isnotalwaysasurjection. III,6 THE DUAL SPACE AND DUAL MODULE 143 Let{xihEI be abasis ofE.For each iEIletfi betheunique functional such thatfi(xj)=Sij(inother words, 1ifi=jand 0ifi=t=j).Such alinear map exists bygeneral properties ofbases (Theorem 4.1). Theorem 6.1. Let Ebe afinite free module over thecommutative ring A, offinite dimension n.Then EVisalsofree, and dim EV=n.If{XI'. . .,xn} isabasis forE,andfiisthefunctional such thatfi(xj)=Sij,thenifl,. . .,fn} isabasis forEV . Proof. LetfE EVand letai=f(Xi) (i=1,..., n).We have f(c}xI+... +cnx n)=clf(xl)+·..+cnf(x n). Hencef=aIfl+· · ·+anfn' and we seethat thefigenerate Ev.Furthermore, theyarelinearly independent, forif bJ+...+b+=OII nJn with biEK,then evaluating theleft-hand side onXiyields b.+.(x.)=0IJii' whence bi=0foralli.This provesour theorem. Given abasis {Xi}(i=1,..., n)asinthetheorem, wecall thebasis {fj} thedual basis. Interms ofthese bases, w,e can expressanelement AofEwith coordinates (aI'. . .,an)' and anelement BofEvwith coordinates (bl,.. .,bn), such that A=alxl+·..+anx n, B=blfl+... +bnfno Then interms ofthese coordinates, we seethat (A,B)=aIb}+... +anbn=A·B istheusual dotproduct ofn-tuples. Corollary 6.2. When Eisfreefinite dimensional, then the map E EVV which toeach XEVassociates thefunctional f (x,f)onEVisanisomorphism ofEonto EVV . Proof. Note that since {fl,. . .,fn} isabasis forEV ,itfollows from the definitions that{xI'. 0 .,xn} isthedual basis inE, soE=Evv . Theorem 6.3. LetU,V,Wbefinite free modules over thecommutative ring A,and let ,\cpOWVUO beanexact sequence ofA-homomorphisms. Then theinduced sequence o HomA(U, A) HomA(V, A) HomA(W, A) 0 144 MODULES III,6 I.e.oUv Vv Wv 0 isalso exact. Proof This isaconsequence ofP2, because afree module isprojective. We now consider properties which have specifically todowith vector spaces, because we aregoing totake factor spaces. So we assume that wedeal with vector spaces over afield K. Let V,V'betwo vector spaces, and suppose givenamapping VxV'-.K denoted by (x,x')1---+(x,x') forxEVandx'EV'.Wecallthemapping bilinear ifforeach xEVthefunction x'1-+(x,x') islinear, andsimilarly foreach x'EV'thefunction x1---+(x,x')is linear. Anelement xEVissaid tobeorthogonal (orperpendicular) toasubset S'ofV'if(x,x')=0forallx'ES'. We make asimilar definition inthe opposite direction. Itisclear that the setofxEVorthogonal toS'isasub- space ofV. Wedefine thekernel ofthebilinear mapontheleft tobethesubspace ofV which isorthogonal toV',andsimilarly forthekernel ontheright. Given abilinear mapasabove, VxV'-.K, letW'beitskernel ontheright and letWbeitskernel ontheleft. Letx'be anelement ofV'.Then x'gives rise toafunctional onV,bytherule x1---+(x,x'), and this functional obviously depends only onthe coset ofx'modulo W'; in other words, ifX'I=x(mod W'), then the functionals x1---+(x,X'I> and x1-+(x,x) areequal. Hence wegetahomomorphism V' VV whose kernel isprecisely W'bydefinition, whence aninjective homomorphism o V'/W' VV. Since allthefunctionals arising from elements ofV'vanish onW, we can view them asfunctionals onV/W, i.e. aselements of(V/W)v. So weactually getan injective homomorphism o V'/W' (V/W)V. One could giveaname tothehomomorphism 9:V' VV III,6 THE DUAL SPACE AND DUAL MODULE 145 such that (x,x')=(x,g(X') forallxEVandx'EV'.However, itwillusually bepossible todescribe itbyan arrow and call ittheinduced map,orthenatural map. Givinganame toit would tend tomake theterminology heavier thannecessary. Theorem 6.4. Let VxV' Kbeabilinear map, letW,W'beitskernels ontheleftandright respectively, and assume that V'/W' isfinite dimensional. Then theinduced homomorphism V'/W' (V/W)v isanisomorphism. Proof. Bysymmetry,wehave aninduced homomorphism V/W (V'/W')V which isinjective. Since dim(V'/W')v=dimV'/W' itfollows thatV/W isfinite dimensional. From theabove injective homomor- phism and theother, namely o V'/W' (V/W)v, wegettheinequalities dim V/W<dimV'/W' and dimV'/W'<dim V/W, whence anequality ofdimensions. Hence ourhomomorphismsaresurjective and inverse toeach other, thereby proving thetheorem. Remark 1. Theorem 6.4 istheanalogue for vector spaces oftheduality Theorem 9.2ofChapter I. Remark 2. Let Abe acommutative ring and letEbeanA-module. Then wemay form two types ofdual: E"=Hom(E, Q/Z), viewing Easanabelian group; EV=HomA(E, A),viewing EasanA-module. Both arecalled dual, andthey usuallyareapplied indifferent contexts. For instance, EVwill beconsidered inChapter XIII, while E"will beconsidered in thetheory ofinjective modules, Chapter XX,4.For anexample ofdual module EVseeExercise 11.Ifbyany chance the two duals arise together and there is need todistinguish between them, then wemay callE"thePontrjagin dual. 146 MODULES III,7 Indeed, inthetheory oftopological groups G,thegroup ofcontinuous homo- morphisms ofGintoR/Z istheclassical Pontrjagin dual, and isclassically denoted byG", soIfind thepreservation ofthatterminology appropriate. Instead ofR/Z one may take other natural groups isomorphic toR/Z. The most common such group isthegroup ofcomplex numbers ofabsolute value 1, which wedenote bySI.The isomorphism withR/Z isgiven bythe map x e27Tix . Remark 3. Abilinear map VxV Kforwhich V'=Viscalled abilinear form. We saythat theform isnon-singular ifthecorresponding maps V' VV and V (V')v areisomorphisms. Bilinear maps and bilinear forms will bestudied atgreater length inChapter XV. See also Exercise 33ofChapter XIII for anice example. 7. MODULES OVER PRINCIPAL RINGS Throughout thissection, weassume that Risaprincipal entire ring. Allmodules are over R,andhomomorphisms areR-homomorphisms, unless otherwise specified. The theorems willgeneralize those proved inChapter Iforabelian groups. Weshall alsopoint out how theproofs ofChapter Ican beadjusted with sub- stitutions ofterminologysoastoyield proofs inthepresentcase. LetFbeafree module over R,with abasis {XJiel. Then thecardinality of Iisuniquely determined, and iscalled thedimension ofF.We recall that this isproved, saybytakingaprime element pinR,andobserving that F/pFisa vector spaceover thefieldR/pR, whose dimension isprecisely thecardinality ofI.We may therefore speak ofthedimension of afree module over R. Theorem 7.1. LetFbeafree module, andMasubmodule. Then Misfree, and itsdimension isless than orequal tothedimension ofF. Proof Forsimplicity,wegive theproof when Fhas afinite basis {Xi}, i=1,..., n.LetMrbetheintersection ofMwith (Xl' ..., xr),themodule generated byXb...,Xr. Then M1=Mn(Xl) isasubmodule of(x1),and is therefore oftype (a 1Xl)with some alER.Hence M1iseither 0orfree, ofdi- mension 1.Assume inductively that Mrisfree ofdimension <r.Let Qbe the setconsisting ofallelements aERsuch that there exists anelement XEM which can bewritten X=blXt+...+brxr +axr+1 III,7 MODULES OVER PRINCIPAL RINGS 147 with biER.Then Qisobviously anideal, and isprincipal, generated saybyan element ar+1.Ifar+1=0,then Mr+ 1=Mrand wearedone with theinductive step. Ifar+1;/=0,let WEMr+1besuch that thecoefficient ofwwith respect to Xr+1isar+1.IfxEMr+1then thecoefficient ofxwith respecttoXr+1is divisible byar+l'and hence there exists CERsuch that x-cwlies inMr. Hence Mr+1=Mr+(w). Ontheother hand, itisclear that Mrn(w)is0,and hence that this sum isdirect, thereby provingour theorem. (For theinfinite case, seeAppendix 2,2.) Corollary 7.2. Let Ebe afinitely generated module and E'asubmodule. Then E'isfinitely generated. Proof We can represent Easafactor module ofafree module Fwith a finite number ofgenerators: IfVI'.. ,,Vnaregenerators ofE,wetake afree module Fwith basis {x1,.. .,xn}and map XionVi.The inverse image ofE'inF isasubmodule, which isfree, andfinitely generated, bythetheorem. Hence E'isfinitely generated. The assertion also follows using simple properties of Noetherian rings and modules. Ifone wants totranslate theproofs ofChapter I,then one makes the following definitions. Afree l-dimensio,nal module over Riscalled infinite cyclic. Aninfinite cyclic module isisomorphic toR,viewed asmodule over itself. Thus every non-zero submodule ofaninfinite cyclic module isinfinite cyclic. Theproof given inChapter Ifortheanalogue ofTheorem 7.1applies without further change. Let Ebeamodule. We saythat Eisatorsion module ifgiven xEE,there exists aER,a;/=0,such that ax =O.Thegeneralization offinite abelian group isfinitely generated torsion module. Anelement xofEiscalled atorsion element ifthere exists aER,a;/=0,such that ax =O. LetEbe amodule. Wedenote byEtor thesubmodule consisting ofalltorsion elements ofE,and call itthetorsion submodule ofE.IfEtor=0,wesaythat Eistorsion free. Theorem 7.3. LetEbefinitely generated. Then E/Etorisfree. There exists afree submodule FofEsuch that Eisadirect sum E=Etor EDF. The dimension ofsuch asubmodule Fisuniquely determined. Proof. We first prove that E/Etor istorsion free. IfxEE,letidenote its residue class mod Etor. Let bER,b=t=0besuch thatbi=o.Then bxEEtop and hence there exists cER, c=t=0,such that cbx=O.Hence xEEtor and i=0,thereby proving thatE/Etoristorsion free. Itisalsofinitely generated. 148 MODULES III,7 Assume now that Misatorsion free module which isfinitely generated. Let {VI'. . .,vn}be amaximal setofelements ofMamongagiven finite setof generators {YI'. . .,Ym} such that{VI'. . .,vn}islinearly independent. IfYis oneofthegenerators, there exist elements a,bb. . .,bnERnotall0,such that ay+btvt +...+bnv n=O. Then a;/=0(otherwise wecontradict thelinear independence ofvt,...,vn). Hence aylies in(Vb...,Vn). Thus foreachj=1,...,mwe can find ajER, aj;/=0,such thatajYjliesin(Vb...,vn). Let a=at...ambetheproduct. Then aM iscontained in(vt,...,vn),and a;/=O.The map X1---+ax isaninjective homomorphism, whose image iscontained inafree module. This image isisomorphictoM,and weconclude from Theorem 7.1that Mis free, asdesired. Togetthesubmodule Fweneed alemma. Lemma 7.4. LetE,E'bemodules, and assume that E'isfree. Letf:E-.E' beasurjective homomorphism. Then there exists afree submodule FofEsuch that therestriction off toFinduces anisomorphism ofFwith E',and such that E=FffiKerf Proof Let{Xaiel beabasis ofE'.For each i,letXibeanelement ofEsuch thatf(xi)=x. LetFbethesubmodule ofEgenerated byalltheelements Xi' iEI.Then one sees atonce that thefamily ofelements {Xi}iel islinearly inde- pendent, and therefore that Fisfree. Given XEE,there exist elements aiER such that f(x)=Laix. Then x-Laixi liesinthekernel off, and therefore E=Kerf+F.Itisclear thatKerf nF=0,and hence that the sum isdirect, thereby proving thelemma. Weapply thelemma tothehomomorphism E E/Etor inTheorem 7.3 to get ourdecomposition E=Etor EDF.The dimension ofFisuniquely determined, because Fisisomorphic toE/Etor foranydecomposition ofEinto adirect sum asstated inthetheorem. The dimension ofthefree module FinTheorem 7.3 iscalled therank ofE. Inorder togetthestructure theorem forfinitely generated modules over R, one canproceed exactly asforabelian groups. Weshall describe thedictionary which allows ustotransport theproofs essentially without change. LetEbeamodule over R.Let xEE.The mapa1-+axisahomomorphism ofRonto thesubmodule generated byx,and thekernel isanideal, which is principal, generated byanelement mER. We saythat misaperiod ofx.We III,7 MODULES OVER PRINCIPAL RINGS 149 note that misdetermined uptomultiplication byaunit(ifm=F0).Anelement cER,c=F0,issaid tobeanexponent forE(resp. forx)ifcE=0(resp. cx =0). Let pbeaprime element. Wedenote byE(p) thesubmodule ofEconsisting ofallelements xhavinganexponent which isapower pr(r>1).Ap-submodule ofEisasubmodule contained inE(p). Weselect once and forallasystem ofrepresentatives fortheprime elements ofR(modulo units). Forinstance, ifRisapolynomial ring inonevariable over afield, wetake asrepresentatives theirreducible polynomials with leading coefficient 1. LetmER, m=FO.Wedenote byEmthekernel ofthemap x1-+mx. Itconsists ofallelements ofEhaving exponentm. Amodule Eissaid tobecyclic ifitisisomorphic toR/(a) for some element aER.Without lossofgenerality ifa=F0,one may assume that aisaproduct of primes inoursystem ofrepresentatives, and then wecould saythat aistheorder ofthemodule. Let rl'. . .,rsbeintegers>1.Ap-module Eissaid tobeoftype (pr1,...,prs) ifitisisomorphic totheproduct ofcyclic modules R/(pri) (i=1,...,s).Ifp isfixed, then one could saythat themodule isoftype (rb...,rs)(relative top). Alltheproofs ofChapter I,8now goover without change. Whenever we argueonthe size of apositive integer m, wehave asimilar argumentonthe number ofprime factors appearing initsprime factorization. Ifwedeal with a prime power pr,we can view theorder asbeing determined byr.The reader can now check that theproofs ofChapter I,8areapplicable. However, weshall develop thetheoryonce again without assuming any knowledge ofChapter I,8.Thus our treatment isself-contained. Theorem 7.5. Let Ebeafinitely generated torsion module =FO.Then Eis thedirect sum E=EBE(p), p taken over allprimes psuch thatE(p) =FO.Each E(p) can bewritten asadirect sum E(p)=R/(pVl)ffi...ffiR/(pVs) with 1< V1<...<vs.The sequence Vb...,Vsisuniquely determined. Proof Let abeanexponent forE,and suppose that a=bcwith (b,c)=(1). Let x,yERbesuch that 1=xb+yc. 150 MODULES III,7 Wecontend that E=EbEt>Ec. Our first assertion then follows byinduction, expressingaasaproduct ofprime powers. Let vEE.Then v=xbv +ycv. Then xbv EEcbecause cxbv =xav =o.Similarly, ycvEEb.Finally EbnEc=0, asone seesimmediately. Hence Eisthedirect sum ofEband Ec. We must now prove that E(p) isadirect sum asstated. IfYb...,Ymare elements ofamodule, weshall saythat theyareindependent ifwhenever wehave arelation a1Y1+...+amYm=0 with aiER,then wemust have aiYi=0foralli.(Observe that independent does not mean linearly independent. )We see atonce thatYl,. ..,Ymareinde- pendent ifandonly ifthemodule (yl'.. .,Ym)hasthedirect sumdecomposition (yl,. . .,Ym)=(yl)Et>...Et>(ym) interms ofthecyclic modules (yi),i==1,...,m. We now have ananalogue ofLemma 7.4formodules havingaprime power exponent. Lemma 7.6. Let Ebeatorsion module ofexponent pr(r>l)forsome prime element p.Let XlEEbe anelement ofperiod pro Let E=E/(x l).Let Yl,...,Ymbeindependent elements ofE.Then foreach ithere exists arepre- sentative YiEEofYi,such that theperiod ofYiisthe same astheperiod ofYi. The elements XbYb...,Ymareindependent. Proof LetYEEhave period pnfor some n>1.Let Ybearepresentative of YinE.Then pnyE(Xl), and hence pny=pScXbCER,p C, for some s<r.Ifs=r,we seethat Yhasthe same periodasy.Ifs<r,then pSCXl hasperiod pr-s, and hence Yhasperiod pn+r-s. We must have n+r-s<r, because prisanexponent forE.Thus weobtain n<s,and we seethat Y-ps-n CXl isarepresentative forY,whose period ispn. Let Yibe arepresentative forYihaving the same period. We prove that XbYl,...,Ymareindependent. Suppose that a,al,...,amERareelements such that aXl +alYl +...+amYm=O. III,7 MODULES OVER PRINCIPAL RINGS 151 Then a1Y1+...+amYm=O. Byhypothesis, wemust have aiYi=0foreach i.Ifpriistheperiod ofYi,then pridivides ai. We then conclude that aiYi=0foreach i,and hence finally that ax1=0,thereby proving thedesired independence. Toget thedirect sum decomposition ofE(p), wefirst note that E(p) is finitely generated. We may assume without loss ofgenerality that E=E(p). Let x1beanelement ofEwhose period prlissuch that r1ismaximal. Let E=E/(x 1).Wecontend that dimEpasvector space over R/pR isstrictly less than dimEp. Indeed, ifYl, ..., Ym arelinearly independent elements ofEp overR/pR, then Lemma 7.6implies that dimEp>m+1because we canalways find anelement of(Xl) having period p,independent ofYb. ..,Ym. Hence dimEp<dimEp.We can prove thedirect sum decomposition byinduction. IfE=F0,there exist elements x2,. ..,Xshaving periods pr2 ,. ..,prsrespectively, such that r2> · · ·>rrByLemma 7.6, there exist representatives X2,. . .,Xr inEsuch that Xihasperiod priand Xl'. ..,Xrareindependent. Since pr1issuch that rlismaximal, wehave rl>r2,and ourdecomposition isachieved. Theuniqueness will beaconsequence ofamore general uniqueness theorem, which westate next. Theorem 7.7. Let Ebe afinitely generated torsion module, E=FO.Then Eisisomorphic toadirect sumofnon-zero factors R/(ql) ffi...ffiR/(qr), where ql,...,qrare non-zero elements ofR,andqllq21.. .Iqr. The sequence ofideals (ql),...,(qr) isuniquely determined bytheabove conditions. Proof. Using Theorem 7.5, decompose Einto adirect sum ofp-submodules, sayE(p1)ffi...ffiE(PI)' and then decompose each E(pJ into adirect sum of cyclic submodules ofperiods P'iij. Wevisualize these symbolicallyasdescribed bythefollowing diagram: E(p1): r11<r12< ... E(p2): r21<r22< ... E(PI): r'1<r'2< ... Ahorizontal row describes thetype ofthemodule with respect totheprime at the left. The exponents rijarearranged inincreasing order for each fixed i=1,...,l.We letqb. ..,qrcorrespond tothecolumns ofthematrix of exponents, inother words q_Prl1pr21Prl1 1- 1 2...I, q_Pr12pr22Prl2 2- 1 2...I, 152 MODULES III,7 The direct sum ofthecyclic modules represented bythefirst column isthen isomorphic toR/{q 1)'because, aswith abelian groups, thedirect sum ofcyclic modules whose periods arerelatively prime isalso cyclic. We have asimilar remark foreach column, and weobserve that ourproof actually orders theqj byincreasing divisibility,aswas tobeshown. Now foruniqueness. Let pbeanyprime, and suppose that E=R/{pb) for some bER,b=1=o.ThenEpisthesubmodule bR/{pb),asfollows atonce from unique factorization inR.But thekernel ofthecomposite map R-.bR -.bR/{pb) isprecisely (p). Thus wehave anisomorphism R/{p) bR/{pb). Let now Ebeexpressedasinthetheorem, asadirect sum ofrterms. An element v=VIEt>...Et> Vr, ViER/{qJ isinEpifandonly ifPVi=0foralli.Hence Episthedirect sum ofthekernel of multiplication bypineach term. ButEpisavector spaceover R/{p), and its dimension istherefore equal tothenumber ofterms R/{qi)such that pdivides qi. Suppose that pisaprime dividing q1,and hence qiforeach i=1,..., r.Let Ehave adirect sum decomposition into dterms satisfying theconditions ofthe theorem, say E=R/{q'l) Et>...Et>R/{q). Then pmust divide atleast roftheelements qj,whence r<s.Bysymmetry, r=s,and pdividesqjforallj. Consider themodule pEe Byapreceding remark, ifwewrite qi=pbi' then pE R/{b 1)Et>...Et>R/{b r), and b1I... Ibr.Some ofthe bimay beunits, but those which are not units determine their principal ideal uniquely, byinduction. Hence if (b1)=...=(bj)=1 but(bj+1)=F(I),then thesequence ofideals (bj-t'1),...,(br) III,7 MODULES OVER PRINCIPAL RINGS 153 isuniquely determined. This provesouruniqueness statement, and concludes theproof ofTheorem 7.7 . The ideals (ql),. . .,(qr)arecalled theinvariants ofE. For oneofthemain applications ofTheorem 7.7 tolinear algebra,seeChapter XV,2. The next theorem isincluded forcompleteness. Itiscalled theelementary divisors theorem. Theorem 7.8. LetFbeafree module over R,and letMbeafinitely generated submodule =t=O.Then there exists abasis CBofF,elements eI'.. .,eminthis basis, and non-zero elements aI'. . .,amERsuch that: (i)The elements aIel'. ..,amemform abasis ofMover R. (ii) Wehave a;Ia;+ Ifori=1,..., m-1. The sequence ofideals (aI),. ..,(am) isuniquely determined bythepreceding conditions. Proof. Write afinite setofgenerators forMaslinear combination ofafinite number ofelements inabasis forF.These elements generateafree submodule offinite rank, and thus itsuffices toprove thetheorem when Fhasfinite rank, which we now assume. We let n=rank(F). The uniqueness isacorollary ofTheorem 7.7. Supposewehave abasis as inthetheorem. Say aI,. . .,asareunits, and socan betaken tobe=1,and as+j=qjwithqllq21. . .Iqrnon-units. Observe that F/M=Fisafinitely generated module over R,having thedirect sum expression r F/M=F=EB(R/qjR)ejEBfree module ofrank n-(r+s) j=I where abar denotes theclass ofanelement ofFmod M.Thus thedirect sum overj=1,...,risthetorsion submodule ofF,whence theelements ql,. .., qrareuniquely determined byTheorem 7.7. We have r+s=m, sotherank ofF/Misn-m,which determines muniquely. Then s=m-risuniquely determined asthenumber ofunits among aI'. . .,am.This proves theuniqueness part ofthetheorem. Next weprove existence. Let Abe afunctional onF,inother words, anelement ofHomR(F, R).We letJ,\=A(M). Then J,\isanideal ofR.Select AIsuch that AI(M) ismaximal inthe setofideals {J,\},that istosay, there isnoproperly larger ideal inthe set{J,\}. Let A.t(M)=(at). Then at =F0,because there exists anon-zero element of M,andexpressing this element interms ofsome basis forFover R,with some non-zero coordinate, wetake theprojectiononthis coordinate togetafunc- tional whose value onMisnot O.Let XtEM besuch that A.t(Xt)=at. For any functional gwemust have g(Xt)E(at)[immediate from themaximality of 154 MODULES III,7 At(M)]. Writing xtinterms ofany basis ofF,we seethat itscoefficients must allbedivisible byat.(Ifsome coefficient isnotdivisible byaI,projectonthis coefficient togetanimpossible functional.) Therefore we can write Xt=atet with some element elEF. Next weprove that Fisadirect sum F=Ret Et>Ker At. Since At(et)=1,itisclear that Ret nKer At=o.Furthermore, given xEF wenote that x-At(x)e tisinthekernel ofAt. Hence Fisthe sum ofthein- dicated submodules, and therefore thedirect sum. We note that KerAlisfree, beingasubmodule ofafree module (Theorem 7.1). We let FI=Ker Al and MI=MnKerAI. We see atonce that M=RXI EBMI. Thus MIisasubmodule ofFIand itsdimension isone less than thedimension ofM.From themaximality condition onAl(M), itfollows atonce that forany functional AonFI'theimage A(M) will becontained inAl(M)(because otherwise, asuitable linear combination offunctionals would yieldanideal larger than (al)). We can therefore complete theexistence proof byinduction. InTheorem 7.8, wecall theideals (al)'. . .,(am) theinvariants ofMinF. For another characterization ofthese invariants, seeChapter XIII, Proposition 4.20. Example. First, seeexamples ofsituations similar tothose ofTheorem 7.8 inExercises 5,7,and 8,and forDedekind rings inExercise 13. Example. Another way toobtain amodule M asinTheorem 7.8 isas amodule ofrelations. Let Wbeafinitely generated module over R,with genera- tors WI'...,wn.Byarelation among {WI'. ..,wn}we mean anelement (a],. . .,an)ERnsuch thatLa;wi=o.The setofsuch relations isasub- module ofRn,towhich Theorem 7.8 may beapplied. Itisalso possible toformulate aproof ofTheorem 7.8byconsidering Mas asubmodule ofRn, andapplying themethod ofrow and column operations to getadesired basis. Inthis context, wemake some further comments which may serve toillustrate Theorem 7.8. We assume that thereader isacquainted with matrices over aring. By row operationswe mean: interchanging two rows; addingamultiple ofone row toanother; multiplyingarowbyaunit inthering. We define column operations similarly. These row and column operations correspond tomultiplication with theso-called elementary matrices inthering. Theorem 7.9. Assume that theelementary matrices inRgenerate GL,iR). Let(xij)be anon-zero matrix with components inR.Then with afinite number ofrow and column oJ}erations, itispossible tobring thematrix to theform III,8 EULER-POINCARE MAPS 155 al 0 o a2o o o o oo oam o o with a1. . .am 0and a1Ia21. . .Iam. We leave theproof forthereader. Either Theorem 7.9 can beviewed as equivalenttoTheorem 7.8, oradirect proof may begiven. Inany case, Theorem 7.9 can beused inthefollowing context. Consider asystem oflinear equations CllXI+...+ClnX n=0 CrlXI+.. ·+CrnX n=o. with coefficients inR.LetFbethesubmodule ofRngenerated bythe vectors X=(Xl'. . .,xn)which aresolutions ofthis system. ByTheorem 7.1, weknow that Fisfree ofdimension<n.Theorem 7.9 can beviewed asprovidinga normalized basis forFinline with Theorem 7.8. Further example. Aspointed outbyPaul Cohen, the row and column method can beapplied tomodules over apower series ring o[[X]], where 0is acomplete discrete valuation ring. Cf.Theorem 3.1ofChapter 5inmyCyclo- tomic Fields Iand II(Springer Verlag, 1990). Forinstance, one could pick0it- self tobe apower series ring k[[T]] inone variable over afield k,but inthe theory ofcyclotomic fields inthe above reference, 0istaken tobethering of p-adic integers. Ontheother hand, George Bergman hasdrawn myattention to P.M.Cohn's "On the structure ofGL-;. ofaring," IHES Publ. Math. No. 30 (1966), giving examples ofprincipal rings where one cannot use row andcolumn operations inTheorem 7.9. 8. EULER-POINCARE MAPS The present section may beviewed asprovidinganexample andapplication oftheJordan-Holder theorem formodules. But aspointed outintheexamples and references below, italso providesanintroduction forfurther theories. Again letAbe aring. We continue toconsider A-modules. Letrbe an abelian group, written additively. Let cpbe arule which tocertain modules associates anelement ofr,subject tothefollowing condition: 156 MODULES III,8 If0-.M' -.M-.Mil -.0isexact, then qJ(M) isdefined ifandonlyifqJ(M') andqJ(M") aredefined, and inthat case, wehave qJ(M)=qJ(M') +qJ(M"). Furthermore qJ(O) isdefined andequal too. Such aruleqJwill becalled anEuler-Poincare mapping onthecategory of A-modules. IfM'isisomorphic toM,then from the exact sequence o-.M' -.M-.0-.0 weconclude that qJ(M') isdefined ifqJ(M) isdefined, and that qJ(M')=qJ(M). Thus ifqJ(M) isdefined for amodule M, qJisdefined onevery submodule and factor module ofM. Inparticular, ifwehave anexact sequence ofmodules M' -.M-.M" and ifqJ(M') and qJ(M") aredefined, then soisqJ(M),asone sees atonce by considering thekernel andimage ofour two maps, andusing thedefinition. Examples. Wecould letA=Z,and letqJbedefined forallfinite abelian groups, and beequal totheorder ofthegroup. The value ofqJisinthemulti- plicative group ofpositive rational numbers. Asanother example, weconsider thecategory ofvector spaces over afield k. WeletqJbedefined forfinite dimensional spaces, and beequal tothedimension. The values ofcparethen intheadditive group ofintegers. InChapter XV weshall seethat thecharacteristic polynomial may becon- sidered asanEuler-Poincare map. Observe that thenatural map ofafinite module into itsimage intheGroth- endieck group defined attheendof4isauniversal Euler-Poincare mapping. We shall developamore extensive theory ofthismapping inChapter XX,3. IfMisamodule (over aring A),then asequence ofsubmodules M =M1::JM2::J...::JMr=0 isalso called afinite filtration, and wecall rthelength ofthefiltration. Amodule Missaid tobesimple ifitdoes notcontain any submodule other than 0and M itself, andifM =FO.Afiltration issaid tobesimple ifeach MdMi+1issimple. The Jordan-Holder theorem asserts that twosimple filtrations ofamodule are equivalent. Amodule Missaid tobeoffinite length ifitis0orifitadmits asimple (finite) filtration. BytheJordan-Holder theorem, thelength ofsuch asimple filtration istheuniquely determined, and iscalled thelength ofthemodule. In thelanguage ofEuler characteristics, theJordan-Holder theorem can be re- formulated asfollows: III,9 THE SNAKE LEMMA 157 Theorem 8.1. LetqJbe arule which toeach simple module associates an element ofacommutative group r,and such thatifM M'then qJ(M)=qJ(M'). ThenqJhas aunique extension toanEuler-Poincare mapping definedonall modules offinite length. Proof Given asimple filtration M=M1 =>M2=>...=>Mr=O wedefine r- 1 qJ(M)=LqJ(Mi/M i+1). i=1 The Jordan-Holder theorem shows immediately that this iswell-defined, and that this extension ofqJisanEule.r-Poincare map. Inparticular, we seethat thelength function isthe Euler-Poincare map taking itsvalues intheadditive group ofintegers, andhaving thevalue 1forany simple module. 9. THE SNAKE LEMMA This section givesavery general lemma, which will beused many times, soweextract ithere. The reader may skip ituntil itisencountered, butalready wegivesome exercises which show how itisapplied: thefive lemma inExercise 15and also Exercise 26.Other substantial applications inthis book will occur inChapter XVI, 3inconnection with the tensor product, and inChapter XX inconnection with complexes, resolutions, and derived functors. Webegin with routine comments. Consider acommutative diagram ofhomo- morphisms ofmodules. M'f )M dJ jd N'h)N Thenfinduces ahomomorphism Ker d' Ker d. Indeed, suppose d'x'=O.Then df(x')=0because df(x')=hd'(x')=O. 158 MODULES III,9 Similarly, hinduces ahomomorphism Coker d' Coker d inanatural wayasfollows. Lety'EN'representanelement ofN'/d'M'. Then hy'mod dMdoes notdependonthechoice ofy'representing thegiven element, because ify"=y'+d'x', then hy"=hy'+hd'x'=hy' +dfx'=hy' mod dM. Thus wegetamap h*:N'/d'M'=Coker d' N/dM=Coker d, which isimmediately verified tobe ahomomorphism. Inpractice, givenacommutative diagramasabove, one sometimes writesf instead ofh,soone writesfforthehorizontal maps both above and below the diagram. This simplifies thenotation, and isnot soincorrect: wemay view M',N' asthe two components ofadirect sum, andsimilarly forM,N.Thenf ismerelyahomomorphism defined onthedirect sum M' EDN'into MEDN. The snake lemma concerns acommutative and exact diagram called asnake diagram: M'fM9M" 0 dJ dj d"j 0)N'f)N)N" 9 Let z"EKer d" .We can construct elements ofN' asfollows. Since 9is surjective, there exists anelement zEMsuch that gz=z".We now move vertically down byd,and take dz.The commutativity d"g=gdshows that gdz=0whence dzisinthekernel of9inN.Byexactness, there exists an element z'EN'such thatfz'=dz.Inbrief, wewrite ,1-1d-1"Z= 00g z. Ofcourse, z'isnotwell defined because ofthechoices made when taking inverse images. However, thesnake lemma will state exactly what goes on. Lemma 9.1. (Snake Lemma). Given asnake diagramasabove, themap b:Ker d"-.Coker d' given bybz"=1-1°dog-1Z"iswelldefined, and wehave anexact sequence dKer d'-.Ker d-.Ker d"-.Coker d'-+Coker d-.Coker d" where themaps besides barethenatural ones. 1I1,10 DIRECT AND INVERSE LIMITS 159 Proof. Itisaroutine verification that the class ofz'mod 1md'isin- dependent ofthechoices made when taking inverse images, whence defining the map b.The proof ofthe exactness ofthe sequence isthen routine, and consists inchasing around diagrams. Itshould becarried out infull detail bythereader who wishes toacquire afeeling forthis type oftriviality. As an example, weshall prove that Ker 5C1mg. whereg*istheinduced maponkernels. Suppose theimage ofz"is0inCoker d'.Bydefinition, there exists u'EM'such that z'=d'u'.Then dz=fz'=fd'u'=dfu' bycommutativity. Hence d(z-fu')=0, and z-fu'isinthekernel ofd.Butg(z-fu')=gz=z".This means that z"is intheimage ofg*,asdesired. All theremainingcases ofexactness will beleft tothereader. The original snake diagram may becompleted bywriting inthekernels and cokernels asfollows (whence the name ofthelemma): Ker d' I M')Ker d I M)Ker d" I M" o o N' I Coker d'N I )Coker dN" I )Coker d" 10. DIRECT AND INVERSE LIMITS We return tolimits, which weconsidered forgroups inChapter I.We now consider linlits inother categories (rings, modules), and wepointoutthat limits satisfyauniversal property, inline with Chapter I, 11. LetI={i}be adirected system ofindices, defined inChapter I, 10.Let C1be acategory, and{Ai}afamily ofobjects inC1.For each pair i,jsuch that 160 MODULES III,10 i<jassume givenamorphism fi..A.-+A' J.I J such that, whenever i<j<k,wehave ft0f=fl andf=ide Such afamily will becalled adirected family ofmorphisms. Adirect limit forthefamily {f}isauniversal object inthefollowing category e.Ob(e) consists ofpairs (A,(fi)) where AEOb(C1) and(fi) isafamily ofmorphisms fi:Ai-+A,iEI,such that foralli<jthefollowing diagram iscommutative: fi.A. J)A.1 Jf\; A (Universal ofcourse means universally repelling.) Thus if(A,(fi)) isthedirect limit, and if(B,(gi)) isanyobject intheabove category, then there exists aunique morphism ({J:A-+Bwhich makes the following diagram commutative: f AA)Aj j B Forsimplicity, oneusually writes A=limA.I' i omitting theffrom thenotation. Theorem 10.1. Direct limits exist inthecategory ofabelian groups, ormore generally inthecategory ofmodules over aring. Proof Let{M i}be adirected system ofmodules over aring. LetMbe their direct sum. LetNbethesubmodule generated byallelements Xij=(.. .,0,x,0,. . .,-flex), 0,. ..) 1I1,10 DIRECT AND INVERSE LIMITS 161 where, for agiven pair ofindices (i,j) withj>i,xijhascomponentxinMi, f(x)inMj,andcomponent 0elsewhere. Then weleave tothereader theveri- fication that thefactor module MjN isadirect limit, where themaps ofMiinto MjN arethenatural ones arising from thecomposite homomorphism Mi-.M-.MjN. Example. LetXbe atopological space, and letxEX.The open neigh- borhoods ofxform adirected system, byinclusion. Indeed, given two open neighborhoods Uand V,then UnVisalso anopen neighborhood contained in both Uand V.Insheaf theory, oneassigns toeach, Uanabelian groupA(U) and foreach pair U::JVahomomorphism h:A(U) A(V) such thatifU::JV::J W then hW0h=h«,. Then thefamily ofsuch homomorphisms isadirected family. The direct limit funA(U) U iscalled thestalk atthepointx .We shall give theformal definition ofasheaf ofabelian groups inChapter XX,6.Forfurther reading, Irecommend atleast two references. First, theself-contained short version ofChapter IIinHartshorne's Algebraic Geometry, Springer Verlag, 1977. (Do alltheexercises ofthatsection, concerning sheaves.) The section isonly five pages long. Second, Irecommend the treatment inGunning's Introduction toHolomorphic Functions ofSeveral Variables, Wadsworth andBrooks/Cole, 1990. We now reverse the arrows todefine inverse limits. We areagain givena directed setIand afamily ofobjects Ai.Ifj>iwe are now givenamorphism f.A.-.A.I. JI satisfying therelations fofl =f' andf:=id, ifj>iand i>k.Asinthedirect case, we can define acategory ofobjects (A,h)withh:A-.Aisuch that foralli,jthefollowing diagram iscom- mutative: (A\A. A.J/1I Auniversal object inthiscategory iscalled aninverse limit ofthesystem (Ai,f). 162 MODULES III,10 Asbefore, weoften saythat A=JimAi i istheinverse limit, omitting theffrom thenotation. Theorem 10.2. Inverse limits exist inthecategory ofgroups, inthecategory ofmodules over aring, and also inthecategory ofrings. Proof. Let{Gi}be adirected family ofgroups, forinstance, and letrbe their inverse limit asdefined inChapter I,10. Letpi: r Gibetheprojection (defined astherestriction from theprojection ofthedirect product, since ris asubgroup ofIIGi).Itisroutine toverify that these data giveaninverse limit inthecategory ofgroups. The same construction also applies tothecategory of rings and modules. Example. Letp beaprime number. For n>mwehave acanonical surjective ring homomorphism f::':Z/pnz Z/pmz. Theprojective limit iscalled thering ofp-adic integers, and isdenoted byZp. For aconsideration ofthisringasacomplete discrete valuation ring,seeExercise 17andChapter XII. Let kbe afield. The power series ringk[[T]] inone variable maybeviewed astheprojective Jimit ofthe factor polynomial rings k[T]/(Tn),where for n>mwehave thecanonical ring homomorphism f;:k[T]/(Tn)k[T]/(Tm). Asimilar remark applies topower series inseveral variables. More generally, letRbe acommutative ring and letJbe aproper ideal. If n>mwehave thecanonical ring homomorphism f::':R/Jn R/Jm. LetRJ=limR/Inbetheprojective limit. Then Rhas anatural homomorphism into RJ.IfRisaNoetherian local ring, then byKrull's theorem (Theorem 5.6 ofChapter X), one knows that nJn={OJ, and sothenatural homorphism ofR initscompletion isanembedding. This construction isapplied especially when Jisthemaximal ideal. Itgivesanalgebraic version ofthenotion ofholomorphic functions forthefollowingreason. Let Rbe acommutative ring and Japroper ideal. Define aJ-Cauchy se- quence {xn} tobeasequence ofelements ofRsatisfying thefollowing condition. Given apositive integer kthere exists Nsuch that foralln,m>Nwehave Xn-XmEJk.Define anull sequence tobe asequence forwhich given kthere exists Nsuch that forall n>Nwehave xnEJk.Define addition andmultipli- III,10 DIRECT AND INVERSE LIMITS 163 cation ofsequences termwise. Then theCauchy sequences form aring e,the null sequences form anideal X,and thefactor ringe/x iscalled theJ-adic completion ofR.Prove these statements asanexercise, and also prove that there isanatural isomorphism e/x=lliTIR/Jn. Thus theinverse limit !i!!!R/Inisalso called theJ-adic completion. SeeChapter XII forthecompletion inthe context ofabsolute values onfields. Examples. Incertain situations one wants todetermine whether there exist solutions ofasystem ofapolynomial equationf(X l'. . .,Xn)=0with coefficients inapower series ring k[T], sayinone variable. One method istoconsider the ring mod (TN), inwhich case thisequation amounts toafinite number ofequations inthecoefficients. Asolution off(X)=0isthen viewed asaninverse limit of truncated solutions. For anearly example ofthis method see[La52], and for anextension toseveral variables [Ar68]. [La52] S.LANG, Onquasi algebraic closure, AnnofMath. 55(1952), pp. 373-390 [Ar68] M.ARTIN, Onthesolutions ofanalytic equations, Invent. Math, 5(1968), pp. 277-291 See also Chapter XII, 7 . InIwasawa theory,one considers asequence ofGalois cyclic extensions Kn over anumber field kofdegree pnwith pprime, and with KnCKn+l. Let Gn betheGalois group ofKnover k.Then one takes theinverse limit ofthegroup rings (Z/pnZ)[G n],following Iwasawa and Serre. Cf. myCyclotomic Fields, Chapter 5.Insuch towers offields, one can also consider theprojective limits ofthemodules mentioned asexamplesattheendof 1.Specifically, consider thegroup ofpn-th roots ofunity pn,and letKn=Q(pn+l),with Ko=Q(p). We let Tp()=!i!!!pn under thehomomorphisms pn+l pngiven by( (p.ThenTp()becomes amodule fortheprojective limits ofthegroup rings. Similarly,one canconsider inverse limits foreach oneofthemodules given intheexamples attheend of 1.(See Exercise 18.) The determination ofthe structure ofthese inverse limits leads tofundamental problems innumber theory andalgebraic geometry. After such examples from real life after basic algebra,wereturn tosome general considerations about inverse limits. Let(Ai'I{)=(Ai) and (Bi,g{)=(Bi)betwo inverse systems ofabelian groups indexed bythe same indexing set. Ahomomorphism (Ai)-+(Bi)isthe obvious thing, namelyafamily ofhomomorphisms hi:Ai-+Bi 164 MODULES 1I1,10 foreach iwhich commute with themaps oftheinverse systems: hj)B. j:A. Iir A.I)B. hiI Asequence o(Ai) (Bi)(Ci) 0 issaid tobeexact ifthecorresponding sequence ofgroups isexact foreach i. Let(An) beaninverse system ofsets, indexed forsimplicity bythepositive integers, with connecting maps Um, n:Am An for m>n. We saythat this system satisfies theMittag-Leffler condition ML ifforeach n, thedecreasing sequence um,n{Am) (m>n)stabilizes, i.e. isconstant for m sufficiently large. This condition issatisfied whenum,nissurjective forallm, n. We note thattrivially, theinverse limit functor isleftexact, inthe sense that given anexact sequence o (An) (Bn) (Cn) 0 then o li.mAn li.mBn li.mCn isexact. Proposition 10.3. Assume that(An) satisfies ML. Given anexact sequence o (An) (Bn)!!.(Cn) 0 ofinverse systems, then o li.mAn JimBn JimCn 0 isexact. Proof Theonly point istoprove thesurjectivityontheright. Let(cn) be anelement oftheinverse limit. Then each inverse image g- l{C n)isacoset of An' soinbijection with An. These inverse images form aninverse system, and theML condition on(An)implies ML on(g-l{C n)).Let Snbethestable subset Sn=()u.n{g- l{C m)). mn III,Ex EXERCISES 165 Then theconnecting maps intheinverse system (Sn) aresurjective, and sothere isanelement (bn)intheinverse limit. Itisimmediate that gmaps this element onthegiven (cn),thereby concluding theproof oftheProposition. Proposition 10.4. Let(C n)beaninverse system ofabeUan groups satisfying ML, and let(um,n)bethesystem ofconnecting maps. Then wehave anexact sequence rIl-urIo-+Jim Cn-+ Cn--. Cn-+O. Proof. For each positive integer Nwehave anexact sequence with afinite product N N o-+lim Cn-+rICn rICn-+o. lnN n=1 n=1 The mapuisthenatural one, whose effect on avector is (0,.. .,0,Cm,0,...,0)1---+(0,...,0,Umm-1Cm,0,...,0)., One seesimmediately that the sequence isexact. The infinite productsarein- verse limits taken over N.The hypothesis implies atonce that ML issatisfied fortheinverse limit ontheleft, and we can therefore apply Proposition 10.3 to conclude theproof. EXERCISES 1.Let Vbeavector spaceover afield K,and letU,Wbesubspaces. Show that dim U+dim W =dim(U +W)+dim(U ()W). 2.Generalize thedimension statement ofTheorem 5.2tofreemodules over acommutative ring. [Hint: Recall how ananalogous statement wasproved forfree abelian groups, and use amaximal ideal instead ofaprime number.] 3.Let Rbe anentire ring containingafield kasasubring, Suppose that Risafinite dimensional vector space over kunder theringmultiplication, Show that Risafield, 4.Direct sums. (a) Prove indetail that theconditions given inProposition 3,2 for asequence to splitareequivalent. Show that asequence 0---7>M' M Mil ---7>0splits if andonly ifthere exists asubmodule NofMsuch that Misequal tothedirect sum 1mfEBN,and thatifthis isthe case, then Nisisomorphic toM".Complete allthedetails oftheproof ofProposition 3,2, 166 MODULES III,Ex (b)Let EandEi(i=1".., m) bemodules over aring. Let 'Pi: Ei Eand .pi:E Eibehomomorphisms having thefollowing properties: .11. 0{(). =idY', 'f" , .pi0qJj=0 ifi:Fj, m LqJi0.pi=ide i=t Show that themapxr-+(.ptJC,..., .pmx)isanisomorphism ofEonto thedirect product oftheEi(i=1,.." m),and that themap (xt,...,xm) qJ1Xt+...+qJmXm isanisomorphism ofthis direct product onto E. Conversely, ifEisequal to adirect product (ordirect sum) ofsubmodules Ei(i=I,, . ,,m),ifwelet'Pibetheinclusion ofEiinE,and .pitheprojection of EonEi,then these maps satisfy theabove-mentioned properties. 5.LetAbeanadditive subgroup ofEuclidean space Rn,and assume that inevery bounded region ofspace, there isonlyafinite number ofelements ofA.Show that Aisafree abelian groupon <ngenerators. [Hint: Induction onthemaximal number of linearly independent elements ofAover R.Let Vb...,Vmbeamaximal setofsuch elements, and letAobethesubgroup ofAcontained intheR-space generated by Vb..,,vm-t.Byinduction, one may assume that any element ofAoisalinear integral combination ofVb ..., Vm-l' Let Sbethe subset ofelements VEAofthe form V=atVt+...+amVmwith real coefficients aisatisfying o<ai<1 o<am<1.ifi=1,...,m-1 Ifvisanelement ofSwith thesmallest am:F0,show that {Vt,...,Vm-hv}isabasis ofAover Z,] Note. The above exercise isapplied inalgebraic number theory toshow that the group ofunits inthering ofintegers ofanumber field modulo torsion isisomorphic toalattice inaEuclidean space. See Exercise 4ofChapter VII. 6.(Artin- Tate). Let Gbe afinite group operatingon afinite setS.For wES,denote 1.wby[w], sothat wehave thedirect sum Z(S)=LZ[w]. WES Define anaction ofGonZ(S) bydefining o'[w]=[o'w] (for wES),andextending 0'toZ(S) bylinearity. LetMbe asubgroup ofZ(S) ofrank #[S]. Show that Mhas aZ-basis {Yw}wes such thatO'Yw=Yaw forall wES.(Cf, myAlgebraic Number Theory, Chapter IX,4,Theorem I.) 7,LetMbe afinitely generated abelian group, Byasemi norm onMwe mean areal- valued function vIvIsatisfying thefollowing properties: III,Ex EXERCISES 167 IvI>0forallvEM; Invl=InIIvifor nEZ; Iv+wi<: IvI+IWIforallv, WEM. Bythekernel oftheseminorm we mean thesubset ofelements vsuch thatIvI=0, (a) LetMo bethe kernel. Show that Mo isasubgroup. IfMo={O}, then the seminorm iscalled anorm. (b) Assume that Mhasrank r.LetVI', . .,vrEMbelinearly independentover Zmod Mo. Prove that there exists abasis {W.,., .,wr}ofM/Mo such that i Iwil<:LIvjl.j=1 [Hint: Anexplicit version oftheproof ofTheorem 7.8gives the result. Without loss ofgenerality,we can asume Mo={O}. LetMI=(V.,, . .,vr), Letdbetheexponent ofM/MI.Then dM has afinite index inMI. Letnj,j bethesmallest positive integer such that there exist integers nj,I'. . .,nj,j_1 satisfying nj,IVI+... +nj,jvj=dWjfor some wjEM. Without lossofgeneralitywemayassume 0<:nj,k<:d-1.Then theelements WI'. . ,,Wrform thedesired basis.] 8,Consider themultiplicative group Q*ofnon-zero rational numbers. For anon-zero rational number x=a/b with a,bEZand(a,b)=1,define theheight h(x)=logmax( IaI,IbI). (a) Show that hdefines aseminorm onQ*, whose kernel consists of+1(the torsion group). (b) LetMIbeafinitely generated subgroup ofQ* ,generated byrational numbers XI', . .,xm'LetMbethesubgroup ofQ*consisting ofthose elements Xsuch that XSEMIfor some positive integers,Show that Misfinitely generated, andusing Exercise 7,find abound fortheseminorm ofasetofgenerators ofMinterms ofthesemi norms ofxI', . .,Xm. Note. The above two exercises areapplied inquestions ofdiophantine approximation, See myDiophantine approximationontoruses, Am. J.Math. 86(1964), pp.521-533, and thediscussion and references Igive inEncy- clopedia ofMathematical Sciences, Number Theory III,Springer Verlag, 1991, pp. 240-243, Localization 9,(a)LetAbe acommutative ring and letMbeanA-module. Let Sbe amultiplicative subset ofA.Define S-I Minamanner analogous tothe one weused todefine S-IA,and show thatS-IM isanS-IA-module. (b)If0 M' M M" 0isanexact sequence, show that the sequence o S-IM' S-IM S-IM" 0isexact. 168 MODULES III,Ex 10.(a)Ifpisaprime ideal, and S=A-pisthecomplement ofpinthering A,then S-IMisdenoted byMp.Show that thenatural map MnMp ofamodule Minto thedirect product ofalllocalizations Mpwhere prangesover allmaximal ideals, isinjective. (b) Show that asequence0 M' M M" 0isexact ifandonly ifthesequence oM Mp M"p0isexact forallprimes p. (c)Let Abeanentire ring and letMbe atorsion-free module. For each prime pof Ashow that thenatural map MMpisinjective. Inparticular AApisinjective, butyoucan seethatdirectly from theimbedding ofAinitsquotient field K, Projective modules over Dedekind rings For the next exercise we assume you have done theexercises onDedekind rings in thepreceding chapter.Weshall seethat forsuch rings,some parts oftheir module theory can bereduced tothe case ofprincipal rings bylocalization. Welet 0be aDedekind ring and Kitsquotient field. 11. LetMbe afinitely generated torsion-free module over o.Prove that Misprojective. [Hint: Given aprime ideal p,thelocalized moduleMpisfinitely generated torsion- free over0p,which isprincipal. ThenMpisprojective,soifFisfinite free over 0, andf:F Misasurjective homomorphism, thenfp:Fp Mphas asplitting gp:Mp Fp,such thatfp0gp=idMp.There exists cpE0such that cpftpand cpgp(M)CF.Thefamily {cp}generates theunit ideal 0(why?),sothere isafinite number ofelements cp,and elements X;E0such that2:x;c p,=1.Let 9=2:x;cp,gp,. Then show that g:M Fgivesahomomorphism such thatfog=idM,] 12.(a)Let a,bbeideals. Show that there isanisomorphism of0-modules aEBboEBab [Hint: First dothis when a,barerelatively prime. Consider thehomomorphism aEBb a+b,and use Exercise 10.Reduce thegeneral case totherelatively primecase byusing Exercise 19ofChapter II.] (b)Let a,bbefractional ideals, andletf:a bbeanisomorphism (ofo-modules, ofcourse). Thenfhasanextension toaK-linear mapfK: K K,Let c=fK(I). Show that b=caand thatfisgiven bythemapping me:X cx(multiplication byc). (c)Let abe afractional ideal. For each bEa-Ithemap mb:a 0isanelement ofthe dual aV.Show that a-I =av=Homo(a,0)under this map, and so aVV =a. 13.(a)LetMbe aprojective finite module over theDedekind ringo.Show that there exist free modules FandF'such that F::)M::)F',andF,F'have the same rank, which iscalled therank ofM. (b) Prove that there exists abasis {e.,.. .,en}ofFand ideals a.,, . ,,ansuch that M=aiel+...+ane n,orinother words, M=EBa;. III,Ex EXERCISES 169 (c) Prove that M=on-I EB afor some ideal a,and that the association M a induces anisomorphism ofKo(0)with thegroup ofideal classes Pic(0).(The group Ko(o) isthegroup ofequivalence classes ofprojective modules defined at theend of4.) Afew snakes 14.Consider acommutative diagram ofR-modules andhomomorphisms such that each row isexact: )M qj)0 M' Ij)M" hj o)N' )N )N" Prove: (a)Iff,hare monomorphisms then gisamonomorphism, (b)Iff, haresurjective, then gISsurjective. (c)Assume inaddition that 0--+M' --+Misexact and that N --+N" --+0isexact. Prove that ifany twooff, g,hareIsomorphisms, then so ISthethud. [Hint:- Use thesnake lemma,] 15. The five lemma. Consider acommutative diagram ofR-modules andhomomorph- isms such that each row isexact: Mt '.j)M2 f,j)M4 14j)M3 1.j)Ms 1,j Nt)N2)N3)N4)Ns Prove: (a)If11issurjective and12,14aremonomorphisms, then/ 3ISamonomorphism, (b)IfIsisamonomorphism and12,14aresurjective, then 13issurjective, [Hint: Use thesnake lemma,] Inverse limits 16. Prove that theinverse limit ofasystem ofsimple groups inwhich thehomomorphisms aresurjective iseither thetrivial group,orasimple group. 17.(a)Let nrange over thepositive integers and letpbe aprime number, Show that the abelian groups An=Z/pnz form aprojective system under thecanonical homomorphism ifn>m,LetZpbeitsinverse limit. Show thatZpmapssur- jectivelyoneach Z/pnz; thatZphas nodivisors of0,and has aunique maximal ideal generated byp.Show thatZpisfactorial, with onlyoneprime, namely p itself. 170 MODULES III,Ex (b)Next consider allideals ofZasformingadirected system, bydivisibility. Prove that !!!!!Z/(a)=nZp, (a)p where thelimit istaken over allideals (a), and theproduct istaken over all pnmes p. 18. (a)Let{An} be aninversely directed sequence ofcommutative rings, and let{M n} beaninversely directed sequence ofmodules, Mnbeingamodule over Ansuch that thefollowing diagram iscommutative: An+1xMn+1 Mn+1 AnXMn Mn The vertical mapsare thehomomorphisms ofthe directed sequence, and the horizontal maps give theoperation oftheringonthemodule. Show that!!!!!Mn isamodule over!!!!! An. (b)LetMbe ap-divisible group. Show thatTp(A)isamodule overZp. (c)LetM,Nbep-divisible groups, Show thatTp(MEBN)=Tp(M)EBTp(N),as modules overZp. Direct limits 19.Let(A;,f)beadirected family ofmodules. Let akEAkfor some k,and suppose that theimage ofakinthedirect limit Aiso.Show that there exists some indexj>ksuch thatf(ak)=O.Inother words whether some element insome group Aivanishes Inthedirect limit canalready beseen within theoriginal data, One way toseethis istouse theconstruction ofTheorem 10.1. 20. LetI,Jbetwo directed sets, and give theproduct IxJtheobvious ordering that (i,j)<(i',j') ifi<i'andj<j'.Let Aijbe afamily ofabelian groups, with homo- morphisms indexed byIxJ,andformingadirected family, Show that thedirect limits limlimAijand limlimAij ij j i exist and areisomorphic inanatural way, State and prove the same result forinverse limi ts. 21. Let(M,f), (M;,g)bedirected systems ofmodules over aring. Byahomomorphism (M;) (M;) one means afamily ofhomomorphisms Ui:M; Miforeach iwhich commute with thef,g.Supposewe aregivenanexact sequence o(MD (M i)(M') 0 ofdirected systems, meaning that foreach i,thesequence oM M. -+M' 0& I I III,Ex EXERCISES 171 isexact. Show that thedIrect limit preserves exactness, that is o hmM hillM; h111M;' 0 ISexact. 22.(a)Let{M;} beafamily ofmodules over aflng. For any module Nshow that Hom(ffi M;,N)=nHom(M i,N) (b)Show that Hom(N, nM;)=nHom(N, M;). 23, Let{M i}beadirected family ofmodules over aring. For anymodule Nshow that 11mHom(N, M;)=Hom(N, JimM;) 24. Show that any module isadirect limit offinitely generated submodules. Amodule Miscalled finitely presented ifthere isanexact sequence FlFoMO where F0,Flare freewith finite bases. Theimage ofF1inF0issaid tobethesubmodule ofrelations, among thefree basis elements ofF0. 25. Show that any module isadirect limit offinitely presented modules (not necessarily submodules). Inother words, given M,there exists adirected system {M;,fJ}with Mi finitely presented forallisuch that M limMi. [Hint: Any finitely generated submodule issuch adirect limit, since aninfinitely generated module ofrelations can beviewed asalimit offinitely generated modules of relations. Make thisprecise togetaproof.] 26, LetEbeamodule over aring. Let{M i}beadirected family ofmodules. IfEisfinitely generated, show that thenatural homomorphism limHom(E, Mi)Hom(E, limMi) ISInjective. IfEisfinitely presented, show that thishomomorphism isanisomorphism. Hint: First prove the statements when Eisfree with finite basis. Then, say Eis finitely presented byanexact sequence F1 F0 E O.Consider thediagram: o )li111Hom(E, Mi) I)limHom(F 0,Mi) I)limHom(F l'Mi) I o )Hom(E, liIIlMi))Hom(F 0'limM;))Hom(F l'limMi) 172 MODULES III,Ex Graded Algebras Let Abeanalgebra over afield k.Byafiltration ofAwe mean asequence ofk- vector spaces Ai(i==0,I,...)such that AocAlcA2c,., and UA;==A, andA;AjcA;+jforalli,j>O.Inparticular, AisanAo-algebra. We then call Aafil- tered algebra, Let Rbe analgebra. We say that Risgraded ifRisadirect sum R==EBR;ofsubspaces such thatR;RjcR;+jforalli,j>o. 27. Let Abe afiltered algebra. Define R;fori>0byR;==A;/A;_I. Bydefinition, A_I=={O}. Let R==EBR;,and R;==gr;(A). Define anatural productonRmaking Rinto agraded algebra, denoted bygr(A), and called theassociated graded algebra. 28. LetA,Bbefiltered algebras, A==UA;and B==UBi.LetL:A-+Bbean(Ao,Bo)- linear map preserving thefiltration, that isL(A;)cB;for alli,andL(ca)== L(c)L(a) for cEAoand aEA;foralli. (a) Show that Linduces an(Ao,Bo)-linear map gr;(L): gr;(A)-+gr;(B) foralli, (b)Suppose thatgr;(L)isanisomorphism foralli.Show that Lisan(Ao,Bo)- isomorphism. 29.Suppose khascharacteristic o.Let nbethe setofallstrictly upper triangular ma- trices ofagiven size nxnover k. (a)For agiven matrix XEn,letDI(X),...,Dn(X) beitsdiagonals,soDI== DI(X)isthemain diagonal, and is0bythedefinition ofn.Let nibethe subset ofnconsisting ofthose matrices whose diagonals DI ,...,Dn-i are O. Thus no=={O}, nlconsists ofallmatrices whose components are 0except possibly for Xnn;n2consists ofallmatrices whose components are 0except possibly those inthelast twodiagonals; and soforth. Show that each n;is analgebra, and itselements arenilpotent (infact the(i+1)-th power ofits elements is0). (b)Let Ubethe setofelements I+Xwith XEn.Show that Uisamulti- plicative group. (c)Let exp betheexponential series defined asusual. Show that exp defines a polynomial function onn(allbut afinite number ofterms are 0when eval- uated on anilpotent matrix), and establishes abijection exp:n-+U. Show that theinverse isgiven bythestandard log series. CHAPTER IV Polynomials This chapter providesacontinuation ofChapter II,3. We prove stan- dard properties ofpolynomials. Most readers will beacquainted with some ofthese properties, especially atthebeginning forpolynomials inone vari- able. However, one ofour purposes istoshow that some ofthese properties also hold over acommutative ring when properly formulated. The Gauss lemma and thereduction criterion forirreducibility will show theimportance ofworking over rings. Chapter IXwill give examples oftheimportance of working over theintegers Zthemselves togetuniversal relations. Ithappens that certain statements ofalgebra areuniversally true. Toprove them, one proves them first forelements ofapolynomial ring over Z,and then one obtains the statement inarbitrary fields (orcommutative ringsasthe case may be)byspecialization. The Cayley-Hamilton theorem ofChapter XV, forinstance, can beproved inthat way. The last section onpower series shows that the basic properties of polynomial rings can beformulated so astohold forpower series rings. I conclude this section with several examples showing theimportance ofpower series invarious parts ofmathematics. 1. BASIC PROPERTIES FOR POLYNOMIALS INONE VARIABLE We start with theEuclidean algorithm. Theorem 1.1. Let Abe acommutative ring, letf,gEA[X] bepoly- nomials in one variable, ofdegrees>0,and assume that the leading 173 174 POLYNOMIALS IV,1 coefficient ofgisaunit InA. Then there exist unique polynomials q,rEA[X] such that f=gq+r anddeg r<deg g. Proof. Write f(X)=anXn+...+ao, g(X)=bdXd+...+bo, where n=degf,d=deg gsothat an,bd=F0and bdisaunit inA.We use induction on n. Ifn=0,and deg g>degf,weletq=0,r=f.Ifdeg g=degf=0,then weletr=0and q=anbi1 . Assume thetheorem proved forpolynomials ofdegree <n(withn>0). We may assume deg g<degf(otherwise, take q=0and r=f). Then f(X)=anbi1Xn-dg(X) +fl(X), where f1(X) hasdegree <n.Byinduction, we can find ql'rsuch that f(X)=anbi1xn-dg(X) +q1(X)g(X) +r(X) and deg r<deg g.Then welet q(X)=anbi1Xn-d+q1(X) toconclude theproof ofexistence for q,r. Asforuniqueness, suppose that f=qlg +r1=q2g +r2 with deg r1<deg gand deg r2<deg g.Subtracting yields (q1-q2)g=r2-r1. Since theleading coefficient ofgisassumed tobeaunit, wehave deg(q1-q2)g=deg(q1-q2)+deg g. Since deg(r2-r1)<deg g,this relation can hold only ifq1-q2=0,I.e. ql=q2,and hence finally r1=r2aswas tobeshown. Theorem 1.2. Let kbe afield. Then thepolynomial ring inone variable k[X] isprincipal. IV,1 BASIC PROPERTIES FOR POLYNOMIALS INONE VARIABLE 175 Proof Let abe anideal ofk[X], and assume Q=Fo.Let gbe an element of Qofsmallest degree>O.Letfbeany element of Qsuch that f=FO.BytheEuclidean algorithm we can find q,rEk[X] such that f=qg+r and degr<deg g.But r=f-qg,whence risin Q.Since ghad minimal degree>0itfollows that r=0,hence that Qconsists ofallpolynomials qg (with qEk[X]). This proves our theorem. ByTheorem 5.2ofChapter IIwe get: Corollary 1.3. The ring k[X] isfactorial. Ifkisafield then every non-zero element ofkisaunit ink,and one sees immediately that the units ofk[X] aresimply theunits ofk.(No polyno- mial ofdegree>1can be aunit because oftheaddition formula for the degree ofaproduct.) Apolynomial f(X)Ek[X] iscalled irreducible ifithasdegree>1,and if one cannot write f(X)asaproduct f(X)=g(X)h(X) with g,hEk[X], and both g,h k.Elements ofkareusually called constant polynomials,sowe can also saythat insuch afactorization, oneof9orhmust beconstant. Apolynomial iscalled monic ifithasleading coefficient 1. Let Abe acommutative ring andf(X}apolynomial inA[X]. Let Abe asubring ofB.Anelement bEBiscalled aroot or azero offinBif f(b)=o.Similarly, if(X) isann-tuple ofvariables, ann-tuple (b)iscalled a zero offiff(b)=o. Theorem 1.4. Let kbe afield andfapolynomial inone variable Xin k[X], ofdegreen>O.Thenfhas atmost nroots ink,andifaisaroot offink,then X-adivides f(X). Proof Suppose f(a)=O.Find q,rsuch that f(X)=q(X)(X-a)+r(X) and deg r<1.Then o=f(a)=r(a). Since r=0orrisanon-zero constant, wemust have r=0,whence X-a divides f(X). Ifat, ..., amaredistinct roots offink,then inductively we see that theproduct (X-at)". (X-am) 176 POLYNOMIALS IV,1 divides j(X), whence m<n,thereby proving thetheorem. The next corollaries give applications ofTheorem 1.4topolynomialfunctions. Corollary 1.5. Let kbe afield and Taninfinite subset ofk.Let f(X)Ek[X] beapolynomialinone variable. Iff(a)=0forall aET,then f=0,i.e.finduces the zero function. Corollary 1.6. Let kbe afield, and letS1' ..., Snbeinfinite subsets ofk. Letf(X l'...,Xn)beapolynomial innvariables over k.Iff(a1' ...,an)=0 forallajESj(i=1,..., n),thenf=o. Proof Byinduction. We have justseen the result istrue for one variable. Let n>2,and write f(X l'...,Xn)=Lh(X l'...,Xn-1)xj j asapolynomial inXnwith coefficients ink[X l'...,Xn-1].Ifthere exists (b1,..., bn-1)E81X...XSn-1 such that for somejwehave h(b 1,...,bn-1);/=0,then f(b 1,.. .,bn-1,Xn) isanon-zero polynomial ink[X n]which takes onthevalue 0fortheinfinite setofelements 8n.This isimpossible. HenceJjinduces the zero function on 81x...X8n-1 forallj,and byinduction wehave Jj=0forallj.Hence f=0,aswas tobeshown. Corollary 1.7. Let kbeaninfinite field andfapolynomial innvariables over k.Iffinduces the zero function onk(n), thenf=o. We shall now consider the case offinite fields. Let kbe afinite field with qelements. Letf(X l'...,Xn)beapolynomial innvariables over k.Write f(X 1,...,Xn)=La(V)X;l...X;". Ifa(v);/=0,werecall that themonomial M(v)(X)occurs infSuppose this is the case, and that inthis monomial M(v)(X),some variable Xioccurs with an exponent Vi>q.We can write X,vi=X9+/lI I' J-l=integer>o. Ifwe now replace xtibyXr+1inthis monomial, then weobtain anew polynomial which gives rise tothe same function asfThe degree ofthis new polynomial isatmost equal tothedegree off IV,1 BASIC PROPERTIES FOR POLYNOMIALS INONE VARIABLE 177 Performing the above operationafinite number oftimes, for allthe monomials occurring infand allthe variables Xl' ...,Xnweobtain some polynomial f*giving rise tothe same function asf,but whose degree in each variable is<q. Corollary 1.8. Let kbe afinite field with qelements. Letfbe a polynomial innvariables over ksuch that thedegree offineach variable is<q.Iffinduces the zero function onken), thenf=o. Proof Byinduction. Ifn=1,then thedegree offis<q,and hencef cannot have qroots unless itisO.The inductive step iscarried outjustas wedidfortheproof ofCorollary 1.6above. Letfbeapolynomial innvariables over thefinite field k.Apolynomial gwhose degree ineach variable is<qwill besaid tobereduced. We have shown above that there exists areduced polynomial f*which gives the same function asfonken). Theorem 1.8 now shows that this reduced polynomial is unique. Indeed, ifgl'g2are reduced polynomials giving the same function, then gl-g2isreduced and gives the zero function. Hence gl-g2=0and gl=g2. We shall give one more application ofTheorem 1.4. Let kbe afield. By amultiplicative subgroup ofkweshall mean asubgroup ofthe group k* (non-zero elements ofk). Theorem 1.9. Let kbe afield and letUbe afinite multiplicative sub- group ofk.Then Uiscyclic. Proof Write Uasaproduct ofsubgroups U(p) foreach prime p,where U(p) isap-group. ByProposition 4.3(vi) ofChapter I,itwill suffice toprove that U(p) iscyclic foreach p.Let abeanelement ofU(p) ofmaximal period prfor some integerr.Then xP" =1forevery element xEU(p), and hence all elements ofU(p) are roots ofthepolynomial Xp" -1. The cyclic group generated byahasprelements. Ifthiscyclic group isnot equal toU(p), then our polynomial has more than prroots, which is impossible. Hence agenerates U(p), and our theorem isproved. Corollary 1.10. Ifkisafinite field, then k*iscyclic. Anelement ,inafield ksuch that there exists anintegern>1such that ,n=1iscalled aroot ofunity, ormore precisely ann-th root ofunity. Thus the setofn-th roots ofunity isthe setofroots ofthepolynomial xn-1. There are atmost nsuch roots, and they obviously form agroup, which is 178 POLYNOMIALS IV,1 cyclic byTheorem 1.9. We shall study roots ofunity ingreater detail later. Agenerator forthegroup ofn-th roots ofunity iscalled aprimitive n-th root ofunity. For example, inthecomplex numbers, e21ti/nisaprimi- tive n-th root ofunity, and then-th roots ofunity areoftype e21tiv/n with 1<v<n. The group ofroots ofunity isdenoted byp.The group ofroots ofunity inafield Kisdenoted byp(K). Afield kissaid tobealgebraically closed ifevery polynomial ink[X] of degree>1has aroot ink.Inbooks onanalysis, itisproved that the complex numbers arealgebraically closed. InChapter Vweshall prove that afield kisalways contained in some algebraically closed field. Ifkis algebraically closed then theirreducible polynomials ink[X] are thepoly- nomials ofdegree1.Insuch acase, theunique factorization ofapolynomial fofdegree>0can bewritten intheform r f(X)=cn(X-i)mi i=l with CEk,c=F0and distinct rootst,...,r. We next developatest when mi>1. Let Abeacommutative ring. We define amap D:A[X] A[X] ofthepolynomial ring into itself. Iff(X)=anXn +...+aowith aiEA,we define thederivative n Df(X)=f'(X)=LvavXv-l =nanXn-t+...+at. v=l One verifies easily that iff,garepolynomials inA[X], then (f+g)'=f'+g', (fg)'=f'g+fg', andifaEA,then (af)'=af'. Let Kbe afield andfanon-zero polynomial inK[X]. Let abe aroot offin K.We can write f(X)=(X-a)mg(X) with some polynomial g(X) relatively prime toX-a(and hence such that g(a) =F0). We call mthemultiplicity ofainf,and say that aisamultiple root ifm>1. IV,1 BASIC PROPERTIES FOR POLYNOMIALS INONE VARIABLE 179 I Proposition 1.11. LetK,fbeasabove. The element aofKisamultiple rootoffifandonlyifitisaroot andf'(a)=o. Proof Factoring fasabove, weget f'(X)=(X-a)mg'(X) +m(X-a)m-l g(x). Ifm>1,then obviously f'(a)=O.Conversely, ifm=1then f'(X)=(X-a)g'(X) +g(X), whence f'(a)=g(a) =FO.Hence iff'(a)=0wemust have m>1,asdesired. Proposition 1.12. LetfEK[X]. IfKhas characteristic 0,andfhas degree>1,thenf'=Fo.Let Khave characteristic p>0andfhave degree>1.Thenf'=0ifandonly if,intheexpression forf(X) given by n f(X)=LayXY , y=l pdivides each integervsuch that ay=Fo. Proof IfKhas characteristic 0,then thederivative ofamonomial ayXY such that v>1and ay=F0isnot zero since itis vayXy-l. IfKhas characteristic p>0,then thederivative ofsuch amonomial is0ifandonly if piv,ascontended. Let Khave characteristic p>0,and letfbewritten asabove, and be such thatf'(X)=O.Then one can write d f(X)=Lb/lXP/l /l=1 withb/lEK. Since thebinomial coefficients()aredivisible bypfor 1<v<P-1we seethat ifKhascharacteristic p,then for a,bEKwehave (a+b)P=aP+bP . Since obviously (ab)P=aPbP ,themap X1---+xP isahomomorphism ofKinto itself, which has trivial kernel, hence is injective. Iterating, weconclude that foreach integer r>1,themap x1---+xP" 180 POLYNOMIALS IV,2 isanendomorphism ofK,called theFrobenius endomorphism. Inductively, if Cl'..., Cnareelements ofK,then (C1+... +cnY'=cf+... +c:. Applying these remarks topolynomials,we seethat forany element aEK wehave (X-a)P'"=xP'"-aP'". IfCEKand thepolynomial xP'"-C has one root ainK,then aP'" =Cand xP'"-C=(X-a)P". Hence our polynomial has preciselyone root, ofmultiplicity pro For In- stance, (X-1)p"=XP'" -1. 2. POLYNOMIALS OVER AFACTORIAL RING Let Abe afactorial ring, and Kitsquotient field. Let aEK, a=FO.We can write aas aquotient ofelements inA,having noprime factor in common. Ifpisaprime element ofA,then we can write a=prb, where bEK, risaninteger, and pdoes not divide the numerator or denominator ofb.Using theunique factorization inA,we see atonce that r isuniquely determined by a,and wecall rthe order of aatp(and write r=ordpa).Ifa=0,wedefine itsorder atptobe 00. Ifa,a'EKand aa' =F0,then ordp(aa')=ordpa+ordpa'. This isobvious. Letf(X) EK[X] beapolynomial inonevariable, written f(X)=ao+a1X+...+anXn . Iff=0,wedefine ordpftobe 00.Iff=F0,wedefine ordpftobe IV,2 POLYNOMIALS OVER AFACTORIAL RING 181 ordpf=min ordpah theminimum being taken over allthose isuch that ai=FO. Ifr=ordpf,wecalluprap-content forf,ifuisany unit ofA.Wedefine the content offtobetheproduct. npordpf , theproduct being taken over allpsuch that ordpI=F0,oranymultiple of this product byaunit ofA. Thus the content iswell defined up to multiplication byaunit ofA.Weabbreviate content bycont. IfbEK,b=F0,then cont(bf)=bcont(f). This isclear. Hence we can write f(X)=e.11(X) where e=cont(f), andfl(X) has content 1.Inparticular, allcoefficients of fllieinA,and their g.c.d. is1.We define apolynomial with content 1tobe aprimitive polynomial. Theorem 2.1. (Gauss Lemma). Let Abe afactorial ring, and letKbe itsquotient field. Letf,gEK[X] bepolynomials inone variable. Then cont(fg)=cont(f) cont(g). Proof. Writing 1=efland g=dg 1where c=cont(f) and d=cont(g), we see that itsuffices toprove: Iff,ghave content 1,then fgalso has content 1,and forthis, itsuffices toprove that foreach prime p,ordp(fg)=O. Let f(X)=anXn+...+ao, g(X)=bmxm +·..+bo,an=F0, bm=F0, bepolynomials ofcontent 1.Let pbe aprime ofA.Itwill suffice toprove that pdoes not divide allcoefficients offg. Let rbethelargest integer such that 0<r<n,ar=F0,and pdoes not divide areSimilarly, letbsbethe coefficient ofgfarthest totheleft, bs=F0,such that pdoes not divide bs. Consider thecoefficient ofxr+s inf(X)g(X). This coefficient isequal to e=arbs +ar+lbs-1+... +ar-lbS+1+... and plarbs. However, pdivides every other non-zero term inthis sum since ineach term there will be some coefficient aitothe left of aror some coefficient bjtotheleftofbs.Hence pdoes not divide e,and our lemma is proved. 182 POLYNOMIALS IV,2 We shall now give another proof forthekey step intheabove argument, namely the statement: Iff,gEA[X] areprimitive (i.e. have content 1)thenfgisprimitive. Proof. We have toprovethat agiven prime pdoes not divide allthe coefficients offg. Consider reduction mod p,namely the canonical homo- morphism A--+A/(p)=A .Denote theimage ofapolynomial byabar, so fH1and g1---+gunder thereduction homomorphism. Then fg=!g. Byhypothesis, 1=F0and g=FO.Since Aisentire, itfollows thatfg =F0,as was tobeshown. Corollary 2.2. Letf(X)EA[X] have afactorization f(X)=g(X)h(X) in K[X]. IfCg=cont(g), Ch=cont(h), and g=Cggl,h=chh l,then f(X)=Cgchgl (X)h 1(X), andCgChisanelement ofA.Inparticular, iff,gEA[X] have content 1, then hEA[X] also. Proof The only thing tobeproved isCgChEA.But cont(f)=CgChcont(g 1hI)=CgCh, whence our assertion follows. Theorem 2.3. Let Abe afactorial ring. Then thepolynomial ringA[X] inonevar_iableisfactorial. Itsprime elements are theprimes ofAandpoly- nomials inA[X] which areirreducible inK[X] and have content 1. Proof LetfEA[X], f=FO.Using theunique factorization inK[X] and thepreceding corollary, we can find afactorization f(X)=c.PI(X)...p,(X) where CEA,and PI' ..., P,arepolynomials inA[X] which areirreducible in [X]. Extracting the!:. contents, wemay assume without loss ofgenerality that the content ofPiis1foreach i.Then c=cont(f) bythe Gauss lemma. This givesusthe existence ofthefactorization. Itfollows that each Pj(X) is irreducible inA[X]. Ifwehave another such factorization, say f(X)=d.ql(X)...qs(X), then from theunique factorization inK[X] weconclude that r=s,and after apermutation ofthefactors wehave Pi=aiqi IV,3 CRITERIA FOR IRREDUCIBiliTY 183 with elements ajEK. Since both Phqiare assumed tohave content 1,it follows that aiinfact liesinAand isaunit. This provesour theorem. Corollary 2.4. Let Abeafactorial ring. Then theringofpolynomials in nvariables A[Xl'.. .,Xn] isfactorial. Itsunits areprecisely theunits of A,and itsprime elements areeither primes ofAorpolynomials which are irreducible inK[X] and have content 1. Proof. Induction. Inview ofTheorem 2.3, when wedeal with polynomials over afactorial ring and having content 1,itisnot necessary tospecify whether such polynomials areirreducible over Aorover thequotient field K. The two notions areequivalent. Remark 1.The polynomial ringK[X 1,...,Xn] over afield Kisnot principal when n>2.For instance, theideal generated byXl'...,Xnisnot princi pal(trivial proof). Remark 2.Itisusually not too easy todecide when agiven polynomial (say inone variable) isirreducible. For instance, thepolynomial X4+4is reducible over therational numbers, because X4+4=(X2-2X+2)(X2+2X+2). Later inthis book we shall giveaprecise criterion when apolynomial xn-aisirreducible. Other criteria aregiven inthe next section. 3. CRITERIA FOR IRREDUCIBiliTY The first criterion is: Theorem 3.1. (Eisenstein's Criterion). Let Abe afactorial ring. Let K beitsquotient field. Letf(X)=anXn +...+aobeapolynomial ofdegree n>1inA[X]. Let pbeaprime ofA,and assume: an=1=0(mod p), ai=0(mod p) ao=1=0(mod p2). Then f(X) isirreducible inK[X].forall i<n, 184 POLYNOMIALS IV,3 Proof Extractingag.c.d. for the coefficients off, we may assume without loss ofgenerality that the content offis1.Ifthere exists a factorization into factors ofdegree>1inK[X], then bythecorollary of Gauss' lemma there exists afactorization inA[X], sayf(X)=g(X)h(X), g(X)=bdXd+...+bo, h(X)=cmXm+...+co, with d,m>1and bdc m=FO.Since boco=aoisdivisible bypbut notp2,it follows that one ofbo,Coisnot divisible byp,say boo Then plc o.Since cmb d=anisnot divisible byp,itfollows that pdoes not divideCm. Let Crbe thecoefficient ofhfurthest totheright such that Cr=1=0(mod p).Then ar=boc r+b1cr-l+.... Since plboc rbut pdivides every other term inthis sum, weconclude that p1ar,acontradiction which provesour theorem. Example. Let abe anon-zero square-free integer =F+1.Then forany integern>1,thepolynomial xn-aisirreducible over Q.The polynomials 3X5-15and 2X10-21areirreducible over Q. There are some cases inwhich apolynomial does notsatisfy Eisenstein's criterion, but asimple transform ofitdoes. Example. Let pbeaprime number. Then thepolynomial f(X)=Xp-l +... +1 isirreducible over Q. Proof Itwill suffice toprove that thepolynomial f(X +1)isirreducible over Q.We note that thebinomial coefficients (p)p! v v!(p-v)!'1<v<P-1, aredivisible byp(because the numerator isdivisible bypand thedenomina- torisnot, and thecoefficient isaninteger). We have (X+I)P-1 XP+pXp-l +...+pXf(X +1)= (X+1)-1= X from which one sees thatf(X+1)satisfies Eisenstein's criterion. Example. LetEbe afield and tanelement ofsome field containing Esuch that tistranscendental over E. Let Kbe the quotient field ofE[t]. IV,3 CRITERIA FOR IRREDUCIBiliTY 185 For any integern>1thepolynomial xn-tisirreducible inK[X]. This comes from thefact that thering A=E[t] isfactorial and that tisaprime init. Theorem 3.2. (Reduction Criterion). LetA,Bbeentire rings, and let ({J:A-+B be ahomomorphism. LetK,Lbethequotient fields ofAand Brespec- tively. LetfEA[X] besuch that ({Jf =F0and deg ({Jf=degfIf({Jf is irreducible inL[X], thenfdoes not have afactorization f(X)=g(X)h(X) with g,hEA[X] and degg, degh>1. Proof. Suppose fhas such afactorization. Then ({Jf=«({Jg)«({Jh). Since deg ({Jg<deg gand deg ({Jh<deg h,our hypothesis implies that we must have equality inthese degree relations. Hence from theirreducibility in L[X] weconclude that gorhisanelement ofA,asdesired. Inthepreceding criterion, suppose that Aisalocal ring, i.e. aring having aunique maximal ideal p,and that pisthe kernel of({J.Then from the irreducibility of({JfinL[X] weconclude theirreducibility offinA[X]. Indeed, any element ofAwhich does not lieinpmust be aunit inA,soour last conclusion intheproofcan bestrengthened tothe statement that gorh isaunit inA. One can also apply thecriterion when Aisfactorial, and inthat case deduce theirreducibility offinK[X]. Example. Let pbe aprime number. Itwill beshown later that XP-X-I isirreducible over thefield Z/pZ. Hence XP-X-I isirreduc- ible over Q.Similarly, X5-5X4-6X-1 isirreducible over Q. There isalso aroutine elementary school test whether apolynomial has a root ornot. Proposition 3.3. (Integral Root Test). Let Abe afactorial ring and K itsquotient field. Let f(X)=anXn+...+aoEA[X]. Let rxEKbe aroot off,with rx=b/dexpressed with b,dEAand b,d relatively prime. Then blao anddla n.Inparticular, iftheleading coefficient anis1,then aroot rxmust lieinAand divides ao. 186 POLYNOMIALS IV,4 We leave theproof tothereader, who should beused tothis one from way back. As anirreducibility test, the test isuseful especially for apolynomial of degree 2or3,when reducibility isequivalent with theexistence ofaroot in thegiven field. 4. HILBERT'S THEOREM This section provesabasic theorem ofHilbert concerning theideals ofa polynomial ring. We define acommutative ring AtobeNoetherian ifevery ideal isfinitely generated. Theorem 4.1. Let Abeacommutative Noetherian ring. Then thepolyno- mial ring A[X] isalso Noetherian. Proof. Let beanideal ofA[X].Leta;consist of0and the setofelements aEAappearingasleading coefficient insome polynomial ao+atX+...+aXi lyingin. Then itisclear thatQiisanideal. (Ifa,bareinQi,then a+bis in Qias one sees bytaking the sum and difference ofthecorresponding polynomials. IfxEA,then xa EQias one sees bymultiplying the corre- sponding polynomial byx.)Furthermore wehave QoCQtCQ2C.", inother words, our sequence ofideals {Qi}isincreasing. Indeed, toseethis multiply theabove polynomial byXtoseethat aEQi+t. Bycriterion (2)ofChapter X, 1,thesequence ofideals {Qi} stops, say at Qr: QoCQtCQ2C...CQr=Qr+t=... . Let aot,..., aonobegenerators forQo, ... ........ .......... ..... art, ..., arn,.begenerators forQr' For each i=0,...,randj=1,..., niletfijbe apolynomial in,ofdegree i,with leading coefficientaij.We contend that thepolynomials /;,jare aset ofgenerators for. Letfbe apolynomial ofdegree din. We shall prove thatfisinthe ideal generated bythe/;,j,byinduction ond.Say d>O.Ifd>r,then we IV,5 PARTIAL FRACTIONS 187 note that theleading coefficients of Xd-rf, Xd-rf, rl'..., rnr generate Qd. polynomialHence there exist elements cl'.. .,cnEAsuch that ther f-Cxd-rf,-... -CXd-rf, 1 r1 n,. rn,. hasdegree <d,and thispolynomial also liesin. Ifd<r,we can subtract alinear combination f-c1h1- ... -cndfdnd togetapolynomial ofdegree <d,also lying in. We note that the polynomial wehave subtracted fromfliesintheideal generated bythehj. Byinduction, we can subtract apolynomial gintheideal generated bythe fijsuch thatf-g=0,thereby proving our theorem. We note that ifcp:A-.Bisasurjective homomorphism ofcommutative rings and AisNoetherian, soisB.Indeed, letbbeanideal ofB,solfJ-1(b) isanideal ofA.Then there isafinite number ofgenerators (ah. . .,an)for cp-1(b), and itfollows since lfJissurjective that b=lfJ(lfJ-l(b)) isgenerated by cp(a 1),...,cp(a n),asdesired. As anapplication, weobtain: Corollary 4.2. Let Abe aNoetherian commutative ring, and letB= A[xI'...,xm] be acommutative ringfinitely generated over A.Then Bis Noetherian. Proof Use Theorem 4.1and thepreceding remark, representing Basa factor ring ofapolynomial ring. Ideals inpolynomial rings will bestudied more deeply inChapter IX. The theory ofNoetherian rings and modules will bedeveloped inChapter X. 5. PARTIAL FRACTIONS Inthis section, weanalyze thequotient field ofaprincipal ring, using the factoriality ofthering. Theorem 5.1. Let Abe aprincipal entire ring, and letPbe asetof representatives for itsirreducible elements. Let Kbethequotient field of A,and let rxEK. For each pEP there exists anelementrxpEAand an integer j(p)>0,such thatj(p)=0for almost allpEP, rxpand pj(P) are 188 POLYNOMIALS IV,5 relatively prime, and _ rxprx- p":ppj(P). Ifwehave another such expression (X=L)' peP P thenj(p)=i(p)forallp,andrxp=PPmod pj(P)forallp. Proof We first prove existence, in aspecialcase. Let a,bberela- tively primenon-zero elements ofA.Then there exists x,YEA such that xa+yb=1.Hence 1 x y- = -+-. ab b a Hence any fraction cjab with cEAcan bedecomposed into asum oftwo fractions (namely cxjb and cyja) whose denominators divide band arespec- tively. Byinduction, itnow follows that anyrxEKhas anexpressionas stated inthetheorem, except possibly for the fact that pmay dividerxp. Canceling thegreatest common divisor yieldsanexpression satisfying allthe desired conditions. Asforuniqueness, suppose that rxhas two expressionsasstated inthe theorem. Let qbeafixed prime inP.Then rxq Pq PP rxp qj(q)- qi(q)= P'tqpi(p)- pj(P). Ifj(q)=i(q)=0,our conditions concerning qare satisfied. Supposeone of j(q) ori(q) >0,sayj(q), and sayj(q)>i(q). Let dbealeast common multiple forallpowers pj(P) and pi(p) such that p=Fq.Multiply theabove equation by dqj(q). Weget d(rx q-qj(q)-i(q)p q)=qj(q)P for some pEA. Furthermore, qdoes not divide d.Ifi(q) <j(q) then q dividesrxq'which isimpossible. Hence i(q)=j(q). We now see that qj(q) dividesrxq-Pq,thereby proving thetheorem. Weapply Theorem 5.1 tothepolynomial ringk[X] over afield k.We letPbethe setofirreducible polynomials, normalized so astohave leading coefficient equal to1.Then Pisasetofrepresentatives foralltheirreduc- ible elements ofk[X]. Intheexpression given for rxinTheorem 5.1, we can now dividerxpbypj(P), i.e. use the Euclidean algorithm, ifdegcxp>degpj(P). We denote thequotient field ofk[X] byk(X), and call itselements rational functions. IV,5 PARTIAL FRACTIONS 189 Theorem 5.2. Let A=k[X] bethepolynomial ring inone variable over a field k.Let Pbethe setofirreducible polynomials ink[X] with leading coefficient1.Then any element fofk(X) has aunique expression 1;,(X)f(X)=L(X)i(P)+g(X), peP p where1;"garepolynomials, 1;,=0ifj(p)=0,1;,isrelatively prime topif j(p) >0,anddeg1;,<degpj(P)ifj(p) >O. Proof. The existence follows atonce from our previous remarks. The uniqueness follows from the fact that ifwe have two expressions, with elements1;,andlfJprespectively, and polynomials g,h,then pj(p) divides 1;,-lfJp'whence1;,-lfJp=0,and therefore1;,=lfJp'g=h. One can further decompose the term1;,lpj(P) byexpanding 1;,according to powers ofp.One can infact dosomething more general. Theorem 5.3. Let kbe afield andk[X] thepolynomial ring inone variable. Letf,gEk[X], and assume deg g>1.Then there exist unique polynomials fo,fl, ...,i1Ek[X] such that degIi<deg gand such that f=fo+ftg+...+i1gd . Proof. We first prove existence. Ifdeg g>degf,then wetake 10=f andIi=0for i>O.Suppose deg g<degfWe can find polynomials q,r with deg r<deg gsuch that f=qg+r, and since deg g>1wehave deg q<degfInductively, there exist polyno- mials ho,hI' ..., hssuch that q=ho+htg+...+hsgS , and hence f=r+hog +...+hsgs+t , thereby proving existence. Asforuniqueness, let f=fo+ftg+...+i1gd=lfJo+lfJtg+·..+lfJmgm betwo expressions satisfying theconditions ofthe theorem. Adding terms 190 POLYNOMIALS IV,6 equal to0toeither side, wemay assume that m=d.Subtracting,weget o=(fo-lpo)+...+(j;,-<Pd)gd. Hence gdivides fo-<Po,and since deg(fo-<Po)<deg gwe seethatfo=<Po. Inductively, take the smallest integer isuch that h=F<Pi(ifsuch iexists). Dividing theabove expression bygiwefind that gdivides Ii-<Piand hence that such icannot exist. This proves uniqueness. We shall call theexpressionforfinterms ofginTheorem 5.3theg-adic expansion offIfg(X)=X,then theg-adic expansion isthe usual expres- sion offasapolynomial. Remark. Insome sense, Theorem 5.2redoes what was done inTheorem 8.1ofChapter IforQ/Z; that is,express explicitlyanelement ofK/Aasa direct sum ofitsp-components. 6. SYMMETRIC POLYNOMIALS Let Abe acommutative ring and lettl'...,tnbealgebraically indepen- dent elements over A.LetXbe avariable over A[tl'...,tnJ. We form the polynomial F(X)=(X-t1)...(X-tn) =xn-Stxn-1+...+(-l)n sn, where eachSi=Si(tl'...,tn)isapolynomial intl'..., tn.Then forinstance Sl=t1+...+tn andSn=t1...tn' The polynomials Sl' ..., Snare called theelementary symmetric polynomials oftl'...,tn' We leave itasaneasy exercise toverify thatSiishomogeneous ofdegree i intl'..., tn. Let (Jbe apermutation oftheintegers (1,..., n).Given apolynomial I(t)EA[t]=A[t 1,.. .,tn],wedefine ulto be u1(tI'...tn)=1(t0'(1),.. .,t0'(n». Ifu, Taretwopermutations, thenuTI=u(Tf)and hence thesymmetric group Gon nletters operatesonthepolynomial ring A[t]. Apolynomial iscalled symmetric iful=Iforall uEG.Itisclear that the setofsymmetric polynomials isasubring ofA[tJ, which contains the constant polynomials IV,6 SYMMETRIC POLYNOMIALS 191 (i.e. Aitself) and also contains theelementary symmetric polynomialssl'...,sn' We shall seebelow that A[sl'...,sn]isthering ofsymmetric polynomials. LetXl' ...,Xnbevariables. Wedefine theweight ofamonomial XVI...XVn1 n to be V1+2V2+...+nvn.We define the weight of apolynomial g(X l'.. .,Xn)tobethemaximum oftheweights ofthemonomials occurring Ing. Theorem 6.1. Letf(t)EA[tl'..., tn]besymmetric ofdegree d.Then there exists apolynomial g(X l'...,Xn)ofweight<dsuch that f(t)=g(Sl, ...,sn). Proof Byinduction on n.The theorem isobvious ifn=1,because Sl=t 1. Assume thetheorem proved forpolynomials inn-1variables. Ifwesubstitute tn=0intheexpression forF(X), wefind (X-t1)...(X-tn-1)X=xn-(Sl)Oxn-l +... +(_1)n-l(sn_l)OX, where (Si)O istheexpression obtained bysubstituting tn=0inSi'We see that (s1)0'...,(Sn-l)Oareprecisely theelementary symmetric polynomials in tl'...,tn-1. We now carry out induction on d.Ifd=0,our assertion istrivial. Assume d>0,and assume our assertion proved forpolynomials ofdegree <d. Let f(tl'..., tn)have degree d. There exists apolynomial g1(Xl'...,Xn-1)ofweight<dsuch that f(t 1,...,tn-I' 0)=gl((Sl)O' ...,(Sn-l)O)' We note thatgl(Sl' ...,Sn-l) hasdegree<dint1,..., tn'Thepolynomial fl(t 1,..., tn)=f(t 1,..., tn)-gl(SI' ...,Sn-l) hasdegree<d(int1,..., tn)and issymmetric. We have fl(tl'.. .,tn-1,0)=o. Hence flisdivisible bytn'i.e.contains tnasafactor. Sinceflissymmetric, itcontains t1...tnasafactor. Hence fl=Snf2(tl'...,tn) for some polynomial f2'which must besymmetric, and whose degreeIS 192 POLYNOMIALS IV,6 <d-n<d.Byinduction, there exists apolynomial g2innvariables and weight<d-nsuch that f2(t 1,..., tn)=g2(S1' ...,sn). Weobtain f(t)=g1(S1' ...,Sn-1) +sng2(S1' ...,sn)' and each term ontheright hasweight<d.This proves our theorem. We shall now prove that theelementary symmetric polynomials S1'''.' Sn arealgebraically independentover A. Ifthey are not, take apolynomial f(X l'...,Xn)EA[X] ofleast degree and notequal to0such that f(sl'·..,Sn)=O. Writefasapolynomial inXnwith coefficients inA[Xl'...,Xn-1], f(X l'...,Xn)=fo(X l'...,Xn-1)+...+h(X l'...,Xn-1)X. Then fo=FO.Otherwise, we can write f(X)=Xnl/1(X) with some polynomial 1/1,and hence snl/1(S1' ...,sn)=O.From this itfollows that I/1(S1' ...,sn)=0,and1/1hasdegree smaller than thedegree off We substitute SiforXiintheabove relation, and get o=fo(sl'...,Sn-1)+...+h(Sl'.. .,Sn-1)S:. This isarelation inA[tl'...,tn],and wesubstitute 0for tninthis relation. Then allterms become 0except thefirst one, which gives o=fO(S1)0, ...,(Sn-1)0), using the same notation asintheproof ofTheorem 6.1. This isanon-trivial relation between theelementary symmetric polynomials intl'...,tn-1,a contradiction. Example. (The Discriminant). Letf(X)=(X-t1)...(X-tn).Con- sider theproduct £5(t)=n(ti-tj). i<j For anypermutation(Jof(1,...,n)we see atonce that £5a(t)=+£5(t). IV,6 SYMMETRIC POLYNOMIALS 193 Hence l5(t)2 issymmetric, and wecall itthediscriminant: Df=D(Sl"'" sn)=n(ti-tj)2. i<j We thus view thediscriminant asapolynomial intheelementary symmetric functions. For acontinuation ofthegeneral theory, see8. We shall now consider special cases. Quadratic case. You should verify that for aquadratic polynomial f(X)=X2+bX+c,one has D=b2-4c. Cubic C9.se. Consider f(X)=X3+aX+b.We wish toprove that D=-4a3-27b2 . Observe first that Dishomogeneous ofdegree 6intl't2'Furthermore, ais homogeneous ofdegree 2and bishomogeneous ofdegree 3.ByTheorem 6.1 weknow that there exists some polynomial g(X 2,X3)ofweight 6such that D=g(a, b). The only monomials X ofweight 6,i.e. such that 2m+3n =6with integers m, n>0,are those forwhich m=3,n=0,or m=0and n=2.Hence g(X 2,X3)=vxl +wxf where v,ware integers which must now bedetermined. Observe that theintegers v,ware universal, inthe sense that for any special polynomial with special values ofa,bitsdiscriminant will begiven byg(a, b)=va3+wb2 . Consider thepolynomial fl(X)=X(X-1)(X +1)=X3-X. Then a=-1, b=0,and D=-va3=-v. But also D=4byusing the definition ofthediscriminant oftheproduct ofthe differences ofthe roots, squared. Hence wegetv=-4. Next consider thepolynomial f2(X)=X3-1. Then a=0,b=-1, and D=2b2=w.But the three roots off2are the cube roots ofunity, namely -1+yC3 -1-yC31,2'2 Using thedefinition ofthediscriminant wefind thevalue D= -27. Hence wegetw= -27. This concludes theproof ofthe formula for the dis- criminant ofthecubic when there isnoX2term. 194 POLYNOMIALS IV,7 Ingeneral, consider acubic polynomial f(X)=X3-StX2+S2X-S3=(X-tt)(X-t2)(X-t3). We find the value ofthediscriminant byreducing this case tothesimpler case when there isnoX2term. Wemake atranslation, and let Y=X-is!so X=Y+iSt=Y+i(tt +t2+t3). Then f(X) becomes f(X)=f*(Y)=y3+aY+b=(Y-ut)(Y-U2)(Y-u3), where a=UtU2+U2U3+UtU3 and b=-UtU2U3, while Ut+U2+U3=O. We have Ui=ti-is! for i=1,2,3, and Ui-uj=ti-tjforalli=Fj,sothediscriminant isunchanged, and you caneasily gettheformula ingeneral. DoExercise 12(b). 7. MASON-STOTHERS THEOREM AND THE abc CONJECTURE Intheearly 80s anew trend ofthought about polynomials started with the discovery ofanentirely new relation. LetI(t) beapolynomial inone variable over thecomplex numbers ifyou wish (analgebraically closed field ofcharac- teristic 0would do).We define no(f)=number ofdistinct roots off. Thus no(f) counts the zeros offbygiving each ofthem multiplicity 1,and no(f)can besmall even though degfislarge. Theorem 7.1(Mason-Stothers, (Mas 84), (Sto 81». Leta(t), b(t), e(t) be relatively prime polynomialssueh that a+b=e.Then maxdeg{ a,b,e}<no(abc)-I. Proof (Mason) Dividing byc,andletting 1=ale, g==blewehave f+g=1, where f,garerational functions. Differentiating wegetf'+g'=0,which werewri teas f' g' ff+gg=0, IV,7 MASON'S THEOREM AND THE abe CONJECTURE 195 sothat b g f'lf- -a-l-- g'lg. Let a(t)=c1n(t-i)mi, b(t)=C2n(t-pj)nj , c(t)=C3n(t-YkYk. Then bycalculus algebraicized inExercise 11(c), weget b f'lf -=-- - a g'lgLmi-Lrk t-i t-Yk Lnj-Lrk t-Pit-Yk Acommon denominator forf'lf andg'lg isgiven bytheproduct No=n(t-i)n(t-Pj)n(t-Yk)' whose degree isno(abc). Observe thatNof'lf and Nog'lg areboth polyno- mials ofdegrees atmost no(abc)-1.From therelation b Nof'lf-= - a Nog'lg' and thefact that a,bare assumed relatively prime,wededuce theinequality inthetheorem. As anapplication, let usprove Fermat's theorem forpolynomials. Thus letx(t), y(t), z(t) berelatively prime polynomials such that one ofthem has degree>1,and such that x(t)n +y(t)n=z(t)n. We want toprove that n<2.BytheMason-Stothers theorem, weget ndeg x=degx(t)n<degx(t) +degy(t)+degz(t)-1, andsimilarly replacing xbyyand zontheleft-hand side. Adding, wefind n(degx+deg y+degz)<3(deg x+deg y+degz)-3. This yieldsacontradiction ifn>3. Asanother application inthe same vein, one has: Davenport's theorem. Letf,gbe non-constant polynomials such that f3-g2=Fo.Then deg(f3-g2)>!degf-1. See Exercise 13. 196 POLYNOMIALS IV,7 One ofthe most fruitful analogies inmathematics isthat between the integers Zand thering ofpolynomials F[t] over afield F.Evolving from theinsights ofMason [Ma 84], Frey [Fr87], Szpiro, and others, Masser and Oesterle formulated the abcconjecture forintegersasfollows. Let mbe a non-zero integer. Define theradical ofmtobe No(m)=np, plm i.e.theproductofalltheprimes dividing m,taken with multiplicity 1. The abcconjecture. Given e>0,there exists apositive number C(e) having thefollowing property. For any non-zero relative prime integers a,b,c such that a+b=c,wehave max(lal, Ibl,Icl)<C(e)N o(abc)l+£. Observe that theinequality says that many prime factors ofa,b,coccur to thefirst power, and that if"small" primes occur tohigh powers, then they have tobecompensated by"large" primes occurring tothefirst power. For instance, onemight consider theequation 2n+1=m. For mlarge, the abcconjecture would state that mhas tobedivisible by large primes tothefirst power. This phenomenon can be seen inthe tables of[BLSTW 83]. Stewart- Tijdeman [ST 86] have shown that itisnecessary tohave the ein theformulation oftheconjecture. Subsequent exampleswere communicated to mebyWojtek Jastrzebowski and Dan Spielmanasfollows. We have togive examples such that forallC>0there exist natural numbers a,b,crelatively prime such that a+b=cand lal>CNo(abc). But trivially, 2nl(32n-1). Weconsider therelations an+bn=Cngiven by 32n-1=Cn' Itisclear that these relations provide thedesired examples. Other examples can beconstructed similarly, since therole of3and 2can beplayed byother integers. Replace 2bysome prime, and 3byaninteger=1mod p. The abcconjecture implies what weshall call the Asymptotic Fermat Theorem. For all nsufficiently large, theequation xn+yn=zn has nosolution inrelatively prime integers =FO. IV,7 MASON'S THEOREM AND THE abe CONJECTURE 197 The proof follows exactly the same patternasforpolynomials, except that wewrite things down multiplicatively, and there isa1+efloating around. The extent towhich the abc conjecture will beproved with an explicit constant C(e) (or sayC(I) tofixideas) yields thecorresponding explicit determination ofthebound for nintheapplication. We now gointo other applications. Hall's conjecture [Ha 71]. Ifu,varerelatively prime non-zero integers such that u3-v2=F0,then lu3-v2 1»luI1/2-t:. The symbol»means that theleft-hand side is>theright-hand side times a constant depending only on e.Again theproof isimmediate from the abc conjecture. Actually, thehypothesis that u, varerelatively prime isnot necessary; thegeneral case can bereduced totherelatively prime case by extracting common factors, and Hall stated hisconjecture inthis more general way. However, healso stated itwithout theepsilon intheexponent, and that does notwork, aswas realized later. Asinthepolynomial case, Hall's conjecture describes how smalllu3-v2 1can be,and the answer isnot toosmall, asdescribed bytheright-hand side. The Hall conjecture can also beinterpretedasgiving abound forintegral relatively prime solutions of v2=u3+bwith integral b. Then wefind lul«IbI2+t:. More generally, inline with conjectured inequalities from Lang-Waldschmidt [La 78], let usfix non-zero integers A,Band let u,v,k,m, nbevariable, with u,vrelatively prime and mv>m+n.Put Aum+Bvn=k. Bytheabcconjecture, one derives easily that (1)m(1+t:) lul«No(k)mn-(m+n) andmn(1+t:) Ivl«No(k)mn-(m+n). From this one gets mn(1+t:) Ikl«No(k)mn-(m+n). The Hall conjecture isaspecialcase after wereplace No(k) with Ikl,because No(k)<Ikl. Next take m=3and n=2,but take A=4and B= -27. Inthis case wewrite D=4u3-27v2 198 POLYNOMIALS IV,7 and weget (2) lul«No(D)2+£and Ivl«No(D)3+£. These inequalities aresupposedtohold atfirst for u, vrelatively prime. Suppose weallow u,vtohave some bounded common factor, say d.Write u=u'd and v=v'd with u',v'relatively prime. Then D=4d3u,3-27d2v'2. Now we canapply inequality (1)with A=4d3and B=-27d2 ,and wefind the same inequalities (2), with the constant implicit inthesign«depending also ond,oron some fixed bound forsuch acommon factor. Under these circumstances, wecallinequalities (2)thegeneralized Szpiro conjecture. The original Szpiro conjecture was stated inamore sophisticated situa- tion, cf.[La 90] for anexposition, and Szpiro's inequalitywas stated inthe form IDI«N(D)6+£, where N(D) isamore subtle invariant, but for our purposes, itissufficient and much easier tousetheradical No(D). The point ofDisthat itoccurs asadiscriminant. The trend ofthoughts inthedirection we arediscussingwas started byFrey [Fr 87], who asso- ciated with each solution ofa+b=cthepolynomial x(x-a)(x +b), which we call theFrey polynomial. (Actually Frey associated the curve defined bytheequation y2=x(x-a)(x +b),formuch deeper reasons, but only thepolynomial on theright-hand side will beneeded here.) The discriminant ofthepolynomial istheproduct ofthe differences ofthe roots squared, and so D=(abc)2. We make atranslation b-a=x+3 togetridofthe x2-term, sothat ourpolynomial can berewritten 3-Y2-Y3, where Y2, Y3are homogeneous ina,bofappropriate weight. The dis- criminant does not change because the roots ofthepolynomial in are IV,7 MASON'S THEOREM AND THE abe CONJECTURE 199 translations ofthe roots ofthepolynomial inx.Then D=4y-27y. The translation with (b-a)/3 introduces asmall denominator. One may avoid this denominator byusing thepolynomial x(x-3a)(x-3b), sothat Y2, Y3then come out tobeintegers, and one canapply thegeneralized Szpiro conjecture tothediscriminant, which then has anextra factor D=36(abc)2. Itisimmediately seen that the generalized Szpiro conjecture implies asymptotic Fermat. Conversely: Generalized Szpiro implies theabcconjecture. Indeed, thecorrespondence (a,b)+-+(Y2,Y3)isinvertible, and hasthe" right" weight. Asimple algebraic manipulation shows that thegeneralized Szpiro estimates onY2, Y3imply thedesired estimates onlal,Ibl.(Do Exercise 14.) From theequivalence between abc and generalized Szpiro, one can use the examples given earlier toshow that theepsilon isneeded intheSzpiro conjecture. Finally, note that thepolynomialcase oftheMason-Stothers theorem and the case ofintegersare notindependent,orspecifically theDavenport theorem and Hall's conjecture arerelated. Examples inthepolynomial case parametrize cases with integers when wesubstitute integers forthevariables. Such examples aregiven in[BCHS 65], oneofthem (due toBirch) being f(t)=t6+4t4+10t2+6 and g(t)=t9+6t7+21tS+35t3+¥t, whence deg(f(t)3-g(t)2)=tdegf+1. This example shows that Davenport's inequality isbest possible, because the degree attains the lowest possible value permissible under the theorem. Substituting large integral values oft=2mod 4gives examples ofsimilarly low values for x3-y2.For other connections ofallthese matters, cf.[La90]. Bibliography [BCHS 65] B.BIRCH, S.CHOWLA, M.HALL, and A.SCHINZEL, On thedifference x3-y2,Norske Vide Selsk. Forrh. 38(1965) pp.65-69 [BLSTW 83] J.BRILLHART, D.H.LEHMER, J.L.SELFRIDGE, B.TUCKERMAN, and S. WAGSTAFF Jr.,Factorization ofbPI+1,b=2,3,5,6,7,10, 11upto high powers, Contemporary Mathematics Vol. 22,AMS, Providence, RI, 1983 [Dav 65] H.DAVENPORT, Onf3(t)-g2(t), Norske Vide Selsk. Forrh. 38(1965) pp.86-87 [Fr87] G.FREY, Links between solutions ofA-B=Candelliptic curves, Number Theory, Lecture Notes 1380, Springer-Verlag, New York, 1989 pp.31-62 200 POLYNOMIALS [Ha 71] [La90] [Ma 84a] [Ma 84b] [Ma 84c] [Si88] [ST86]IV,8 M. HALL, The diophantine equationx3-y2=k,Computers and Number Theory, ed.byA.O.L.Atkin and B.Birch, Academic Press, London 1971 pp.173-198 S.LANG, Old and new conjectured diophantine inequalities, Bull. AMS Vol. 23No.1 (1990) pp.37-75 R.C.MASON, Equationsover function fields, Springer Lecture Notes 1068 (1984), pp. 149-157; inNumber Theory, Proceedings ofthe Noordwijkerhout,1983 R.C.MASON, Diophantine equations over function fields, London Math. Soc. Lecture Note Series Vol. 96,Cambridge University Press, Cambridge, 1984 R.C.MASON, The hyperelliptic equation over function fields, Math. Proc. Cambridge Phi/os. Soc. 93(1983) pp.219-230 J.SILVERMAN, Wieferich's criterion and the abcconjecture, Journal of Number Theory 30(1988) pp.226-237 C.L.STEWART and R.TUDEMAN, OntheOesterle-Masser Conjecture, Mon. Math. 102(1986) pp.251-257 Seeadditional references attheendofthechapter. 8. THE RESULTANT Inthis section, we assume that the reader isfamiliar with determinants. The theory ofdeterminants will becovered later. The section can beviewed asgiving further examples ofsymmetric functions. Let Abe acommutative ring and letVo,...,vn,Wo,..., Wmbealge- braically independent over A.Weform twopolynomials: Iv(X)=voXn+...+vn, gw(X)=woXm+...+Wm. We define theresultant of(v,w),orofIv,gw,tobethedeterminant m WoW1...WmVoV1...Vn VoV1...Vn nVoV1...Vn WoW1...Wm WoW1...Wm l y m+n The blank spacesaresupposed tobefilled with zeros. IV,8 THE RESULTANT 201 Ifwesubstitute elements (a)=(ao,...,an)and (b)=(bo,...,bm)inAfor (v)and (w)respectively inthe coefficients offvand gw' then weobtain polynomials faand gbwith coefficients inA,and wedefine their resultant to bethedeterminant obtained bysubstituting (a)for (v)and (b)for(w) inthe determinant. We shall write theresultant offv,gwintheform Res(fv' gw) or R(v, w). The resultant Res(fa, gb)isthen obtained bysubstitution of(a),(b)for(v),(w) respectively. We observe that R(v, w)isapolynomial with integer coefficients, i.e. we may take A=Z.Ifzisavariable, then R(zv, w)=zmR(v, w) and R(v, zw)=znR(v, w) asone sees immediately byfactoring out zfrom thefirst mrows (resp. the last nrows) inthe determinant. Thus Rishomogeneous ofdegreeminits first setofvariables, and homogeneous ofdegree ninitssecond setof variables. Furthermore, R(v, w)contains themonomial vmwn o m with coefficient 1,when expressedasasum ofmonomials. Ifwesubstitute 0for Voand Wointheresultant, weobtain 0,because the first column ofthedeterminant vanishes. Let uswork over theintegers Z.Weconsider thelinear equations Xm-tfv(X)=voxn+m-t +Vtxn+m-2+... +vnXm-t Xm-2fv(X)=voxn+m-2 +... +vXm-2 n fv(X)=voXn+... +Vn xn-tgw(X)=woxn+m-t+wtxn+m-2+...+wmXn-t xn-2gw(X)= woxn+m-2+...+wmXn-2 gw(X)=woxm +...+wm. Let Cbethecolumn vector ontheleft-hand side, and let Co, ..., Cm+n bethecolumn vectors ofcoefficients. Our equationscan bewritten C=xn+m-tco+...+1.Cm+n. ByCramer's rule, applied tothelast coefficient which is=1, R(v, w)=det(C o,...,Cm+n)=det(C o,...,Cm+n-t'C). 202 POLYNOMIALS IV,8 From this we see that there exist polynomials CPv,wandt/lv,winZ[v,w][X] such that 'P,v,w/v+t/1v,wgw=R(v,w)=Res(fv, fw). Note that R(v, w)EZ[v, w]butthat thepolynomials ontheleft-hand side involve thevariable X. IfA.:Z[v, w]-.Aisahomomorphism into acommutative ring Aand we letA.(v)=(a),A.(w)=(b),then CPa,bfa +t/la,bgb=R(a, b)=Res(h, fb). Thus from theuniversal relation oftheresultant over Zweobtain asimilar relation forevery pair ofpolynomials, inany commutative ring A. Proposition 8.1. Let Kbe asubfield of afield L,and leth, gbbe polynomials inK[X] havingacommon root inL.Then R(a, b)=o. Proof Iffa()=gb()=0,then wesubstitute forXintheexpression obtained forR(a, b)and find R(a, b)=o. Next, weshall investigate therelationship between the resultant and the roots ofourpolynomials fv,gw. We need alemma. Lemma 8.2. Leth(X l'.. .,Xn)be apolynomial in nvariables over the integers Z.Ifhhas thevalue 0when wesubstitute XIforX2and leave the other Xifixed (i=F2),then h(X l'...,Xn)isdivisible byXI-X2in Z[X 1,...,Xn]. Proof Exercise forthereader. Let vo, tl'..., tn'WO,Ul'..., Urnbealgebraically independent over Zand form thepolynomials fv=vo(X-t1)...(X-tn)=voXn+... +vn, gw=wo(X-u1)...(X-urn)=woxm +... +wrn. Thus welet Vi=(-l)ivosi(t) and wj=(-l)iwosj(u). We leave tothereader the easy verification that vo,VI' ..., vn,WO, WI' ..., Wm arealgebraically independent over Z. Proposition 8.3. Notation being asabove, wehave n m Res(fv' gw)=vO'w8nn(ti-uj). i=1j=1 IV,8 THE RESULTANT 203 Proof. Let Sbetheexpression ontheright-hand side oftheequality in the statement oftheproposition. Since R(v,w)ishomogeneous ofdegreeminitsfirst variables, and homogeneous ofdegreeninitssecond variables, itfollows that R=vow8h(t, u) where h(t,u)EZ[t, u]. ByProposition 8.1, the resultant vanishes when we substitute tiforUj(i=1,...,nand j=1,...,m),whence bythelemma, view- ing Rasanelement ofZ[vo,wo,t,u]itfollows that Risdivisible byti-Uj for each pair (i,j). Hence Sdivides RinZ[v o,wo,t,u], because ti-Ujis obviouslyaprime inthat ring, and different pairs (i,j) give rise todifferent prImes. From theproduct expression forS,namely (1)n m S=vow8f1f1(ti-uj), i=1j=1 weobtain n n m f1g(t i)=w8f1f1(ti-uj), i=l i=1j=1 whence (2)n S=Vof1g(tJ. i=l Similarly, (3)m S=(-1)nmw8f1f(uj). j=1 From (2) we seethat Sishomogeneous and ofdegreenin(w), and from (3) we seethat Sishomogeneous and ofdegreemin(v). Since Rhasexactly the same homogeneity properties, and isdivisible byS,itfollows that R=cSfor some integerc.Since both Rand Shave amonomial vow occurring in them with coefficient 1,itfollows that c=1,and ourproposition isproved. We also note that thethree expressions found for Sabove now give usa factorization ofR.We also getaconverse forProposition 8.1. Corollary 8.4. Letfa'gbbepolynomials with coefficients inafield K,such that aob o=F0,and such that fa, gbsplit infactors ofdegree1inK[X]. Then Res(h, gb)=0ifandonlyifhand gbhave aroot incommon. Proof. Assume that theresultant iso.If h=ao(X-1)". (X-n)' gb=bo(X-PI)...(X-Pn)' isthefactorization ofh,gb,then wehave ahomomorphism 204 POLYNOMIALS IV,8 Z[vo, t,wo,u]-+K such that Vo1---+ao,Wo1---+bo,ti1---+(Xi'and uj1---+Pjforalli,j.Then o=Res(la, gb)=aO'bonn«(Xi-Pj), ij whence fa,Ibhave aroot incommon. The converse has already been proved. We deduce one more relation fortheresultant inaspecialcase. LetIvbe asabove, Iv(X)=voXn+... +Vn=vo(X-t1)...(X-tn). From (2) weknow that ifI:isthederivative ofIv,then (4) Res(lv' I:)=v8-1nI'(t i). i Using theproduct rule fordifferentiation, wefind: I:(X)=Lvo(X-t1)...(X-ti)...(X-tn), i I:(t i)=VO(t i-t1)...(ti-ti)...(ti-tn), where aroof over aterm means that this term istobeomitted. We define thediscriminant ofIvtobe D(lv)=D(v)=(_1)n(n-l)/2 v5n-2n(ti-tj). i#:j Proposition 8.5. LetIvbe asabove and have algebraically independent coefficients over Z.Then (5) Res(lv, I:)=v5n-1n(ti-tj)=(-1)n(n-l)/2 voD(lv)' i#:j Proof One substitutes theexpression obtained forI:(t i)into theprod- uct(4). The result follows atonce. When wesubstitute 1for Vo,wefind that thediscriminant aswedefined itinthepreceding section coincides with thepresent definition. Inparticular, wefind anexplicit formula forthediscriminant. The formulas inthespecial case ofpolynomials ofdegree 2and 3will begivenasexercises. Note that thediscriminant can also bewritten astheproduct D(lv)=v5n-2n(ti-tj)2. i<j Serre once pointed out tomethat thesign (_1)n(n-l)/2 was missing inthe first edition ofthis book, and that this sign error isquite common inthe literature, occurringasitdoes invan derWaerden, Samuel, and Hilbert (but not inhiscollected works, corrected byOlga Taussky); onthe other hand thesign iscorrectly given inWeber's Algebra, Vol. I,50. For acontinuation ofthis section, seeChapter IX,3and4. IV,9 POWER SERIES 205 9. POWER SERIES LetXbe aletter, and letGbethemonoid offunctions from the set{X} tothenatural numbers. IfvEN, wedenote byXVthefunction whose value atXisv.Then Gisamultiplicative monoid, already encountered when we discussed polynomials. Itselements areXO ,Xl,X2 ,...,Xv,... . Let Abe acommutative ring, and letA[[X]] bethe setoffunctions from Ginto A,without any restriction. Then anelement ofA[[X]] may be viewed asassigning toeach monomial Xv acoefficient avEA. We denote this element by 00 LavXv. v=O The summation symbol isnot asum, ofcourse, but weshall write theabove expression also intheform aoXO+atXt+... and wecall itaformal power series with coefficients inA,inone variable. We call ao,al,...itscoefficients. Given two elements ofA[[X]], say 00 LavXvand v=O00 Lb/lX/l, /l=O wedefine their product tobe 00 LCiXi i=O where Ci=Lavb/l. v+/l=i Just aswith polynomials,one defines their sum tobe 00 L(av+bv)Xv . v=O Then we seethat the power series form aring, theproof being the same as forpolynomials. One can also construct the power series ring inseveral variables A[[X t,...,Xn]] inwhich every element can beexpressed intheform La(V)X:l...X;"=La(v)M(v)(X l,...,Xn) (v) with unrestricted coefficientsa(v)inbijection with then-tuples ofintegers (Vt, ..., vn)such that Vi>0foralli.Itisthen easy toshow that there isan isomorphism between A[[X l,...,Xn]] and therepeated power series ring A[[Xt]]...[[X n]]. We leave this asanexercise forthereader. 206 POLYNOMIALS IV,9 The next theorem will give ananalogue ofthe Euclidean algorithm for power series. However, instead ofdealing with power series over afield, itis important tohave somewhat more general coefficients forcertain applica- tions, sowehave tointroduce alittle more terminology. Let Abearing andIanideal. We assume that 00 nIV ={O}. v=l We can view the powers IVasdefining neighborhoods of0inA,and we can transpose theusual definition ofCauchy sequence inanalysis tothissituation, namely: wedefine asequence {an} inAtobeCauchy ifgiven some power IV there exists aninteger Nsuch that forallm,n>Nwehave am-anEIV. Thus IVcorresponds tothegivenEofanalysis. Then wehave the usual notion ofconvergence ofasequence toanelement ofA.One says that Ais complete inthe/-adic topology ifevery Cauchy sequence converges. Perhaps the most important example ofthis situation iswhen Aisalocal ring and I=misitsmaximal ideal. By acomplete local ring, one always means alocal ring which iscomplete inthem-adic topology. Let kbeafield. Then thepower series ring R=k[[X l,...,Xn]] in nvariables issuch acomplete local ring. Indeed, let mbethe ideal generated bythevariables Xl' ...,Xn. Then Rim isnaturally isomorphic to thefield kitself, somisamaximal ideal. Furthermore, any power series of theform f(X)=Co-fl(X) withCoEk,Co=F0andfl(X)Emisinvertible. Toprove this, one may first assume without loss ofgenerality that Co=1.Then (1-fl(X))-l=1+fl(X) +fl(X)2 +fl(X)3 +... gives theinverse. Thus we seethat mistheunique maximal ideal and Ris local. Itisimmediately verified that Riscomplete inthe sense wehave just defined. The same argument shows that ifkisnot afield butCoisinvertible ink,then again f(X) isinvertible. Again letAbe aring. We may view thepower series ring innvariables (n>1)asthering ofpower series inone variable Xnover thering ofpower series inn-1variables, that iswehave anatural identification A[[Xl' ...,Xn]]=A[[Xl' ...,Xn-l]][[X n]]. IfA=kisafield, theringk[[X l,...,Xn-l]]isthen acomplete local ring. More generally, if0isacomplete local ring, then thepower series ring o[[X]] isacomplete local ring, whose maximal ideal is(m,X)where mis themaximal ideal of o.Indeed, ifapower series LavXv has unit constant IV,9POWER SERIES 207 term aoE0*,then thepower series isaunit ino[[X]],because first, without loss ofgenerality,wemay assume that ao=1,and then wemay invert 1+h with hE(m,X)bythegeometric series 1-h+h2-h3+.,. . In anumber ofproblems, itisuseful toreduce certain questionsabout power series inseveral variables over afield toquestions about powerseries inone variable over the more complicated ring asabove. We shall now apply thisdecomposition totheEuclidean algorithm forpower series. Theorem 9.1. Let 0beacomplete local ring with maximal ideal m.Let 00 f(X)=LaiXi i=O be apower series ino[[X]] (one variable), such that notallailieinm. Say ao, ..., an-1Em,and anE0*isaunit. Given gE0[[X]]we can solve theequation g=qf+r uniquely with qEo[[X]],rEo[X], anddeg r<n-1. Proof (Manin). Let rxand tbetheprojections onthebeginning and tailend oftheseries, given by n-l rx:LbiXi1--+LbiXi=bo+b1X+...+bn-1xn-l, i=O 00 t:LbiXil--+ LbiXi-n=bn+bn+1X+bn+2X2+.... i=n Note that t(hXn)=hforany hEo[[X]];and hisapolynomial ofdegree <nifandonly ift(h)=O. The existence ofq,risequivalent with thecondition that there exists q such that t(g)=t(qf). Hence ourproblem isequivalent with solving t(g)=t(qrx(f») +t(qt(f)Xn)=t(qrx(f») +qt(f). Note thatt(f) isinvertible. Put Z=qt(f). Then the above equation is equivalent with (rx(f»)(rx(f»)-r(g)=-rZ -r(f)+Z=I+-r0 -r(f)Z. Note that rx(f) -r0 -r(f):o([X]]-+mo[[X]], because rx(f)/t(f)Emo[[X]]. We can therefore invert tofind Z,namely 208 POLYNOMIALS IV,9 ((X(f))-l Z=I+.0 .(f).(g), which proves both existence and uniqueness and concludes theproof. Theorem 9.2. (Weierstrass Preparation). The power seriesfinthepre- vious theorem can bewritten uniquely intheform f(X)=(Xn+bn-1xn-1+...+bo)u, where biEm,and uisaunit ino[[X]]. Proof. Write uniquely Xn=qf+r, bytheEuclidean algorithm. Then qisinvertible, because q=CO+CI X+.", f=...+anXn+..., sothat 1=coa n(mod m), and thereforeCoisaunit ino.Weobtain qf=xn-r,and f=q-l(Xn-r), with r=0(mod m). This proves theexistence. Uniqueness isimmediate. The integerninTheorems 9.1and 9.2iscalled theWeierstrass degree off, and isdenoted bydeg wf.We seethat apower series not allofwhose coeffi- cients lieinmcan beexpressedasaproduct ofapolynomial having thegiven Weierstrass degree, times aunit inthe power series ring. Furthermore, all the coefficients ofthepolynomial except theleading one lieinthemaximal ideal. Such apolynomial iscalled distinguished, oraWeierstrass polynomial. Remark. Irather like the useoftheEuclidean algorithm intheproof of the Weierstrass Preparation theorem. However, one can also giveadirect proof exhibiting explicitly the recursion relations which solve forthe coeffi- cients ofu, asfollows. Write u=LCiXi. Then we have tosolve the equations boco=ao, boc1+b1Co=a1, bOcn-1+...+bn-1Co=an-l, boc n+...+Co=an, boCn+1+...+C1=an+1, IV,9 POWER SERIES 209 Infact, the system ofequations has aunique solution mod mrfor each positive integer r,after selecting Cotobe aunit, say Co=1.Indeed, from thefirst nequations (from 0to n-1)we seethat bo,..., bn-1areuniquely determined tobe 0mod m. Then Cn'cn+1,...are uniquely determined mod mbythesubsequent equations. Now inductively, suppose we have shown that thecoefficients bi'cjareuniquely determined mod mr .Then one sees immediately that from theconditions ao, ..., an-1=0mod mthefirst n equations define biuniquely mod mr+l because all bi=0mod m. Then thesubsequent equations defineCjmod mr+l uniquely from the values of bimod mr+l andCjmod mr .The unique system ofsolutions mod mrforeach rthen defines asolution intheprojective limit, which isthecomplete local rIng. We now have allthetools todeal with unique factorization inoneimportant case. Theorem 9.3. Let kbeafield. Then k[[X.,...,Xn]] isfactorial. Proof. Letf(x)=f(X 1,. ..,Xn)Ek[[X]] be =f=.O.After makingasufficiently general linear change ofvariables (when kisinfinite) x.="c..Y. with c..EkI L.J I}} I}' wemayassume without loss ofgenerality thatf(O,. . .,0,xn)=f=.o.(When kis finite, one has tomake anon-linear change, cf.Theorem 2.1ofChapter VIII.) Indeed, ifwewritef(X)=fd(X)+higher terms, wherefd(X) isahomogeneous polynomial ofdegree d>0,then changing thevariables asabove preserves the degree ofeach homogeneous component off,and since kisassumed infinite, thecoefficientsCijcan betaken sothat infact each power Yf(i=1,..., n) occurs with non-zero coefficient. We now proceed byinduction onn.LetRn=k[[X 1,...,Xn]] bethepower series innvariables, and assume byinduction that Rn-Iisfactorial. ByTheorem 9.2,writef=9Uwhere uisaunit and 9isaWeierstrass polynomial inRn-I[X n]. ByTheorem 2.3,Rn-I[X n]isfactorial, and sowe can write 9asaproduct of irreducible elements 91'. . .,9rERn-I [Xn],sof=91· ··9ru,where thefactors 9iareuniquely determined uptomultiplication byunits. This proves theexistence of afactorization. As touniqueness, suppose fisexpressedas aproduct of irreducible elements inRn,f=fl...fs.Thenfq(O,...,0,xn)=f=.0foreach q=1,.. .,s,sowe canwritefq=hquwhere uisaunit andhqisaWeierstrass polynomial, necessarily irreducible inRn-I[X n].Thenf=9U=nhqnu with 9and allhqWeierstrass polynomials. ByTheorem 9.2, we must have 9=nhq,and since Rn-I [Xn]isfactorial, itfollows that thepolynomials hq arethe same asthepolynomials 9i,uptounits. This proves uniqueness. Remark. As waspointed out tomebyDan Anderson, Iincorrectly stated inaprevious printing thatif()isafactorial complete local ring, then()[[X]] isalso factorial. This assertion isfalse, asshown bytheexample k(t)[[X., X2,X3]]/(Xr+xi+X) 210 POLYNOMIALS IV,9 due toP.Salmon, Su unproblema post daP.Samuel, Atti Acad. Naz. Lincei Rend. Cl. Sc. Fis. Matern. 40(8) (1966) pp.801-803. Itistrue thatif()isa regular local ring inaddition tobeing complete, then()[[X]] isfactorial, butthis isadeeper theorem. The simple proof Igave forthepower series over afield isclassical. Ichose theexposition in[GrH 78]. Theorem 9.4. IfAisNoetherian, thenA[[X]] isalso Noetherian. Proof Our argument will be amodification oftheargument used inthe proof ofHilbert's theorem forpolynomials. We shall consider elements of lowest degree instead ofelements ofhighest degree. Let beanideal ofA[[X]]. We letQibethe setofelements aEAsuch that aisthecoefficient ofXiinapower series aXi+terms ofhigher degree lyingin. Then Qiisanideal ofA,andQiCQi+l (theproof ofthis assertion being the same asforpolynomials). The ascending chain ofideals stops: QoCQ1CQ2C...CQr=Qr+ 1=... Asbefore, letaij(i=0,...,rand j=1,...,ni)begenerators forthe ideals Qi,and letfijbepower series inAhaving aijasbeginning coefficient. Given fE21,starting with aterm ofdegree d,say d<r,we can find elements Ct,. . .,CndEAsuch that f-c1hl- ... -cndhnd starts with aterm ofdegree>d+1.Proceeding inductively, wemayas- sume that d>r.We then use alinear combination f-C(d)Xd-rf,-... -C(d)xd-rf, 1 rl n,. rn,. togetapower series starting with aterm ofdegree>d+1.Inthis way, if westart with apower series ofdegree d>r,then itcan beexpressedas a linear combination offrt,. . .,frnrbymeans ofthecoefficients 00 00 g1(X)=Lcv)xv-r ,...,gn,.(X)=LC:)xv-r , v=d v=d and we seethat thefijgenerate our ideal, aswas tobeshown. Corollary 9.5.IfAisaNoetherian commutative ring, or afield, then A[[X l'...,Xn]]isNoetherian. Examples. Power series inone variable are atthe core ofthetheory of functions ofone complex variable, and similarly forpower series inseveral variables inthehigher-dimensionalcase. Seeforinstance [Gu 90]. Weierstrass polynomialsoccur inseveral contexts. First, theycan beused toreduce questions about power series toquestions about polynomials, in studying analyticsets. See forinstance [GrH 78], ChapterO.Inanumber- IV,9 POWER SERIES 211 theoretic context, such polynomials occur ascharacteristic polynomials in the Iwasawa theory ofcyclotomic fields. Cf.[La90], starting with Chapter 5. Power series can also beused asgenerating functions. Suppose that to each positive integernwe associate anumber a(n). Then thegenerating function isthe power series La(n)tn .Insignificant cases, itturns out that this function representsarational function, and itmay be amajor result to prove that this isso. For instance inChapter X,6 we shall consider aPoincare series, associated with thelength ofmodules. Similarly, intopology,consider a topological space Xsuch that itshomology groups (say)arefinite dimen- sional over afield kofcoefficients. Let hn=dimHn(X, k),where Hnisthe n-th homology group. The Poincare series isdefined tobethegenerating serIes Px(t)=Lhntn . Examples arise inthetheory ofdynamical systems. One considers a mapping T:X-.Xfrom aspace Xinto itself, and weletNnbethenumber offixed points ofthen-th iterate Tn =ToT 0...0T(ntimes). The generat- ing function isLNntn .Because ofthe number ofreferences Igive here, I list them systematically atthe end ofthe section. See first Artin-Mazur [ArM 65]; aproof byManning of aconjecture ofSmale [Ma 71]; and Shub's book [Sh 87], especially Chapter 10,Corollary 10.42 (Manning's theo rem). For anexample inalgebraic geometry, let Vbe analgebraic variety defined over afinite field k.LetKnbetheextension ofkofdegreen(ina given algebraic closure). Let Nnbethenumber ofpoints ofVinKn. One defines the zeta function Z(t)asthepower series such that Z(O)=1and 00 Z'/Z(t)=LNntn-1 . n=l Then Z(t)isarational function (F.K.Schmidt when thedimension ofVis1, and Dwork inhigher dimensions). For adiscussion and references tothe literature, seeAppendix CofHartshorne [Ha77]. Finallywemention thepartition function p(n), which isthe number of waysapositive integercan beexpressedasasum ofpositive integers. The generating function was determined byEuler tobe 00 00 1+Lp(n)tn=n(1-tn)-l. n=l n=l Seeforinstance Hardy andWright [HardW 71],Chapter XIX. The generat- ing series for thepartition function isrelated tothe power series usually expressed interms ofavariable q,namely 212 POLYNOMIALS IV,9 00 00 &=qn(1-qn)24=Lt(n)qn, n=l n=l which isthegenerating series fortheRamanujan function t(n). The power series for &isalso theexpansion of afunction inthetheory ofmodular functions. For anintroduction, see Serre's book ESe73], last chapter, and books onelliptic functions, e.g. mine. We shall mention oneapplication of thepower series for&intheGalois theory chapter. Generating power series also occur inK-theory, topological andalgebraic geometric, asinHirzebruch's formalism fortheRiemann-Roch theorem and itsextension byGrothendieck. SeeAtiyah [At67], Hirzebruch [Hi66], and [FuL 86]. Ihave extracted some formal elementary aspects having directly todowith power series inExercises 21-27, which can beviewed asbasic examples. See also Exercises 31-34 ofthe next chapter. [ArM 65] [At67] [FuL 85] [GrH 78] [Gu 90] [HardW 71] [Hart 77] [Hi66] [La90] [Ma 71] ESe73] [Sh87]Bibliography M.ARTIN and B.MAZUR, Onperiodic points, Ann. Math. (2)81(1965) pp.89-99 M.ATIYAH, K-Theory, Addison-Wesley 1991 (reprinted from the Ben- jamin Lecture Notes, 1967) W. FULTON and S.LANG, Riemann-Roch Algebra, Springer-Verlag, New York, 1985 P.GRIFFITHS and J.HARRIS, Principles ofAlgebraic Geometry, Wiley- Interscience, New York, 1978 R.GUNNING, Introduction toHolomorphic Functions ofSeveral Vari- ables, Vol. II:Local Theory, Wadsworth andBrooks/Cole, 1990 G.H.HARDY and E.M.WRIGHT, AnIntroduction totheTheory of Numbers, Oxford University Press, Oxford, UK, 1938-1971 (several editions) R.HARTSHORNE, Algebraic Geometry, Springer-Verlag, New York, 1977 F.HIRZEBRUCH, Topological Methods inAlgebraic Geometry, Springer- Verlag, New York, 1966 (translated and expanded from theoriginal German, 1956) S.LANG, Cyclotomic Fields, IandII,Springer-Verlag, New York, 1990, combined edition oftheoriginal editions, 1978, 1980 A.MANNING, Axiom Adiffeomorphisms have rational zeta functions, Bull. Lond. Math. Soc. 3(1971) pp.215-220 J.P.SERRE, ACourse inArithmetic, Springer-Verlag, New York, 1973 M.SHUB, Global Stability ofDynamical Systems, Springer-Verlag, New York, 1987 IV,Ex EXERCISES 213 EXERCISES 1.Let kbeafield andf(X)Ek[X]anon-zero polynomial. Show that thefollowing conditions areequivalent: (a)The ideal(f(X)) isprime. (b)The ideal(f(X))ismaximal. (c)f(X) isirreducible. 2.(a)State and prove theanalogue ofTheorem 5.2fortherational numbers. (b)State and prove theanalogue ofTheorem 5.3forpositive integers. 3.Letfbe apolynomial inone variable over afield k.LetX,Ybetwo variables. Show that ink[X, Y] wehave a"Taylor series" expansion II f(X +Y)=f(X) +Llpi(X) yi , i=l where lpi(X) isapolynomial inXwith coefficients ink.Ifkhas characteristic 0, show that D'.f(X)lpi(X)= .,. l. 4.Generalize thepreceding exercise topolynomials inseveral variables (introduce partial derivatives and show that afinite Taylor expansion exists for apolynomial inseveral variables). 5.(a)Show that thepolynomials X4+1and X6+X3+1areirreducible over the rational numbers. (b)Show that apolynomial ofdegree 3over afield iseither irreducible orhas a root inthefield. IsX3-5X2+1irreducible over therational numbers? (c)Show that thepolynomial intwo variables X2+y2-1isirreducible over therational numbers. Isitirreducible over thecomplex numbers? 6.Prove theintegral root testof3. 7.(a)Let kbe afinite field with qelements. Letf(X l'...,XII) be apolynomial in k[X] ofdegree dand assume f(O,..., 0)=O.An element (a1,..., all)Ek(lI) such thatf(a)=0iscalled azero off.Ifn>d,show thatfhas atleast one other zero ink(II). [Hint: Assume thecontrary, and compare thedegrees of thereduced polynomial belonging to 1-f(X)Q-1 and (1-Xl-1)...(1-X:-1).The theorem isdue toChevalley.] (b)Refine theabove results byproving that thenumber Nofzeros offink(lI) is =0(mod p),arguingasfollows. Letibeaninteger>1.Show that '". {q-1=-1 ifq-1divides i,x' = XEk 0 otherwise. Denote thepreceding function ofibytjJ(i). Show that 214 POLYNOMIALS IV,Ex N=L(1-f(x)q-l) xeken) and foreach n-tuple (iI'...,in)ofintegers>0that LXl... xn="'(il)..."'(in). xek(n) Show that both terms inthe sum forNabove yield 0mod p.(The above argument isdue toWarning.) (c)Extend Chevalley's theorem torpolynomials fl, ...,f,.ofdegrees d1,..., dr respectively, innvariables. Ifthey have noconstant term and n>Ldi,show that they have anon-trivial common zero. (d)Show that anarbitrary function f:k(n) kcan berepresented by apoly- nomial. (Asbefore, kisafinite field.) 8.Let Abe acommutative entire ring and Xavariable over A.Let a,bEAand assume that aisaunit inA. Show that the map X....-+ aX+bextends to a unique automorphism ofA[X] inducing theidentity onA.What isthe inverse automorphism? 9.Show that every automorphism ofA[X] isofthetype described inExercise 8. 10. LetKbe afield, andK(X) thequotient field ofK[X]. Show that every automorphism ofK(X) which induces theidentityonKisoftype aX+b X....-+ eX+d with a,b,e,dEKsuch that (aX +b)/(eX +d)isnot anelement ofK, or equivalently, ad-be:FO. 11.Let1beacommutative entire ring and letKbeitsquotient field. We show here that some formulas from calculus have apurely algebraic setting. Let D:A--.A be aderivation, that isanadditive homomorphism satisfying the rule for the derivative ofaproduct, namely D(xy)=xDy +yDx for x,YEA. (a)Prove that Dhas aunique extension toaderivation ofKinto itself, and that this extension satisfies therule /)yDx-xDyD(x y= 2 Y for x,YEA and y:Fo.[Define theextension bythisformula, prove that itis independent ofthechoice ofx,ytowrite thefraction x/y, and show that it isaderivation having theoriginal value onelements ofA.] (b)Let L(x)=Dx/x for xEK*. Show that L(xy)=L(x) +L(y). The homo- morphism Liscalled thelogarithmic derivative. (c)Let Dbethestandard derivative inthepolynomial ringk[X] over afield k. LetR(X)=ef1(X-CXi)miwithCXiEk,eEk,and miEZ,soR(X) isarational IV,Ex EXERCISES 215 function. Show that m. R'/R=LI .X-ex.I 12.(a)Iff(X)=aX2+bX+c,show that thediscriminant offisb2-4ac. (b)Iff(X)=aoX3+atX2+a2X+a3,show that thediscriminant offis aa-4aoa-4aa3-27aa +18aoata2a3. (c)Letf(X)=(X-tt)...(X-tll).Show that II Df=(_1)"("-1)/2 f1f'(t i). i=1 13.Polynomials will betaken over analgebraically closed field ofcharacteristic O. (a)Prove Davenport's theorem. Letf(t), g(t) bepolynomials such thatf3-g2:Fo.Then deg(f3-g2)>1degf+1. Orput another way, leth=f3-g2and assume h:Fo.Then degf<2deg h-2. Todothis, first assume f,9relatively prime andapply Mason's theorem. In general, proceedasfollows. (b)Let A,B,f,9bepolynomials such that Af,Bg arerelatively prime :FO.Let h=Af3 +Bg2 .Then degf<deg A+deg B+2deg h-2. This follows directly from Mason's theorem. Then starting with f,9not necessarily relatively prime, start factoring out common factors until no longer possible, toeffect thedesired reduction. When Ididit,Ineeded todo this step three times, sodon't stop until you getit. (c)Generalize (b)tothe case offm-g"forarbitrary positive integer exponents mand n. 14.Prove that thegeneralized Szpiro conjecture implies theabcconjecture. 15.Prove that theabcconjecture implies thefollowing conjecture: There areinfinitely many primes psuch that 2p-1 =1=1mod p2.[Cf. thereference [Sit88] and[La90] attheend of7.] 16.Let wbe acomplex number, and let c=max(l, Iwl). Let F,Gbe non-zero polynomials inone variable with complex coefficients, ofdegrees dand d'respec- tively, such that IFI,IGI>1.Let Rbetheir resultant. Then IRI<Cd+d'[IF(w)1 +IG(w)l] IFld'IGld(d +d,)d+d'. (We denote byIFIthemaximum oftheabsolute values ofthecoefficients ofF.) 17.Let dbe aninteger>3.Prove the existence ofanirreducible polynomial of degree dover Q,having precisely d-2real roots, and apair ofcomplex conjugate roots. Use thefollowing construction. Let b1,..., bd-2bedistinct 216 POLYNOMIALS IV,Ex integers, and let abeaninteger>O.Let g(X)=(X2+a)(X-b1)...(X-bd-1)=Xd+Cd_1Xd-1+...+Co. Observe thatCiEZforalli.Let pbeaprime number, and let Pgll(X)=g(X) +diI P sothat gllconverges tog(i.e. the coefficients ofgllconverge tothe coefficients ofg). (a)Prove that gllhasprecisely d-2real roots for nsufficiently large. (You may use abitofcalculus, orusewhatever method youwant.) (b)Prove that gllisirreducible over Q. Integral-valued polynomials 18.LetP(X) EQ[X] be apolynomial inone variable with rational coefficients. It may happen that P(n)EZforallsufficiently large integersnwithout necessarily P having integer coefficients. (a)Give anexample ofthis. (b)Assume that Phas the above property. Prove that there are integers Co, C1, ..., Crsuch that P(X)=CO()+Cl(,X l)+ooo+c" where (X )=X(X-1)...(X-r+1)r r! isthebinomial coefficient function. Inparticular, P(n) EZforall n.Thus we maycall Pintegral valued. (c)Letf:Z--.Zbe afunction. Assume that there exists anintegral valued polynomial Qsuch that thedifference functionfdefined by (f)(n)=f(n)-f(n-1) isequal toQ(n) forall nsufficiently large. Show that there exists anintegral- valued polynomial Psuch thatf(n)=P(n) forall nsufficiently large. Exercises onsymmetric functions 19.(a)LetXl'...' XII bevariables. Show that any homogeneous polynomial in Z[Xl'...,XII] ofdegree >n(n-1)liesintheideal generated bytheelemen- tary symmetric functions Sl, ..., SII. (b)With the same notation show that Z[X l'...,XII] isafreeZ[Sl'''.' SII] module with basis themonomials x(r) =Xl...X;" with 0<ri<n-i. IV,Ex EXERCISES 217 (c)LetXl'... ,X"and Y1,..., Ymbetwo independent sets ofvariables. Let s1,...,srIbetheelementary symmetric functions ofXands,..., sthe elementary symmetric functions ofY(using vector vector notation). Show thatZ[X, Y]isfree over Z[s,s']with basis x(r)y(q), and theexponents (r),(q) satisfying inequalitiesasin(b). (d)LetIbeanideal inZ[s, s']. Let Jbetheideal generated byIinZ[X, Y]. Show that JnZ[s,s']=I. 20,Let Abeacommutative ring. Let tbeavariable. Let m f(t)=Laiti i=Oand" g(t)=Lbiti i=O bepolynomials whose constant terms are ao=bo=1.If f(t)g(t)=1, show that there exists aninteger N(=(m+n)(m +n-1))such that any mono- mial arl...arn 1 " withLjrj>Nisequal toO.[Hint: Replace the a'sand b'sbyvariables. Use Exercise 19(b) toshow that any monomial M(a) ofweight> Nlies intheideal I generated bytheelements k Ck=Laibk-i i=O (lettingao=bo=1).Note that Ckisthek-th elementary symmetric function of the m+nvariables (X,Y).] [Note: For some interesting contexts involving symmetric functions, see Cartier's talk attheBourbaki Seminar, 1982-1983.] A-rings Thefollowing exercises start atrain ofthought which will bepursued inExercise 33ofChapter V;Exercises 22-24 ofChapter XVIII; and Chapter XX,3. These originated toalarge extent inHirzebruch's Riemann- Roch theorem and itsextension byGrothendieck who defined A-rings ingeneral. LetKbeacommutative ring. Byl-operations we mean afamily ofmappings Ai:K K foreach integer i>0satisfying therelations forallxEK: AO(X)=1, A1(x)=x, and forallintegers n>0,and x,yEK, " A"(X +y)=LAi(X)A"-i(y). i=O 218 POLYNOMIALS IV,Ex The reader will meet examples ofsuch operations inthechapteronthe alternat- ing and symmetric products, but the formalism ofsuch operations depends only onthe above relations, and so can bedeveloped here inthe context offormal power series. Given aA.-operation, inwhich case wealso say that Kisal-ring, wedefine thepower series 00 A.t(x)=LA.i(X)ti . i=O Prove thefollowing statements. 21.The mapx....-+ A.t(x) isahomomorphism from the additive group ofKinto the multiplicative group ofpower series 1+tK[[t]]whose constant term isequal to 1.Conversely, any such homomorphism such that A.t(x)=1+xt+higher terms gives rise toA.-operations. 22.Let s=at+higher terms be apower series inK[[t]] such that aisaunit inK. Show that there isapower series t=g(s)=Lbisiwith biEK. Show that any power series f(t)EK[[t]]can bewritten intheform h(s)for some other power series with coefficients inK. Given aA.-operationonK,define thecorresponding Grotbendieck power series Yt(x)=A.t/(1-t)(x)=A.s(x) where s=t/(1-t).Then themap x....-+ Yt(x) isahomomorphismasbefore. We define yi(x) bytherelation Yt(x)=Lyi(X)ti . Show that Ysatisfies thefollowing properties. 23.(a)For every integern>0wehave II y"(X +y)=Lyi(X)y"-i(y). .i=O (b)Yt(l)=1/(1-t). (c)Yt(-1)=1-t. 24.Assume that A.iU =0fori>1.Show: (a)Yt(u-1)=1+(u-l)t. 00 (b)Yt(1-u)=L(1-U)iti . i=O 25.Bernoulli numbers. Define the Bernoulli numbers Bkasthe coefficients Inthe powersenes t00tk F(t)= t=LBk-. e-1 k=O k! IV,ExEXERCISES 219 Ofcourse, et=LtPlIn! isthestandard power series with rational coefficients tin!. Prove: (a)Bo=1,Bt=-1, B2=t. (b)F(-t)=t+F(t), and Bk=0ifkisodd 1. 26.Benoulli pol.ynomials.Define the Bernoulli polynomials Ba:(X) bythe power senes expanS10n tetX 00tk F(t,X)= t=LBk(X)-. e-1 k=O k! Itisclear that Bk=Bk(O),sotheBernoulli numbers arethe constant terms ofthe Bernoulli polynomials. Prove: (a)Bo(X)=1,B1(X)=X-1, B2(X)=x2-X+t. (b)For each positive integer N, Bk(X)=Nk-tNfBk(X+a ). a=O N (c)Bk(X)=Xk-1kXk-t +lower terms. tk (d)F(t,X+1)-F(t,X)=teXt =tLXk,. k. (e)Bk(X +1)-Bk(X)=kXk-tfork>1. 27.Let Nbe apositive integer and letfbe afunction onZ/NZ. Form the power senes N-1te(a+X)t Ff(t, X)=Lf(a) Nt. a=O e-1 Following Leopoldt, define thegeneralized Bernoulli polynomials relative tothe function fby 00tk Ff(t, X)= kf:OBk,f(X)kr Inparticular, the constant term ofBk,f(X)isdefined tobethe generalized Bernoulli numberBk,f=Bk,f(O)introduced byLeopoldt incyclotomic fields. Prove: (a)Ff(t,X+k)=ektFf(t, X). (b)Ff(t,X+N)-Ff(t, X)=(eNt-I)Ff(t,X). 1N-1 (c)k[Bk,f(X +N)-Bk.f(X)]=aof(a)(a +xl-1 . (d)Bk.J(X)=itoe)B;,fxn-i =Bk,f+kBk-t.fX+...+kB1,fXk-1+BO,fXk . Note. The exercises onBernoulli numbers and polynomials aredesigned not only togive examples for the material inthe text, but toshow how this material leads into majorareas ofmathematics: intopology and algebraic geometry centering 220 POLYNOMIALS IV,Ex around Riemann-Roch theorems; analytic and algebraic number theory,asinthe theory ofthe zeta functions and thetheory ofmodular forms, cf.my Introduction toModular Forms, Springer-Verlag, New York, 1976, Chapters XIV and XV; my Cyclotomic Fields, IandII,Springer-Verlag, New York, 1990, Chapter 2,2;Kubert- Lang's Modular Units, Springer-Verlag, New York, 1981; etc. Further Comments, 1996-2001. Iwas informed byUmberto Zannier that what has been called Mason's theorem wasproved three years earlier byStothers [Sto 81], Theo- rem 1.1. Zannier himself haspublished some results onDavenport's theorem [Za95], without knowing ofthepaper byStothers, usingamethod similar tothat ofStothers, andrediscovering some ofStothers' results, butalso going beyond, Indeed, Stothers uses the"Belyi method" belonging toalgebraic geometry, and increasingly appearingasa fundamental tool. Mason gaveavery elementary proof, accessible atthebasic level of algebra. An even shorter and very elegant proof oftheMason-Stothers theorem was given byNoah Snyder [Sny 00]. Iam much indebted toSnyder forshowing methat proof before publication, and Ireproduced itin[La99b]. But Irecommend looking at Snyder's version, [La99b] S.LANG, Math Talks forUndergraduates, Springer Verlag 1999 [Sny 00] N.SNYDER, Analternate proof ofMason's theorem, Elemente der Math. 55 (2000) pp.93-94 [Sto 81] W,STOTHERS, Polynomial identities andhauptmoduln, Quart. J.Math. Oxford (2)32(1981) pp.349-370 [Za95] U.ZANNIER, OnDavenport's bound forthedegree off3-g2and Riemann's existence theorem, Acta Arithm. LXXI.2 (1995) pp. 107-137 Part Two ALGEBRAIC EQUATIONS This part isconcerned with the solutions ofalgebraic equations, inone orseveral variables. This isthe recurrent theme inevery chapter ofthis part, and welay the foundations for allfurther studies concerning such equations. Given asubring Aof aring B,and afinite number ofpolynomials i1' ...,ininA[Xl'...,Xn],we areconcerned with then-tuples (b1,.. .,bn)EB(n) such that h(b 1,..., bn}=0 for i=1,..., r.For suitable choices ofAand B,this includes thegeneral problem ofdiophantine analysis when A,Bhave an"arithmetic" structure. We shall study various cases. Webegin bystudying roots ofonepolyno- mial inone variable over afield. We prove the existence ofanalgebraic closure, andemphasize therole ofirreducibility. Next westudy the group ofautomorphisms ofalgebraic extensions ofa field, both intrinsically and as agroup ofpermutations ofthe roots of a polynomial. We shall mention some major unsolved problems along the way. Itisalso necessary todiscuss extensions ofaring, togive thepossibil- ityofanalyzing families ofextensions. The ground work islaid inChapter VII. InChapter IX, we come tothe zeros ofpolynomials inseveral variables, essentially over algebraically closed fields. But again, itisadvantageous to 221 222 ALGEBRAIC EQUATIONS PART TWO consider polynomials over rings, especially Z,since inprojective space, the conditions that homogeneous polynomials have anon-trivial common zero can begiven universally over Zinterms oftheir coefficients. Finally weimpose additional structures like those ofreality, ormetric structures given byabsolute values. Each one ofthese structures gives rise to certain theorems describing the structure ofthe solutions ofequationsas above, andespecially proving theexistence ofsolutions inimportantcases. CHAPTER V Algebraic Extensions Inthis first chapter concerning polynomial equations, weshow that given apolynomial over afield, there always exists some extension ofthe field where thepolynomial has aroot, and weprove theexistence ofanalgebraic closure. We make apreliminary study ofsuch extensions, including the automorphisms, and wegive algebraic extensions offinite fields asexamples. 1. FINITE AND ALGEBRAIC EXTENSIONS Let Fbe afield. IfFisasubfield ofafield E,then wealso saythat Eis anextension field ofF.We may view Easavector space over F,and wesay that Eisafinite orinfinite extension ofFaccordingasthedimension ofthis vector space isfinite orinfinite. Let Fbe asubfield ofafield E.Anelement rxofEissaid tobealgebraic over Fifthere exist elements ao, ..., an(n>1)ofF,not allequal to0,such that ao+a1rx+...+anrxn=O. If rx=F0,and rxisalgebraic, then we can always find elements aiasabove such that ao=F0(factoring out asuitable power ofrx). LetXbe avariable over F.We can also saythat rxisalgebraic over Fif thehomomorphism F[X] E 223 224 ALGEBRAIC EXTENSIONS V,1 which istheidentity onFand maps Xon rxhas anon-zero kernel. Inthat case the kernel isanideal which isprincipal, generated byasingle polyno- mial p(X), which wemay assume hasleading coefficient 1.We then have an isomorphism F[X]j(p(X)) F[rx], and since F[rx] isentire, itfollows that p(X) isirreducible. Having normal- ized p(X)sothat itsleading coefficient is1,we seethat p(X) isuniquely determined byrxand will becalled THE irreducible polynomial of rxover F. We sometimes denote itbyIrr( rx,F,X). An extension EofFissaid tobealgebraic ifevery element ofEis algebraic over F. Proposition 1.1. Let Ebe afinite extension ofF. Then Eisalgebraic over F. Proof Let rxEE, rx=Fo.The powers ofrx, 12 n ,rx,rx,..., rx, cannot belinearly independent over Fforallpositive integers n,otherwise thedimension ofEover Fwould beinfinite. Alinear relation between these powers shows that rxisalgebraic over F. Note that the converse ofProposition1.1isnot true; there exist infinite algebraic extensions. We shall see later that the subfield ofthecomplex numbers consisting ofallalgebraic numbers over Qisaninfinite extension ofQ. IfEisanextension ofF,wedenote by [E:F] thedimension ofEasvector space over F.Itmay beinfinite. Proposition 1.2. Let kbeafield andFeE extension fields ofk.Then [E:k]=[E:F][F:k]. If{Xi}ielisabasis forFover kand{Yj}jeJisabasis for Eover F,then {XiYj}(i,j)elxJisabasis for Eover k. Proof Let ZEE.Byhypothesis there exist elements rxjEF,almost all rxj=0,such that Z=LrxjYjo jeJ ForeachjEJthere exist elements bjiEk,almost allofwhich areequal to0, such that V,1 FINITE AND ALGEBRAIC EXTENSIONS 225 rxj=LbjiX i, ieI and hence z= b..x,y.i..Ji..J JI Irji This shows that{xiYj}isafamily ofgenerators forEover k.We must show that itislinearly independent. Let{cij}beafamily ofelements ofk,almost allofwhich are0,such that c..x. y.=0IJ IJ. ji Then foreachj, c..x, =0i..J IJ I i because the elementsYjarelinearly independent over F.Finally Cij=0for each ibecause {Xi}isabasis ofFover k,thereby proving ourproposition. Corollary 1.3. The extension Eofkisfinite ifandonlyifEisfiniteover Fand Fisfiniteover k. Aswith groups,wedefine atower offields tobeasequence F1cF2C...cFn ofextension fields. The tower iscalled finite ifandonly ifeach step isfinite. Let kbe afield, Eanextension field, and rxEE.We denote byk(rx) the smallest subfield ofEcontaining both kand rx.Itconsists ofallquotients f(rx)/g(rx), where f,garepolynomials with coefficients inkand g(rx) =FO. Proposition 1.4. Let rxbealgebraic over k.Then k(rx)=k[rx], and k(rx) is finiteover k.The degree [k(rx):k]isequal tothedegree ofIrr(rx, k,X). Proof Letp(X)=Irr(rx, k,X). Letf(X)Ek[X] besuch that f(rx) =FO. Then p(X) does not divide f(X), and hence there exist polynomials g(X), h(X)Ek[X] such that g(X)p(X) +h(X)f(X)=1. From this wegeth(rx)f(rx)=1,and we seethat f(rx) isinvertible ink[rx]. Hence k[rx] isnotonlyaring but afield, and must therefore beequal to k(rx). Let d=degp(X). The powers 1d-l ,rx,...,rx arelinearly independent over k,forotherwise suppose ao+alrx+... +ad_1rxd-1=0 226 ALGEBRAIC EXTENSIONS V,1 with aiEk,not allai=O.Letg(X)=ao+·..+ad-lXd-1 .Then g=F0and g()=O.Hence p(X) divides g(X), contradiction. Finally, letf(a.)Ek[], where f(X)Ek[X]. There exist polynomials q(X), r(X)Ek[X] such that degr<dand f(X)=q(X)p(X) +r(X). Thenf()=r(), and we seethat1,,..., d-l generate k[]asavector space over k.This proves ourproposition. Let E,Fbeextensions ofafield k.IfEand Farecontained insome field Lthen wedenote byEF the smallest subfield ofLcontaining both Eand F,and call itthecompositum ofEand F,inL.IfE,Fare notgivenas embedded inacommon field L,then wecannot define thecompositum. Let kbe asubfield ofEand let l'...,nbeelements ofE.We denote by k(l' ...,n) thesmallest subfield ofEcontaining kand1' ...,n'Itselements consist of allquotients f(l'.. .,n) g(l' ...,n) where f,garepolynomials innvariables with coefficients ink,and g(l'...,n) =FO. Indeed, the setofsuch quotients forms afield containing kand l'...,n. Conversely, any field containing kand 1,...,n must contain these quotients. We observe that Eisthe union ofallitssubfields k(l'"'' n)as (1' ...,n)ranges over finite subfamilies ofelements ofE.We could define the compositum ofanarbitrary subfamily ofsubfields ofafield Lasthe smallest subfield containing allfields inthefamily. We saythat Eisfinitely generated over kifthere isafinite family ofelements l'...,nofEsuch that E=k(l'...,n)' We seethat Eisthecompositum ofallitsfinitely generated subfields over k. Proposition 1.5. Let Ebe afinite extension ofk.Then Eisfinitely generated. Proof. Let{1'.." n}be abasis ofEasvector space over k.Then certainly E=k(l' ...,n)' V,1 FINITE AND ALGEBRAIC EXTENSIONS 227 IfE=k(rxl'...,rxn)isfinitely generated, and Fisanextension ofk,both F,Econtained inL,then EF =F(rx 1,..., rxn), and EF isfinitely generated over F.We often draw thefollowing picture: EF/""F E""/ k Lines slanting upindicate aninclusion relation between fields. We also call the extension EFofFthetranslation ofEtoF,oralso thelifting ofEto F. Let rxbealgebraic over the field k.Let Fbe anextension ofk,and assume k(rx),Fboth contained insome field L.Then rxisalgebraic over F. Indeed, theirreducible polynomial for rxover khas afortiori coefficients in F,and gives alinear relation forthepowers of rxover F. Suppose that wehave atower offields: kck(rx 1)Ck(rx 1,rx2)c...ck(rx 1,..., rxn), each one generated from thepreceding field by asingle element. Assume that eachrxiisalgebraic over k,i=1,..., n.As aspecial case ofourpreceding remark, wenote that rxi+l isalgebraic over k(rx 1,..., rxi).Hence each step of thetower isalgebraic. Proposition 1.6. Let E=k(rx 1,..., rxn)beafinitely generated extension of afield k,and assumerxialgebraic over kfor each i=1,..., n.Then Eis finite algebraic over k. Proof From theabove remarks, weknow that Ecan beobtained asthe end ofatower each ofwhose steps isgenerated by one algebraic element, and istherefore finite byProposition 1.4. We conclude that Eisfinite over k byCorollary 1.3,and that itisalgebraic byProposition 1.1. Letebe acertain class ofextension fields FeE. We shall saythat eis distinguished ifitsatisfies thefollowing conditions: (1)LetkeF cEbeatower offields. The extension kcEisineifand only ifkeF isineandFeE isine. (2)IfkcEisine,ifFisany extension ofk,and E,Fare both contained insome field, then FcEFisine. (3)IfkeF and kcEareineand F,Earesubfields ofacommon field, then kcFE isine. 228 ALGEBRAIC EXTENSIONS V,1 The diagrams illustrating ourpropertiesare asfollows: E I F I k (1)EF /"'F E",/ k (2)EF/ E F/ k (3) These lattice diagrams offields areextremely suggestive inhandling exten- sion fields. We observe that (3)follows formally from the first two conditions. Indeed, one views EF over kasatower with steps keF cEF. As amatter ofnotation, itisconvenient towrite ElF instead ofFeE to denote anextension. There can benoconfusion with factor groups since we shall never usethenotation ElF todenote such afactor group when Eisan extension field ofF. Proposition 1.7. The class ofalgebraic extensions isdistinguished, and so istheclass offinite extensions. Proof: Consider first the class offinite extensions. We have already proved condition (1). Asfor(2), assume that Elk isfinite, and letFbeany extension ofk.ByProposition 1.5there exist elements al,..., anEEsuch that E=k(a l,...,an)' Then EF =F(al' ...,an), and hence EFIFisfinitely generated byalgebraic elements. Using Proposition 1.6 weconclude that EFIF isfinite. Consider next theclass ofalgebraic extensions, and let kcFcE be atower. Assume that Eisalgebraic over k.Then afortiori, Fis algebraic over kand Eisalgebraic over F.Conversely, assume each step in the tower tobealgebraic. Let aEE.Then asatisfies anequation anan+...+ao=0 with aiEF,not allai=O.Let Fo=k(a n,...,ao)'Then Foisfinite over kby Proposition 1.6,and aisalgebraic over Fo. From the tower kcFo=k(a n,...,ao)c:Fo(a) and the fact that each step inthis tower isfinite, weconclude that Fo(a) is finite over k,whence aisalgebraic over k,thereby proving that Eisalgebraic over kandproving condition (1)foralgebraic extensions. Condition (2)has already been observed tohold, i.e. anelement remains algebraic under lifting, and hence sodoes anextension. V,2 ALGEBRAIC CLOSURE 229 Remark. Itistrue thatfinitely generated extensions form adistinguished class, but one argument needed toprove part of(1)can becarried outonly with more machinery than wehave atpresent. Cf.thechapterontranscen- dental extensions. 2. ALGEBRAIC CLOSURE Inthis and the next section weshall deal with embeddings ofafield into another. We therefore define some terminology. Let Ebeanextension ofafield Fand let u:F-+L be anembedding (i.e. aninjective homomorphism) ofFinto L.Then u induces anisomorphism ofFwith itsimage uF, which issometimes written FlI .Anembedding tofEinLwill besaid tobeover uiftherestriction oft toFisequal tou.We also saythat textends u.Ifuistheidentity then we saythat tisanembedding ofEover F. These definitions could bemade inmore general categories, since they depend only ondiagrams tomake sense: E inej F )LET) L in\Ie FT )L jid (J Remark. Letf(X)EF[X] be apolynomial, and let rxbe aroot offin E.Sayf(X)=ao+...+anXn .Then o=f(a)=ao+ala+·..+anan . Iftextends uasabove, then we seethat trxisaroot offllbecause o=t{f(rx))=ag+ar(trx) +...+a:(trx)n. Here wehave written allinstead ofu(a). This exponential notation is frequently convenient and will beused again inthesequel. Similarly, we write FlIinstead ofu(F) oruF. In our study ofembeddings itwill also beuseful tohave alemma concerning embeddings ofalgebraic extensions into themselves. For this we note that ifu:E-+Lisanembedding over k(i.e.inducing theidentity onk), then ucan beviewed asak-homomorphism ofvector spaces, because both E,Lcan beviewed asvector spacesover k.Furthermore uisinjective. 230 ALGEBRAIC EXTENSIONS V,2 Lemma 2.1. Let Ebeanalgebraic extension ofk,and let (1:E-+Ebean embedding ofEintoitself over k.Then (1isanautomorphism. Proof. Since (1isinjective, itwill suffice toprove that (1issurjective. Let rxbeanelement ofE,letp(X) beitsirreducible polynomial over k,and letE' bethesubfield ofEgenerated byallthe roots ofp(X) which lieinE.Then E'isfinitely generated, hence isafinite extension ofk.Furthermore, (1must maparoot ofp(X) on aroot ofp(X), and hence (1maps E'into itself. We can view (1as ak-homomorphism ofvector spaces because (1induces the identity onk.Since (1isinjective, itsimage (1(£') isasubspace ofE'having the same dimension [E' :k].Hence u(E')=E'. Since aEE',itfollows that aisintheimage ofu,and our lemma isproved. Let E,Fbeextensions ofafield k,contained insome bigger field L.We can form therIng E[F] generated bytheelements ofFover E.Then E[F]= F[E], and EF isthequotient field ofthisring. Itisclear that theelements of E[F] can bewritten intheform a1b1+...+anb n with aiEEand biEF.Hence EFisthefield ofquotients ofthese elements. Lemma 2.2. Let E1,E2beextensions ofafield k,contained insome bigger field E,and let (1beanembedding ofEinsome field L.Then (1(E 1E2)=(1(E 1)(1(E 2). Proof Weapply(1toaquotient ofelements oftheabove type, say (a1b1+...+anb n)arbr +... +a:b: (1 a;b;+...+ab= aab;a+.. ·+a;:b;:, and seethat theimage isanelement of(1(E 1)(1(E 2).Itisclear that theimage (1(E 1E2)is(1(E 1)(1(E 2). Let kbe afield, f(X)apolynomial ofdegree>1ink[X]. We consider theproblem offinding anextension Eofkinwhichfhas aroot. Ifp(X) is anirreducible polynomial ink[X] which divides j(X), then any root ofp(X) will also be aroot of,f(X),so we may restrict ourselves toirreducible polynomials. Letp(X) beirreducible, and consider thecanonical homomorphism (1:k[X]-+k[X]j(p(X)). Then (1induces ahomomorphism onk,whose kernel is0,because every nonzero element ofkisinvertible ink,generates theunit ideal, and 1does not lieinthekernel. Let betheimage ofXunder (1,i.e. =(1(X) isthe residue class ofXmod p(X). Then pa()=pa(xa}=(p(x))a=o. V,2 ALGEBRAIC CLOSURE 231 Hence isaroot ofpel,and assuch isalgebraic over uk. We have now found anextension ofuk,namely uk() inwhich pelhas aroot. With aminor set-theoretic argument, weshall have: Proposition 2.3. Let kbe afield andfapolynomial ink[X] ofdegree >1.Then there exists anextension Eofkinwhichfhas aroot. Proof We may assume thatf=pisirreducible. We have shown that there exists afield Fand anembedding u:k--.F such that pelhas aroot inF.Let Sbe asetwhose cardinality isthe same asthat ofF-uk(=thecomplement ofukinF)and which isdisjoint from k.Let E=kuS. We can extend u:k-+Ftoabijection ofEonF.We now define afield structure onE.Ifx,yEEwedefine xy=u-1(u(x)u(y»), x+y=u-1(u(x) +u(y»). Restricted tok,our addition and multiplication coincide with thegiven addition andmultiplication ofouroriginal field k,and itisclear that kisa subfield ofE. We let rx=U-1(). Then itisalso clear that p(rx)=0,as desired. Corollary 2.4. Let kbe afield and letf1, ...,f"bepolynomials ink[X] ofdegrees>1.Then there exists anextension Eofkinwhich each hhas aroot, i=1,..., n. Proof Let£1beanextension inwhich f1has aroot. We may view f2 asapolynomial over E1.Let E2beanextension ofE1inwhich f2has a root. Proceeding inductively, ourcorollary follows atonce. We define afield Ltobealgebraically closed ifevery polynomial inL[X] ofdegree>1has aroot inL. Theorem 2.5. Letkbeafield. Then there exists analgebraically closed field containing kasasubfield. Proof We first construct anextension Elofkinwhich every polyno- mial ink[X] ofdegree>1has aroot. One canproceedasfollows (Artin). Toeach polynomial fink[X] ofdegree>1weassociate aletter Xfand we letSbethe setofallsuch letters Xf(sothat Sisinbijection with the setof polynomials ink[X] ofdegree>1).Weform thepolynomial ringk[S], and contend that theideal generated byallthepolynomials f(Xf)ink[S] isnot theunit ideal. Ifitis,then there isafinite combination ofelements inour ideal which isequal to1: glfl(Xfl)+...+gnf,,(X fn)=1 232 ALGEBRAIC EXTENSIONS V,2 with giEk[S]. For simplicity, write Xiinstead ofXfi.The polynomials gi will involve actually onlyafinite number ofvariables, sayXl' ...,XN(with N>n).Our relation then reads n Lgi(X 1,...,XN)Ii{X i)=1. i=l Let Fbe afinite extension inwhich each polynomial fl'...,f"has aroot, sayrxiisaroot ofIiinF,fori=1,..., n.Let rxi=0fori>n.Substitute rxi forXiinour relation. We get0=1,contradiction. Let mbe amaximal ideal containing theideal generated byallpolyno- mials f(Xf)ink[S]. Then k[S]jm isafield, and wehave acanonical map u:k[S]-+k[S]jm. For anypolynomial fEk[X] ofdegree>1,thepolynomial fahas aroot in k[S]jm, which isanextension ofuk. Using the same type ofset-theoretic argumentasinProposition 2.3, weconclude that there exists anextension E1ofkinwhich every polynomial fEk[X] ofdegree>1has aroot inE1. Inductively, we can form asequence offields £1cE2CE3C...cEn... such that every polynomial inEn[X] ofdegree>1has aroot inEn+l. Let E betheunion ofallfields En'n=1,2,.... Then Eisnaturallyafield, forif x,yEEthen there exists some nsuch that x,yEEn' and we can take the product orsum xy orx+yinEn. This isobviously independent ofthe choice of nsuch that x,yEEn' and defines afield structure on E.Every polynomial inE[X] has itscoefficients insome subfield En'hence aroot in En+1,hence aroot inE,asdesired. Corollary 2.6. Let kbe afield. There exists anextension k8which is algebraic over kandalgebraically closed. Proof Let Ebeanextension ofkwhich isalgebraically closed and let k8betheunion ofallsubextensions ofE,which arealgebraic over k.Then k8isalgebraic over k.IfrxEEand rxisalgebraic over k8then rxisalgebraic over kbyProposition 1.7.Iffisapolynomial ofdegree>1ink8[X], then fhas aroot rxinE,and rxisalgebraic over k8 .Hence rxisink8and k8is algebraically closed. We observe that ifLisanalgebraically closed field, andfEL[X] has degree>1,then there exists cELand rxl'...,rxnELsuch that f(X)=c(X-rx1)...(X-rxn). Indeed, fhas aroot rx1inL,sothere exists g(X)EL[X]such that f{X)=(X-rx1)g(X). Ifdeg g>1,we can repeat this argument inductively, and express fas a V,2 ALGEBRAIC CLOSURE 233 product ofterms (X-ai)(i=1,...,n)and anelement cEL.Note that cis theleading coefficient off,i.e. f(X)=cX" +terms oflower degree. Hence ifthecoefficients offlieinasubfield kofL,then cEk. Let kbe afield and (1:k-+Lanembedding ofkinto analgebraically closed field L.We areinterested inanalyzing theextensions of (1toalgebraic extensions Eofk.We begin byconsidering thespecialcase when Eis generated byone element. Let E=k(a) where aisalgebraic over k.Let p(X)=Irr(a, k,X). Letpbearoot ofpClinL.Given anelement ofk(a)=k[a], we can write it intheform f(a) with some polynomial f(X)Ek[X]. We define anextension of (1bymapping f(a)t--+fCl(/3). This isinfact well defined, i.e.independent ofthechoice ofpolynomial f(X) used toexpress our element ink[a]. Indeed, ifg(X) isink[X] and such that g(a)=f(a), then (g-f)(a)=0,whence p(X) divides g(X)-f(X). Hence pCl(X) divides gCl(X)-fCl(X), and thus gCl(P)=fCl(/3). Itisnow clear that our map isahomomorphism inducing(1onk,and that itisanextension of (1to k(a). Hence weget: Proposition 2.7. The number ofpossible extensions of(1tok(a) is<the number ofroots ofp,and isequal tothenumber ofdistinct roots ofp. This isanimportant fact, which weshall analyze more closely later. For the moment, we are interested inextensions of (1toarbitrary algebraic extensions ofk.Wegetthem byusing Zorn's lemma. Theorem 2.8. Let kbe afield, Eanalgebraic extension ofk,and (1:k-+Lanembedding ofkinto analgebraically closed field L. Then there exists anextension of(1to anembedding ofEinL.IfEis algebraically closed and Lisalgebraic over (1k, then any such extension of (1isanisomorphism ofEonto L. Proof Let Sbethe setofallpairs (F,t)where Fisasubfield ofE containing k,and tisanextension of (1toanembedding ofFinL.If(F,t) and (F',t') are such pairs, wewrite (F,t)<(F' ,t')ifFcF'andt'IF=t. Note that Sisnot empty [itcontains (k,(1)], and isinductively ordered: If {(fi,ti)}isatotally ordered subset, weletF=Uliand define tonFtobe equal totioneach lieThen (F,t)isanupper bound forthetotally ordered subset. Using Zorn's lemma, let(K, A.)beamaximal element inS.Then A.is anextension of(1,and wecontend that K=E.Otherwise, there exists aEE, 234 ALGEBRAIC EXTENSIONS V,2 rxrtK.Bywhat we saw above, ourembeddingA.has anextension toK(rx), thereby contradicting themaximality of(K, A.).This proves that there exists anextension of (1toE.We denote this extension again by(1. IfEisalgebraically closed, and Lisalgebraic over (1k, then (1Eis algebraically closed and Lisalgebraic over (1E,hence L=(1E. As acorollary, wehave acertain uniqueness for an"algebraic closure" of afield k. Corollary 2.9. Let kbeafield and letE,E'bealgebraic extensions ofk. Assume that E,E' arealgebraically closed. Then there exists an iso- morphism t:E-+E' ofEonto E'inducing theidentity onk. Proof Extend theidentity mapping onktoanembedding ofEinto E' andapply thetheorem. We see that analgebraically closed and algebraic extension ofkis determined up to anisomorphism. Such anextension will becalled an algebraic closure ofk,and wefrequently denote itbyk8 .Infact, unless otherwise specified, we use thesymbol k8only todenote algebraic closure. Itisnow worth while torecall thegeneral situation ofisomorphisms and automorphisms ingeneral categories. Let Cibe acategory, and A,Bobjects inCi.We denote byIso(A, B)the setofisomorphisms ofAon B.Suppose there exists atleast one such isomorphism(1:A-+B,with inverse (1-1: B-+A.IfqJisanautomorphism of A,then (10qJ:A-+Bisagainanisomorphism. If1/1isanautomorphism of B,then1/10(1:A-+Bisagainanisomorphism. Furthermore, the groups ofautomorphisms Aut(A) andAut(B) areisomorphic, under themappings qJ(10qJ0(1-1, (1-1 01/10(1+-11/1, which are inverse toeach other. The isomorphism(10qJ0(1-1 ISthe one which makes thefollowing diagram commutative: A(J )B l 1<10O<1-1 A )B (J We have asimilar diagram for(1-1 01/10(1. Let t:A-+Bbeanother isomorphism. Then t-10(1isanautomorphism ofA,and t0(1-1 isanautomorphism ofB.Thus twoisomorphisms differ by anautomorphism (ofAorB). We seethat thegroup Aut(B) operates onthe V,3 SPLITTING FIELDS AND NORMAL EXTENSIONS 235 setIso(A, B) ontheleft, and Aut(A) operatesonthe setIso(A, B) onthe right. We also seethat Aut(A) isdetermined uptoamapping analogous 'to a conjugation. This isquite different from the type ofuniqueness given by universal objects inacategory. Such objects have only theidentity auto- morphism, and hence aredetermined uptoaunique isomorphism. This isnot the case with thealgebraic closure ofafield, which usually has alarge amount ofautomorphisms. Most ofthischapter and the next is devoted tothestudy ofsuch automorphisms. Examples. Itwill beproved later inthis book that thecomplex numbers arealgebraically closed. Complex conjugation isanautomorphism ofC. There aremany more automorphisms, but theother automorphisms=1=ide are notcontinuous .We shall discuss other possible automorphisms inthechapter ontranscendental extensions. The subfield ofCconsisting ofallnumbers which arealgebraicover Qisanalgebraic closure QaofQ.Itiseasy toseethat Qa isdenumerable. Infact, prove thefollowingasanexercise: Ifkisafield which isnotfinite, then anyalgebraic extension ofkhas the same cardinalityask. Ifkisdenumerable, one can first enumerate allpolynomials ink,then enumerate finite extensions bytheir degree, andfinally enumerate thecardi- nality ofanarbitrary algebraic extension. We leave thecounting details as exerCIses. Inparticular, Q8 =FC.IfRisthefield ofreal numbers, then R8=C. Ifkisafinite field, then algebraic closure k8ofkisdenumerable. We shall infact describe ingreat detail the nature ofalgebraic extensions of finite fields later inthischapter. Not allinteresting fields are subfields ofthecomplex numbers. For instance, one wants toinvestigate thealgebraic extensions of afield C(X) where Xisavariable over C.The study ofthese extensions amounts tothe study oframified coverings ofthesphere (viewed asaRiemann surface), and infact one hasprecise information concerning the nature ofsuch extensions, because one knows thefundamental group ofthesphere from which afinite number ofpoints has been deleted. We shall mention this example again later when wediscuss Galois groups. 3. SPLITTING FIELDS AND NORMAL EXTENSIONS Let kbe afield and letfbe apolynomial ink[X] ofdegree>1.Bya splitting field Koffweshall mean anextension Kofksuch thatfsplits into linear factors inK,i.e. 236 ALGEBRAIC EXTENSIONS V,3 f(X)=c(X-1)...(X-n) withiEK,i=1,..., n,and such that K =k(1' ...,n)isgenerated byall the roots off Theorem 3.1. LetKbeasplitting field ofthepolynomial f(X)Ek[X]. If Eisanother splitting field off,then there exists anisomorphism u:E-.K inducing theidentity onk.IfkeKe k8 ,where k8isanalgebraic closure ofk,then any embedding ofEink8inducing theidentity onkmust be an isomorphism ofEonto K. Proof LetK8beanalgebraic closure ofK.Then K8isalgebraic over k,hence isanalgebraic closure ofk.ByTheorem 2.8 there exists an embedding u:E-+K8 inducing theidentity onk.We have afactorization f(X)=c(X-P1)...(X-Pn) with PiEE,i=1,..., n.The leading coefficient cliesink.Weobtain f(X)=fCl(X)=c(X-UP1).,.(X-u/3n). We have unique factorization inK8[X]. Sincefhas afactorization f(X)=c(X-1)...(X-n) inK[X], itfollows that (U/31' ...,uPn) differs from(1'"'' n)byapermuta- tion. From this weconclude that UPi EKfor i=1,..., nand hence that uE cK.But K =k(1' ...,n)=k(UP1' ...,u/3n), and hence uE =K,because E=k(P1'...,Pn). This proves our theorem. We note that apolynomial f(X) Ek[X] always has asplitting field, namely thefield generated byitsroots inagiven algebraic closure k8ofk. LetIbe asetofindices and let{h}ielbe afamily ofpolynomials in k[X], ofdegrees>1.Byasplitting field forthisfamily weshall mean an extension Kofksuch that every hsplits inlinear factors inK[X], and Kis generated byallthe roots ofallthepolynomials h,iEI.Inmost applica- tions wedeal with afinite indexing setI,but itisbecoming increasingly important toconsider infinite algebraic extensions, and soweshall deal with them fairly systematically. One should also observe that theproofs weshall give forvarious statements would not besimpler ifwerestricted ourselves to thefinite case. Let k8beanalgebraic closure ofk,and letKibe asplitting field ofhin k8 .Then thecompositum ofthe Kiisasplitting field for our family, V,3 SPLITTING FIELDS AND NORMAL EXTENSIONS 237 since the two conditions definingasplitting field areimmediately satisfied. Furthermore Theorem 3.1extends atonce totheinfinite case: Corollary 3.2. Let Kbe asplitting field for thefamily {};}iel and letE beanother splitting field. Any embedding ofEinto K8inducing the identity onkgives anisomorphism ofEonto K. Proof Let the notation be asabove. Note that Econtains aunique splitting field Eiof};and Kcontains aunique splitting field Kiof};. Any embedding(JofEinto K8must map Eionto KibyTheorem 3.1, and hence maps Einto K.Since Kisthecompositum ofthefields Ki,our map(Jmust send Eonto Kand hence induces anisomorphism ofEonto K. Remark. IfIisfinite, and ourpolynomialsare11' ...,fn'then asplit- ting field for them isasplitting field for the single polynomial f(X)= fl(X)...fn(X) obtained bytaking theproduct. However, even when dealing with finite extensions only, itisconvenient todeal simultaneously with sets ofpolynomials rather than asingle one. Theorem 3.3. Let Kbe analgebraic extension ofk,contained inan algebraic closure k8ofk.Then thefollowing conditions areequivalent: NOR 1.Every embedding ofKink8over kinduces anautomorphism ofK. NOR 2.Kisthesplitting field ofafamily ofpolynomials ink[X]. NOR 3.Every irreducible polynomial ofk[X] which has aroot inK splits into linear factors inK. Proof. Assume NOR 1.Let rxbe anelement ofKand letPa(X) beits irreducible polynomial over k.Letpbe aroot ofPClink8 .There exists an isomorphism ofk(rx) onk(P) over k,mappingrxonp.Extend this iso- morphism toanembedding ofKink8 .This extension isanautomorphism(J ofKbyhypothesis, hence (Jrx =pliesinK.Hence every root ofPClliesinK, andPClsplits inlinear factors inK[X]. Hence Kisthesplitting field ofthe family {pCI} CIeKas rxranges over allelements ofK,and NOR 2issatisfied. Conversely, assume NOR 2,and let{};}ielbethefamily ofpolynomials ofwhich Kisthesplitting field. Ifrxisaroot ofsome };inK,then forany embedding(JofKink8over kweknow that (Jrxisaroot of};. Since Kis generated bythe roots ofallthepolynomials /;"itfollows that (Jmaps K into itself. We now apply Lemma 2.1toconclude that (Jisanautomorphism. Our proof that NOR 1implies NOR 2also shows that NOR 3is satisfied. Conversely,assume NOR 3.Let (Jbe anembedding ofKink8 over k.Let rxEKand letp(X) beitsirreducible polynomialover k.If(Jis anembedding ofKink8over kthen (Jmapsrxon aroot pofp(X), and by hypothesis plies inK. Hence (Jrxlies inK,and (Jmaps Kinto itself. By Lemma 2.1,itfollows that (Jisanautomorphism. 238 ALGEBRAIC EXTENSIONS V,3 Anextension Kofksatisfying thehypotheses NOR 1,NOR 2,NOR 3 will besaid tobenormal. Itisnot true that theclass ofnormal extensions is distinguished. For instance, itiseasily shown that anextension ofdegree 2 isnormal, but the extension Q(.y2)oftherational numbers isnot normal (thecomplex roots ofX4-2are notinit),and yetthis extension isobtained bysuccessive extensions ofdegree 2,namely E=Q()=>F=>Q, where F=Q(),=v0-and E=F(). Thus atower ofnormal extensions isnotnecessarily normal. However, we still have some oftheproperties: Theorem 3.4. Normal extensions remain normal under lifting. If K =>E=>kand Kisnormal over k,then Kisnormal over E.IfKl'K2 arenormal over kand arecontained insome field L,then K1K2isnormal over k,and soisK1nK2. Proof Forour first assertion, letKbenormal over k,letFbeany extension ofk,and assume K,Farecontained insome bigger field. Let (Jbe anembedding ofKF over F(inFa). Then (Jinduces theidentity onF,hence onk,and byhypothesis itsrestriction toKmaps Kinto itself. We get (KFY'=KCIFCI=KF whence KF isnormal over F. Assume that K =>E=>kand that Kisnormal over k.Let (Jbe an embedding ofKover E.Then (Jisalso anembedding ofKover k,and our assertion follows bydefinition. Finally, ifK1,K2arenormal over k,then foranyembedding(JofK1K2 over kwehave (J(K 1K2)=(J(K1)(J(K 2) and our assertion again follows from thehypothesis. The assertion concern- ingtheintersection istrue because (J(K InK2)=(J(KI)n(J(K 2). We observe that ifKisafinitely generated normal extension ofk,say K =k(I'...,n)' and P1,.'" Pnare therespective irreducible polynomials ofl' ...,nover kthen Kisalready thesplitting field ofthe finite family PI'.'" Pn' We shall investigate later when Kisthesplitting field of asingle irreducible polynomial. v, SEPARABLE EXTENSIONS 239 4. SEPARABLE EXTENSIONS Let Ebeanalgebraic extension ofafield Fand let a:F L beanembedding ofFinanalgebraically closed field L.Weinvestigate more closely extensions ofatoE.Any such extension ofamaps Eon asubfield ofLwhich isalgebraic over aF. Hence for our purposes,weshall assume that Lisalgebraic over aF,hence isequal toanalgebraic closure ofaF. Let S(Ibethe setofextensions ofatoanembedding ofEinL. Let L'beanother algebraically closed field, and let t:F L'be an embedding. We assume asbefore that L'isanalgebraic closure oftF. ByTheorem 2.8, there exists anisomorphismA.:L L'extending the map t00'-1applied tothefield aF. This isillustrated inthefollowing diagram: L' (A.L E(1. I tF ( F )aF T (1 We let S1:bethe setofembeddings ofEinL'extending t. If0'* ES(Iisanextension ofatoanembedding ofEinL,then A.00'*is anextension ofttoanembedding ofEinto L',because fortherestriction to Fwehave A.00'* =t00'-1 0a=t. Thus A.induces amapping from S(Iinto S1:' Itisclear that the inverse mapping isinduced byA.-1,and hence that S(I' S1:areinbijection under the mappIng 0'* 1---+A.00'*. Inparticular, thecardinality ofS(I' S1:isthe same. Thus this cardinality depends only ontheextension ElF, and will bedenoted by [E:F]s. We shall call ittheseparable degree ofEover F.Itismostly interesting when ElF isfinite. Theorem 4.1. Let E::JF::Jkbeatower. Then [E:k]s=[E:F]s[F: k]s. Furthermore, ifEisfinite over k,then[E:k]sisfinite and 240 ALGEBRAIC EXTENSIONS v, [E:k]s<[E:k]. The separable degree isatmost equal tothedegree. Proof. Let 0':k-+Lbeanembedding ofkinanalgebraically closed field L.Let{O'i}iel bethefamily ofdistinct extensions of(JtoF,and foreach i,let {tij}bethefamily ofdistinct extensions of(JitoE.Bywhat we saw before, each(Jihasprecisely [E:F]sextensions toembed dings ofEinL.The setof embeddings {tij}contains precisely [E:F]s[F: k]s elements. Any embedding ofEinto Lover (Jmust beone ofthetij,and thus we seethat thefirst formula holds, i.e. wehave multiplicativity intowers. Astothesecond, let usassume that Elk isfinite. Then we can obtain E asatower ofextensions, each step being generated byone element: kck(a 1)ck(a1' (2)c...ck(a 1,..., ar)=E. Ifwedefine inductively F"+1=F,,(a,,+1) then byProposition 2.7, [F,,(a"+l):F,,]s<[F,,(a"+l):1;]. Thus ourinequality istrue ineach step ofthe tower. Bymultiplicativity, it follows that theinequality istrue fortheextension Elk,aswas tobeshown. Corollary 4.2. Let Ebefiniteover k,and E::JF::Jk.Theequality [E:k]s=[E:k] holds ifand onlyifthecorresponding equality holds ineach step ofthe tower, i.e.forElF andFlk. Proof. Clear. Itwill beshown later (and itisnotdifficult toshow) that[E:k]sdivides the degree [E:k]when Eisfinite over k.We define [E:k]i tobethe quotient,sothat [E:k]s[E: k]i=[E:k]. Itthen follows from themultiplicativity oftheseparable degree and ofthe degree intowers that thesymbol [E:k]iisalso multiplicative intowers. We shall deal with itatgreater length in6. Let Ebeafinite extension ofk.We shall saythat Eisseparable over kif [E:k]s=[E:k]. An element aalgebraic over kissaid tobeseparable over kifk(a) is separable over k.We seethat this condition isequivalent tosaying that the irreducible polynomial Irr(a, k,X)has nomultiple roots. Apolynomial f(X)Ek[X] iscalled separable ifithas nomultiple roots. v, SEPARABLE EXTENSIONS 241 Ifaisaroot of aseparable polynomial g(X) Ek[X] then the irreducible polynomial ofaover kdivides gand hence rxisseparable over k. We note thatifkeF CKand aEKisseparable over k,then aisseparable over F.Indeed, iffisaseparable polynomial ink[X] such thatf(a)=0,then falso hascoefficients inF,and thus aisseparable over F .(We may saythat a separable element remain's separable under lifting.) Theorem 4.3. Let Ebeafinite extension ofk.Then Eisseparable over k ifandonlyifeach element ofEisseparable over k. Proof Assume Eisseparable over kand let aEE.We consider the tower kck(a)cE. ByCorollary 4.2, we must have [k(a): k]=[k(a): k]swhence aisseparable over k.Conversely,assume that each element ofEisseparableover k.We can write E=k(al'. ..,an) where each a;isseparable over k.We consider the tower kck(a 1)ck(a 1,(2)c." ck(a 1,...,an). Since each aiisseparable over k,each aiisseparable over k(aI'.. .,ai-I) for i>2.Hence bythe tower theorem, itfollows that Eisseparable over k. We observe that our last argument shows: IfEisgenerated by afinite number ofelements, each ofwhich isseparable over k,then Eisseparable over k. Let Ebe anarbitrary algebraic extension ofk.We define Etobe separable over kifevery finitely generated subextension isseparable over k,i.e., ifevery extension k(aI'...,an) with a1,...,anEEisseparable over k. Theorem 4.4. Let E/ be analgebraic extension ofk,generated bya family ofelements {ai}iel. Ifeach aiisseparable over kthen Eis separable over k. Proof Every element ofEliesinsome finitely generated subfield k(ai ,...,a:),1 .n and asweremarked above, each such subfield isseparable over k.Hence every element ofEisseparable over kbyTheorem 4.3, and this concludes theproof. Theorem 4.5. Separable extensions form adistinguished class ofexten- sions. 242 ALGEBRAIC EXTENSIONS v, Proof Assume that Eisseparableover kand letE::J F::Jk.Every element ofEisseparable over F,and every element ofFisanelement ofE, soseparable over k.Hence each step inthe tower isseparable. Conversely, assume that E::JF::Jkissome extension such that E/Fisseparable and F/k isseparable. IfEisfinite over k,then we can useCorollary 4.2. Namely,we have anequality oftheseparable degree and thedegree ineach step ofthetower, whence anequality forEover kbymultiplicativity. IfEisinfinite, let rxEE.Then rxisaroot ofaseparable polynomial f(X) with coefficients inF. Let these coefficients bean,..., ao. Let Fo= k(a n,..., ao).Then Foisseparable over k,and rxisseparable over Fo. We now deal with thefinite tower kcFocFo(rx) and we therefore conclude that Fo(rx) isseparableover k,hence that rx isseparable over k. This proves condition (1) inthe definition of "distinguished." Let Ebeseparable over k.Let Fbeany extension ofk,and assume that E,Fareboth subfields ofsome field. Every element ofEisseparable over k, whence separable over F.Since EF isgenerated over Fbyalltheelements ofE,itfollows that EF isseparableover F,byTheorem 4.4. This proves condition (2)inthedefinition of"distinguished," and concludes theproof of our theorem. Let Ebeafinite extension ofk.The intersection ofallnormal extensions Kofk(in analgebraic closure E8)containing Eisanormal extension ofk which contains E,and isobviously the smallest normal extension ofk containing E.If0'1' ..., O'nare thedistinct embeddings ofEinE8 ,then the extension K =(0'1E)(0'2E)...(O'nE), which isthecompositum ofallthese embeddings, isanormal extension ofk, because for any embedding ofit,say t,we canapply ttoeach extension O'iE. Then (to'1' ...,to'n) isapermutation of(0'1' ..., O'n)and thus tmaps K into itself. Any normal extension ofkcontaining Emust contain O'iE for each i,and thus thesmallest normal extension ofkcontaining Eisprecisely equal tothecompositum (0'1E)...(0'nE). IfEisseparable over k,then from Theorem 4.5 and induction we conclude that thesmallest normal extension ofkcontaining Eisalso separ- able over k. Similar results hold for aninfinite algebraic extension Eofk,taking an infinite compositum. v, SEPARABLE EXTENSIONS 243 Inlight ofTheorem 4.5, thecompositum ofallseparable extensions ofa field kinagiven algebraic closure kaisaseparable extension, which will be denoted bykSorksep ,and will becalled theseparable closure ofk.As a matter ofterminology, ifEisanalgebraic extension ofk,and (Jany embedding ofEinkaover k,then wecall (JE aconjugate ofEinka .We can say that thesmallest normal extension ofkcontaining Eisthecompositum of alltheconjugates ofEinEa. Let rxbealgebraic over k.If(Jl'..., (Jrarethedistinct embeddings ofk(rx) into kaover k,then wecall (J1rx,...,(Jrrx theconjugates of rxin/(8. These elements aresimply thedistinct roots oftheirreducible polynomial of rxover k.The smallest normal extension ofkcontaining one ofthese conjugatesis simply k((J1rx,...,(Jrrx). Theorem 4.6. (Primitive Element Theorem). Let Ebeafinite extension ofafield k.There exists anelement rxEEsuch that E=k(rx)ifandonly ifthere exists onlyafinite number offields Fsuch that kc:FeE. IfE isseparable over k,then there exists such anelement rx. Proof Ifkisfinite, then weknow that themultiplicative group ofEis generated by one element, which will therefore also generate Eover k.We assume that kisinfinite. Assume that there isonlyafinite number offields, intermediate between kand E.Let rx,pEE. As cranges over elements ofk,we canonly have afinite number offields oftype k(rx+cP). Hence there exist elements cl' C2Ekwith Cl=FC2such that k(rx+c1P)=k(rx+c2P). Note that rx+clP and rx+c2Pareinthe same field, whence sois(cl-C2)P, and hence soisp.Thus rxisalso inthat field, and we seethat k(rx,P)can be generated byone element. Proceeding inductively, ifE=k(rx l,..., rxn)then there will exist elements C2,..., CnEksuch that E=k() where =rxl+C2rx2 +...+cnrx n.This proves half ofour theorem. Conversely, assume that E=k(rx) for some rx,and letf(X)=Irr(rx, k,X). LetkeF cE.LetgF(X)=Irr(rx, F,X). Then gFdivides f.We have unique factorization inE[X], and any polynomial inE[X] which has leading coefficient 1and divides f(X) isequal toaproduct offactors (X-rxi)where at,...,anarethe roots offinafixed algebraic closure. Hence there isonlya finite number ofsuch polynomials. Thus wegetamapping FgF from the setofintermediate fields into afinite setofpolynomials. Let Fobe 244 ALGEBRAIC EXTENSIONS V,5 thesubfield ofFgenerated over kbythecoefficients ofgF(X). Then gFhas coefficients inFoand isirreducible over Fosince itisirreducible over F. Hence thedegree of rxover Foisthe same asthedegree of rxover F.Hence F=Fo. Thus our field Fisuniquely determined byitsassociated poly- nomials gF, and our mapping istherefore injective. This proves the first assertion ofthetheorem. As tothe statement concerning separable extensions, using induction, wemay assume without loss ofgenerality that E=k(rx,P)where rx,pare separable over k.Let0'1'''.' O'nbethedistinct embeddings ofk(rx,P)inka over k.Let P(X)=n(O'irx+XO'iP- O'jrx-XO'jP).ij Then P(X) isnot the zero polynomial, and hence there exists CEksuch that P(c) =FO.Then theelements O'i(rx +cP)(i=1,...,n)aredistinct, whence k(rx+cP) hasdegree atleast nover k.But n=[k(rx, P):k],and hence k(rx,P)=k(rx+cP), asdesired. IfE=k(rx), then wesaythat rxisaprimitive element ofE(over k). 5. FINITE FIELDS We have developed enough general theorems todescribe the structure of finite fields. This isinteresting foritsown sake, and also gives usexamples forthegeneral theory. Let Fbe afinite field with qelements. As wehave noted previously,we have ahomomorphism Z-+F sending1on 1,whose kernel cannot be0,and hence isaprincipal ideal generated byaprime number psince Z/pZ isembedded inFand Fhas no divisors ofzero. Thus Fhascharacteristic p,and contains afield isomorphic toZ/pZ. We remark that Z/pZ has noautomorphisms other than theidentity. Indeed, any automorphism must map1on 1,hence leaves every element fixed because 1generates Z/pZ additively. Weidentify Z/pZ with itsimage inF.Then Fisavector space over Z/pZ, and this vector space must be V,5FINITE FIELDS 245 finite since Fisfinite. Let itsdegree be n.Let W1,..., W"be abasis forF over Z/pZ. Every element ofFhas aunique expression oftheform alWI+···+a"w" with aiEZ/pZ. Hence q=p". Themultiplicative group F*ofFhas order q-1.EveryaEF*satisfies theequation xq-t =1.Hence every element ofFsatisfies theequation f(X)=xq-X =o. This implies that thepolynomial f(X) has qdistinct roots inF,namely all elements ofF.Hencefsplits into factors ofdegree1inF,namely xq-X =n(X-a). a.eF Inparticular, Fisasplitting field forf.But asplitting field isuniquely determined uptoanisomorphism. Hence ifafinite field oforder p"exists, it isuniquely determined, up to anisomorphism,asthesplitting field of Xp" -Xover Z/pZ. As amatter ofnotation, wedenote Z/pZ byFp.Let nbeaninteger>1 and consider thesplitting field of Xp"-X=f(X) inanalgebraic closure F;.We contend that thissplitting field isthe setof roots off(X) inF;.Indeed, leta,pberoots. Then (a+P)P"-(a+P)=aP"+PP"-a-p=0, whence a+pisaroot. Also, (aP)P"-ap=aP"pP"-ap=ap-ap=0, and apisaroot. Note that 0,1are roots off(X). IfP#=0then (P-1 )P"_p-1=(Pp")-1_p-l=0 sothatp-t isaroot. Finally, (-p)P"-(-13)=(-I)P"pP" +p. Ifpisodd, then(-I)P"=-1and we seethat-pisaroot. Ifpiseven then -1=1(inZ/2Z) and hence-p=13isaroot. This proves our contention. The derivative off(X) is f'(X)=p"XP"-1-1=-1. Hence f(X) has nomultiple roots, and therefore has p"distinct roots in F;.Hence itssplitting field has exactly p"elements. We summarize our results: 246 ALGEBRAIC EXTENSIONS V,5 Theorem 5.1. For each prime pand each integer n>1there exists afinite field oforder p"denoted byFpn, uniquely determined asasubfield ofan algebraic closure F;.Itisthesplitting field ofthepolynomial Xpn-X, and itselements are the roots ofthis polynomial. Every finite field is isomorphic toexactly onefield Fpn. Weusually write p"=qandFqinstead ofFpn. Corollary 5.2. LetFqbe afinite field. Let nbe aninteger>1.In a given algebraic closure F:,there exists one andonly one extension ofFqof degree n,and this extension isthefield Fqn. Proof Let q=pm. Then q"=pm". The splitting field ofxqn-Xis precisely Fpmnand has degreemn over ZjpZ. SinceFqhas degree mover ZjpZ, itfollows thatFqnhasdegreenoverFq.Conversely, any extension of degreenoverFqhasdegreemn overFpand hence must beFpmn.This proves ourcorollary. Theorem 5.3. Themultiplicative group ofafinite field iscyclic. Proof. This hasalready been proved inChapter IV,Theorem 1.9. We shall determine allautomorphisms ofafinite field. Let q=pnand letFqbethefinite field with qelements. We consider the Frobenius mapping cp:Fq-+Fq such that cp(x)=xp .Then cpisahomomorphism, and itskernel is0sinceFq isafield. Hence cpisinjective. SinceFqisfinite, itfollows thatcpis surjective, and hence thatcpisanisomorphism. We note that itleavesFp fixed. Theorem 5.4. The group ofautomorphisms ofFqiscyclic ofdegree n, generated by cpo Proof. Let Gbethe group generated by cpo We note that cpn=id because cp"(x)=xpn=xforallxEFq.Hence nisanexponent forcpoLet d betheperiod ofcp,sod>1.We have cpd(X)=xPdforallxEFq.Hence each xEFqisaroot oftheequation Xpd -X=o. This equation has atmost pdroots. Itfollows that d>n,whence d=n. There remains tobeproved that Gisthegroup ofallautomorphisms of Fq. Any automorphism ofFqmust leaveFpfixed. Hence itisanauto- V,6 INSEPARABLE EXTENSIONS 247 morphism ofFqoverFp. By Theorem 4.1, the number ofsuch auto- morphisms is<n.HenceFqcannot have any other automorphisms except forthose ofG. Theorem 5.5. Let m, nbeintegers>1.Then inanyalgebraic closure of FP'thesubfield Fp'Iiscontained inFp'"ifandonlyifndivides m.Ifthat isthe case, letq=pn, and letm=nd. Then Fp'"isnormal andseparable over Fq' and thegroup ofautomorphisnls ofF pmover Fqiscyclic oforder d,generated bycpn. Proof. Allthe statements aretrivial consequences ofwhat hasalready been proved andwill beleft tothereader. 6. INSEPARABLE EXTENSIONS This section isofafairly technical nature, and can beomitted without impairing theunderstanding ofmost ofthe rest ofthebook. Webegin with some remarks supplementing those ofProposition 2.7. Letf(X)=(X-)mg(X) be apolynomial ink[X], and assume X- does not divide g(X). We recall that miscalled themultiplicity of inf Weaythat isamultiple root offifm>1.Otherwise, wesaythat isa simple root. Proposition 6.1. Let bealgebraic over k, Ek8 ,and let f(X)=Irr(, k,X). Ifchar k=0,then allroots offhave multiplicity1(fisseparable). If char k=p>0, then there exists aninteger Jl>0such that every rootoffhasmultiplicity pll. Wehave [k(): k]=pll[k(): k]s, andp/lisseparable over k. Proof Let l'...,rbethedistinct roots offinkaand let=1.Let mbethemultiplicity of inf.Given 1<i<r,there exists anisomorphism (1:k()-.k(i) over ksuch that (1 =i.Extend (1toanautomorphism ofk8and denote 248 ALGEBRAIC EXTENSIONS V,6 this extension also by(1.Sincefhas coefficients inkwehave f(1=fWe note that r f(X)=IT(X-ua)mj j=1 ifmjisthemultiplicity ofajinf.Byunique factorization, weconclude that mi=m1and hence that allmiareequal tothe same integerm. Consider thederivative f'(X). Iffandf'have aroot incommon, then rx isaroot of apolynomial oflower degree than degf.This isimpossible unless degf'=-00, inother words, f'isidentically O.Ifthecharacteristic is0,this cannot happen. Hence iffhasmultiple roots, we areincharacteris- ticp,andf(X)=g(XP) for some polynomial g(X)Ek[X]. Therefore rxPisa root ofapolynomial gwhose degree is<degfProceeding inductively, we take the smallest integer Jl>0such that rxp/l isthe root of aseparable polynomial ink[X], namely thepolynomial hsuch that f(X)=h(XP"). Comparing thedegree offand g,weconclude that [k(rx):k(rxP)]=p. Inductively, wefind [k(rx):k(rxP")]=pll. Since hhas roots ofmultiplicity 1,weknow that [k(rxp/l): k]s=[k(rxp/l): k], and comparing thedegree offand thedegree ofh,we seethat the num- ber ofdistinct roots offisequal tothe number ofdistinct roots ofh. Hence [k(rx):k]s=[k(rxP"):k]s. From this our formula forthedegree follows bymultiplicativity, and our proposition isproved. We note that the roots ofhare p/l p/lrx1,..., rxr· Corollary 6.2. For anyfinite extension Eofk,theseparable degree [E:k]sdivides thedegree [E:k]. The quotient is1ifthecharacteristic is 0,and apower ofpifthecharacteristic isp>O. Proof Wedecompose Elk into atower, each step being generated by one element, and apply Proposition 6.1,together with themultiplicativity of our indices intowers. IfElK isfinite, wecall thequotient V,6 INSEPARABLE EXTENSIONS 249 [E:k] [E:k]s theinseparable degree (ordegree ofinseparability), and denote itby[E:k]ias in4. We have [E:k]s[E:k]i=[E:k]. Corollary 6.3. Afinite extension isseparable ifandonlyif[E:k]i=1. Proof Bydefinition. Corollary 6.4IfE=>F=>kare twofinite extensions, then [E:k]i=[E:F]i[F:k]i' Proof. Immediate byTheorem 4.1. We now assume throughout that kisafield ofcharacteristic p>O. Anelement rxalgebraic over kissaid tobepurely inseparable over kif there exists anintegern>0such that rxp"liesink. Let Ebe analgebraic extension ofk.We contend that thefollowing conditions areequivalent: P.Ins. 1.We have [E:k]s=1. P.Ins. 2.Every element rxofEispurely inseparable over k. P.Ins. 3. ForeveryrxEE,theirreducible equation of rxover kisoftype Xp" -a=0with some n>0and aEk. P.Ins. 4.There exists asetofgenerators {rxi}ie1ofEover ksuch that each rxiispurely inseparable over k. Toprove theequivalence, assume P.Ins. 1.Let rxEE.ByTheorem 4.1, weconclude that [k(rx): k]s=1.Letf(X)=Irr(rx, k,X). Thenfhasonly one root since [k(rx):k]s isequal tothe number ofdistinct roots off(X). Let m=[k(rx): k]. Then degf=m,and the factorization offover k(rx) isf(X)=(X-rx)m. Write m=pnrwhere risaninteger prime top.Then f(X)=(XP"-rxP"t =xp"r-rrxP"Xp"(r-l) +lower terms. Since thecoefficients off(X) lieink,itfollows that rrxp" 250 ALGEBRAIC EXTENSIONS V,6 lies ink,and since r=F0(ink),then rxP"lies ink.Let a=rxp".Then (Xis aroot ofthepolynomial XP"-a,which divides f(X). Itfollows that f(X)=XP"-a. Essentially the same argumentasthepreceding one shows that P.Ins. 2 implies P.Ins. 3.Itistrivial that thethird condition implies thefourth. Finally, assume P.Ins. 4.Let Ebe anextension generated bypurely inseparable elements rxi(iEI).Any embedding ofEover kmaps rxion aroot of h(X)=Irr(rx i,k,X). Buth(X) divides some polynomial Xp" -a,which hasonly one root. Hence anyembedding ofEover kistheidentity oneach (Xi'whence theidentity on E,and weconclude that[E:k]s=1,asdesired. An extension satisfying the above four properties will becalled purely inseparable. Proposition 6.5. Purely inseparable extensions form adistinguished class ofextensions. Proof The tower theorem isclear from Theorem 4.1, and thelifting property isclear from condition P.Ins. 4. Proposition 6.6. Let Ebe analgebraic extension ofk.Let Eo bethe compositum ofallsubfields FofEsuch that F::J kand Fisseparable over k.Then Eo isseparable over k,and Eispurely inseparable over Eo. Proof Since separable extensions form adistinguished class, weknow that Eoisseparable over k.Infact, Eoconsists ofallelements ofEwhich areseparable over k.ByProposition 6.1,givenrxEEthere exists apower of p,say pnsuch that rxP"isseparable over k.Hence Eispurely inseparable over Eo,aswas tobeshown. Corollary 6.7.If analgebraic extension Eofkisboth separable and purely inseparable, then E=k. Proof Obvious. Corollary 6.8. Let Kbenormal over kand letKo beitsmaximal separa- blesubextension. Then Ko isalso normal over k. Proof Let (Jbeanembedding ofKoinK8over kand extend (Jtoan embedding ofK.Then (Jisanautomorphism ofK.Furthermore, (JKois separable over k,hence iscontained inKo, since Ko isthemaximal separa- blesubfield. Hence (JKo=Ko,ascontended. V,6 INSEPARABLE EXTENSIONS 251 Corollary 6.9. Let E,Fbetwofinite extensions ofk,and assume that Elk isseparable, Flk ispurely inseparable. Assume E,Fare subfields ofa common field. Then [EF: F]=[E:k]=[EF: k]s, [EF: E]=[F:k]=[EF: k]i. Proof. Thepicture isasfollows: EF p;/ E Fks The proof isatrivial juggling ofindices, using thecorollaries ofProposition 6.1. We leave itasanexercise. Corollary 6.10. Let EPdenote thefield ofallelements xP ,xEE.Let E beafinite extension ofk.IfEPk =E,then Eisseparable over k.IfEis separable over k,then EP"k =Eforall n>1. Proof. LetEobethemaximal separable subfield ofE.Assume EPk =E. Let E=k(rx 1,..., rxn).Since Eispurely inseparable over Eothere exists m such that rxr'"EEofor each i=1,..., n.Hence EP'" cEo. But EP'"k =E whence E=Eoisseparable over k.Conversely, assume that Eisseparable over k.Then Eisseparableover EPk. Since Eisalso purely inseparable over EPk weconclude that E=EPk. SimilarlywegetE=Epnkforn:>1,aswas tobeshown. Proposition 6.6shows that any algebraic extension can bedecomposed into atower consisting ofamaximal separable subextension and apurely inseparable step above it.Usually, one cannot reverse the order ofthe tower. However, there isanimportantcase when itcan bedone. Proposition 6.11. LetKbenormal over k.LetGbeitsgroup ofautomorphisms over k.Let[(Gbethefixed field ofG (see Chapter VI,1). Then KG ispurely inseparableover k,andKisseparableover KG.IfKoisthemaximal separa- blesubextension ofK,then K==KGKoand KonKG==k. Proof Let rxEKG. Let tbe anembedding ofk(rx) over kinKaand extend ttoanembedding ofK,which wedenote also byt.Then tisan automorphism ofKbecause Kisnormal over k.Bydefinition, trx =rxand hence tistheidentity onk(rx). Hence [k(rx): k]s=1and rxispurely in- separable. Thus KG ispurely inseparable over k.The intersection ofKo 252 ALGEBRAIC EXTENSIONS V,6 and KG isboth separable andpurely inseparableover k,and hence isequal tok. Toprove that Kisseparableover KG, assume first that Kisfinite over k,and hence that Gisfinite, byTheorem 4.1. Let rxEK.Let 0'1' ..., O'rbe a maximal subset ofelements ofGsuch that theelements 0'1rx,..., O'rrx aredistinct, and such that a}istheidentity, and rxisaroot ofthepolynomial r f(X)=n(X-O'irx). i=1 For any tEGwenote that fT:=fbecause tpermutes the roots. We note thatfisseparable, and that itscoefficients areinthefixed field KG. Hence rx isseparable over KG. The reduction oftheinfinite case tothefinite case is done byobserving that everyrxEKiscontained in some finite normal subextension ofK.We leave thedetails tothereader. We now have thefollowing picture: K KKG/0,,- Ko "-KG KonKG=k ByProposition 6.6,Kispurely inseparable over Ko, hence purely insepara- ble over KoKG. Furthermore, Kisseparable over KG, hence separableover KoKG. Hence K=KoKG, thereby provingourproposition. We seethat every normal extension decomposes into acompositum of apurely inseparable and aseparable extension. We shall define aGalois ex- tension inthe next chapter tobe anormal separable extension. Then Ko isGalois over kand thenormal extension isdecomposed into aGalois and a purely inseparable extension. The group Giscalled theGalois group ofthe extension Klk. Afield kiscalled perfect ifkP=k.(Every field ofcharacteristic zero is also called perfect.) Corollary 6.12. Ifkisperfect, then every algebraic extension ofkis separable, and every algebraic extension ofkisperfect. Proof Every finite algebraic extension iscontained inanormal exten- sion, and weapply Proposition 6.11 togetwhat wewant. V,Ex EXERCISES 253 EXERCISES 1.Let E=Q(), where isaroot oftheequation 3+2+ +2=0, Express (2+ +1)(2 +)and(-1)-1 intheform a2 +b+c with a,b,CEQ. 2.Let E=F() where isalgebraicover F,ofodd degree. Show that E=F(2). 3.Let and f3betwo elements which arealgebraic over F.Letf(X)=Irr(, F,X) andg(X)=Irr(f3, F,X). Suppose that degfand deg garerelatively prime. Show that gisirreducible inthepolynomial ringF()[X]. 4,Let bethe real positive fourth root of2.Find allintermediate fields inthe extension Q() ofQ. 5.If isacomplex root ofX6+X3+1,find allhomomorphisms a:Q()-+c. [Hint: The polynomial isafactor ofX9-1.] 6.Show that.j2+J3isalgebraic over Q,ofdegree 4. 7.Let E,Fbetwo finite extensions ofafield k,contained inalarger field K.Show that [EF: k]<[E:k][F:k]. If[E:k]and[F:kJarerelatively prime, show that one has anequality sign in theabove relation. 8.Letf(X) Ek[X] be apolynomial ofdegreen.Let Kbeitssplitting field. Show that[K:k] divides n! 9.Find thesplitting field ofXp8 -lover thefield Z/pZ. 10.Let beareal number such that4 =5. (a)Show thatQ(i2) isnormal over Q. (b)Show thatQ( +i)isnormal overQ(i2). (c)Show thatQ( +i)isnot normal over Q. 11.Describe thesplitting fields ofthefollowing polynomialsover Q,and find the degree ofeach such splitting field. (a)X2-2 (b)X2-1 (c)X3-2 (d)(X3-2)(X2-2) (e)X2+X+1 (f)X6+X3+1 (g)X5-7 12.Let Kbe afinite field with pnelements. Show that every element ofKhas a unique p-th root inK. 254 ALGEBRAIC EXTENSIONS V,Ex 13.Ifthe roots ofamonic polynomial f(X) Ek[X] insome splitting field aredistinct, and form afield, then char k=pandf(X)=Xpn-Xfor some n>1. 14.Let char K =p.LetLbe afinite extension ofK,and suppose [L:K]prime to p.Show that Lisseparableover K. 15.Suppose char K =p.Let aEK.Ifahas nop-th root inK,show that Xp" -ais irreducible inK[X] forallpositive integersn. 16.Let char K =p.Let (1bealgebraic over K.Show that (1isseparable ifandonly ifK((1)=K((1P") forallpositive integersn. 17.Prove that thefollowing twopropertiesareequivalent: (a)Every algebraic extension ofKisseparable. (b)Either char K =0,orchar K=pand every element ofKhas ap-th root in K. 18.Show that every element ofafinite field can bewritten asasum oftwo squares inthat field. 19.Let Ebe analgebraic extension ofF.Show that every subring ofEwhich contains Fisactuallyafield, Isthis necessarily true ifEisnotalgebraic over F? Prove orgiveacounterexample. 20,(a)Let E=F(x) where xistranscendental over F.LetK:FFbe asubfield ofE which contains F.Show that xisalgebraic over K. (b)Let E=F(x). Let y=f(x)jg(x) be arational function, with relatively prime polynomials f,gEF[x]. Let n=max(deg f,degg).Supposen>1.Prove that [F(x):F(y)]=n. 21. Let Z+ bethe setofpositive integers, and Aanadditive abelian group. Let f:Z+ Aand g:Z+ Abemaps. Suppose that foralln, f(n)=Lg(d). din LetJ.1.betheMobius function (cf.Exercise 12ofChapter II). Prove that g(n)=LJ.1.(njd)f(d). din 22. Let kbe afinite field with qelements. Letf(X)Ek[X] beirreducible. Show that f(X) divides xq"-Xifand only ifdegfdivides n.Show themultiplication form ula xq"-X=f1f1fd(X), dinfdirr where the inner product isover allirreducible polynomials ofdegree dwith leading coefficient 1.Counting degrees, show that qn=Ldt/J(d), din where t/J(d)ISthe number ofirreducible polynomials ofdegree d.Invert by V,Ex EXERCISES 255 Exercise 21and find that nt/J(n)=LJ.l(d)qn/d. din 23.(a)Let kbe afinite field with qelements. Define the zeta function Z(t)=(I-t)-lrI(1-tdegp)-l, P where pranges over allirreducible polynomials p=p(X) ink[X] with leading coefficient 1.Prove that Z(t) isarational function and determine this rational function. (b)Let 1tq(n)bethenumber ofprimes pasin(a)ofdegree<n.Prove that qqm 1t(m)'" - qq-Imfor m 00. Remark. This istheanalogue oftheprime number theorem innumber theory, but itisessentially trivial inthepresent case, because the Riemann hypothesis is trivially verified. Things get more interesting fast after this case. Consider an equation y2=x3+ax+bover afinite fieldFqofcharacteristic :F2,3,and having qelements. Assume-4a3-27b2:F0,inwhich case the curve defined by thisequation iscalled anelliptic curve. Define Nnby Nn-I=number ofpoints (x,y)satisfying theabove equation with x,yEFq" (the extension ofFqofdegree n). Define the zeta function Z(t) tobetheunique rational function such that Z(O)=I and Z'/Z(t)=LNntn-1 . Afamous theorem ofHasse asserts that Z(t) isarational function oftheform (I-t)(l-t)Z(t)=, (1-t)(1-qt) where isanimaginary quadratic number (not real, quadratic over Q), isits complex conjugate, and =q,soII=ql/2. SeeHasse, "Abstrakte Bergrundung der komplexen Multiplikation und Riemannsche Vermutung inFunktionen- korpern," Abh. Math. Sem. Univ. Hamburg 10(1934) pp.325-348. 24.Let kbe afield ofcharacteristic pand lett,ubealgebraically independentover k.Prove thefollowing: (a)k(t,u)hasdegree p2over k(tP ,uP). (b)There exist infinitely many extensions between k(t,u)andk(tP ,uP). 25. Let Ebe afinite extension ofkand letpr=[E:k]i. We assume that the characteristic isp>o.Assume that there isnoexponent pSwith s<rsuch that Epsk isseparableover k(i.e., such that pS isseparableover kfor each inE), Show that Ecan begenerated by one element over k.[Hint: Assume first that Eispurely inseparable.] 256 ALGEBRAIC EXTENSIONS V,Ex 26. Let kbe afield, f(X)anirreducible polynomial ink[X], and letKbe afinite normal extension ofk.Ifg,haremonic irreducible factors off(X) inK[X], show that there exists anautomorphismuofKover ksuch that 9=her.Give anexample when this conclusion isnotvalid ifKisnotnormal over k. 27.LetXl' ..., XIIbealgebraically independent over afield k.Let ybealgebraic over k(x)=k(x l,...,XII). LetP(X II+I)bethe irreducible polynomial ofyover k(x). Let <p(x) bethe least common multiple ofthedenominators ofthecoefficients of P.Then thecoefficients of<p(x)P areelements ofk[x]. Show that thepolynomial f(X I'...,XII +I)=<p(X I'...,XII)P(X II+I) isirreducible over k,asapolynomial inn+1variables. Conversely, letf(X I'...,XII+I)be anirreducible polynomialover k.Let XI'...,XIIbealgebraically independentover k.Show that f(x l,.. .,XII'Xn+1) isirreducible over k(x l,...,XII). Iffisapolynomial in nvariables, and (b)=(bl,...,bll)isann-tuple of elements such thatf(b)=0,then wesaythat (b)isazero off.We saythat (b)is non-trivial ifnotallcoordinates biareequal too. 28.Letf(X l'...,XII) be ahomogeneous polynomial ofdegree 2(resp. 3)over afield k.Show that iffhas anon-trivial zero inanextension ofodd degree (resp. degree 2)over k,thenfhas anon-trivial zero ink. 29.Letf(X, Y)beanirreducible polynomial intwo variables over afield k.Let tbe transcendental over k,and assume that there exist integers m, n:F0and elements a,bEk,ab:F0,such thatf(at", btm)=o.Show that after inverting possibly Xor and uptoaconstant factor,fisoftype Xmy"-C with some CEk. The answer tothefollowing exercise isnotknown. 30.(Artin conjecture). Letfbe ahomogeneous polynomial ofdegree din nvari- ables, with rational coefficients. Ifn>d,show that there exists aroot ofunity', and elements XI' ..., XIIEQ[,] notall0such thatf(x l,...,XII)=O. 31.Difference equations. Let UI'..., Udbeelements ofafield K.We want tosolve forinfinite vectors (xo,XI'.. .,XII'...)satisfying (*) XII=UIXII-I +...+UdXII-d for n>d. Define thecharacteristic polynomial ofthesystem tobe Xd-(UIXd-1+...+ud)=f(X). V,Ex EXERCISES 257 Supposecxisaroot off. (a)Show that XII=cx"(n>0)isasolution of(*). (b)Show that the setofsolutions of(*)isavector space ofdimension d. (c)Assume that the characteristic polynomial has ddistinct roots cxl,..., CXd. Show that thesolutions (cx), ...,(cx;) form abasis forthespace ofsolutions. (d)Let XII=blcx+...+bdcx; for n>0,show how tosolve forbl,..., bdinterms of CXI,..., CXdand Xo, ...,Xd-l. (Use theVandermonde determinant.) (e)Under theconditions of(d),letF(T)=LXIIT". Show that F(T) representsa rational function, and give itspartial fraction decomposition. 32.Let d=2forsimplicity. Given ao,ai' u,v,w,tEK,wewant tofind thesolutions ofthesystem all=uall-1-vtall-2-t"w for n>2. Let CXI,CX2bethe root ofthecharacteristic polynomial, that is 1-uX+vtX2=(I-cxIX)(1-cx2X). Assume thatCXI,CX2aredistinct, and also distinct fromt.Let 00 F(X)=LallX". 11=0 (a)Show that there exist elements A,B,CofKsuch that ABC F(X)= + + . l-cxlX l-cx2Xl-tX (b)Show that there isaunique solution tothedifference equation given by all=Acx +Bcx; +Ct" for n>o. (Tosee anapplication ofthis formalism tomodular forms, asinthework of Manin, Mazur, and Swinnerton-Dyer, cf.myIntroduction toModular Forms, Springer-Verlag, New York, 1976, Chapter XII,2.) 33.Let Rbearing which we assume entire forsimplicity. Let g(T)=Td-ad-l Td-l- ... -ao beapolynomial inR[T], and consider theequation Td=ao+aIT+...+ad-l Td-l . Let Xbearoot ofg(T). (a)For anyintegern>dthere isarelation X" =aO,1I+al,lIx +...+ad_I,lIxd-1 with coefficientsai,jinZ[ao,...,ad-I]cR. (b)LetF(T)ER[T] beapolynomial. Then F(x)=ao(F) +al(F)x +...+ad-l(F)Xd-1 where thecoefficients ai(F) lieinRanddepend linearlyonF. 258 ALGEBRAIC EXTENSIONS V,Ex (c)LettheVandermonde determinant be 1Xl 1 X2V(x 1,...,Xd)="-1Xl d-lX2=fl(Xj-Xi). i<j 1Xdd-lXd Suppose that theequation g(T)=0has droots and that there isafactoriza- tion d g(T)=fl(T-Xi). i=1 Substituting Xifor Xwith i=1,...,dandusing Cramer's rule ontheresulting system oflinear equations, yields aj(F)=Aj(F) where istheVandermonde determinant, andAlF)isobtained byreplacing thej-th column byt(F(x1),...,F(x,,)),so 1Xl F(x 1) 1X2 F(x 2) Aj(F)=d-lXl d-lX2 1 Xd F(x d)d-lXd IfA#0then we can write aiF)=iF)/. Remark. IfF(T) isapower series inR[[T]] andifRisacomplete local ring, with Xl' ..., Xdinthemaximal ideal, and x=Xifor some i,then we can evaluate F(x) because the series converges. The above formula forthe coefficientsaj(F) remains valid. 34.LetXl' ..., Xdbeindependent variables, and letAbethering d Q[[Xl,..., Xd]][T]/fl (T-Xi). i=l Substituting some XiforTinduces anatural homomorphism qJiofAonto Q[[Zl, ...,Xd]]=R, and themapZi-+(CPl(z),...,tpd(Z») gives anembedding ofAinto theproduct ofR with itself dtimes. Let kbeaninteger, and consider theformal power series d(T-x.)eT-Xid F(T)=ekTfl T-xI=ekTflh(T-Xi) i=l ei-I i=l where h(t)=tet/(et-1).Itisaformal power series inT,T-Xl' ..., T-Xd. Under substitution ofsomeXjforTitbecomes apower series inXjandXj-Xi' and thus converges inQ[[Xl, ..., xd]]. V,Ex EXERCISES 259 (a)Verify that d F(T)=ao(F) +...+ad_1(F)Td-1modn(T-Xi) i=l where ao(F), ..., ad-1(F) EQ[[Xb ..., xd]],and that theformula given inthe preceding exercise for these coefficients interms ofVandermonde determi- nants isvalid. (b)Show that ad-1(F)=0if-(d-1)<k<0andad-l(F)=1ifk=O. Remark. The assertion in(a)isasimple limit. The assertion in(b)isafact which has been used intheproof oftheHirzebruch-Grothendieck-Riemann- Roch theorem and asfar asIknow there was nosimple known proof until Roger Howe pointed outthat itcould bedone bytheformula ofthepreceding exercise asfollows. We have 1Xld-2F(Xl) Xl V(x 1,...,xn)ad-l (F)= 1 Xdd-2F(Xd)Xd Furthermore, F(x.)=ekxjn(Xj-xn)eXj-Xn .J n:#:jeXj-Xn -1 We usetheinductive relation ofVandermonde determinants V(xl'...,Xd)=V(xl'...,j'.. .,Xd)(-1)d-jn(xj-xn). ":#:j Weexpand thedeterminant for ad-1(F)according tothelast column toget d1 ad-1(F)=Le(k+d-l)xj n x x. j=l n:#:jej-en Using theinductive relation backward, and replacing XibyeXiwhich wedenote byYifortypographical reasons, weget 1Yld-2y+d-l Yl d-2y;+d-l YdV(Yl' ...,Yd)ad-l (F)= 1Yd Ifk:F0then two columns ontheright arethe same, sothedeterminant isO.If k=0then weget the Vandermonde determinant ontheright,soad-l(F)=1. This proves thedesired value. CHAPTER VI Galois Theory This chapter contains the core ofGalois theory.We study the group of automorphisms ofafinite (and sometimes infinite) Galois extension atlength, andgive examples, such ascyclotomic extensions, abelian extensions, and even non-abelian ones, leading into thestudy ofmatrix representations oftheGalois group and their classifications .We shall mention anumber offundamental unsolved problems, the most notable ofwhich iswhether givenafinite group G,there exists aGalois extension ofQhaving this groupasGalois group. Three surveys give recent points ofview onthose questions and sizeable bibliographies: B,MATZA T,Konstruktive Galoistheorie, Springer Lecture Notes 1284, 1987 B.MATZA T,Uber dasUmkehrproblem derGaloisschen Theorie, lahrsbericht Deutsch. Mat.-Verein. 90(1988), pp, 155-183 J.P.SERRE, Topics inGalois theory, course atHarvard, 1989, Jones and Bartlett, Boston 1992 More specific references will begiven inthe text attheappropriate moment concerning this problem and theproblem ofdetermining Galois groups over specific fields, especially therational numbers. 1. GALOIS EXTENSIONS LetKbeafield and letGbeagroup ofautomorphisms ofK.Wedenote byKGthesubset ofKconsisting ofallelements xEKsuch that x(J =xforall aEG.Itisalso called thefixed field ofG.Itisafield because ifx,YEKGthen (x+y)(J=x(J+y(J=x+Y 261 262 GALOIS THEORY VI,1 forall (JEG,andsimilarly, one verifies that Kisclosed under multiplication, subtraction, andmultiplicative inverse. Furthermore, KGcontains 0and 1, hence contains theprime field. Analgebraic extension Kofafield kiscalled Galois ifitisnormal and separable. Weconsider Kasembedded inanalgebraic closure. The group of automorphisms ofKover kiscalled theGalois group ofKover k,and isdenoted byG(K/k), GK1k ,Gal(K/k), orsimply G.Itcoincides with the setofembeddings ofKinJ(8-over k. For theconvenience ofthereader, weshall now state themain result ofthe Galois theory forfinite Galois extensions. Theorem 1.1. LetKbeafinite Galois extension ofk,with Galois group G. There isabijection between the setofsubfields EofKcontaining k,and the setofsubgroups HofG, given byE=KH .Thefield EisGalois over kifand onlyifHisnormal inG,andifthat isthecase, then themap(J1---+ (JIEinduces anisomorphism ofGjHonto theGalois group ofEover k. Weshall give theproofs stepbystep, and asfar aspossible,wegive them for infinite extensions. Theorem 1.2. LetKbeaGalois extension ofk.Let GbeitsGalois group. Then k=KG.IfFisanintermediate field, keF cK,then KisGalois over F.The map F1---+G(KjF) from the setofintermediate fields into the setofsubgroups ofGisinjective. Proof Let rxEKG. Let (Jbeanyembedding ofk(rx) inK8 ,inducing the identity onk.Extend (Jtoanembedding ofKinto K8 ,and callthis extension (J also. Then (Jisanautomorphism ofKover k,hence isanelement ofG.By assumption,(Jleaves rxfixed. Therefore [k(rx):k]s=1. Since rxisseparableover k,wehave k(rx)=kand rxisanelement ofk.This proves ourfirst assertion. LetFbe anintermediate field. Then Kisnormal andseparable over Fby Theorem 3.4 and Theorem 4.5ofChapter V.Hence KisGalois over F.IfH= G(K/F) then bywhat weproved above weconclude that F=KH .IfF,F' are intermediate fields, andH=G(K/F), H'=G(K/F'), then F=KHand F' =KH'. IfH =H'weconclude that F=F',whence our map F1---+G(KjF) isinjective, thereby provingour theorem. VI, 1 GALOIS EXTENSIONS 263 Weshall sometimes call thegroup G(K/F) ofanintermediate field thegroup associated with F.We saythat asubgroup HofGbelongs toanintermediate field FifH =G(K/F). Corollary 1.3. LetK/k beGalois with group G.Let F,F'betwo inter- mediate fields, and letH,H'bethesubgroups ofGbelonging toF,F'respec- tively. Then HnH'belongs toFF'. Proof Every element ofHnH'leaves FF'fixed, and every element ofG which leaves FF'fixed also leaves Fand F'fixed and hence lies inHnH'. This proves our assertion. Corollary 1.4. Let thenotation beasinCorollary 1.3.Thefixedfield ofthe smallest subgroup ofGcontaining H,H'isFnF'. Proof Obvious. Corollary 1.5. Let the notation be asinCorollary 1.3. Then FcF'if andonlyifH'cH. Proof IfFcF'and (JEH'leaves F'fixed then (Jleaves Ffixed, so (Jlies inH.Conversely, ifH'cHthen thefixed field ofHiscontained inthefixed field ofH', soFcF'. Corollary 1.6. Let Ebeafinite separable extension ofafield k.LetKbe thesmallest normal extension ofkcontaining E.Then Kisfinite Galois over k.There isonlyafinite number ofintermediate fields Fsuch thatkeF cE. Proof We know that Kisnormal and separable, and Kisfinite over k since we saw that itisthefinite compositum ofthefinite number ofconjugates ofE.The Galois group ofK/k hasonlyafinite number ofsubgroups. Hence there isonlyafinite number ofsubfields ofKcontaining k,whence afortioria finite number ofsubfields ofEcontaining k. Ofcourse, thelast assertion ofCorollary 1.6hasbeen proved inthepreceding chapter, but wegetanother proof here from another point ofview. Lemma 1.7. Let Ebeanalgebraic separable extension ofk.Assume that there isanintegern>1such that every element rxofEisofdegree<nover k. Then Eisfinite over kand[E:k]<n. Proof Let rxbeanelement ofEsuch that thedegree [k(rx):k]ismaximal, say m<n.Wecontend that k(rx)=E.Ifthis isnottrue, then there exists an element pEE such that prtk(rx), and bytheprimitive element theorem, there exists anelementYEk(rx,P)such that k(rx,P)=key). But from thetower kck(rx)ck(rx,P) we seethat [k(rx, P):k]>mwhence yhasdegree> mover k,contradiction. 264 GALOIS THEORY VI, 1 Theorem 1.8. (Artin). LetKbeafield and letGbeafinite group ofauto- morphisms ofK,oforder n.Let k=KGbethefixed field. Then Kisafinite Galois extension ofk,and itsGalois group isG.Wehave[K:k]=n. Proof. Let rxEKand letU1, ..., Urbeamaximal setofelements ofGsuch that U1rx,..., Urrxare distinct. If! EGthen (!U 1rx,...,!Urrx)differs from (U1rx,. ..,urrx) byapermutation, because! isinjective, and every !Uirx isamong the set{U1rx,..., urrx}; otherwise this setisnotmaximal. Hence rxisaroot of thepolynomial r f{X)=n(X-Uirx), i= 1 and forany!EG,ff=f.Hence thecoefficients offlieinKG =k.Further- more,fisseparable. Hence every element rxofKisaroot ofaseparable polynomial ofdegree<nwith coefficients ink.Furthermore, this poly- nomial splits inlinear factors inK. Hence Kisseparable over k,isnormal over k,hence Galois over k.ByLemma 1.7, wehave [K:k]<n.The Galois group ofKover khasorder<[K:k] (byTheorem 4.1ofChapter V), and hence Gmust bethefull Galois group. This proves allour assertions. Corollary 1.9. LetKbeafinite Galois extension ofkand letGbeitsGalois group. Then every subgroup ofGbelongs tosome subfield Fsuch that keF cK. Proof. LetHbeasubgroup ofGand letF=KH .ByArtin's theorem we know that KisGalois over Fwith group H. Remark. When Kisaninfinite Galois extension ofk,then thepreceding corollary isnot true any more. This shows that some counting argument must beused intheproof ofthefinite case. Inthepresent treatment, wehave used anold-fashioned argument. The reader can look upArtin's own proof in hisbook Galois Theory. Intheinfinite case, one defines theKrull topologyon theGalois group G(cf. exercises 43-45), and Gbecomes acompact totally disconnected group. The subgroups which belong totheintermediate fields are theclosed subgroups. The reader maydisregard theinfinite caseentirely through- out ourdiscussions without impairing understanding. Theproofs intheinfinite case areusually identical with those inthefinite case. The notions ofaGalois extension and aGalois group aredefined completely algebraically. Hence they behave formally under isomorphisms the way one expects from objects inanycategory. Wedescribe this behavior more explicitly inthepresent case. LetKbeaGalois extension ofk.Let A.:K-.A.K VI,1 GALOIS EXTENSIONS 265 be anisomorphism. Then AKisaGalois extension ofAk. K). )AK k)AkA Let GbetheGalois group ofKover k.Then themap 0'1---+A.oO'OA.-1 givesahomomorphism ofGinto theGalois group ofAK over Ak,whose inverse isgiven by A.-10t0A. t. Hence G(AK/ Ak) isisomorphictoG(K/k) under theabove map. We may write G(lK/A.k)A=G(K/k) or G(A.K/A.k)=A.G(K/k)A. -1, where theexponentA.is"conjugation," O'A =A.-100'0A.. There isnoavoiding thecontravariance ifwewish topreserve therule (O'A)W=O'AW when wecompose mappingsA.and w. Inparticular, letFbeanintermediate field, keF cK,and letA.:F-+A.F beanembedding ofFinK,which we assume isextended toanautomorphism ofK.Then A.K =K.Hence G(K/A.F)A=G(K/F) and G(K/A.F)=A.G(K/F)A.- 1. Theorem 1.10. LetKbeaGalois extension ofkwith group G.Let Fbea subfield, keF cK,and letH =G(K/F). Then Fisnormal over kifand onlyifHisnormal inG.IfFisnormal over k,then therestriction map0'1---+ 0'IF 266 GALOIS THEORY VI,1 isahomomorphism ofGonto theGalois group ofFover k,whose kernel isH. Wethus have G(F/k) G/H. Proof Assume Fisnormal over k,and letG'beitsGalois group. The restriction map(J (JIFmaps Ginto G',and bydefinition, itskernel isH. Hence Hisnormal inG.Furthermore, any element! EG'extends toanem- bedding ofKinKa ,which must beanautomorphism ofK, sotherestriction map issurjective. This proves thelast statement. Finally,assume that Fisnot normal over k.Then there exists anembeddingA.ofFinKover kwhich isnot anautomorphism, i.e. A.F =FF.Extend A.toanautomorphism ofKover k. The Galois groups G(K/A.F) andG(K/F) areconjugate, and they belong to distinct subfields, hence cannot beequal. Hence Hisnotnormal inG. AGalois extension K/k issaid tobeabelian (resp. cyclic) ifitsGalois group G isabelian (resp. cyclic). Corollary 1.11. LetK/k beabelian (resp. cyclic). IfFisanintermediate field, keF cK,then FisGalois over kand abelian (resp. cyclic). Proof This follows atonce from thefact that asubgroup ofanabelian groupisnormal, and afactor group ofanabelian (resp. cyclic) group isabelian (resp. cyclic). Theorem 1.12. LetKbeaGalois extension ofk,letFbeanarbitrary exten- sion and assume thatK,Fare subfields afsome other field. Then KF isGalois over F,andKisGalois over KnF.LetHbetheGalois group ofKF over F, and GtheGalois group ofKover k.If(JEHthen therestriction of(JtoKis inG,and themap O'I-+(JIK gives anisomorphism ofHontheGalois group ofKover KnF. Proof Let (JEH.The restriction of (JtoKisanembedding ofKover k, whence anelement ofGsince Kisnormal over k.The map(J1-+ (JIKisclearlya homomorphism. If(JIKistheidentity, then (Jmust betheidentity ofKF (since every element ofKF can beexpressedasacombination ofsums, products, andquotients ofelements inKandF). Hence ourhomomorphism(J1-+ (JIKis injective. LetH'beitsimage. Then H'leaves KnFfixed, andconversely, ifan element rxEKisfixed under H', we seethat rxisalso fixed under H,whence rxEFand rxEKnF.Therefore KnFisthefixed field. IfKisfinite over k, oreven KFfinite over F,then byTheorem 1.8, weknow that H'istheGalois group ofKover KnF,and thetheorem isproved inthat case. (Intheinfinite case, one must add theremark that fortheKrull topology, our mapu ulKiscontinuous, whence itsimage isclosed since Hiscompact. See Theorem 14.1;Chapter I,Theorem 10.1;and Exercise 43.) VI, 1 GALOIS EXTENSIONS 267 Thediagram illustrating Theorem 1.12 isasfollows: /KFFK/ KnF k Itissuggestive tothink oftheopposite sides ofaparallelogramasbeing equal. Corollary 1.13. LetKbeafinite Galois extension ofk. LetFbeanarbitrary extension ofk.Then [KF:F]divides [K:k]. Proof Notation beingasabove, weknow that theorder ofHdivides the order ofG,soour assertion follows. Warning. The assertion ofthecorollary isnotusually valid ifKisnot Galois over k.For instance, let lJ.=.j2bethereal cube root of2,let(be a cube root of1,(=I1,say -1+13(= v-J 2' and letP=(rx. Let E=Q(P). Since Piscomplex and rxreal, wehave Q(P) =IQ(rx). Let F=Q(rx). Then EnFisasubfield ofEwhose degreeover Qdivides 3. Hence thisdegree is3or1,and must be 1since E=IF.But EF=Q(rx, P)=Q(rx, ()=Q(rx,J=3 ). Hence EFhasdegree2over F. Theorem 1.14. LetKland K2beGalois extensions ofafield k,with Galois groups G1and G2respectively. Assume KbK2aresubfields ofsome field. Then KlK2 isGalois over k.Let GbeitsGalois group. Map G-.G1XG2 byrestriction, namely a1---+(aIKl' aIK 2). This map isinjective. IfK1nK2=kthen themap isanisomorphism. 268 GALOIS THEORY VI,1 Proof Normality andseparabilityarepreserved intaking thecompositum oftwofields, soK1K2isGalois over k.Our map isobviouslyahomomorphism ofGinto G1xG2.Ifanelement (JEG induces theidentity onK1and K2 then itinduces theidentity ontheir compositum,soourmap isinjective. Assume that K1nK2=k.According toTheorem 1.12, givenanelement (J1EG1there exists anelement (JoftheGalois group ofK1K2over K2which induces(J1on K1.This (Jisafortiori inG,and induces theidentity onK2.Hence G1x{e2} iscontained intheimage ofourhomomorphism (where e2istheunit element of G2).Similarly, {e 1}xG2iscontained inthis image. Hence their product is contained intheimage, and their product isprecisely G1XG2.This proves Theorem 1.14. K1K2/ K1 K2/ K1nK2 k Corollary 1.15. Let Kb...,Kn beGalois extensions ofkwith Galois groups Gb...,Gn.Assume that Ki+1n(K 1...Ki)=kfor each i=1,..., n-1.Then theGalois group ofK1...Knisisomorphic tothe product G1X... xGninthenatural way. Proof Induction. Corollary 1.16. LetKbe afinite Galois extension ofkwith group G,and assume that Gcan bewritten asadirect product G=G1X... xGn.Let Kibethefixed field of G1X... x{I}x... xGn where thegroup with 1element occurs inthei-thplace. Then KiisGalois over k,andKi+1n(K 1...K;)=k.Furthermore K =K1...Kn. Proof ByCorollary 1.3,thecompositum ofallKibelongs totheintersection oftheir corresponding groups, which isclearly theidentity. Hence thecompos i- turn isequal toK.Each factor ofGisnormal inG,soKiisGalois over k.By Corollary 1.4,theintersection ofnormal extensions belongs totheproduct of their Galois groups, and itisthen clear that Ki+1n(K 1...Ki)=k. VI,2 EXAMPLES AND APPLICATIONS 269 Theorem 1.17. Assume allfields contained insome common field. (i)IfK,Lareabelian over k,soisthecomposite KL. (ii)IfKisabelian over kand Eisany extension ofk, then KE isabeUan over E. (iii)IfKisabelian over kandK ::JE :::>kwhere Eisanintermediatefield, then Eisabelian over kandKisabelian over E. Proof Immediate from Theorems 1.12 and 1.14. Ifkisafield, thecomposite ofallabelian extensions ofkinagiven algebraic closure kdiscalled themaximum abelian extension ofk,and isdenoted bykab . Remark onnotation. We have used systematically thenotation: ka=algebraic closure ofk; kS=separable closure ofk; kab=abelian closure ofk=maximal abelian extension. Wehave replaced other people's notation k(and mine aswell inthefirstedition) with kainorder tomake thenotation functorial with respect totheideas. 2. EXAMPLES AND APPLICATIONS Letkbe afield andf(X)aseparable polynomial ofdegree>1ink[X]. Let f(X)=(X-1)...(X-n) beitsfactorization inasplitting field Kover k.Let GbetheGalois group ofK over k.Wecall GtheGalois group offover k.Then theelements ofGpermute theroots off Thus wehave aninjective homomorphism ofGinto thesymmetric group Snon nelements. Not every permutation need begiven byanelement ofG.We shall discuss examples below. Example 1.Quadratic extensions. Let kbe afield and aEk.Ifaisnot asquare ink,then thepolynomial X2 -ahas noroot inkand istherefore irreducible. Assume char k=t=2.Then thepolynomial isseparable (because 2=t=0), andifaisaroot, then k(a) isthesplitting field, isGalois, and its Galois group iscyclic oforder 2. Conversely, given anextension Kofkofdegree 2,there exists aEksuch that K=k(a) and a2=a.This comes from completing thesquare and thequadratic formula asinelementary school. The formula isvalid aslongasthecharacteristic ofkis =t=2. 270 GALOIS THEORY VI,2 Example 2. Cubic extensions. Let kbe afield ofcharacteristic =t=2or 3.Let f(X)=X3+aX+b. Any polynomial ofdegree 3can bebrought into this form bycompleting the cube. Assume thatfhas noroot ink.Thenfis irreducible because anyfactoriza- tion must have afactor ofdegree1.Let abe aroot off(X). Then [k(a): k]=3. LetKbethesplitting field. Since char k=t=2,3,fisseparable. Let Gbethe Galois group. Then Ghasorder 3or6since Gisasubgroup ofthesymmetric group S3.Inthesecond case, k(a) isnotnormal over k. There isaneasy way totest whether theGalois group isthefullsymmetric group. We consider thediscriminant. Ifell' el2,el3are thedistinct roots of f(X),welet b=(ell-el2)(el2-el3)(elt-el3) and =b2 . IfGistheGalois group and (JEGthen (J(b)=+b.Hence (Jleaves fixed. Thus Llisintheground field k,and inChapter IV,6, wehave seen that Ll=-4a3 -27b2 . The setof (]"inGwhich leave 5fixed isprecisely the setofeven permutations. Thus Gisthesymmetric group ifandonly ifLlisnot asquare ink.We may summarize theabove remarks asfollows. Letf(X) beacubic polynomial ink[X], and assume char k=t=2,3.Then: (a)fisirreducible over kifandonlyiffhas noroot ink. (b) Assume firreducible. Then theGalois group off isS3ifand onlyifthe discriminant offisnot asquare ink.Ifthediscriminant isasquare, then theGalois group iscyclic oforder 3,equal tothealternating group A3as apermutation ofthe roots off. Forinstance, consider f(X)=X3-X+1 over therational numbers. Any rational root must be 1or-1,and sof(X) is irreducible over Q.The discriminant is-23,and isnot asquare. Hence the Galois group isthesymmetriroup.Thesplitting field contains asubfield of degree 2,namely k(8)=k(V Ll). Ontheother hand, letf(X)=X3-3X+1.Thenfhas noroot inZ,whence noroot inQ, sofisirreducible. The discriminant is81,which isasquare,so theGalois group iscyclic oforder 3. Example 3. We consider the polynomial f(X)=X4-2over the rationals Q.Itisirreducible byEisenstein's criterion. Let elbe areal root. VI,2 EXAMPLES AND APPLICATIONS 271 Let i=.J=l.Then +rxand +irxarethefour roots off(X), and [Q(a):Q]=4. Hence thesplitting fieldofj(X) is K =Q(rx, i). The field Q(rx) nQ(i) hasdegree1or2over Q.Thedegree cannot be2otherwise iEQ(rx), which isimpossible since rxisreal. Hence thedegree is1.Hence ihas degree 2over Q(rx) and therefore [K:Q]=8.The Galois group off(X) has order 8. There exists anautomorphism! ofKleaving Q(rx) fixed, sending ito-i, because KisGalois over Q(rx),ofdegree 2.Then!2 =ide Q(rx, i)=Ky Q(rx) Q(i) Q Bythemultiplicativity ofdegrees intowers, we seethat thedegreesare as indicated inthediagram. Thus X4-2isirreducible over Q(i). Also, Kis normal over Q(i). There exists anautomorphism(1ofKover Q(i)mapping the root aofX4 -2totheroot ia.Then one verifies atonce that 1,U,u2 ,U3are distinct and (14 =ideThus (1generatesacyclic group oforder 4.Wedenote it by«(1). Since! rt«(1)itfollows that G=«(1,!)isgenerated by(1and! because «(1) hasindex 2.Furthermore, one verifies directly that !(1=(13!, because this relation istrue when applied to rxand iwhich generate Kover Q. This gives usthestructure ofG.Itisthen easy toverify that thelattice ofsub- groups isasfollows: G 221:-:--- 2 3/'"U .)U'I '/(i'U'U.,(1,.) (1,U2.)1'U(1'U.)(1,u3.) (1) 272 GALOIS THEORY VI,2 Example 4. Let kbeafield and lett1,...,tnbealgebraically independent over k.LetK =k(tl'...,tn).Thesymmetric group Gon nletters operates on Kbypermuting (tl'...,tn)and itsfixed field isthefield ofsymmetric functions, bydefini tion thefield ofthose elements ofKfixed under G.Let Sl'...,Snbethe elementary symmetric polynomials, and let n f(X)=n(X-t;). i=1 Up toasign, thecoefficients offare Sb...,Sn.We letF=KG .Wecontend that F=k(sl'.. .,sn).Indeed, k(sl'...,Sn)CF. Ontheother hand, Kisthesplitting field off(X), and itsdegreeover Fisn!. Itsdegree over k(Sb...,sn)is<n!and hence wehave equality, F=k(S1'...,sn). The polynomial f(X) above iscalled thegeneral polynomial ofdegreen. We have just constructed aGalois extension whose Galois groupisthesym- metric group. Using theHilbert irreducibility theorem, one can construct aGalois extension ofQwhose Galois group isthesymmetric group. (Cf. Chapter VII, end of2, and[La83], Chapter IX.) Itisunknown whether givenafinite group G,there exists aGalois extension ofQwhose Galois group isG.Byspecializing para- meters, Emmy Noether remarked that onecould prove thisifone knew that every field Esuch that Q(Sb...,Sn)CECQ(tb...,t n) isisomorphic to afield generated bynalgebraically independent elements. However, matters are not sosimple, because Swan proved that thefixed field of acyclic subgroup ofthesymmetric group isnotnecessarily generated by algebraically independent elements over k[Sw69], [Sw 83]. Example 5. We shall prove that thecomplex numbers arealgebraically closed. This will illustrate almost allthetheorems wehave proved previously. We usethefollowing properties ofthereal numbers R:Itisanordered field, every positive element isasquare, and every polynomial ofodd degree inR[X] has aroot inR.Weshall discuss ordered fields ingeneral later, and our argu- ments apply toany ordered field having theabove properties. Let i=yCl" (inother words aroot ofXl+1).Every element inR(i) has asquare root. Ifa+biER(i), a,bER,then the square root isgiven by c+di,where 2a+Ja2+b2 2-a+Ja2+b2 c = 2and d= 2. Each element ontheright ofourequalities ispositive and hence has asquare root inR,Itisthen trivial todetermine thesign ofcand dsothat (c+di)2=a+bi. VI,2 EXAMPLES AND APPLICATIONS 273 Since Rhascharacteristic 0,every finite extension isseparable. Every finite extension ofR(i) iscontained inanextension Kwhich isfinite and Galois over R.We must show that K =R(i). Let GbetheGalois group over Rand letH be a2-Sylow subgroup ofG.LetFbeitsfixed field. Counting degrees and orders, wefind that thedegree ofFover Risodd. Bytheprimitive element theorem, there exists anelement rxEFsuch that F=R(rx).Then rxistheroot of anirreducible polynomial inR[X] ofodd degree. This canhappen only ifthis degree is1.Hence G=Hisa2-group. We now seethat KisGalois over R(i). Let G1beitsGalois group. Since G1 isap-group (with p=2),ifG1isnot thetrivial group, then G1has asubgroup G2ofindex 2.LetFbethefixed field ofG2.Then Fisofdegree 2over R(i); it isaquadratic extension. But we saw that every element ofR(i) has asquare root, and hence that R(i) has noextensions ofdegree 2.Itfollows that G1isthe trivial group and K =R(i), which iswhat wewanted. (The basic ideas oftheabove proofwere already inGauss. The variation oftheideas which wehave selected, makingaparticularly efficient useofthe Sylow group, isdue toArtin.) Example 6. Letf(X) beanirreducible polynomial over thefield k,and assume thatfisseparable. Then theGalois group Gofthesplitting field is representedasagroup ofpermutations ofthe nroots, where n=degf When- ever one has acriterion forthis group tobethefullsymmetric group Sn,then one can seeifitapplies tothisrepresentation ofG.Forexample, itisaneasy exercise (cf. Chapter I,Exercise 38) that forpprime, Spisgenerated by [123. · ·p]and anytransposition.We then have thefollowing result. Letf(X) beanirreducible polynomial with rational coefficients andofdegree pprime. Iffhasprecisely two nonreal roots inthecomplex numbers, then the Galois group off isSp. Proof The order ofGisdivisible byp,and hence bySylow's theorem, G contains anelement oforder p.Since Gisasubgroup ofSpwhich hasorder p!, itfollows that anelement oforder pcan berepresented byap-cycle [123· · ·p] after asuitable ordering oftheroots, because any smaller cycle has order less than p,sorelatively prime top.But thepair ofcomplex conjugate roots shows that complex conjugation induces atransposition inG.Hence thegroup isall ofSp. Aspecificcase iseasily given. Drawing thegraph of f(X)=XS-4X+2 shows thatfhas exactly three real roots, soexactly twocomplex conjugateroots. Furthermore fisirreducible over QbyEisenstein's criterion, sowe canapply thegeneral statement proved above toconclude that theGalois group off over QisSs.See also Exercise 17ofChapter IV. 274 GALOIS THEORY VI,2 Example 7. Thepreceding example determines aGalois group byfinding some subgroups passing toanextension field oftheground field. There are other possible extensions ofQrather than thereals, forinstance p-adic fields which will bediscussed later inthis book. However, instead ofpassing toan extension field, itispossible tousereduction mod p.For our purposes here, we assume thefollowing statement, which will beproved inChapter VII, theorem 2.9. Letf(X)EZ[X] be apolynomial with integral coefficients, and leading coefficient1.Let pbe aprime number. Letl(X)=f(X) mod pbethe polynomial obtained byreducing thecoefficients mod p.Assume thatfhas nomultiple roots inanalgebraic closure ofFp.Then there exists abijection (b...,n)1-+(b. ..,n) oftheroots off onto those of1,and anembedding oftheGalois group ofJasa subgroup oftheGalois group o.ff,which givesanisomorphism oftheaction of those groups onthe setofroots. The embedding will bemade precise inChapter VII, but here wejust want to use this result tocompute Galois groups. For instance, consider X5-X-I over Z.Reducing mod 5shows that thispolynomial isirreducible. Reducing mod 2gives theirreducible factors (X2+X+1)(X3+X2+1)(mod 2). Hence theGalois group over therationals contains a5-cycle and aproduct ofa 2-cycle and a3-cycle. The third power oftheproduct ofthe2-cycle and3-cycle isa2-cycle, which isatransposition. Hence theGalois group contains atrans- position and thecycle [123.· ·p],which generate Sp(cf. theexercises ofChapter Ionthesymmetric group). Thus theGalois group ofX5-X-I isSp. Example 8. Thetechnique ofreducing mod primes togetlotsofelements inaGalois groupwas used bySchur todetermine theGalois groups ofclassical polynomials [Schur 31]. Forinstance, Schur proves that theGalois group over Qofthefollowing polynomials over Qisthesymmetric group: n (a)f(X)=Lxm1m!(inother words, thetruncated exponential series), if m=O nisnotdivisible by4.Ifnisdivisible by4,hegets thealternating group. (b)Let Hm(X)=(-l)mex2/2:;m(e-X2/2) bethem-th Hermite polynomial. Put H2n(X)=KO)(X2) and H2n+1(X)=XK1)(X2). Then theGalois group ofK<j)(X) over Qisthesymmetric group Snfori=0, 1,providedn>12.Theremaining cases were settled in[Schulz 37]. VI,2 EXAMPLES AND APPLICATIONS 275 Example 9.This example isaddressed tothose who know something about Riemann surfaces andcoverings. Let tbetranscendental over the com- plex numbers C,and letk=C(t). The values oftinC,or00,correspond tothe points oftheGauss sphere S,viewed asaRiemann surface. LetP1,...,Pn+1be distinct points ofS.The finite coverings ofS-{PI,...,Pn-I}areinbijection with certain finite extensions ofC(t), those which are unramified outside PI,...,Pn-I. LetKbetheunion ofallthese extension fields correspondingto such coverings, and letnn)bethefundamental group of S-{Pb...,Pn+l}. Then itisknown thatn\n)isafreegroup on ngenerators, and has anembedding intheGalois group ofKover C(t), such that thefinite subfields ofKover C(t) areinbijection with thesubgroups ofn\n)which areoffinite index. Given a finite group Ggenerated bynelements (11' ..., (1nwe can find asurjective homomorphism nn)Gmapping thegenerators ofnin)on(11,...,an. LetH bethekernel. Then Hbelongs toasubfield KHofKwhich isnormal over C(t) and whose Galois group isG.Inthelanguage ofcoverings, Hbelongs toa finite covering of S-{P I,. ..,Pn+I}. Over thefield C(t)one can useanalytic techniques todetermine theGalois group. The Galois group isthecompletion of afree group,asproved by Douady [Dou 64]. For extensions tocharacteristic p,see[Pop 95]. Afunda- mental problem istodetermine theGalois group over Q(t), which requires much deeper insight into thenumber theoretic nature ofthis field. Basic con- tributions were made byBelyi [Be80], [Be83], who also considered thefield Q(Jl)(t), where Q(Jl) isthefield obtained byadjoining allroots ofunity tothe rationals. Belyi proved that over this latter field, essentially alltheclassical fi- nite groupsoccur asGalois groups. Seealso Conjecture 14.2 below. For Galois groupsover Q(t), see the survey [Se88], which contains a bibliography. One method iscalled therigidity method, first applied byShih [Shi 74], which Isummarize because itgives examples ofvarious notions defined throughout this book. Theproblem istodescend extensions ofC(t) with agiven Galois group Gtoextensions ofQ(t) with the same Galois group. Ifthisextension isKover Q(t), one also wants the extension toberegular over Q(see the definition inChapter VIII, 4). Togiveasufficient condition, weneed some definitions. LetGbeafinite group with trivial center. LetCI'C2,C3beconjugacy classes. Let P=P(C I'C2,C3)bethe setofelements (91,92,93)EC1XC2XC3 such that 919293=1.LetP'bethe subset ofPconsisting ofallelements (9., 92'93)EPsuch that Gisgenerated by91'92'93.We saythat thefamily (C., C2,C3)isrigid ifGoperates transitivelyonP', andP'isnotempty. 276 GALOIS THEORY VI,3 We define aconjugacy class CofGtoberational ifgiven gECand a positive integersrelatively primetotheorder ofg,then g5EC.(Assuming that thereader knows theterminology ofcharacters defined inChapter XVIII, this condition ofrationality isequivalenttothecondition that every character Xof Ghas values intherational numbers Q.) One then has thefollowing theorem, which iscontained intheworks ofShih, Fried, Belyi, Matzat andThompson. Rigidity theorem. Let Gbe afinite group with trivial center, and let C}, C2,C3beconjugacy classes which arerational, and such that thefamily (C}, C2,C3)isrigid. Then there exists aGalois extension ofQ(t) with Galois group G(and such that theextension isregular over Q). [Be80] [Be83] [Dou 64] [La83] [Pop 95] [Se88] [Shi 74] [Sw69] [Sw 83]Bibliography G.BELYI, Galois extensions ofthemaximal cyclotomic field, lzv. Akad. Nauk SSR 43(1979) pp. 267-276 (=Math. USSR lzv. 14(1980), pp. 247-256 G.BEL YI,Onextensions ofthemaximal cyclotomic field havingagiven classical Galois group, J.reine angew. Math. 341 (1983), pp. 147-156 A.DOUADY, Determination d'un groupe deGalois, C.R,Acad. Sci. 258 (1964), pp.5305-5308 S.LANG, Fundamentals ofDiophantine Geometry. Springer Verlag 1983 F.PoP, Etale Galois covers ofaffine smooth curves, Invent. Math. 120 (1995), pp.555-578 J.-P.. SERRE, Groupes deGalois surQ,Seminaire Bourbaki, 1987-1988 Asterisque 161-162, pp.73-85 R.-Y.SHIH, Ontheconstruction ofGalois extensions offunction fields and number fields, Math. Ann. 207(1974), pp.99-120 R.SWAN, Invariant rational functions and aproblem ofSteenrod, Invent. Math. 7(1969), pp. 148-158 R.SWAN, Noether's problem inGalois theory, Emmy Noether inBryn Mawr, J.D.Sally and B.Srinivasan, eds., Springer Verlag, 1983, pp. 40 3. ROOTS OF UNITY Let kbe afield. Byaroot ofunity (ink)weshall mean anelement' Ek such that ,n=1for some integer n>1.Ifthecharacteristic ofkisp,then the equation Xpm=1 hasonly oneroot, namely 1,and hence there isnopm-th root ofunity except1. VI,3 ROOTS OFUNITY 277 Let nbeaninteger>1and notdivisible bythecharacteristic. Thepolynomial Xn-1 isseparable because itsderivative isnXn-1 =F0,and theonly root ofthederiva- tive is0,sothere isnocommon root. Hence inkathepolynomial xn-1has n distinct roots, which are roots ofunity. They obviously form agroup,and we know that every finite multiplicative group in afield iscyclic (Chapter IV, Theorem 1.9). Thus thegroup ofn-th roots ofunity iscyclic. Ageneratorfor this groupiscalled aprimitive n-th root ofunity. IfJlndenotes thegroup ofalln-th roots ofunity inkaand m,narerelatively prime integers, then Jlnzn JimXJln. This follows because Jim, Jlncannot have any element incommon except 1, and becauseJlmJln consequently has mnelements, each ofwhich isanmn-th root ofunity. Hence JlmJln=Jlmn' and thedecomposition isthat ofadirect product. As amatter ofnotation, toavoid double indices, especially intheprime power case, wewrite J1[n] forJ1n.Soifpisaprime, J1[pr] isthegroup of pr-th roots ofunity. Then J1[pOO] denotes the union ofallJ1[pr] for all positive integersr.See the comments in 14. Letkbeany field. Let nbenotdivisible bythecharacteristic p.Let(= (nbe aprimitive n-th root ofunity inka .Let (fbeanembedding ofk«() inka over k.Then (a()n=a«(n)= 1 sothata(isann-th root ofunity also. Hence a(=(ifor some integer i=i(a), uniquely determined mod n.Itfollows that amaps k«() into itself, and hence that k«() isnormal over k.IfTisanother automorphism ofk«() over kthen aT(=(i(G)i(f). Since aand Tare automorphisms, itfollows that i(a) and i(T) areprime ton (otherwise, a(would have aperiod smaller than n).Inthis waywegetahomo- morphism oftheGalois group Gofk(()over kinto themultiplicative group (ZjnZ)* ofintegers prime ton,mod n.Ourhomomorphism isclearly injective since ;(a) isuniquely determined by amod n,and the effect ofaonk«() is determined byitseffect on(.Weconclude that k«() isabelian over k. We know that theorder of(ZjnZ)* isqJ(n). Hence thedegree [k«(): k] divides qJ(n). For aspecific field k,thequestion arises whether theimage ofGK({)/Kin (Z/nZ)* isallof(Z/nZ)* .Looking at K=RorC,one sees that this isnot always the case. We now giveanimportant example when itisthe case. 278 GALOIS THEORY VI,3 Theorem 3.1. Let(beaprimitive n-th rootofunity. Then [Q«():Q]=qJ(n), wherecpistheEuler function. The map u i(u) gives anisomorphism GQ«()/Q (Z/nZ)*. Proof. Letj'(X)betheirreducible polynomial of(over Q.Then f(X) divides xn-1,sayxn-1=f(X)h(X), where bothf, hhave leading coefficient 1.BytheGauss lemma, itfollows thatJ: hhave integral coefficients. Weshall now prove that ifpisaprime number notdividing n,then (Pisalso arootoff Since (Pisalso aprimitive n-th root ofunity, and since anyprimitive n-th root of unitycan beobtained byraising (toasuccession ofprime powers, with primes notdividing n,thiswillimply that alltheprimitive n-th roots ofunity areroots ofJ: which must therefore have degree>qJ(n), and hence precisely qJ(n). Suppose (Pisnot arootoff Then (Pisaroot ofh,and(itself isaroot ofh(XP). Hencef(X) divides h(XP), and we canwrite h(XP)=f(X)g(X). Sincefhasintegral coefficients and leading coefficient 1,we seethat 9has integral coefficients. Since aP-a(mod p)foranyinteger a,weconclude that h(XP)=h(X)P (mod p), and hence h(X)P=f(X)g(X) (mod p). Inparticular, ifwedenote byfand hthepolynomials inZ/pZ obtained by reducing fand hrespectively mod p,we seethatJand Iiare notrelatively prime, i.e.have afactor incommon. Butxn-T=.f(X)Ii(X), and hence xn-Thasmultiple roots. This isimpossible,asone sees bytaking thede- rivative, and our theorem isproved. Corollary 3.2. Ifn,marerelative prime integers>1,then Q«(n) nQ«(m)=Q. Proof We note that (nand (mareboth contained inQ((mn) since (:Zn isa primitive m-th root ofunity. Furthermore, (m(nisaprimitive mn-th root of unity. Hence Q«(n)Q«(m)=Q('mn). Our assertion follows from themultiplicativity qJ(mn)=qJ(m)qJ(n). Suppose that nisaprime number p(having nothing todowith thecharacter- istic). Then XP-1=(X-1)(XP-1+...+1). VI,3 ROOTS OFUNITY 279 Anyprimitive p-th root ofunity isaroot ofthesecond factor ontheright ofthis equation. Since there areexactly p-1primitive p-th roots ofunity,we con- clude that these roots areprecisely the roots of XP-1+. ..+1. We saw inChapter IV,3that thispolynomial could betransformed into anEisenstein polynomialover the rationals. This gives another proof that [Q«(p): Q]=p-1. Weinvestigate more closely thefactorization ofxn-1,and suppose that we areincharacteristic 0forsimplicity. We have Xn -1=IT(X-(), ( where theproduct istaken over alln-th roots ofunity. Collect together allterms belonging toroots ofunity having the same period. Let <I>d(X)= IT (X-,) period (=d Then Xn -1=IT<I>d(X). din We seethat <I>}(X)=X-I, and that <l>n(X)=Xn -1 IT <I>d(X) din d<n From this wecancompute<I>(X)recursively, and we seethat<l>n(X) isapolynomial inQ[X] because wedivide recursively bypolynomials having coefficients inQ. All ourpolynomials have leading coefficient 1,sothat infact <I>n(X) hasinteger coefficients byTheorem 1.1ofChapter IV. Thus ourconstruction isessentially universal and would hold over any field (whose characteristic does notdivide n). We call <l>n(X) then-th cyclotomic polynomial. The roots of<l>nareprecisely theprimitive n-th roots ofunity, and hence deg <l>n=cp(n). From Theorem 3.1 weconclude that <I>nisirreducible over Q,and hence <l>n(X)=Irr«(n, Q,X). 280 GALOIS THEORY VI,3 We leave theproofs ofthefollowing recursion formulas asexercises: 1.IfPisaprime number, then <l>p(X)=Xp-l +Xp-2 +... +1, and for anintegerr>1, r-l <l>pr(X)=<l>p(XP ). 2.Let n=p;l...Pbeapositive integer with itsprime factorization. Then <l>n(X)=<l>Pl''Ps(XPP-1 pS-l). 3.Ifnisodd >1,then <l>2n(X)=<I>n(-X). 4.Ifpisaprime number, notdividing n,then _<I>n(XP)<Ppn(X)- <Pn(X). Ontheother hand, ifpin, then <I>pn(X)=<I>n(XP). 5.We have <I>n(X)=IT(Xn/d -1)JL(d). din Asusual, J1istheMobius function: {o ifnisdivisible byp2for some prime p, J1(n)=(-I)r ifn=PI...Prisaproduct ofdistinct primes, 1 ifn=1. As anexercise, show that LJL(d)={Iifn=I, dln 0ifn>1. Example. Inlight ofExercise 21ofChapter V,wenote that theassociation n <l>n(X)can beviewed as afunction from thepositive integers into the multiplicative group ofnon-zero rational functions. Themultiplication formula xn-1=n<I>d(X)can therefore beinverted bythegeneral formalism of convolutions. Computations ofanumber ofcyclotomic polynomials show that forlow values ofn,they have coefficients equal to0or+1.However, Iam indebted toKeith Conrad forbringing tomyattention anextensive literature on thesubject, starting with Bang in1895. Iinclude only thefirst and last items: A.S.BANG, OmLigningen <l>m(X)=0,NytTidsskrift forMatematik (B) 6(1895), pp.6-12 H.L.MONTGOMERY and R.C.VAUGHN, The order ofmagnitude ofthem-th coef- ficients ofcyclotomic polynomials, Glasgow Math, J.27(1985), pp, 143-159 VI,3 ROOTS OFUNITY 281 Inparticular, if<I>n(X)=anjXj,define L(j)=log maxnIanjI.Then Montgomery andVaughn prove that .1/2 .1/2J«LU«J (logj)1/4(logJ)1/4 where thesign« means that theleft-hand side isatmost apositive constant times theright-hand side forj00.Bang also points out that <l>lOS(X)isa cyclotomic polynomial ofsmallest degree having coefficients =1=0or+1:the coefficient ofX7and X41is-2(all others are0or+1). If(isann-th root ofunity and(=F1,then 1-(n 1y yn-101-(=++...+ =. This istrivial, butuseful. LetFqbethefinite field with qelements, qequal toapower oftheoddprime number p.Then F:has q-1elements and isacyclic group. Hence wehave theindex (F::F:2)=2. Ifvisanon-zero integer notdivisible byp,let ()={-ifv=x2(mod p)for some x, ifv=1=x2(mod p)forallx. This isknown asthequadratic symbol, anddepends onlyontheresidue class ofvmod p. From ourpreceding remark, we seethat there are asmany quadratic residues asthere arenon-residues mod p. Theorem 3.3. Let(beaprimitive p-th rootofunity, and let s=()c, the sumbeing taken over non-zero residue classes mod p.Then S2=(/)P. Every quadratic extension ofQ iscontained inacyclotomic extension. Proof. The last statement follows atonce from theexplicit expression of +pasasquare inQ«(), because thesquare root ofaninteger iscontained inthe 282 GALOIS THEORY VI,4 field obtained byadjoining thesquare root oftheprime factors initsfactoriza- tion, and alsoJ=1.Furthermore, fortheprime 2,wehave (1+i)2=2i.We now prove our assertion concerning S2. Wehave S2=L()(J1 )(V+Jl=L(VJ1 )(V+Jl. v,JlP P v,JlP As vranges over non-zero residue classes, sodoesVJ1forany fixedJ1,and hence replacingvbyvJ1yields S2 =L(VJ12 )(Jl(V+1)=L()(Jl(V+1) V,Jl P V,11 P =L()(O+L()L(I1(V+ 1). 11 P v*-l P 11 But 1+(+...+(P-I=0,and the sum ontheright overJ1consequently yields-1.Hence S2=()(P-1)+(-1) L() P v*-1P =p(/)-G) =p(/). asdesired. We seethatQ(JP)iscontained inQ«(,J=1 )orQ«(), dependingonthe sign ofthequadratic symbol with-1.Anextension ofafield issaid tobe cyclotomic ifitiscontained inafield obtained byadjoining roots ofunity. We have shown above that quadratic extensions ofQarecyclotomic. A theorem ofKronecker asserts that every abelian extension ofQiscyclotomic, buttheproof needs techniques which cannot becovered inthis book. 4. LINEAR INDEPENDENCE OF CHARACTERS Let Gbeamonoid andKafield. Byacharacter ofGinK(inthischapter), weshall mean ahomomorphism X:G-+K* ofGinto themultiplicative group ofK.The trivial character isthehomo- VI,4 LINEAR INDEPENDENCE OFCHARACTERS 283 morphism taking the constant value I.Functions /;:G-+Karecalled linearly independent over Kifwhenever wehave arelation aIfl+. ,.+anin=0 with aiEK,then allai=O. Examples. Characters will occur invarious contexts inthis book. First, thevarious conjugate embeddings ofanextension field inanalgebraicclosure can beviewed ascharacters. These arethecharacters which most concern usin thischapter. Second, weshall meet characters inChapter XVIII, when weshall extend the next theorem toamore general kind ofcharacter inconnection with group representations. Next, one meets characters inanalysis. Forinstance, given aninteger m,the functionf: R/Z C*such thatf(x)=e21Timxisacharacter onR/Z. Itcan be shown that allcontinuous homomorphisms ofR/Z into C* areofthis type. Similarly, givenareal number y,thefunction x.....-+ e21Tixy isacontinuous character onR,anditisshown inFourier analysis that allcontinuous characters ofabsolute value 1onRareofthis type. Further, letXbeacompact space and letRbethering ofcontinuous complex- valued functions onX.LetR*bethegroup ofunits ofR.Then given xEXthe evaluation mapff(x) isacharacter ofR*into C*.(Actually, this evaluation map isaring homomorphism ofRonto C.) Artin found aneat way ofexpressingalinear independence property which covers allthese cases, aswell asothers, inthefollowing theorem [Ar44]. Theorem 4.1. (Artin). Let Gbeamonoid and Kafield. LetX.,. ..,Xn bedistinct characters ofGinK.Then they arelinearly independentover K. Proof One character isobviously linearly independent. Suppose that we have arelation aIXl+...+anXn=0 with a;EK,notallO.Take such arelation with nassmall aspossible. Then n>2,and noaiisequal toO.SinceXI'X2aredistinct, there exists ZEGsuch that Xl(Z) =FX2(Z). For allxEGwehave alXl(xz) +...+anXn(xz)=0, and sinceXiisacharacter, alXl (Z)Xl +. ..+anXn(z)Xn=O. Divide byXl(Z) and subtract from ourfirst relation. The term alXI cancels, and wegetarelation (a2X2(Z)-a2)X2+...=O. XI(z) 284 GALOIS THEORY VI,5 The first coefficient isnot0,and this isarelation ofsmaller length than ourfirst relation, contradiction. Asanapplication ofArtin's theorem, one can consider the case when Kisa finite normal extension ofafield k,and when thecharacters aredistinct auto- morphisms (11'...,(1nofKover k,viewed ashomomorphisms ofK*into K*. This special case had already been considered byDedekind, who, however, expressed thetheorem inasomewhat different way, considering thedeterminant constructed from(1iWjwhere wjisasuitable setofelements ofK,andproving in amore complicated way thefactthat thisdeterminant isnot o.The formulation given above and itsparticularly elegant proofaredue toArtin. Asanother application,wehave: Corollary 4.2. Let ab...,anbedistinct non-zero elements ofafield K.If aI,. . .,anare elements ofKsuch thatforallintegersv>0wehave a1a+...+ana=0 then ai=0foralli. Proof Weapply thetheorem tothedistinct homomorphisms V1---+aY I ofZ?;.o into K* . Another interesting application will begivenasanexercise (relative in- variants). 5. THE NORM AND TRACE Let Ebeafinite extension ofk.Let[E:kJs=r,and let pJl=[E: kJi ifthecharacteristic isp>0,and 1otherwise. Let(11,...,(1rbethedistinct embeddings ofEinanalgebraic closure kaofk.Ifaisanelement ofE,we define itsnorm from Etoktobe NE/k(a)=N[(rx.)= vO/1vrx.P"=(01Uvrx.YE:kJi . Similarly,wedefine thetrace r TrE/k(a)=Trf(Cl)=[E:kJiL(1va . v=1 The trace isequal to0if[E:kJi>1,inother words, ifElk isnotseparable. VI,5 THE NORM AND TRACE 285 Thus ifEisseparable over k,wehave Nf(a)=naa (1 where theproduct istaken over thedistinctembeddings ofEink8over k. Similarly, ifElk isseparable, then Trf(a)=Lua. (1 Theorem 5.1. LetElk beafinite extension. Then the norm Nt isamulti- plicative homomorphism ofE*into k*and the trace isanadditive homo- morphism ofEinto k.IfE =>F =>kisatower offields, then thetwo maps are transitive, inother words, Nf=N[0N: and Trf=Trf0Tr:. IfE=k(a),andf(X)=Irr(a, k,X)=xn+an_1Xn-1+...+ao,then N(a)(a)=(-l)naoand Tr(a)(a)= -an-1. Proof For the first assertion, we note that aPI-L isseparable over kif pJ1=[E:kJi. Ontheother hand, theproduct r nuvaPI-L v= 1 isleft fixed under any isomorphism into k8because applying such aniso- morphism simply permutes thefactors. Hence thisproduct must lieinksince aPI-Lisseparable over k.Asimilar reasoning applies tothe trace. For thesec0!ld assertion, let{Ti}bethefamily ofdij?ctembeddings ofF into k8over k. Extend each Tjtoanautomorphism ofk8,and denote this extension byTjalso. Let{u;} bethefamily ofembeddings ofEink8over F. (Without loss ofgenerality,wemay assume that Eck8.)Ifuisanembedding ofEover kink8 ,then for some j,Tj-1aleaves Ffixed, and hence7:}1U =Uifor some i.Hence u=7:jUiand consequently thefamily {TjUi} gives alldistinct embeddings ofEinto k8over k.Since theinseparability degree ismultiplicative intowers, our assertion concerning thetransitivity ofthe norm and trace is obvious, because wehave already shown thatNmaps Einto F,andsimilarly forthe trace. Suppose now that E=k(a). Wehave f(X)=«X-a1)...(X-ar))[E:k], ifab. . .,ararethedistinct roots off Looking attheconstant termoffgivesus theexpression forthenorm, andlooking atthe next tohighest term givesusthe expression forthe trace. We observe that the trace isak-linear map ofEinto k,namely Trf(ca)=cTrf(a) 286 GALOIS THEORY VI,5 forall aEEand cEk.This isclear since cisfixed under every embedding of Eover k.Thus the trace isak-linear functional ofEinto k.Forsimplicity, wewrite Tr =Trt. Theorem 5.2. Let Ebeafinite separable extension ofk.Then Tr: E-.kis anon-zero functional. The map (x,y) Tr(xy) ofExE-.kisbilinear, andidentifies Ewith itsdual space. Proof. That Trisnon-zero follows from thetheorem onlinear indepen- dence ofcharacters. For each xEE,themap Trx:E-.k such that Trx(Y)=Tr(xy) isobviouslyak-linear map, and themap XJ-+Trx isak-homomorphism ofEinto itsdual space EV .(We don't write E*forthe dual space because we use the star todenote themultiplicative group ofE.) IfTrxisthe zero map, then Tr(xE)=O.Ifx=F0then xE=E.Hence the kernel ofxJ-+Trx iso.Hence wegetaninjective homomorphism ofEinto thedual space E.Since these spaces have the same finite dimension, itfollows that wegetanisomorphism. This proves ourtheorem. Corollary 5.3. Let WI,...,Wnbeabasis ofEover k.Then there exists a basis W/ I,.. .,WofEover ksuch thatTr(Wiwj)=bij. Proof The basis W/ I,...,wisnone other than thedual basis which we defined when weconsidered thedual space ofanarbitrary vector space. Corollary 5.4. Let Ebeafinite separable extension ofk,and let (JI'...,(Jn bethedistinct setofembeddings ofEinto k8over k.Let WI'...,Wnbeele- ments ofE.Then the vectors I=«(JIWb...,(JIWn), n=«(JnWb..., (Jnwn) arelinearly independent over EifWI'...,wnformabasis ofEover k. Proof Assume that Wb...,Wnform abasis ofElk. Let aI'...,anbeele- ments ofEsuch that aII+...+ann=O. Then we seethat aI(JI+...+an(Jn VI,5 THE NORM AND TRACE 287 applied toeach one ofWI'...,Wngives thevalue O.But 0'l'...,0'narelinearly independentascharacters ofthemultiplicative group E*into k8*.Itfollows that rxi=0fori=1,...,n,and our vectors arelinearly independent. Remark. Incharacteristic 0,one sees much more trivially that the trace is notidentically o.Indeed, ifCEkand c=F0,then Tr(c)=ncwhere n=[E:k], and n=FO.This argument also holds incharacteristic pwhen nisprimetop. Propoition 5.5. Let E=k(rx) beaseparable extension. Let f(X)=Irr( rx,k,X), and letf'(X) beitsderivative. Let ({()IX)=Po+P.X +...+Pn_. xn-. with PiEE.Then thedual basis of1,rx,...,rxn-1is Po Pn-l f'(rx)'...' f'(rx). Proof Let rx1,...,rxnbethedistinct roots off Then if(X) IX=X' i=1(X-rxi)f'(rxi)for 0<r<n-1. To seethis, letg(X) bethedifference oftheleft- andright-hand side ofthis equality. Then ghasdegree<n-1,and has nroots rxf,...,rxn.Hence gis identically zero. Thepolynomialsf(X) (X-(Xi)f'(rxi)rxI areallconjugate toeach other. Ifwedefine the trace ofapolynomial with coefficients inEtobethepolynomial obtained byapplying the trace tothe coefficients, then [f(X)rxr ]rTr (X_IX)f'(IX)=X · Looking atthecoefficients ofeach power ofXinthisequation,we seethat (i/3j)_Tr IX f'(IX)-bij, thereby provingourproposition. Finallyweestablish aconnection with determinants, whose basic properties we now assume. Let Ebe afinite extension ofk,which weview as afinite dimensional vector spaceover k.For each aEEwehave thek-linear map 288 GALOIS THEORY VI,6 multiplication bya, ma: E Esuch that ma(x)=ax. Then wehave thedeterminant det(m a),which can becomputedasthedeterminant ofthematrix Marepresenting mawith respect toabasis. Similarlywehave the trace Tr(m a),which isthe sum ofthediagonal elements ofthematrix Ma. Proposition 5.6. LetEbeafinite extension ofkand letaEE.Then det(m a)=NE/k(a) and Tr(m a)=TrE/k(a). Proof. Let F=k(a). If[F:k]=d,then 1,a,..., -I isabasis for Fover k.Let{Wb...'wr}be abasis for Eover F.Then{aiwj} (i=0,. ..,d-1;j=1,...,r)isabasis forEover k.Let f(X)=Xd+ad_I Xd-I+ . . .+ao betheirreducible polynomial ofaover k.Then NF/k(a)=(-1)dao ,andbythe transitivity ofthe norm, wehave NE/k(a)=NF/k(a)r. The reader canverify directlyontheabove basis thatNF/k(rx)risthedeterminant ofma onF,and then that NF/k(a)disthedeterminant ofma onE,thus concluding theproof forthedeterminant. The trace ishandled exactly inthe same way, except thatTrE/k(a)=r·TrF/k(a). The trace ofthematrix forma onFisequal to-ad-I. From this the statement identifying the two traces isimmediate, asit was forthe norm. 6. CYCLIC EXTENSIONS We recall that afinite extension issaid tobecyclic ifitisGalois and its Galois group iscyclic. The determination ofcyclic extensions when enough roots ofunity areintheground field isbased onthefollowing fact. Theorem 6.1. (Hilbert's Theorem 90). LetK/k becyclic ofdegreen with Galois group G.Let (Jbe agenerator ofG.Let {3EK. The norm N:({3)=N(fJ) isequaltoIifandonlyifthere exists anelement rx=F0inK such that f3=rx/(Jrx. Proof Assume such anelement rxexists. Taking the norm of{3weget N(rx)jN((Jrx). Butthenorm istheproduct over allautomorphisms inG.Inserting (Jjust permutes these automorphisms. Hence the norm isequal to1. Itwill beconvenient touse anexponential notation asfollows. Ift,t'EG and EKwewrite T+t'=tt'. VI,6 CYCLIC EXTENSIONS 289 ByArtin's theorem oncharacters, themap given by id+fJu+pi+au2+...+pi+a+...+a"-2 Un-I onKisnotidentically zero. Hence there exists f}EKsuch that theelement rx=e+pea +pi+aea2+.. .+pi+a+...+a"- 20a"-1 isnotequal toO.Itisthen clear that prxa= rxusing thefact that N(P)=1,and hence that when weapplyutothelast term inthesum, weobtain f}.Wedivide byrxatoconclude theproof. Theorem 6.2. Let kbeafield,naninteger >0prime tothecharacteristic ofk,and assume that there isaprimitive n-th rootofunity ink. (i)LetKbeacyclic extension ofdegreen.Then there exists rxEKsuch that K =k(rx), and rxsatisfies anequation Xn-a=0for some aEk. (ii)Conversely, letaEk.Let rxbearootofxn-a.Then k(rx) iscyclic over k,ofdegree d,din, and rxdisanelement ofk. Proof Let(beaprimitive n-th root ofunity ink,and letK/k becyclic with groupG. LetubeageneratorofG. WehaveN((-l)=((-I)n=1.ByHilbert's theorem 90,there exists rxEKsuch that urx =(rx. Since (isink,wehave uirx =(irxfori=1,..., n.Hence theelements (irx are ndistinct conjugates of rx over k,whence [k(rx):k]isatleast equal ton.Since [K:k]=n,itfollows that K =k(rx). Furthermore, u(rxn)=u(rx)n=((rx)n= rxn . Hence rxnisfixed under u,hence isfixed under each power ofu,hence isfixed under G.Therefore rxnisanelement ofk,and welet a= rxn .This proves the first part ofthetheorem. Conversely, let aEk.Let abe aroot ofxn-a.Then a(i isalso aroot for each i=1,. . .,n,and hence allroots lieink(a) which istherefore normal over k.Allthe roots aredistinct sok(a) isGalois over k.Let GbetheGalois group. Ifuisanautomorphism ofk(rx)/k then urxisalso aroot ofxn-a.Hence urx =warxwhere Waisann-th root ofunity, notnecessarily primitive. The map u1---+Waisobviouslyahomomorphism ofGinto thegroup ofn-th roots ofunity, and isinjective. Since asubgroup ofacyclic group iscyclic,weconclude that Giscyclic, oforder d,anddin. The image ofGisacyclic group oforder d. Ifuisagenerator ofG,then ClJuisaprimitive dth root ofunity. Now weget u(rxd)=(urx)d=(Warx)d=ad. Hence rxdisfixed under u,and therefore fixed under G.Itisanelement ofk,and our theorem isproved. 290 GALOIS THEORY VI,6 We now pass totheanalogue ofHilbert's theorem 90incharacteristic pfor cyclic extensions ofdegree p. Theorem 6.3. (Hilbert's Theorem 90,Additive Form). Letkbeafield and K/k acyclic extension ofdegreenwith group G.Let ()beagenerator ofG. Let {3EK.The trace Trt({3) isequal to0ifandonlyifthere exists anelement rxEKsuch that {3= rx-arx. Proof Ifsuch anelement rxexists, then we seethat the trace is0because the trace isequal tothe sum taken over allelements ofG,andapplyingaper- mutes these elements. Conversely, assume Tr(f3)=O.There exists anelement (JEKsuch that Tr((J)=FO.Let a;=Tre)[pea +(P+up)ea2+...+(P+up+...+u"-2p)ean- ']. From thisitfollows atonce that {3= rx-arx. Theorem 6.4. (Artin-Schreier) Let kheafield ofcharacteristic p. (i)LetKbeacyclic extension ofkofdegree p.Then there exists r:1EKsuch that K=k(r:1) and r:1satisfies anequation XP-X-a=0with some aEk. (ii)Conversely, givenaEk,thepolynomial f(X)=XP-X-aeither has one root ink,inwhich case allitsroots are ink,oritisirreducible. In thislatter case,ifrxisaroot then k(r:1)iscyclic ofdegree pover k. Proof LetK/k becyclic ofdegree p.Then Trf(-1)=0(itisjust the sum of-1with itself ptimes). Let abe agenerator oftheGalois group. Bythe additive form ofHilbert's theorem 90,there exists rxEKsuch that arx-rx=1, orinother words, arx = rx+1.Hence airx = rx+iforallintegers i=1,...,p and rxhaspdistinct conjugates. Hence [k(rx):k]>p.Itfollows that K=k(rx). We note that a(rxP-rx)=a(rx)P-a(r:1)=(rx+I)P-(rx+1)=r:1P-rx. Hence rxP-rxisfixed under a,hence itisfixed under the powers ofa,and therefore under G.Itliesinthefixed field k.Ifwelet a= rxP-rxwe seethat ourfirst assertion isproved. Conversely, let aEk.If rxisaroot ofXP-X-athen rx+iisalso a root for i=1,...,p.Thus f(X) has pdistinct roots. Ifone root lies ink then allroots lieink.Assume that noroot lies ink.We contend that the VI,7 SOLVABLE AND RADICAL EXTENSIONS 291 polynomialisirreducible. Suppose that f(X)=g(X)h(X) with g,hEk[X] and 1<deg g<p.Since p f(X)=n(X-(l-i) i=1 we seethatg(X) isaproduct over certain integers i.Let d=deg g.The co- efficient ofXd-1in9isasum ofterms -(a +i)taken over precisely dintegers i.Hence itisequal to-do: +jfor some integer j.But d=1=0ink,and hence 0:liesink,because thecoefficients of9lieink,contradiction. Weknow therefore thatf(X) isirreducible. All roots lieink(a), which istherefore normal over k. Since f(X) has nomultiple roots, itfollows that k«(l) isGalois over k.There exists anautomorphism aofk«(l) over ksuch that a(l =(l+ 1(because (l+ 1 isalso aroot). Hence thepowersaiofagive ai(l =(l+ifori=1,...,pand aredistinct. Hence theGalois group consists ofthese powers and iscyclic, there byproving thetheorem. Forcyclic extensions ofdegree pr,seetheexercises onWitt vectors and the bibliographyattheend of8. 7. SOLVABLE AND RADICAL EXTENSIONS Afinite extension Elk(which weshall assume separable forconvenience) is said tobesolvable iftheGalois group ofthesmallest Galois extension Kofk containing Eisasolvable group. This isequivalent tosaying that there exists a solvable Galois extension Lofksuch that kcEeL. Indeed, we have kcEeKe Land G(Klk) isahomomorphic image ofG(Llk). Proposition 7.1. Solvable extensionsformadistinguished class ofextensions. Proof LetElk besolvable. LetFbeafield containing kand assume E,f' aresubfields ofsome algebraically closed field. LetKbeGalois solvable over k, and EcK.Then KF isGalois over FandG(KFIF) isasubgroup ofG(Klk) byTheorem 1.12. Hence EFIFissolvable. Itisclear that asubextension ofa solvable extension issolvable. Let E::JF::Jkbeatower, and assume thatElF issolvable and FIkissolvable. LetKbeafinite solvable Galois extension ofk containing F.Wejustsaw thatEKIK issolvable. LetLbeasolvable Galois extension ofKcontaining EK. Ifaisanyembedding ofLover kinagiven algebraic closure, then aK=Kand hence aLisasolvable extension ofK.We letMbethecompositum ofallextensions aLforallembeddingsaofLover k. 292 GALOIS THEORY VI,7 Then MisGalois over k,and istherefore Galois over K.The Galois group of Mover Kisasubgroup oftheproduct nG(uLIK) (1 byTheorem 1.14. Hence itissolvable. We have asurjective homomorphism G(Mlk)-.G(Klk) byTheorem 1.10. Hence theGalois group ofM/k has a solvable normal subgroup whose factor group issolvable. Itistherefore solvable. Since EcM,ourproofiscomplete. EK/ E K/ F k Afinite extension Fofkissaid tobesolvable byradicals ifitisseparable and ifthere exists afinite extension Eofkcontaining F,andadmittingatower decomposition k=EoCE1CE2C... CEm=E such that each step E;+liE; isoneofthefollowing types: 1.Itisobtained byadjoiningaroot ofunity. 2.Itisobtained byadjoiningaroot ofapolynomial xn-awith aEEiand nprime tothecharacteristic. 3.Itisobtained byadjoiningaroot ofanequation XP-X-awith aEEiifPisthecharacteristic >o. One can see atonce that the class ofextensions which are solvable by radicals isadistinguished class. Theorem 7.2. Let Ebeaseparable extension ofk.Then Eissolvable by radicals ifandonlyifElk issolvable. Proof Assume that Elk issolvable, and letKbe afinite solvable Galois extension ofkcontaining E.Let mbetheproduct ofallprimes unequal tothe characteristic dividing thedegree [K:k],and letF=k«() where (isaprimitive m-th root ofunity. Then FIkisabelian. WeliftKover F.Then KF issolvable over F.There isatower ofsubfields between Fand KF such that each step is cyclic ofprime order, because every solvable group admits atower ofsub- VI,8 ABELIAN KUMMER THEORY 293 groups ofthe same type, and we can useTheorem 1.10. ByTheorems 6.2and 6.4, weconclude that KF issolvable byradicals over F,and hence issolvable byradicals over k.This proves thatElk issolvable byradicals. KF/ K Fk/ Conversely, assume thatElk issolvable byradicals. For anyembedding(J ofEinE8over k,theextension (JElk isalso solvable byradicals. Hence the smallest Galois extension KofEcontaining k,which isacomposite ofEand itsconjugates issolvable byradicals. Let mbetheproduct ofallprimes unequal tothecharacteristic dividing thedegree [K:k]andagain letF=k(,)where' isaprimitive m-th root ofunity. Itwill suffice toprove that KF issolvable over F,because itfollows then that KF issolvable over kand hence G(Klk) issolvable because itisahomomorphic image ofG(KF Ik). ButKFIFcan bedecomposed into atower ofextensions, such that each step isprime degree and ofthetype described inTheorem 6.2 orTheorem 6.4, and thecorresponding root ofunity isinthefield F.Hence KFIF issolvable, and ourtheorem isproved. Remark. One could modify ourpreceding discussion bynotassuming separability. Then one must deal with normal extensions instead ofGalois extensions, and one must allow equations XP-ainthesolvability byradicals, with pequal tothecharacteristic. Then westill have thetheorem corresponding toTheorem 7.2. Theproof isclear inview ofChapter V,6. For aproof that every solvable group isaGalois group over therationals, I refer toShafarevich [Sh54], aswell ascontributions ofIwasawa [Iw53]. [lw53] K.IWAsAwA, Onsolvable extension ofalgebraic number fields, Ann. ofMath. 58(1953), pp.548-572 [Sh54] I.SHAF AREVICH, Construction offields ofalgebraic numbers with given solvable Galois group, lzv. Akad. Nauk SSSR 18(1954), pp. 525-578 (Amer. Math. Soc. Transl. 4(1956), pp.185-237) 8. ABELIAN KUMMER THEORY Inthis section weshall carry out ageneralization ofthetheorem concerning cyclic extensions when theground field contains enough roots ofunity. Let kbe afield and mapositive integer. AGalois extension Kofkwith group Gissaid tobeofexponentmif(Jm = 1forall (JEG. 294 GALOIS THEORY VI,8 We shall investigate abelian extensions ofexponentm.We first assume that misprime tothecharacteristic ofk,and that kcontains aprimitive m-th root ofunity. Wedenote byPmthegroup ofm-th roots ofunity. We assume that allouralgebraic extensions inthis section arecontained inafixed algebraic closure ka . Let aEk.The symbol al/", (or)isnotwell defined. Ifrxm=aand(is anm-th root ofunity, then «(rx)"'=aalso. We shall use thesymbol al/mto denote any such element rx,which will becalled anm-th root ofa.Since the roots ofunity areintheground field, weobserve that thefield k(rx) isthe same nomatter which m-th root rxofaweselect. Wedenote this field byk(al/m). Wedenote byk*m thesubgroup ofk*consisting ofallm-th powers ofnon- zero elements ofk.Itistheimage ofk*under thehomomorphismx1---+xm . LetBbeasubgroup ofk*containing k*m. Wedenote byk(BI/m)orKBthe composite ofallfields k(al/m)with aEB.Itisuniquely determined byBasa subfield ofka . Let aEBand let rxbeanm-th root ofa.Thepolynomial Xm-asplits into linear factors inKB,and thus KBisGalois over k,because this holds forall aEB.Let GbetheGalois group. Let UEG.Then urx =W(lrx for some m-th root ofunity W(IEPmCk*. The map UJ-+W(I isobviouslyahomomorphism ofGinto Pm' i.e.fort,UEGwehave turx=WtW(lrx=W(lWtrx. We may write W(1=urx/rx. This root ofunity W(1isindependent ofthechoice ofm-th root ofa,forifrx'isanother m-th root, then rx'=(rxfor some (EPm' whence urx'/rx'=(urx/(rx=urx/rx. Wedenote W(Iby(u,a). The map (u,a)1---+(u,a) gIves usamap GxB-+Pm. Ifa,bEBand rxm=a,pm=bthen (rx{3)m=aband u(rxf3)/rx{3=(urx/rx)(u{3/{3). Weconclude that themap above isbilinear. Furthermore, ifaEk*mitfollows that <u,a)=1. Theorem 8.1. Letkbeafield,maninteger> 0prime tothecharacteristic of k,and assume that aprimitive m-th rootofunity liesink.LetBbeasubgroup ofk*containing k*m and letKB=k(BI/m).Then KBisGalois, and abelian ofexponentm.Let GbeitsGalois group. We have abilinear map GxB-+Pm given by (u,a)J-+(u,a). VI,8 ABELIAN KUMMER THEORY 295 IfUEGand aEB,and rxm=athen <u,a)=urx/rx. The kernel ontheleftis1 and thekernel ontheright isk*m. The extension KB/k isfinite ifandonlyif (B :k*m) isfinite. Ifthat isthe case, then B/k*m=G", and inparticularwehave theequality [K B:k]=(B:k*m). Proof Let uEG.Suppose (u,a)= 1forallaEB.Then forevery gener- ator rxofKBsuch that rxm=aEBwehave urx=rx.Hence uinduces theidentity onKBand thekernel ontheleft is1.Let aEBand suppose (u,a)=1forall uEG.Consider thesubfield k(a1/m)ofKB.Ifa1/misnot ink,there exists an automorphism ofk(a11m)over kwhich isnot theidentity. Extend this auto- morphism toKB,and call this extension u.Then clearly (u,a) =F1.This proves ourcontention. Bytheduality theorem ofChapter I,9we seethat Gisfinite ifandonly ifB/k*m isfinite, and inthat case wehave theisomorphism asstated, sothat inparticular theorder ofGisequal to(B:k*m), thereby proving thetheorem. Theorem 8.2. Notation being asinTheorem 8.1, the map B KBgives a bijection ofthe setofsubgroups ofk*containing k*mand theabeUan extensions ofkofexponent m. Proof LetBb B2besubgroups ofk*containing k*m. IfB1cB2then k(B}/m)ck(B/m). Conversely, assume that k(B}/m)ck(B/m). We wish to prove B1cB2.LetbEB1.Then k(b1/m)ck(B/m) andk(b1/m)iscontained in afinitely generated subextension ofk(B/m). Thus wemay assume without loss ofgenerality that B2/k*m isfinitely generated, hence finite. LetB3bethesub- group ofk*generated byB2and b.Then k(B/m)=k(B/m) and from what we saw above, thedegree ofthis field over kisprecisely (B2:k*m) or (B 3:k*m). Thus these two indices areequal, and B2=B3. This proves that B1CB2. We now have obtained aninjection ofour setofgroups Binto the setof abelian extensions ofkofexponent m.Assume finally that Kisanabelian extension ofkofexponent m.Any finite subextension isacomposite ofcyclic extensions ofexponentmbecause any finite abelian group isaproduct of cyclic groups, and we canapply Corollary 1.16. ByTheorem 6.2, every cyclic extension can beobtained byadjoininganm-th root. Hence Kcan beobtained byadjoiningafamily ofm-th roots, say m-th roots ofelements {bj}jeJwith bjEk*. Let Bbethesubgroup ofk*generated byallbjand k*m. Ifb' =bam with a,bEkthen obviously k(b'1/m)=k(b11m). Hence k(B1/m)=K, asdesired. 296 GALOIS THEORY VI,8 When wedeal with abelian extensions ofexponent pequal tothechar- acteristic, then wehave todevelopanadditive theory, which bears the same relationship toTheorems 8.1and 8.2 asTheorem 6.4bears toTheorem 6.2. Ifkisafield, wedefine theoperator by (x)=xP-x for xEk.Then isanadditive homomorphism ofkinto itself. Thesubgroup (k) plays the same role asthesubgroup k*m inthemultiplicative theory, whenever misaprime number. Thetheory concerningapower ofpisslightly more elaborate and isdue toWitt. We now assume khascharacteristic p.Aroot ofthepolynomialXP-X-a with aEkwill bedenoted by KJ-1a.IfBisasubgroup ofkcontaining k weletKB=k(-1B)bethefield obtained byadjoining-1atokforallaEB. Weemphasize thefact that Bisanadditive subgroup ofk. Theorem 8.3. Let kbeafield ofcharacteristic p.The map B1---+k(-1B) isabijection between subgroups ofkcontaining kand abelian extensions of kofexponent p.Let K =KB=k(-1B),and letGbeitsGalois group. If(1EGand aEB,and(X=a,let«(1,a)= (1(X-(X.Then wehave abilinear map GxB Z/pZ given by «(1,a) «(1,a). The kernel ontheleft is1and thekernel ontheright istJk. The extension KB/k isfinite ifandonlyif(B:k) isfinite andifthat isthe case, then [K B:k]=(B:k). Proof. Theproofisentirely similar totheproof ofTheorems 8.1and 8.2. Itcan beobtained byreplacing multiplication byaddition, andusing the"-th root" instead ofanm-th root. Otherwise, there isnochange inthewording of theproof. The analogous theorem forabelian extensions ofexponent pnrequires Witt vectors, and will bedeveloped intheexercises. Bibliography [Wi 35] E,WIlT, Der Existenzsatz fur abelsche Funktionenkorper, J,relne angew. Math. 173(1935), pp,43-51 [Wi 36] E.WITT, Konstruktion von galoisschen Korpern der Charakteristik pmit vorgegebener Gruppe derOrdung pf,J.reine angew. Math. 174(1936), pp. 237-245 [Wi 37] E,WITT, Zyklische Korper undAlgebren derCharakteristik pvom Grad pn. Struktur diskret bewerteter perfekter Korper mit vollkommenem Restklas- senkorper derCharakteristik p,J.reine angew. Math. 176(1937), pp, 126- 140 VI,9THE EQUATION Xn-a=0297 9. THE EQUATION Xn-B=0 When theroots ofunityarenotintheground field, theequation xn-a=0 isstillinteresting but alittle more subtle totreat. Theorem 9.1. Letkbeafield and naninteger>2.Let aEk,a=FO.Assume thatforallprime numbers psuch that pin wehave artkP ,andif41nthen art-4k4 .Then Xn-aisirreducible ink[X]. Proof Our first assumptionmeans that aisnot ap-th power ink.We shall reduce our theorem tothe case when nisaprime power, byinduction. Write n=prmwith pprime tom,and podd. Let m xm-a=n(X-v) v= 1 bethefactorization ofxm-ainto linear factors, and say=1.Substituting XprforXweget m xn-a=xprm-a=n(Xpr -v). v=l We may assume inductively thatxm-aisirreducible ink[X]. We contend that isnot ap-th power ink(). Otherwise,=pP,PEk(). Let Nbethe norm fromk() tok.Then -a=(-l)mN()=(-l)mN(pp)=(-l)mN(p)P. Ifmisodd, aisap-th power, which isimpossible. Similarly, ifmiseven and p isodd, wealso getacontradiction. This provesourcontention, because mis prime top.Ifweknow our theorem forprime powers, then weconclude that Xpr -isirreducible overk(). IfAisaroot ofXp"-r:1then kck(r:1)ck(A) givesatower, ofwhich thebottom step hasdegreemand thetopstep hasdegree proItfollows that Ahasdegreenover kand hence that Xn-aisirreducible. We now suppose that n=prisaprime power. Ifpisthecharacteristic, let beap-th root ofa.Then XP-a=(X-)P and hence Xpr -a=(XPr-1 -)Pifr>2.Byanargument even more trivial than before, we see that aisnot ap-th power ink(a), hence inductively )(pr-I-aisirreducible over k(a). Hence)(pr-aisirreducible over k. Suppose that pisnot thecharacteristic. Wework inductively again, and let bearoot ofXP-a. Supposeaisnot ap-th power ink.We claim that XP-aisirreducible. Otherwise aroot aofXP-ageneratesanextension k(a) ofdegree d<P and aP=a,Taking the norm from k(a) tokwegetN(a)P=ad. Since dis prime top,itfollows that aisap-th power ink,contradiction. 298 GALOIS THEORY VI,9 Let r>2.Welet rx= rx 1.We have P XP-a=n(X-rxv)\'=1 and p )(pr-a=IT(Xpr-I-av). v=I Assume that rxisnot ap-th power ink(rx). LetAbearoot ofXpr-1 -rx.Ifp isodd then byinduction, Ahasdegree pr-lover k(rx), hence hasdegree prover kand we aredone. Ifp=2,supposerx=-4p4 with PEk(rx). LetNbethe norm from k(a) tok.Then -a=N(a)=16N({3)4,so-aisasquare ink.Since p=2wegetv=I Ek(a) and a=(v=I 2(32)2,acontradiction. Hence again byinduction, wefind that Ahasdegree prover k.We therefore assume that a=/3Pwith some /3Ek(a), and derive the consequences. Taking the norm from k(rx) tokwefind -a=(-1)P N(rx)=(-I)PN(PP)=(-l)PN(fJ)P. Ifpisodd, then aisap-th power ink,contradiction. Hence p=2,and -a=N(P)2 isasquare ink.Write -a=b2with bEk.Since aisnot asquare inkwecon- clude that-1isnot asquare ink.Let i2= -1.Over k(i) wehavethefactoriza- tion X2r_a=X2"+b2=(X2r-1+ib)(X2r-1 -ib). Each factor isofdegree 2r-1and weargue inductively. IfX2r-1+ibisreducible over k(i)then +ibisasquare ink(i) orliesin-4(k(i))4. Ineither case, +ibisa square ink(i),say +ib =(e+di)2=c2+2edi-d2 with c,dEk.Weconclude that c2=d2orc=+d,and+ib=2cdi=+2c2;. Squaring givesacontradiction, namely a=-b2=-4c4 . We now conclude byunique factorization that X2r+b2cannot factor in k[X], thereby provingour theorem. The conditions ofourtheorem arenecessary because X4t4b4=(X2+2bX +2b2)(X2-2bX +2b2). Ifn=4mand aE-4k4then xn-aisreducible. VI,9 THE EQUATION Xn-a=0299 Corollary 9.2. Let kbeafield and assume that aEk,a=F0,and that aisnot ap-th power for some prime p.Ifpisequal tothecharacteristic, orifpisodd, thenfor every integer r> 1thepolynomial Xpr -aisirreducible over k. Proof The assertion islogically weaker than theassertion ofthetheorem. Corollary 9.3. Let kbeafield and assume that thealgebraic closure kaofk isoffinite degree >lover k.Then ka=k(i) where i2= -1,and khas characteristic o. Proof. We note that k8isnormal over k.Ifk8isnotseparable over k,so char k=p>0,then /(!lispurely inseparable over some subfield ofdegree> 1(byChapter V,6), and hence there isasubfield Econtaining k,and anelement aEEsuch that XP-aisirreducible over E.ByCorollary 9.2, k8cannot beof finite degree over E.(The reader may restrict hisorherattention tocharacteristic oifChapter V,6wasomitted.) We may therefore assume that kaisGalois over k.Let k1=k(i). Then ka isalso Galois over k1.Let GbetheGalois group ofka/kI.Suppose that there isaprime number pdividing theorder ofG,and letHbeasubgroup oforder p. LetFbeitsfixed field. Then [ka :F]=p.Ifpisthecharacteristic, then Exercise 29attheend ofthechapter willgive thecontradiction. We may assume that p isnot thecharacteristic. Thep-th roots ofunity =F 1arethe roots ofapoly- nomial ofdegree<p-1(namely Xp-1+...+1),and hence must lieinF. ByTheorem 6.2, itfollows that kaisthesplitting field ofsome polynomial XP-awith aEF.The polynomial XP2 -aisnecessarily reducible. Bythe theorem, wemust have p=2and a=-4b4with bEF.This implies ka=F(al/2)=F(i). But weassumed iEk1,contradiction. Thus wehave proved/(!l=k(i). Itremains toprove that char k=0,andfor this Iuse anargument shown tomebyKeith Conrad. Wefirst show that asum ofsquares inkisasquare. Itsuffices toprove this for asum oftwo squares, and inthis case wewrite anelement x+iyEk(i)=k8asasquare. x+iy=(u+iv)2, x,y,u,vEk, and then x2+y2=(u2+v2)2.Then toprove khascharacteristic 0,wemerely observ thatifthecharacteristic is>0,then-1isafinite sum 1+ ....+1, whence asquare bywhat wehave just shown, but k8=k(i), sothis concludes theproof. Corollary 9.3 isdue toArtin; see[Ar24], given attheendofChapter XI. Inthatchapter, much more will beproved about thefield k. Example 1. Let k=Qand letGQ=G(Qa/Q). Then theonly non-trivial torsion elements inGQhave order 2.Itfollows from Artin's theory (asgiven inChapter XI) that allsuch torsion elements areconjugate inGQ.One uses Chapter XI, Theorems 2.2, 2.4, and2.9.) 300 GALOIS THEORY VI,9 Example 2. Let kbe afield ofcharacteristic notdividingn.Let aEk, a =t=0and letKbethesplitting field ofXn -a.Let abe one root of xn -a,and let(be aprimitive n-th root ofunity. Then K=k(a, ()=k(a, fLn). We assume thereader isacquainted with matrices over acommutative ring. Let uEGK/k.Then (ua)n=a,sothere exists some integer b=b(u) uniquely determined mod n,such that u(a)=ar'(u). Since uinduces anautomorphism ofthecyclic group fLn' there exists aninteger d(u)relatively primetonanduniquely determined mod nsuch that u(() (d(u). LetG(n) bethesubgroup ofGL2(Z/nZ)consisting ofallmatrices M=G)with bEZ/nZ and dE(Z/nZ)*. Observe that#G(n)=ncp(n). We obtain aninjective map 0" M(O")=(b()d(»)ofGK1k=-+G(n), which isimmediately verified tobe aninjective homomorphism. The question arises, when isitanisomorphism? The next theorem givesananswer over some fields, applicable especially totherational numbers. Theorem 9.4. Letkbeafield. Let nbeanoddpositive integer prime tothe characteristic, and assume that[k(fLn):k]=cp(n). Let aEk,and suppose that for each prime pintheelement aisnot ap-th power ink.LetKbethesplitting field ofxn -aover k.Then the above homomorphism u M(u) isan isomorphism ofGK/kwith G(n). The commutator group isGal(K/k(fLn))'so k(fLn) isthemaximal abelian subextension ofK. Proof. This isaspecialcase ofthegeneral theory of 11,and Exercise 39, taking into account therepresentation ofGK/kinthegroup ofmatrices. One need onlyuse thefact that theorder ofGK/kisncp(n), according tothat exercise, and so#(G K/k)=#G(n), soGK/k=G(n). However, weshall givenanindependent proofasanexample oftechniques ofGalois theory.We prove thetheorem by induction. Suppose first n=pisprime. Since[k(fLp):k]=p-1isprime top,it follows that ifais aroot ofXP-a,then k(a) nk(fLp)=kbecause [k(a):k]=p.Hence [K :k]=p(p-1), soGK/k=G(p). Adirect computation of acommutator ofelements inG(n) forarbitraryn shows that the commutator subgroup iscontained inthegroup ofmatrices G).bEZ/nZ. VI,9 THE EQUATION Xn-a=0301 and somust bethatsubgroup because itsfactor group isisomorphicto(ZjnZ)* under theprojectiononthediagonal. This proves thetheorem when n=p. Now letpin and write n=pm. Then [k(J1m):k]=cp(m), immediately from thehypothesis that [k(J1n):k]=cp(n). Let abe aroot ofxn -a,and let f3=aPeThen f3isaroot ofxm -a,andbyinduction we canapply thetheorem toxm -a.The field diagram isasfollows. /k(a,J.tn) k(l3, J.tn) k(J.tn)/k(a) k(l3fP'k Since ahasdegree pm over k,itfollows that acannot have lower degree than pover k(f3),so[k(a):k(f3)]=pandXP -(3isirreducible over k(f3).Weapply thefirst part oftheproof toXP-f3over k(f3). The property concerning the maximal abelian subextension ofthesplitting field shows that k(a)nk(f3, J1n)=k(f3). Hence [k(a, J1n):k(f3, J1n)]=p.Byinduction, [k(f3, J1n):k(J1n)]=m,again because ofthemaximal abelian subextension ofthesplitting field ofxm -a over k.This proves that[K:k]=ncp(n), whence GK/k=G(n), andthecommutator statement hasalready been proved. This concludes theproof ofTheorem 9.4. Remarks. When niseven, there are some complications, because for instance Q(V2) iscontained inQ(J18),sothere aredependence relations among thefields inquestion. The non-abelian extensions, asinTheorem 9.4, areof intrinsic interest because they constitute thefirst examples ofsuch extensions that come tomind, buttheyarose inother important contexts. For instance, Artin used them togiveaprobabilistic model forthedensity ofprimes psuch that 2(say)iaprimitive root mod p(that is,2generates thecyclic group (ZjpZ)*. Instead of2hetook any non-square integer=t=+1.Atfirst, Artin did notrealize explicitly the above type ofdependence, and socame toananswer that was offbysome factor insome cases. Lehmer discovered thediscrepancy bycomputations. AsArtin then said, one has tomultiply bythe"obvious" factor which reflects thefield dependencies. Artin never published hisconjecture, but thematter isdiscussed indetail byLang-Tate intheintroduction tohiscollected papers (Addison- Wesley, Springer Verlag). Similar conjectural probabilistic models were constructed byLang-Trotter in connection with elliptic curves, and more generally with certain p-adic repre- sentations oftheGalois group, in"Primitive pointsonelliptic curves", Bull. AMS 83No.2 (1977), pp.289-292; and[LaT 75](end of 14). Forfurther comments onthep-adic representations ofGalois groups,see 14 and 15. 302 GALOIS THEORY VI,10 10. GALOIS COHOMOLOGY Let Gbeagroup and Aanabelian group which wewrite additively forthe general remarks which wemake, preceding our theorems. Let usassume that Goperates onA,bymeans ofahomomorphism G-.Aut(A). Byal-cocycle of GinAone means afamily ofelements {rxa} aEGwithrxaEA,satisfying therelations rxu+urxr= rxar forallu,tEG.If{rxa}aEG and {Pa}aEG arel-cocycles, then we can add them to getal-cocycle {rxa+Pa} aEG.Itisthen clear that l-cocycles form agroup, denoted byZl(G, A).Byal-coboundary ofGinAone means afamily ofele- ments {rxa}aEG such that there exists anelement pEA forwhichrxa=up-P forall uEG.Itisthen clear that aI-coboundary isal-cocycle, and that the l-coboundaries form agroup, denoted byB1(G,A). The factor group Zl(G, A)jB1(G,A) iscalled thefirst cohomology group ofGinAand isdenoted byH1(G,A). Remarks. Suppose Giscyclic. Let TrG: A AbethehomomorphismaLu(a). UEG Let ybe agenerator ofG.Let(I-y)A bethesubgroup ofAconsisting ofall elements a-y(a) with aEA.Then (I-y)A iscontained inkerTrG. The reader willverifyasanexercise that there isanisomorphism kerTrG/(1-y)A=H1(G, A). Then the next theorem for acyclic group isjust Hilbert's Theorem 90of6. Cf.also thecohomology ofgroups, Chapter XX, Exercise 4,for aneven more general context. Theorem 10.1. LetKjk be afinite Galois extension with Galois group G. Then }'or theoperation ofGonK* wehave H1(G,K*)=I,andfor the operation ofGontheadditive group ofKwehave H1(G,K)=O.Inother words, thefirst cohomology group istrivial inboth cases. Proof. Let {rxa}aEG beal-cocycle ofGinK*. Themultiplicative cocycle relation reads a rxarxt=rxat. VI,10 GALOIS COHOLOLOGY 303 Bythelinear independence ofcharacters, there exists (JEKsuch that theelement P=LC<r!«(J) TeG is=FO.Then up=LU!«(J)=LC<arC<; lut«(J) reG reG =C<;1LC<aTU!((J)=C<a-1p. reG Weget C<a=PluP, andusing P-1instead ofPgives what wewant. For theadditive part ofthetheorem, wefind anelement (JEKsuch that the trace Tr((J)isnotequal toO.Given al-cocycle {C<a} intheadditive group ofK, welet 1 P= Tr(O)tGIXt.(0). Itfollows atonce thatC<a=P-up,asdesired. The next lemma will beapplied tothenon-abelian Kummer theory ofthe next section. Lemma 10.2. (Sah). Let Gbeagroup and letEbeaG-module. Lettbein the center ofG.Then H1(G,E)isannihilated bythemap x1-+tX-xonE. Inparticular, ifthis map isanautomorphism ofE,then H1(G,E)=O. Proof. Letfbeal-cocycle ofGinE.Then f(u) =f(1:u1:-1)=f(1:) +1:(f(ut-1) =f(1:) +1:[/(u) +uf(t-1)]. Therefore !f(u)-f(u)= -utf(1:- 1)-f(t). Butf(l)=f(l) +f(l)impliesf(l)=0,and o=f(l)=f(!1:- 1)=f(t) +1:f(t-1). This shows that(!-l)f(u)=(u-l)f(1:),sofisacoboundary. This proves thelemma. 304 GALOIS THEORY VI, 11 11. NON-ABELIAN KUMMER EXTENSIONS We areinterested inthesplitting fields ofequations xn-a=0when the n-th roots ofunity are notcontained intheground field. More generally,we want toknow roughly (or aspreciselyaspossible) theGalois group ofsimul- taneous equations ofthis type. For this purpose, weaxiomatize thepattern ofproof toanadditive notation, which infact makes iteasier toseewhat is gOIng on. We fix aninteger N>1,and weletMrangeover positive integers divid- ingN.We letPbethe setofprimes dividing N.We letGbeagroup, and let: A=G-module such that theisotropy group ofany element ofAisoffinite index inG .We also assume that Aisdivisible bytheprimes piN, that is pA=A forallpEP. r=finitely generated subgroup ofAsuch thatrispointwise fixed byG. We assume that ANisfinite. Then risalsofinitely generated. Note that 1-r::JAN. N Example. For our purposes here, the above situation summarizes the properties which hold inthefollowing situation. LetKbeafinitely generated field over therational numbers, oreven afinite extension oftherational numbers. WeletAbethemultiplicative group ofthealgebraic closure Ka .WeletG=GK betheGalois group Gal(KajK). We letrbeafinitely generated subgroup of themultiplicative group K*. Then alltheabove properties aresatisfied. We seethat AN=J1Nisthegroup ofN-throots ofunity. The group written r inadditive notation iswritten r1/Ninmultiplicative notation. Next wedefine theappropriate groups analogous totheGalois groups of Kummer theory, asfollows. For any G-submodule BofA,welet: G(B)=image ofGinAut(B), G(N)=G(A N)=image ofGinAut(A N), H(N)=subgroup ofGleaving ANpointwise fixed, Hr(M, N)(forMIN)=image ofH(N) inAut(r} VI, 11 NON-ABELIAN KUMMER EXTENSIONS 305 Then wehave anexact sequence: o-+Hr<M, N)-+G(r+AN)-+G(N)-+O. Example. Inthe concrete case mentioned above, the reader will easily recognize these various groupsasGalois groups. Forinstance, letAbethe multiplicative group. Then wehave thefollowing lattice offield extensions with corresponding Galois groups: G(r1/MpN)K( r1/M )}JlN'1Hr(M, N) K(PN) }1 G(N) K Inapplications,wewant toknow how much degeneracy there iswhen wetrans- lateK(PM' r1/M)over K(PN) with MIN. This isthe reason weplay with the pair M,Nrather than asingle N. Let usreturn toageneral Kummer representationasabove. We arein- terested especially inthat part of(ZINZ)* contained inG(N), namely thegroup ofintegersn(mod N)such that there isanelement En]inG(N) such that [n]a=na forall aEAN. Such elements arealways contained inthe center ofG(N), and are called homotheties. Write N =f1pn(p) Let Sbeasubset ofP.We want tomake some non-degeneracy assumptions about G(N). Wecall Sthespecial set. There isaproduct decomposition (ZINZ)*=n(Zlpn(p)z)*. piN If21Nwesuppose that 2ES.For each pESwesuppose that there isaninteger c(p)=pf(p)withf(p)>1such that G(A )nV n(Zlpn(p)z)* , N:::::> e(p)X peS p,S whereVe(P)isthesubgroup ofZ(pn(p») consisting ofthose elements =1mod c(P). 306 GALOIS THEORY VI, 11 Theproduct decompositionontheright isrelative tothedirect sum decom- position AN=EBApn(p).piN The above assumption will becalled thenon-degeneracy assumption. The integers c(p) measure theextent towhich G(A N)isdegenerate. Under thisassumption,weobserve that [2] EG(A M)ifMIN and Misnotdivisible byprimes ofS; [1+c]EG(A M)ifMINand Misdivisible only byprimes ofS, where c=c(S)=nc(p). peS We can then use[2]-[1]=[1]and [1+c]-[1]=[c]inthe context of Lemma 10.2, since [1]and[c] areinthe center ofG. For any Mwedefine c(M)=nc(p). plM peS Define 1r' = -rnAG N and theexponent e(r'/r)=smallest positive integeresuch that er' cr. Itisclear that degeneracy intheGalois group Hr(M, N)defined above can arise from lots ofroots ofunity intheground field, oratleast degeneracy in theGalois group ofroots ofunity; and also ifwelook atanequation XM-a=0, from thefact that aisalready highly divisible inK.This second degeneracy would arise from theexponent e(r'/r),ascan beseen bylooking attheGalois group ofthedivisions ofr.The next theorem shows that these aretheonly sources ofdegeneracy. We have theabelian Kummer pairing forMIN, Hr(M, N)xr/Mr-.AM given by (t,x)1-+ty-y, where yisany element such thatMy=x.The value ofthepairing isindepen- VI,11 NON-ABELIAN KUMMER EXTENSIONS 307 dent ofthechoice ofy.Thus for xEr,wehave ahomomorphism CfJx:Hr(M, N)-.AM such that CfJx(t)=ty-y, where My=x. Theorem 11.1. LetMIN. LetCfJbethehomomorphism CfJ:r-.Hom(Hr(M, N),AM) and letrq)beitskernel. LeteM(r)=g.c.d. (e(r'/r), M). Under the non- degeneracy assumption,wehave c(M)eM(r)r q>cMr. Proof Let xErand suppose CfJx=O.LetMy=x.For (JEGlet Ya=(JY-y. Then {Ya} isa1-cocycle ofGinAM' andbythehypothesis thatCfJx=0,this cocycle depends only ontheclass of (Jmodulo thesubgroup ofGleaving the elements ofANfixed. Inother words, wemay view {Ya}asacocycle ofG(N) in AM. Let c=c(N). ByLemma 10.2, itfollows that {cYa} splitsasacocycle of G(N) inAM. Inother words, there exists toEAMsuch that cYa=(Jto-to, and thisequation infact holds for (JEG.Let tbesuch that ct =to.Then c(JY-cy=(Jct-cy, whence c(y-t)isfixed byall (JEG,and therefore liesin r.Therefore N e(r'/r)c(y-t)Er. Wemultiply both sides byMand observe thatcM(y-t)=cMy=cx. This shows that c(N)e(r'/r)rqJcMr. Since r/Mr hasexponent M, wemay replace e(r'/r) bythegreatestcommon divisor asstated inthetheorem, and we canreplace c(N) byc(M) toconclude theproof. Corollary 11.2. Assume that Misprime to2(r':r)and isnotdivisible by anyprimes ofthespecial setS.Then wehave aninjection CfJ:r/Mf-.Hom(Hr<M, N),AM). 308 GALOIS THEORY VI,12 Ifin addition risfree with basis {af,...,ar},and weletCPi=CPa;,then themap Hr(M, N) A<;j given by !-.(CPt(!),. . .,CPr(i)) isinjective. IfAM iscyclic oforder M,this map isanisomorphism. Proof Under thehypotheses ofthecorollary, wehave c(M)= 1and CM(r)=1inthetheorem. Example. Consider the case ofGalois theory when Aisthemultiplicative group ofKa. Let af,...,arbeelements ofK*which aremultiplicatively inde- pendent. They generateagroupasinthecorollary. Furthermore, AM=PM iscyclic, sothecorollary applies. IfMisprime to2(r':r)and isnotdivisible byanyprimes ofthespecialsetS,wehave anisomorphism cp:r/Mr-.Hom(Hr(M, N),PM). 12. ALGEBRAIC INDEPENDENCE OF HOMOMORPHISMS Let Abeanadditive group, and letKbeafield. LetAt,...,An:A-.Kbe additive homomorphisms. We shall say that At, ..., Anarealgebraically dependent (over K) ifthere exists apolynomial !(X t,..., Xn)in K[X t,. . .,Xn] such that forallxEAwehave f(At(X),..., An(X))=0, but such thatfdoes not induce the zero function onK(n), i.e. onthedirect product ofKwith itself ntimes. Weknow that with each polynomial we can associate aunique reductXl polynomial giving the same function. IfKis infinite, the reduced polynomial isequal tofitself. Inour definition ofde- pendence, wecould aswell assume thatfisreduced. Apolynomial f(X b...,Xn)will becalled additive ifitinduces anadditive homomorphism ofK(n) into K. Let(Y)=(Yt,...,) bevariables inde- pendent from (X). Let g(X, Y)=f(X +Y)-f(X)-f(Y) where X+Yisthecomponentwise vector addition. Then thetotal degree of gviewed asapolynomial in(X) with coefficients inK[Y]isstrictly less than thetotal degree off, andsimilarly, itsdegree ineach Xiisstrictly less than the degree offineach Xi. One sees this easily byconsidering thedifference of monomials, VI, 12 ALGEBRAIC INDEPENDENCE OFHOMOMORPHISMS 309 M(v)(X+Y)-M(v)(X)-M(v)(Y) =(X t+Yt)VI. ..(X n+)Vn-X'11...Xn-Y'11...Y". Asimilar assertion holds forgviewed asapolynomial in(Y)with coefficients in K[X]. Iffisreduced, itfollows that gisreduced. Hence iffisadditive, itfollows that gisthe zero polynomial. Example. LetKhave characteristic p.Then inonevariable, themap 1---+apm for aEKand m>1isadditive, and given bytheadditive polynomial aXpm. Weshall seelater that this isatypical example. Theorem 12.1. (Artin). LetAb...' An:A-.Kbeadditive homomorph- isms ofanadditive group into afield. Ifthese homomorphisms arealge- braically dependent over K,then there exists anadditive polynomial f(Xb.. .,Xn)=F0 inK[X] such that f(At (x),...,An(X))=0 forallxEA. Proof Letf(X)=f(X b..., Xn)EK[X] be areduced polynomial of lowest possible degree such thatf=F0but forall xEA,f(i\(x))=0,where i\(x) isthe vector (AI(x), ..., An(X)). Weshall prove thatfis additive. Letg(X, Y)=f(X +Y)-f(X)-f(Y). Then g(i\(x), i\(y))=f(i\(x +y))-f(i\(x))-f(i\(y))=0 forallx,YEA. Weshall prove that ginduces the zero function onK(n) XK(n). Assume otherwise. Wehave two cases. Case 1.We have g(,i\(y))=0for all EK(n) and allYEA. By hypothesis, there exists'EK(n) such thatg(', Y)isnotidentically O.Let P(Y)=g(', Y). Since thedegree ofgin(Y) isstrictly smaller than thedegree off,wehave acontradiction. Case 2.There exist'EK(n) and y'EAsuch thatg(', i\(y')) =Fo.Let P(X)=g(X,i\(y')). Then Pisnotthe zero polynomial, butP(i\(x))=0forall xEA,againacontradiction. 310 GALOIS THEORY VI, 12 Weconclude that ginduces the zero function onK(n) XK(n), which proves what wewanted, namely thatfisadditive. We now consider additive polynomialsmore closely. Letfbe anadditive polynomial innvariables over K,and assume thatfis reduced. Let /;(X;)=f(O,. ..,Xi'...,0) with Xiinthei-thplace, and zeros intheother components. Byadditivity, it follows that f(Xl'...,Xn)=fl(X1)+...+fn{Xn) because thedifference oftheright-hand side and left-hand side isareduced polynomial taking the value 0onK(n). Furthermore, each/;isanadditive polynomial inone variable. We now study such polynomials. Letf{X) beareduced polynomial inonevariable, which induces alinear map ofKinto itself. Suppose that there occurs amonomial arXrinfwith coefficient ar=Fo.Then themonomials ofdegreerin g{X, Y)=f{X +Y)-f{X)-f{Y) aregiven by ar{X +y)r-arXr-aryr. We have alreadyseen that gisidenticallyO.Hence theabove expression is identically O.Hence thepolynomial {X+y)r_xr_yr isthe zero polynomial. Itcontains the term rxr-1Y.Hence ifr>1,our field must have characteristic pand risdivisible byp.Write r=pnls where sis prime top.Then o={X+yy-xr_yr ={Xpm +ypm)s_{Xpm)s_(ypm)s. Arguingasbefore, weconclude that s=1. Hence iffisanadditive polynomial inonevariable, wehave m f{X)=LavXPv, v=o with avEK.Incharacteristic 0,theonly additive polynomials inone variable areoftype aXwith aEK. Asexpected,wedefine A.t,...,A.ntobealgebraically independent if,whenever fisareduced polynomial suchthatf{i\(x»=0forallxEK,thenfis the zero polynomial. VI, 12 ALGEBRAIC INDEPENDENCE OFHOMOMORPHISMS 311 Weshall apply Theorem 12.1 tothe case when A.1,...,A.nareautomorphisms ofafield, and combine Theorem 12.1 with thetheorem onthelinear indepen- dence ofcharacters. Theorem 12.2. LetKbeaninfinite field, and let (Jl'...,(Jnbethedistinct elements ofafinite group ofautomorphisms ofK. Then (J1,...,(Jnarealge- braically independent over K. Proof (Artin). Incharacteristic 0,Theorem 12.1 and the linear inde- pendence ofcharacters show that our assertion istrue. Letthecharacteristic bep>0,and assume that (J1,...,(Jnarealgebraically dependent. There exists anadditive polynomial f(X 1,...,X n)inK[X] which is reduced, j'=F0,and such that f((J1(x),...,(Jn(x))=0 forallxEK.Bywhat we saw above, we canwrite this relation intheform n m LLair(Ji(X)pr=0 i=1 r=1 forallxEK,and with notallcoefficients airequal toO.Therefore bythelinear independence ofcharacters, theautomorphisms {up ,.r }.h. 1 d 1 WIt I=,..., n an r=,..., m cannot bealldistinct. Hence wehave (Jpr =(Jl!s, J with either i=Fjorr=Fs.Sayr<s.For allxEKwehave (Ji(X)pr=(Jj(x)ps. Extracting p-th roots incharacteristic pisunique. Hence () ()PS_r (pS-r ) (JiX =(Jjx =(Jjx forallxEK.Let (J=(J;I(Ji.Then s-r (J(x)=xP forallxEK.Taking (In =idshows that Pn(s-r)x=x forallxEK.Since Kisinfinite, this can hold only ifs=r.But inthat case, (Ji=(Jj,contradicting thefact that westarted with distinct automorphisms. 312 GALOIS THEORY VI, 13 13. THE NORMAL BASIS THEOREM Theorem 13.1. LetK/k beafinite Galois extension ojdegreen.Let (J1,...,(Jn betheelements oftheGalois group G.Then there exists anelement WEK such that (J1W,...,(JnWform abasis ofKover k. Proof. We prove this here only when kisinfinite. The case when kis finite can beproved later bymethods oflinear algebra,asanexercise. For each (JEG,letX(1beavariable, and lettu.t=X(1-1t. LetXi=X(1j.Let f{X b.. .,Xn)=det(t uj .(1j). Thenfisnotidentically 0,asone sees bysubstituting1forXidand 0forX(1if (J=FideSince kisinfinite,fis reduced. Hence thedeterminant will not be0for allxEKifwesubstitute (Ji{X) forXiinf. Hence there exists WEKsuch that det{(Ji-1(Jj{W))=FO. Suppose ab...,anEkaresuch that a1(J1(w)+...+an(Jn{w)=o. Apply (Ji-1tothis relation foreach i=1,..., n.Since ajEkwegetasystem of linear equations, regarding the ajasunknowns. Since thedeterminant ofthe coefficients is=F0,itfollows that a. =0Jforj=1,..., n and hence that Wisthedesired element. Remark. Interms ofrepresentationsasinChapters IIIandXVIII, the normal basis theorem says that therepresentation oftheGalois grouponthe additive group ofthefield istheregular representation. One may also say that Kisfree ofdimension lover thegroup ring k[G]. Such aresult may beviewed asthefirst step inmuch more subtle investigations having todowith algebraic number theory. LetKbe anumber field (finite extension ofQ)and let 0Kbe itsring ofalgebraic integers, which will bedefined inChapter VII, 1.Then one may askfor adescription of 0KasaZ[G]module, which isamuch more difficult problem. For fundamental work about thisproblem,seeA.Frohlich, Galois Module Structures ofAlgebraic Integers, Ergebnisse derMath. 3Folge Vol. 1,Springer Verlag (1983). See also thereference [CCFT 91]given atthe end ofChapter III, 1. VI, 14 INFINITE GALOIS EXTENSIONS 313 14. INFINITE GALOIS EXTENSIONS Althoughwehave already givensome ofthebasic theorems ofGalois theory already forpossibly infinite extensions, thenon-finiteness did notreally appear inasubstantial way. We now want todiscuss itsrole more extensively. LetK/kbe aGalois extension with group G.For each finite Galois subex- tension F, we have the Galois groups GKIF and GFlk.Put H=GKIF. Then Hhasfinite index, equal to#(G Flk)=[F:k].This just comes asaspecial case ofthegeneral Galois theory.We have acanonical homomorphism G G/H=GFlk. Therefore by the universal property ofthe Inverse limit, we obtain a homomorphism G limG/H, HEft where thelimit istaken forHinthefamily ofGaloisgroups GKIFasabove. Theorem 14.1. Thehomomorphism G limG/H isanisomorphism. Proof. First thekernel istrivial, because iflTisinthekernel, then lTrestricted toevery finite subextension ofKistrivial, and soistrivial onK.Recall that an element oftheinverse limit isafamily {lTH} with lTHEG/H,satisfyingacertain compatibility condition. This compatibility condition means that wemay define anelement ITofGasfollows. Let aEK.Then aiscontained insome finite Galois extension FCK.LetH=Gal(K/F). Let ua=uHa. Thecompatibility condition means that lTHa isindependent ofthechoice ofF.Then itisimmediately verified that ITisanautomorphism ofKover k,which maps toeachlTHinthe canonical map ofGinto G/H.Hence themap G lim. G/Hissurjective, thereby proving thetheorem. Remark. For thetopological interpretation,seeChapter I,Theorem 10. 1, and Exercise 43. Example. Let J1[pOC] betheunion ofallgroups ofroots ofunity J1[pn], where pisaprime and n=1,2,... rangesover thepositive integers. Let K=Q(J1[pOC]). Then Kisanabelian infinite extension ofQ.LetZpbethering ofp-adic integers, andZ;thegroup ofunits. From 3, weknow that(Z/pnz)* isisomorphictoGal(Q(J1[pn]/Q)). These isomorphismsarecompatible inthe tower ofp-th roots ofunity,soweobtain anisomorphism Z; Gal(Q(J1[pOO]/Q)). 314 GALOIS THEORY VI,14 Towers ofcyclotomic fields have been extensively studied byIwasawa. Cf. asystematic exposition andbibliography in[La90]. For other types ofrepresentations inagroup GL2(Zp),see Serre [Se68], [Se72], Shimura [Shi 71], andLang-Trotter [LaT 75]. One general framework inwhich therepresentation ofGalois groupsonroots ofunity can beseen has todowith commutative algebraic groups, starting with ellipticcurves. Specif- ically, consider anequation y2=4x3 -g2x-g3 with g2, g3EQand non-zero discriminant:=g-27g=t=o.The setof solutions together with apointatinfinity isdenoted byE.From complex analysis (orbypurely algebraic means), one sees thatifKisanextension ofQ,then the setofsolutions E(K) with x,yEKand00form agroup, called thegroup of rational points ofEinK.One isinterested inthetorsion group, sayE(Qa)tor of points inthealgebraic closure, orfor agiven prime p,inthegroup E(Qa)[pr] and E(Qa)[pOC]. As anabelian group, there isanisomorphism E(Qa)[pr]=(Zfprz)x(Zfprz), sotheGalois group operatesonthepoints oforder prvia arepresentation in GL2(Zfprz), rather thanGLt(Zfprz)=(Zfprz)* inthe case ofroots ofunity. Passing totheinverse limit, one obtains arepresentation ofGal(QafQ)=GQ inGL2(Zp).One ofSerre's theorems isthat theimage ofGQinGL2(Zp)isa subgroup offinite index, equal toGL2(Zp)forallbut afinite number ofprimes p,ifEnd C(E)=Z. More generally, using freely thelanguage ofalgebraic geometry, when Ais acommutative algebraic group, saywith coefficients inQ,then one may consider itsgroup ofpoints A(Qa)top and therepresentation ofGQin asimilar way. Developing thenotions todeal with these situations leads intoalgebraic geometry. Instead ofconsidering cyclotomic extensions ofaground field, one may also consider extensions ofcyclotomic fields. The following conjecture isdue to Shafarevich. See thereferences attheendof7. Conjecture 14.2. Letko=Q(J1) bethecompositum ofallcyclotomic exten- sions ofQinagiven algebraic closure Qa. Let kbeafinite extension ofko. Let Gk=Gal(Qafk). Then Gkisisomorphic tothecompletion ofafree group oncountably many generators. IfGisthefree group, then werecall that thecompletion isthe inverse limit lim GfH,taken over allnormal subgroups Hoffinite index. Readers should view thisconjectureasbeing inanalogytothesituation with Riemann surfaces, asmentioned inExample 9of2.Itwould beinteresting toinvestigate theextent towhich theconjecture remains valid ifQ(J1) isreplaced byQ(A(Qa)tor)' where Aisanelliptic curve. For some results about free groups occurringasGalois groups,see also Wingberg [Wi91]. VI, 15 [La90] [LaT 75] [Se68] [Se72] [Shi 71] [Wi 91]THE MODULAR CONNECTION 315 Bibliography S.LANG, Cyclotomic Fields IandII,Second Edition, Springer Verlag, 1990 (Combined edition from thefirst editions, 1978, 1980) S,LANG and H,TROTTER, Distribution ofFrobenius Elements inGL2-Extensions oftheRational Numbers, Springer Lecture Notes 504 (1975) J.-P.SERRE, Abelian l-adic Representations andElliptic Curves, Benjamin,1968 J.-P. SERRE, Proprietes galoisiennes despoints d'ordre fini des courbes ellip- tiques, Invent, Math. 15(1972), pp.259-331 G.SHIMURA, Introduction tothearithmetic theory ofAutomorphic Functions, Iwanami Shoten and Princeton University Press, 1971 K.WINGBERG, OnGalois groups ofp-closed algebraic number fields with restricted ramification, I,J.reine angew. Math. 400 (1989), pp. 185-202; andII,ibid., 416(1991), pp. 187-194 15. THE MODULAR CONNECTION This final section givesamajor connection between Galois theory and the theory ofmodular forms, which has arisen since the 1960s. One fundamental question iswhether givenafinite group G,there exists a Galois extension KofQwhose Galois group isG.InExercise 23youwill prove this when Gisabelian. Already inthenineteenth century, number theorists realized thebigdifference between abelian and non-abelian extensions, and started understanding abelian extensions. Kronecker stated and gave what aretoday considered incomplete arguments that every finite abelian extension ofQiscontained insome extension Q((),where (isaroot ofunity. Thedifficulty layinthepeculiarities ofthe prime 2.The trouble was fixed byWeber attheend ofthenineteenth century. Note that thetrouble with 2has been systematic since then. Itarose inArtin's conjecture about densities ofprimitiveroots asmentioned intheremarks after Theorem 9.4. Itarose intheGrunwald theorem ofclass field theory (corrected byWang, cf.Artin- Tate [ArT 68], Chapter 10). Itarose inShafarevich' sproof that givenasolvable group, there exists aGalois extension ofQhaving that groupasGalois group, mentioned attheend of7. Abelian extensions ofanumber field Fareharder todescribe than over the rationals, and thefundamental theory givingadescription ofsuch extensions is called class field theory (see the above reference). Ishall giveonesignificant example exhibiting theflavor. Let RFbethering ofalgebraic integers inF.It can beshown that RFisaDedekind ring. (Cf. [La70], Chapter I,6,Theorem 2.)Let Pbe aprime ideal ofRF.Then Pnz=(p)for some prime number p. 316 GALOIS THEORY VI,15 Furthermore, RF/Pisafinite field with qelements. Let Kbe afinite Galois extension ofF.Itwill beshown inChapter VII that there exists aprime Qof RKsuch that QnRF=P.Furthermore, there exists anelement FrQEG=Gal(K/F) such thatFrQ(Q)=Qand forall aERKwehave FrQa=a'lmod Q. WecallFrQaFrobenius element intheGalois group Gassociated with Q.(See Chapter VII, Theorem 2.9.) Furthermore, forallbut afinite number ofQ,two such elements areconjugatetoeach other inG.We denote anyofthem byFrp. IfGisabelian, then there isonlyone element Frp intheGalois group. Theorem 15.1. There exists aunique finite abelian extension KofFhaving thefollowing property. IfPI' P2are prime ideals of RF,then Frpi=Frp2ifandonlyifthere isanelement aofKsuch that aPI=P2. Inasimilar but more complicated manner, one can characterize allabelian extensions ofF.This theory isknown asclass field theory, developed byKro- necker, Weber, Hilbert, Takagi, andArtin. The main statement concerning the Frobenius automorphismasabove isArtin' sReciprocity Law. Artin- Tate's notes giveacohomological account ofclass field theory. MyAlgebraic Number Theory givesanaccount following Artin's firstproof dating back to1927, with later simplifications byArtin himself. Both techniques arevaluable toknow. Cyclotomic extensions should beviewed inthelight ofTheorem 15. 1 .Indeed, letK=Q((),where (isaprimitive n-th root ofunity. For aprime ptn,we have theFrobenius automorphism Frp,whose effect on(is Frp«()=(P.Then Frp1=Frp2ifandonly ifPI=P2mod n. Toencompass both Theorem 15.1 and thecyclotomiccase inoneframework, one has toformulate theresult ofclass field theory forgeneralized ideal classes, notjust theordinaryones when two ideals areequivalent ifandonly ifthey differ multiplicatively byanon-zero field element. See myAlgebraic Number Theory for adescription ofthese generalized ideal classes. The non-abelian case ismuch more difficult. Ishall indicate brieflyaspecial case which givessome oftheflavor ofwhat goes on. The problem istodofor non-abelian extensions what Artin didforabelian extensions. Artin went asfar assaying that theproblemwas not togive proofs but toformulate what was to beproved. Theinsight ofLanglands and others inthesixties shows thatactually Artin was mistaken. Theproblem liesinboth. Shimura made several computations inthis direction involving "modular forms" [Sh66]. Langlands gaveanumber ofconjectures relating Galois groups with "automorphic forms", which showed that the answer layindeeper theories, whose formulations, letalone their proofs, were difficult. Great progresswas made intheseventies bySerre andDeligne, who provedafirst case ofLangland's conjecture [DeS 74]. VI, 15 THE MODULAR CONNECTION 317 The study ofnon-abelian Galois groups occurs viatheir linear "representa- tions". Forinstance, letlbe aprime number. We can askwhether GLn(F/),or GL2(F[),orPGL2(F[)occurs asaGalois group over Q,and"how". Theproblem istofind natural objectsonwhich theGaloisgroup operatesasalinear map, such that wegetinanatural wayanisomorphism ofthis Galois group with one ofthe above linear groups. The theories which indicate inwhich direction to find such objectsaremuch beyond thelevel ofthis course, and lieinthetheory ofmodular functions, involving both analysis andalgebra, which form aback- ground forthenumber theoretic applications. Again Ipick aspecialcase togive theflavor. LetKbe afinite Galois extension ofQ,with Galois group G=Gal(K/Q). Let p:G GL2(F[) be ahomomorphism ofGinto thegroup of2x2matrices over thefinite field F[for some prime l.Such ahomomorphismiscalled arepresentation ofG. From elementary linear algebra, if M=e) isa2x2matrix, wehave itstrace and determinant defined by tr(M)=a+dand detM=ad-bc. Thus we can take the trace and determinant trp(u)and detp(u)for uEG. Consider theinfinite product with avariable q: oc oc Il(q)=qf1(1-qn)24=Lanqn.n=I n=I The coefficients anareintegers, and al=1. Theorem 15.2. For each prime lthere exists aunique Galois extension Kof Q,with Galois group G,and aninjective homomorphism p:G GL2(F[) having thefollowing property. For allbut afinite number ofprimes p,ifapis thecoefficient ofqPinIl(q), then wehave trp(Fr p)=apmodland detp(Fr p)=pIl mod l. Furthermore, forallprimes l=f=.2,3,5,7,23,691, theimage p(G) inGL2(F/) consists ofthose matrices MEGL2(F/) such that detMisaneleventh power inFf. 318 GALOIS THEORY VI,15 The above theorem was conjectured bySerre in1968 [Se68]. Aproof of theexistence asinthefirst statement wasgiven byDeligne [De68]. The second statement, describing howbigtheGalois group actually isinthegroup ofmatrices GL2(F/) isdue toSerre andSwinnerton-Dyer [Se72], [SwD 73]. The point ofIl(q) isthatifweput q=e27Tiz ,where Zisavariable inthe upper half-plane, then Ilisamodular form ofweight 12.Fordefinitions and an introduction, seethelastchapter of[Se73], [La73], [La76], and thefollowing comments. Thegeneral result behind Theorem 15.2 formodular forms ofweight >2was given byDeligne [De 73]. Forweight 1,itisdue toDeligne-Serre [DeS 74].We summarize thesituation asfollows. LetNbe apositive integer. ToNweassociate thesubgroups r(N) Cr}(N) Cro(N) ofSL2(Z)defined bytheconditions for amatrix a=(:)ESL2(Z): aEr(N) ifandonly ifa=d=1mod Nand b==c=0mod N; aEr}(N) ifandonly ifa=d=1mod Nand c=0mod N; aEro(N) ifandonly ifc=0mod N. Letf be afunction ontheupper half-plane Sj={zEC,Im(z) >O}.Let k be aninteger. For y=(:)ESL2(R), definef0[')']k (anoperationontheright) by az+bf0[')']k(z)=(cz+d)-'l(')'z) where')'Z= cz+d. Letrbe asubgroup ofSL2(Z)containing r(N). Wedefinefto bemodular of weight konrif: Mk1.fisholomorphiconSj; Mk2.fisholomorphicatthecusps, meaning that forall aESL2(Z), the function f0[a]k has apower series expansion 00 f0[a]k(z)=Lane27Tinz/N; n=O Mk3.Wehavefo [')']k=ffor all')'Er. One says thatfiscuspidal ifinMk2thepower series has azero; that is,the power starts with n>1. VI, 15 THE MODULAR CONNECTION 319 Suppose thatfis modular ofweight konr(N). Thenfis modular onr)(N) ifandonly iff(z+1)=f(z), orequivalently fhas anexpansion oftheform ::x; f(z)=f::x;(qz)=Lanqn where q=qz=e2mz . n=O This power series iscalled theq-expansion off. Suppose fhasweight konr}(N). IfYEro(N) and yisthe above written matrix, thenf0[Y]k depends onlyontheimage ofdin(Z/NZ)*, and wethen denote f0[Y]k byf0[d]k. Let e:(Z/NZ)* C* be ahomomorphism (also called aDirichlet character). One says that eisodd ife(-1)=-1, and even ife(-1)=1.One says thatfismodular oftype (k,e)onro(N) iffhas weight konr}(N), and f0[d]k=e(d)f forall dE(Z/NZ)*. Itispossibletodefine analgebra ofoperatorsonthespace ofmodular forms ofgiven type. This requiresmore extensive background, and Irefer thereader to[La76]for asystematic exposition. Among allsuch forms, itisthen possible todistinguishsome ofthem which areeigenvectors forthis Hecke algebra, or, asone says, eigenfunctions forthisalgebra. One may then state theDeligne- Serre theorem asfollows. Letf=t=0beamodular form oftype (1 ,e)onro(N),sofhasweight 1.Assume that eisodd. Assume thatfisaneigenfunction oftheHecke algebra, with q- expansion fx=Lanqn, normalized sothat a}=1.Then there exists aunique finite Galois extension KofQwith Galois group G,and arepresentation p:G GL2(C) (actually aninjective homomorphism), such thatfor all primes p%Nthecharacteristic polynomial ofp(Frp)is X2 -apX+e(p). Therepresentation pisirreducible ifandonlyiffiscuspidal. Note that therepresentation phasvalues inGL2(C). For extensive work ofSerre and hisconjectures concerning representations ofGalois groups inGL2(F)when Fisafinite field, see[Se87]. Roughly speaking, thegeneral philosophy started byaconjecture ofTaniyama-Shimura and theLanglands conjectures isthat everything insight is"modular". Theorem 15.2 and theDeligne-Serre theorem areprototypes ofresults inthisdirection. For"modular" representationsinGL2(F), when Fisafinite field, Serre's conjectures have been proved, mostly byRibet [Ri90]. As aresult, following anidea ofFrey, Ribet also showed how the Taniyama-Shimura conjecture implies Fermat's lasttheorem [Ri90b]. Note that Serre's conjectures that certain representations inGL2(F) aremodular imply the Taniyama-Shimura conjecture. 320 GALOIS THEORY [ArT 68] [De68] [De 73] [DeS 74] [La70] [La73] [La76] [Ri90a] [Ri90b] [Se68] [Se72] [Se73] [Se87] [Shi 66] [Shi 71] [SwD 73]VI,Ex Bibliography E.ARTIN andJ, TATE, Class Field Theory, Benjamin-Addison-Wesley,1968 (reprinted byAddison-Wesley, 1991) P,DELIGNE, Formes modulaires etrepresentations l-adiques, Seminaire Bour- baki 1968-1969, exp, No. 355 P,DELIGNE, Formes modulaires etrepresentationsdeGL(2), Springer Lecture Notes 349 (1973), pp,55-105 P,DELIGNE and J.P.SERRE, Formes modulaires depoids 1,Ann, Sci. ENS 7(1974), pp.507-530 S,LANG, Algebraic Number Theory, Springer Verlag, reprinted from Addison- Wesley (1970) S.LANG, Elliptic functions, Springer Verlag, 1973 S,LANG, Introduction tomodular forms, Springer Verlag, 1976 K,RIBET, Onmodular representations ofGal( Q/Q)arising from modular forms, Invent. Math. 100(1990), pp.431-476 K,RIBET, From theTaniyama-Shimura conjecture toFermat's lasttheorem, Annales delaFac, des Sci. Toulouse (1990), pp. 116-139 J.-P.SERRE, Uneinterpretation des congruences relatives alafonction de Ramanujan, Seminaire Delange-Pisot-Poitou, 1967-1968 J.-P. SERRE, Congruencesetformes modulaires (d'apres Swinnerton-Dyer), Seminaire Bourbaki, 1971-1972 J,-Po SERRE, Acourse inarithmetic, Springer Verlag, 1973 J.-P. SERRE, Sur lesrepresentations modulaires dedegre2deGal( Q/Q), Duke Math. j,54(1987), pp. 179-230 G.SHIMURA, Areciprocity law innon-solvable extensions, J.reine angew. Math. 221(1966), pp.209-220 G.SHIMURA, Introduction tothearithmetic theory ofautomorphic functions, Iwanami Shoten and Princeton University Press, 1971 H,P.SWINNERTON-DYER, Onl-adic representations and congruences for coefficients ofmodular forms, (Antwerp conference) Springer Lecture Notes 350 (1973) EXERCISES 1.What istheGalois group ofthefollowing polynomials? (a)X3-X-lover Q. (b)X3-10over Q. (c)X3-10overQ(J2), (d)X3-10over Q(J=3 ), (e)X3-X-lover Q(J=23 ). (f)X4-5over Q,Q(J5), Q(j-=5 ),Q(i). (g)X4-awhere aISanyinteger #0,#+ 1and ISsquare free, Over Q. VI,Ex EXERCISES 321 (h)X3-awhere aisanysquare-free integer>2.Over Q. (i)X4+2over Q,Q(i), (j)(X2-2)(X2-3)(X2-5)(X2-7)over Q. (k) Let PI' .,., Pnbedistinct prime numbers. What isthe Galois group of (X2-PI)...(X2-Pn)over Q? (I)(X3-2)(X3-3)(X2-2)over Q(J-3). (m)xn-t,where tistranscendental over thecomplex numbers Cand nisa positive integer. Over C(t), (n)X4-t,where tisasbefore. Over R(t), 2.Find theGalois groups over Qofthefollowing polynomials. (a)X3+X+ 1 (b)X3-X+ 1 (g)X3+X2-2X-1 (c)X3+2X+ 1 (d)X3-2X+ 1 (e)X3-X-I (f)X3-12X +8 3.Letk=C(t) bethefield ofrational functions inone variable. Find theGalois group over kofthefollowing polynomials: (a)X3+X+t (b)X3-X+t (c)X3+tX+ 1 (d)X3-2tX+t (e)X3-X-t (f)X3+t2X-t3 4.Let kbe afield ofcharacteristic =1=2,Let cEk,ctt.k2 ,Let F=k(\!'";;), Let a=a+b\!'";; with a,bEkand notboth a,b=O.Let E=F(). Prove that thefollowing conditions areequivalent. (1) EisGalois over k, (2) E=F(W), where a'=a-b\!'";;. (3)Either aa'=a2 -cb2Ek2orcaa' Ek2 . Show that when these conditions aresatisfied, then Eiscyclic over kofdegree 4if andonly ifcaa' Ek2 , 5.Let kbe afield ofcharacteristic =1=2,3,Letf(X), g(X)=X2-cbeirreducible polynomialsover k,ofdegree 3and 2respectively. Let Dbethediscriminant off. Assume that [k(DI/2):k]=2and k(DI/2)=1=k(CIl2). Let abe aroot offand (3aroot of9inanalgebraic closure. Prove: (a)Thesplitting field offgover khasdegree 12, (b)Let)'=a+(3,Then [k()'):k]=6. 6,(a)Let Kbecyclicover kofdegree 4,and ofcharacteristic =1=2.Let GKlk=(a), Let Ebetheunique subfield ofKofdegree 2over k.Since [K :E]=2,there exists aEKsuch that a2='YEEand K=E(a). Prove that there exists ZEEsuch that zaz= -1,aa=za, z2=a)'/)'. (b)Conversely, letEbe aquadratic extension ofkand letGElk=(T), Let zEE be anelement such that ZTZ= -1.Prove that there exists )'EEsuch that z2=T)'/)'. Then E=k()').Let a2=)',and letK=k(a). Show that Kis Galois, cyclic ofdegree 4over k,Let abe anextension of TtoK.Show that aisanautomorphism ofKwhich generates GKlk ,satisfyinga2a= -aand aa= +za.Replacingzby-zoriginally ifnecessary,one can then have aa=za, 322 GALOIS THEORY VI,Ex 7.(a) Let K=Q() where aEZ, a<O.Show that Kcannot beembedded ina cyclic extension whose degreeover Qisdivisible by4. (b) Letf(X)=X4+30X2+45. Let abe aroot ofF.Prove thatQ(a)iscyclic of degree 4over Q. (c)Letf(X)=X4+4x2+2.Prove thatf isirreducible over Qand that theGalois group iscyclic. 8.Letf(X)=X4+aX2+bbeanirreducible polynomialover Q,with roots +ex,+(J, andsplitting field K. (a)Show thatGal{K/Q) isisomorphic toasubgroup ofDs(the non-abelian group oforder 8other than thequaternion group), and thus isisomorphic tooneofthe following: (i)Z/4Z (ii)Z/2ZxZ/2Z (iii) Ds. (b)Show that thefirst case happens ifandonly if a {3 {3- aEQ. Case (ii)happens ifandonly ifa{3EQora2-{32EQ.Case (iii) happens otherwise. (Actually, in(ii), the case a2-{32EQcannot occur. Itcorresponds toasubgroup DsCS4which isisomorphictoZ/2Z xZ/2Z, but isnot transitive on{I,2,3,4}). (c)Find thesplittingfield KinCofthepolynomial X4-4X2-1. Determine theGalois group ofthissplitting field over Q,and describe fully thelattices ofsubfields andofsubgroups oftheGalois group. 9.LetKbe afinite separable extension ofafield k,ofprime degree p.Let ()EKbe such that K =k{(}), and let(}b...,(}pbetheconjugates of(}over kinsome algebraIc closure. Let (}=0t.If(}2Ek(0),show that KisGalois and infactcyclicover k, 10.Letf{X)EQ[X] beapolynomial ofdegree n,and letKbeasplitting fieldoffover Q, Suppose thatGal(K/Q) isthesymmetric group Snwith n>2. (a)Show thatfis irreducible over Q. (b)Ifexisarootoff, show that theonly automorphism ofQ(ex) istheidentity. (c)Ifn>4,show that exnQ. 11.Apolynomial f(X) issaid tobereciprocal ifwhenever r:xisaroot, then I/ISalso a root. Wesuppose thatfhascoefficients inareal subfield kofthecomplex numbers. If jisirreducible over k,and has anonreal root ofabsolute value 1,show thatjIS reciprocal ofeven degree. 12.What istheGalois groupover therationals ofX5-4X+2? 13.What istheGalois groupover therationals ofthefollowing polynomials: (a)X4+2X2+X+3 (b)X4+3X3-3X-2 (c)X6+22X5-9X4+12X3-37X2-29X-15 [Hint: Reduce mod 2,3,5.] 14.Prove that givenasymmetric group Sn,there eXIsts apolynomialf(X)EZ[X] with leading coefficient 1whose Galois groupover QisSn.[Hint: Reducing mod 2,3,5, show that there exists apolynomial whose reductions aresuch that theGalois group VI,Ex EXERCISES 323 contaIns enough cycles togenerate SII. Use theChInese remainder theorem, also to beable toapply Eisenstein's criterion.] 15. LetK/kbe aGalois extension, and letFbeanintermediate field between kand K. LetHbethesubgroup ofGal(K/k) mapping Finto itself. Show that Histhenormal- izer ofGal(K/F) inGal(K/k). 16. LetK/k be afinite Galois extension with group G.Let aEKbesuch that {aa}UEGisanormal basis, For each subset SofGletS(a)=2: UESaa .LetHbe a subgroup ofGand letFbethefixed field ofH.Show that there exists abasis ofF over kconsisting ofelements oftheform S(a), Cyclotomic fields 17.(a)Let kbe afield ofcharacteristic t2n, for some oddinteger n>1,andlet' be aprimitive n-th root ofunity, ink,Show that kalso contains aprimitive 2n-th root ofunity. (b)Letkbeafinite extension oftherationals. Show that there isonlyafinite number ofroots ofunity ink. 18. (a) Determine which roots ofunity lieinthefollowing fields: Q(i), Q(v=2 ), Q(v2), Q(Y=3 ),Q(V3), Q(v=5 ). (b)For which integersmdoes aprimitive m-th root ofunity have degree 2over Q? 19. Let(beaprimitive n-th root ofunity. LetK=Q((). (a)Ifn=pr(r>1)isaprime power, show thatNK/Q(l-()=p. (b)Ifniscomposite (divisible byatleast twoprimes) thenNK/Q(l-,)=1. 20. LetI(X)EZ[X] be anon-constant polynomial with integer coefficients. Show that thevalues I(a) with aEZ+ aredivisible byinfinitely many primes. [Note: This istrivial. Amuch deeper question iswhether there areinfinitely many asuch that/(a) isprime, There arethree necessary conditions: The leading coefficient ofIispositive. Thepolynomial isirreducible. The setofvalues I(Z+) has nocommon divisor> 1. Aconjecture ofBouniakowski [Bo 1854] states that these conditions aresufficient. The conjecturewas rediscovered later and generalized toseveral polynomials by Schinzel [Sch 58]. Aspecialcase istheconjecture that X2+1represents infinitely many primes. For adiscussion ofthegeneral conjecture and aquantitative version givingaconjectured asymptotic estimate, seeBateman and Horn [BaH 62]. Also see thecomments in[HaR 74]. More precisely, letI.,. ..,Irbepolynomials with integer coefficients satisfying the first two conditions (positive leading coefficient, irre- ducible). Let I=II.·.Ir betheir product, and assume thatIsatisfies thethird condition. Define: 7T(f)(X)=number ofpositive integersn<xsuch that/l(n),...,Ir(n)areallprimes. (We ignore thefinite number ofvalues ofnforwhich some li(n) isnegative.) The 324 GALOIS THEORY VI,Ex Bateman-Horn conjecture isthat x 1T(f)(X)-(d 1.. .dr)-IC(f)I(10;t)'dt, o where C(f)=9{(1-)-r(1-)}, theproduct being taken over allprimes p,andNf(p)isthenumber ofsolutions of thecongruence f(n)==0mod p. Bateman and Horn show that theproduct converges absolutely. When r=1and f(n)=an+bwith a,brelatively prime integers,a>0,then one gets Dirichlet's theorem that there areinfinitely many primes inanarithmetic progression, together with theDirichlet density ofsuch primes. [BaH 62] P.T.BATEMAN and R.HORN, Aheuristic asymptotic formula concerning thedistribution ofprime numbers, Math. Compo 16(1962) pp.363-367 [Bo 1854] V,BOUNIAKOWSKY, Sur lesdiviseurs numeriques invariables des fonc- tions rationnelles entieres, Memoires sc.math. etphys. T.VI(1854- 1855) pp.307-329 [HaR 74] H.HALBERSTAM and H.-E. RICHERT, Sieve methods, Academic Press, 1974 [Sch 58] A.SCHINZEL and W. SIERPINSKI, Sur certaines hypotheses concernant lesnombres premiers, Acta Arith. 4(1958) pp. 185-208 21. (a)Let abe anon-zero integer, paprime,napositive integer, and pn.Prove that pI<l>n(a) ifandonly ifahasperiodnin(Z/pZ)*. (b)Againassume p,rnProve that pI<l>n(a) for some aEZifandonly ifp= 1 mod n.Deduce from this that there areinfinitely many primes==1mod n, a specialcase ofDirichlet's theorem fortheexistence ofprimes inanarithmetic progression. 22. LetF=Fpbetheprime field ofcharacteristic p.LetKbethefield obtained from Fbyadjoining allprimitive I-th roots ofunity, forallprime numbers I=1=p.Prove that Kisalgebraically closed. [Hint: Show thatifqisaprime number, and ran integer>1,there exists aprime Isuch that theperiod ofpmod Iisqr,byusing thefollowing oldtrick ofVan derWaerden: Let Ibe aprime dividing thenumber qr1 bP- (r- 11r- 1 2= qr-1 1=pq-l)q-+q(pq-l)q-+... +q. p- IfIdoes notdivide pqr-1 -1,we aredone. Otherwise, I=q.But inthat case q2does notdivide b,and hence there exists aprime I#-qsuch that Idivides b.Then thedegree ofF(,,) over Fisqr,soKcontains subfields ofarbitrary degree over F.] 23. (a)Let Gbe afinite abelian group. Prove that there exists anabelian extension of Qwhose Galois group isG. VI,Ex EXERCISES 325 (b)Let kbe afinite extension ofQ,and letGbe afinite abelian group.Prove that there exist infinitely many abelian extensions ofkwhose Galois groupisG. 24. Prove that there areinfinitely many non-zero integers a,b =1=0such that -4a3-27b2isasquareinZ, 25. Let kbe afield such that every finite extension iscyclic. Show that there exists an automorphismuofkaover ksuch that kisthefixed field ofu. 26. Let Qabe afixed algebraic closure ofQ.Let Ebe amaximal subfield ofQanot containing \12 (such asubfield exists byZorn's lemma). Show that every finite extension ofEiscyclic. (Your proof should work taking anyalgebraic irrational number instead of\12.) 27. Let kbe afield, kaanalgebraic closure, and uanautomorphism ofkaleaving k fixed. LetFbethefixed field ofu,Show that every finite extension ofFiscyclic, (The above two problemsareexamples ofArtin, showing how todig holes inan algebraically closed field,) 28. Let Ebeanalgebraic extension ofksuch that every non-constant polynomial f(X) ink[X] has atleast one root inE,Prove that Eisalgebraically closed. [Hint: Discuss theseparable andpurely inseparablecases separately, and use theprimitive element theorem. ] 29.(a)LetKbeacyclic extension ofafield F,with Galois group Ggenerated by(1.Assume that the characteristic isp,and that[K:F]=pm-1for some integerm>2. Let (Jbeanelement ofKsuch thatTr:({J)=1.Show that there exists anelement (1inKsuch that (1(1-(1={JP-(J. (b) Prove that thepolynomialXP-X-exisirreducible inK[X]. (c)Ifeisaroot ofthIspolynomial, prove that F(e) isaGalois, cyclic extension of degree pmofF,and that itsGalois groupisgenerated byanextension (1*of (1 such that (1*(e)=()+(J. 30. Let Abeanabelian group and letGbeafinite cyclic group operatingonA[bymeans ofahomomorphism G Aut(A)]. Let (1be agenerator ofG.We define the trace TrG=Tr onAbyTr(x)=L!x. Let ATrdenote the kernel ofthe trace, and let reG (1-(1)A denote thesubgroup ofAconsIsting ofallelements oftype y-(1Y, Show that HleG,A) ATr/(1-(1)A, 31. LetFbe afinite field and Kafinite extension ofF.Show that the norm N:and the trace Tr:aresurjective (asmaps from Kinto F). 32. LetEbeafinite separable extension ofk,ofdegreen.Let W =(w t".,,wn)beelements ofE.Let (11', . ,,(1nbethedistinct embeddings ofEInkaover k,Define thedis- criminant ofWtobe DE/k(W)=det(UiWj)2 , Prove: (a)IfV=(V.,, . .,vn)isanother setofelements ofEand C=(cij)isamatrix ofelements ofksuch that Wi=2:cijvj,then DE/k(W)=det(C)2D E/k(V). 326 GALOIS THEORY VI,Ex (b)The discriminant isanelement ofk. (c)Let E=k(rx) andletf(X)=Irr(rx,k, X). Letrxl'...'rxnbethe roots off and sayrx= rxl.Then n f'(rx)=n(rx-rx). j=2 Show that DElk(l, rx,...,rxn- 1)=(_l)"(n-l)/2Nf(f'(rx». (d)Letthenotation beasin(a). Show thatdet(Tr(wiw)=(det(O"iw)2. [Hint: Let Abethematrix (0";w).Show that tAA isthematrix (Tr(wiw),] Rational functions 33, LetK =C(x) where xistranscendental over C,andlet' be aprimitive cube root of unity inC.Let 0"betheautomorphism ofKover Csuch that O"X='x.Let! bethe automorphism ofKover Csuch that !X =X-I. Show that a3=1=il and Ta=a-1T. Show that thegroup ofautomorphisms Ggenerated byaand Thas order 6and the subfield FofKfixed byGisthefield C(y) where y=x3+x-3 . 34. Give anexample ofafield Kwhich isofdegree 2over two distinct subfields Eand F respectively, but such that Kisnotalgebraic over EnF. 35. Let kbeafield and Xavariable over k.Let (X)=f(X) qJg(X) be arational function ink(X), expressedasaquotient oftwopolynomials f,gwhich arerelatively prime. Define thedegree ofqJtobemax(deg f,degg). Let Y=qJ(X). (a)Show that thedegree ofqJisequal tothedegree ofthefield extension k(X) over key) (assuming yk),(b)Show that every automorphism ofk(X) over kcan berepresented byarational function({Jofdegree 1,and istherefore induced byamap aX+bX eX+d with a,b,e,dEkand ad-be#-o.(c)Let Gbethegroup ofautomorphisms ofk(X) over k.Show that Gisgenerated bythefollowing automorphisms: !b:X X+b, O"a:XaX (a#-0), XX-l with a,bEk. 36. Letkbeafinite field with qelements. LetK =k(X) betherational field inone variable. Let Gbethegroup ofautomorphisms ofKobtained bythemappings aX+bX eX+d VI,Ex EXERCISES 327 with a,b,c,dinkand ad-bc#-O.Prove thefollowing statements: (a)The order ofGisq3-q. (b)The fixed field ofGisequal tokey) where (xq2 _X)q+1 y= (xq-X)q2+1. (c)LetHIbethesubgroup ofGconsisting ofthemappings Xt--+aX+bwith a#-O.The fixed field ofHIisk(T) where T=(xq-X)q- 1. (d)LetH2bethesubgroup ofHIconsisting ofthemappings X X+bwith bEk.The fixed field ofH2isequal tok(Z) where Z=xq-X. Some aspects ofKummer theory 37. Let kbe afield ofcharacteristic 0,Assume that foreach finite extension Eofk,the index (E*:E*n) isfinite forevery positive integern.Show that foreach positive integer n,there exists onlyafinite number ofabelian extensions ofkofdegree n. 38. Let a#-0,#-+1be asquare-free integer. For each prime number p,letKpbe thesplitting field ofthepolynomial XP-aover Q.Show that[K p:Q]=p(p-1). Foreach square-free integerm>0,let Km=nKp plm bethecompositum ofallfields Kpforpim.Let dm=[Km: Q]bethedegree ofKm over Q.Show that ifmisodd then dm=ndp,andifmiseven, m=2nthen d2n=dn plm or2dnaccordingasva isorisnotinthefield ofm-th roots ofunity Q('m). 39. LetKbeafield ofcharacteristic 0forsimplicity. Letrbeafinitely generated subgroup ofK*. LetNbeanoddpositive integer. Assume that foreach prime piN wehave r=rl/pnK, and also thatGal(K(PN)/K) Z(N)*. Prove thefollowing. (a)f/fN=f/(fnK*N)=fK*N/K*N. (b) LetKN=K(PN). Then rnKN=rN . [Hint: Ifthese two groups arenotequal, then for some prime piN there exists anelement aErsuch that a=bPwith bEKNbutbK. Inother words, aisnot ap-th power inKbutbecomes ap-th power inKN. The equation xP-aisirreducible over K.Show that bhasdegree pover K(p p), and that K(pp,a11P)isnot abelian over K, so al/phasdegree poverK(pp). Finish theproof yourself.] 328 GALOIS THEORY VI,Ex (c)Conclude that thenatural Kummer map f/fNHom(Hr<N), J1N) isanisomorphism. (d)LetGr(N)=Gal(K(fl/N ,J1N)/K). Then the commutator subgroup ofGr(N) isHr<N), and inparticular Gal(KN/ K)isthemaximal abelian quotient of Gr<N). 40. LetKbeafield and paprime number notequal tothecharacteristic ofK.Letfbea finitely generated subgroup ofK*, and assume thatfisequal toitsown p-division group inK,that isifZEKand zPEf,then zEf.Ifpisodd, assume thatJ1pcK,and ifp=2,assume that J14cK.Let (f:fP)=pr+I . Show thatflIp isitsown p-divislon group inK(fl/p),and [K(f1/pm): K]=pm(r+1) forallpositive integersm. 41. Relative invariants (Sato). Letkbeafield andKanextension ofk.Let Gbeagroup ofautomorphisms ofKover k,and assume that kisthefixed field ofG.(We donot assume that Kisalgebraic over k.)Byarelative invariant ofGinKweshall mean an element PEK,P#-0,such that foreach UEGthere exists anelement l(u)Ekfor which pCT =X(u)P, Since uisanautomorphism,wehave X(u)Ek*. We saythat the map X:G k*belongs toP,and callitacharacter, Prove thefollowing statements: (a)The map Xabove isahomomorphism. (b)Ifthe same character Xbelongs torelative invariants Pand Qthen there exists CEk*such that P=cQ. (c)The relative invariants form amultiplicative group, which wedenote byI. Elements PI'. ,.,PmofIarecalled multiplicatively independent mod k*if their images inthefactor group l/k* aremultiplicatively independent, i.e.if given integersvI'...,Vmsuch that p"l..,p"m =CEk* 1m' then VI= ...= Vm=o. (d)IfPI'...,Pm aremultiplicatively independent mod k*prove that they are algebraically independentover k,[Hint: Use Artin's theorem oncharacters,] (e)Assume that K =k(X l'...,Xn)isthequotient field ofthepolynomial ring k[X h... ,Xn]=k[X], and assume that Ginduces anautomorphism ofthe polynomial ring. Prove: IfF1(X)and F2(X)arerelative invariant polynomials, then their g.c.d. isrelative invariant. IfP(X)=F1(X)/F 2(X) isarelative invariant, and isthequotient oftworelatively prime polynomials, then F1(X) and F2(X)are relative invariants. Prove that the relative invariant poly- nomials generate I/k*. Let Sbethe setofrelative invariant polynomials which cannot befactored into aproduct oftwo relative invariant polynomials of degrees>1.Show that theelements ofS/k*aremultiplicatively independent, and hence thatl/k* isafree abelian group. [Ifyou know about transcendence degree, then using (d)youcanconclude that this group isfinitely generated.] VI,Ex EXERCISES 329 42, Letf(z) bearational function with coefficients inafinite extension oftherationals. Assume that there areinfinitely many roots ofunity, suchthatf(') isaroot ofunity. Show that there exists anintegernsuchthatf(z)=cz"forsome constant c(which isin fact aroot ofunity). This exercise can begeneralizedasfollows: Letrobe afinitely generated multi- plicative group ofcomplex numbers. Letrbethegroup ofallcomplex numbers ')' such that ymliesinr0for some integerm#-O.Letf(z) bearational function with complex coefficients such that there exist infinitely many yErforwhichf(y) liesinr. Then again,f(z)=cz"for some cand n.(Cf. Fundamentals ofDiophantine Geometry.) 43. LetK/k be aGalois extension. We define theKrull topology onthe group G(K/k)=Gbydefiningabase foropensets toconsist ofallsets aHwhere aEG and H=G(K/F)for some finite extension Fofkcontained inK. (a)Show thatifone takes only those sets aHforwhich Fisfinite Galois over kthen one obtains another base forthe same topology. (b)Theprojective limit!i!!!G/Hisembedded inthedirect product limG/H nG/H.HH Give thedirect product theproduct topology. ByTychonoff's theorem in elementary point settopology, thedirect product iscompact because itisa direct product offinite groups, which arecompact (and ofcourse alsodiscrete). Show that theinverselimit!!!!!.G/Hisclosed intheproduct, and istherefore compact. (c)Conclude thatG(K/k) iscompact. (d)Show that every closed subgroup offinite index inG(K/k) isopen. (e) Show that the closed subgroups ofG(K/k) areprecisely those subgroups which areoftheform G(K/ F)for some extension Fofkcontained inK. (f)LetHbeanarbitrary subgroup ofGand letFbethefixed field ofH.Show thatG(K/F)istheclosure ofHinG. 44. Let kbe afield such that every finite extension iscyclic, andhaving one extension of degreenforeach integern.Show that theGalois group G=G(k8jk)istheinverse limit limZjmZ,asmZ ranges over allideals ofZ,ordered byinclusion. Show that thislimit isisomorphic tothedirect product ofthelimits nlimZ/pnz=nZppn-+oop taken over allprime numbers p,inother words, itisisomorphic totheproduct ofall p-adic integers. 45. Let kbe aperfect field and kaitsalgebraic closure. Let aEG(k8/k) beanelement ofinfinite order, and suppose kisthefixed field ofa.For each prime p,letKpbe thecomposite ofallcyclic extensions ofkofdegreeapower ofp. (a) Prove that kaisthecomposite ofallextensionsKp. (b)Prove that eitherKp=k,orKpisinfinite cyclicover k.Inother words, Kp cannot befinite cyclic over kand =1=k. (c)Suppose k8=Kpfor some prime p,sokaisaninfinite cyclic tower of p-extensions. Let ube ap-adic unit, uEZ;such that udoes notrepresent arational number. Define aU, and prove that a,aU arelinearly independent 330 GALOIS THEORY VI,Ex over Z,i.e. thegroup generated byaand aUisfree abelian ofrank 2.In particular {a} and{a,aU} have the same fixed field k. Witt vectors 46. LetXl'X2,. ..be asequence ofalgebraically independent elements over theintegers Z.For each integern:>1define x(n) =LdX:;/d. din Show that Xncan beexpressed interms ofX(d)fordin, with rational coefficients. Using vector notation, wecall(xI'X2,...)theWitt components ofthe vector x, and call(xCI), x(2),...)itsghost components. Wecall xaWitt vector. Define thepower series fx{t)=n(1-xntn). nl Show that d -t-logfx(t)=Lx(n)tn . dtnI [By logf(t)wemeanf'(t)/f(t) iff(t) isapower series, and thederivativef'(t) istaken dt formally.] Ifx,yare two Witt vectors, define their sum andproduct componentwise with respect totheghost components, i.e. (xty)(n)=xcn)tyen). What is(x+Y)n? Well, show that !x(t)!y(t)=0(1 +(x+Y)ntn)=!x+y(t). Hence (x+Y)nisapolynomial with integer coefficients inXI'y.,..., xn,Yn.Also show that !xy(t)=n(1-xr;/dy/etm)de/m d.e I where mistheleast common multiple ofd,eand d,erange over allintegers:>1.Thus (xY)n isalso apolynomial inx.,Yl ..., xn,Ynwith integer coefficients. The above arguments aredue toWitt (oral communication) and differ from those ofhisoriginal paper. IfAisacommutative ring, then takingahomomorphic image ofthepolynomial ring over Zinto A,we seethat we can define addition andmultiplication ofWitt vectors with components inA,and that these Witt vectors form aring W(A). Show that Wisafunctor, i.e.that anyringhomomorphism qJofAinto acommutative ring A' induces ahomomorphism W(qJ): W(A)-+W(A'). VI,ExEXERCISES 331 47. Letpbe aprime number, and consider theprojection ofW(A) onvectors whose componentsareindexed byapower ofp.Now use thelog tothe base ptoindex these components,sothat wewrite Xninstead ofxp".Forinstance, Xonow denotes what was Xlpreviously. For aWitt vector x=(Xo,Xl'.. .,X n,. ..)define Yx =(0,Xo,Xb...) and Fx =(xg,xf,...). Thus Yisashifting operator. We have V0F=F 0Y.Show that (Yx)(n)=px(n-l) and xCn)=(Fx)(n-l) +pnxn. Also from thedefinition, wehave x(n)=x"+pX"-l+...+Pn.xn. 48. Letkbeafield ofcharacteristic p,andconsider W(k). Then Visanadditive endomorph- ismofW(k), andFisaringhomomorphism ofW(k) into itself. Furthermore, ifXEW(k) then pX=VFx. Ifx,yEW(k), then (Vix)(Yjy)=yi+j(FPjx. FPiy).For aEkdenote by{a}theWitt vector (a,0,0,,..).Then we canwrite symbolically 00 x=Lyi{x;}. i=O Show that ifxEW(k) and Xo#-0then xisaunit inW(k). Hint: One has l-x{xo1}=Vy and then 00 00 x{xo I}L(Vy)i=(1-Vy)L(Vy)i=1. o 0 49. Let nbeaninteger>1and paprime number again. Letkbeafield ofcharacteristic p. Let J.t;,(k) bethering oftruncated Witt vectors (xo,...,Xn-1)with components ink. Weview Jt;,(k)asanadditive group. IfxEJt;,(k), define fcJ(x)=Fx-x.ThenfcJisa homomorphism. IfKisaGalois extension ofk,and uEG(K/k), and XEWn(K)we can define (JXtohave component «(Jxo,...,(Jxn-1).Prove theanalogue ofHilbert's Theorem 90forWitt vectors, and prove that thefirstcohomology group istrivial. (One takes avector whose trace isnot0,and finds acoboundary the same wayasintheproof ofTheorem 10.1). 50.IfxEJt;,(k), show that there exists EJ.t;,(k) such that fcJ()=x.Dothisinductively, solving first forthefirst component, and then showing that avector (0,(11'...,(1n-1)is intheimage offcJifandonly if«(1.,.,,,(1n-1)isintheimage offcJ.Prove inductively that if,'EJ.t;,(k') for some extension k'ofkandiffcJ=fcJ' then-'isavector with components intheprime field. Hence thesolutions offcJ=xforgivenxEJt;,(k) alldiffer bythe vectors with components intheprime field, and there arepnsuch vectors. Wedefine k()=k(0'..., n-1), 332 GALOIS THEORY VI,Ex orsymbolically, k(p-1x), Prove that itisaGalois extension ofk,and show that thecyclic extensions ofk,of degree p",areprecisely those oftypek(p-1x)with avector xsuch that Xopk. 51.Develop theKummer theory forabelian extensions ofkofexponent p"byusing (k). Inother words, show that there isabijection between subgroups Bof(k) containing p(k) and abelian extensions asabove, given by BKB where KB=k(p-lB). Allofthis isdue toWitt, cf.thereferences attheendof8, especially [Wi 37]. The proofsarethe same, mutatis mutandis, asthose given for theKummer theory inthe text. Further Progress and directions Major progresswas made inthe 90sconcerning some problems mentioned inthe chapter. Foremost was Wiles's proof ofenough oftheShimura- Taniyama conjecture to imply Fermat's Last Theorem [Wil 95],[TaW 95]. [TaW 95] R.TAYLOR and A.WILES, Ring-theoretic properties orcertain Hecke alge- bras, Annals ofMath. 141(1995) pp,553-572 [Wil 95] A.WILES, Modular elliptic curves and Fermat's last theorem, Annals. of Math. 141(1995) pp.443-551 Then aproof ofthecomplete Shimura- Taniyama conjecturewasgiven in[BrCDT0I]. [BrCDT 01] C.BREUIL, B.CONRAD, F,DIAMOND, R.TAYLOR, Onthemodularity ofel- lipticcurves over Q:Wild 3-adic exercises, J.Arner, Math. Soc. 14(2001) pp.843-839 Inaquite different direction, Neukirch started thecharacterization ofnumber fields bytheir absolute Galois groups [Ne 68],[Ne 69a], [Ne69b], and proved itforGalois extensions ofQ.His results were extended and hissubsequent conjectures were proved byIkeda and Uchida [Ik77],[Uch 77],[Uch 79],[Uch 81]. These results were extended tofinitely generated extensions ofQ(function fields) byPop [Pop 94], who has amore extensive bibliography onthese and related questions ofalgebraic geometry. For these references, seethebibliography attheendofthebook. CHAPTER VII Extensions ofRings Itisnotalways desirable todeal only with field extensions. Sometimes one wants toobtain afield extension byreducingaring extension modulo aprime ideal. This procedureoccurs inseveral contexts, and sowe areled togive the basic theory ofGalois automorphismsover rings, looking especially athow the Galois automorphisms operateonprime ideals ortheresidue class fields. The two examples given after Theorem 2.9 show theimportance ofworkingover rings, togetfamilies ofextensions intwo very different contexts. Throughout thischapter, A,B,Cwill denote commutative rings. 1. INTEGRAL RING EXTENSIONS InChapters Vand VI wehave studied algebraic extensions offields. For a number ofreasons, itisdesirable tostudy algebraic extensions ofrings. For instance, given apolynomial with integer coefficients, sayX5-X-1, one can reduce thispolynomial mod pforanyprime p,and thus getapoly- nomial with coefficients inafinite field. Asanother example, consider the polynomial XnXn- 1+Sn-1 +.. .+So whereSn_l'. ..,Soarealgebraically independent over afield k.This poly- nomial hascoefficients ink[so,...,Sn-1]andbysubstituting elements ofkfor So,...,Sn-lone obtains apolynomial with coefficients ink.One can then get 333 334 EXTENSION OFRINGS VII, 91 information about polynomials bytakingahomomorphism ofthering in which they have their coefficients. This chapter isdevoted toabrief description ofthebasic facts concerning polynomialsover rings. LetMbeanA-module. We saythat Misfaithful if,whenever aEAissuch that aM=0,then a=O.We note that Aisafaithful module over itself since Acontains aunit element. Furthermore, ifA =t=0,then afaithful module over Acannot betheO-module. LetAbe asubring ofB.Let aEB.Thefollowing conditions areequivalent: INT 1.The element aisaroot ofapolynomial Xn+an-1xn-1+. ..+ao with coefficients aiEA,anddegreen>1.(The essential thing here isthat theleading coefficient isequal to1.) INT 2. Thesubring A[a] isafinitely generated A-module. INT 3. There exists afaithful module over A[a] which isafinitely gener- ated A-module. We prove theequivalence. Assume INT 1.Letg(X) be apolynomial InA[X] ofdegree>Iwith leading coefficient 1such that g(a)=O.If f(X)EA[X] then f(X)=q(X)g(X) +r(X) with q,rEA[X] and degr<deg g.Hence f(a)=r(a), and we seethat if deg g=n,then 1,a,...,an-1aregenerators ofA[a] asamodule over A. Anequation g(X)=0with gasabove, such that g(a)=0iscalled an integral equation for aover A. Assume INT 2.We letthemodule beA[a] itself. Assume INT 3,and letMbethefaithful module over A[a] which isfinitely generated over A,saybyelements WI'...,Wn.Since aM cMthere exist ele- mentsaijEAsuch that aw1=allWI+. ..+a1nWn, aWn=an 1WI+.. .+annWn. TransposingaWb.. .,aWn totheright-hand side ofthese equations, we con- clude that thedeterminant a-all a-a22 -a..I) d= -a..I) a-ann VII,1 INTEGRAL RING EXTENSIONS 335 issuch that dM=o.(This will beproved inthechapter when wedeal with determinants.) Since Misfaithful, wemust have d=O.Hence rxisaroot of thepolynomial det(X bij-aij), which givesanintegral equation for rxover A. Anelement rxsatisfying the three conditions INT 1,2,3iscalled integral over A. Proposition 1.1. Let Abeanentire ring and Kitsquotient field. I£t rxbe algebraic over K. Then there exists anelement c=F0inAsuch that crxis integral over A. Proof. There exists anequation n n-l0 anrx+an-1rx +...+ao= with aiEAand an#O.Multiply itbya- 1.Then (anrx)n+...+aoa:-1=0 isanintegral equation for anrx over A.This proves theproposition. LetACBbesubrings ofacommutative ring C,andletaEC.Ifaisintegral over Athen aisafortiori integral over B.Thus integrality ispreserved under lifting. Inparticular,aisintegral over anyring which isintermediate between Aand B. Let Bcontain Aasasubring. We shall saythat Bisintegral over Aifevery element ofBisintegral over A. Proposition 1.2. IfBisintegral over Aandfinitely generated asanA-algebra, then Bisfinitely generated asanA-module. Proof. We may prove thisbyinduction onthenumber ofring generators, and thus wemay assume that B=A[rx] for some element rxintegralover A,by consideringatower AcA[rx 1]cA[rx brxl]C...cA[rxl'...,rxn]=B. But wehave alreadyseen that our assertion istrue inthat case, thisbeing part ofthedefinition ofintegrality. Just aswedidforextension fields, one may define aclass eofextension rings AcBtobedistinguished ifitsatisfies theanalogous properties, namely: (1)Let AcBcCbe atower ofrings. The extension AcCisineif andonly ifAcBisineand BcCisine. (2)IfAcBisine,ifCisany extension ring ofA,and ifB,Careboth subrings of some ring, then CcB[C] isine.(We note that B[C]=C[B] isthesmallest ring containing both Band C.) 336 EXTENSION OFRINGS VII,1 Aswith fields, wefindformallyasaconsequence of(1)and(2)that (3)holds, namely: (3)IfAcBandAce are ine,and B,Care subrings ofsome ring, then AcB[C] isine. Proposition 1.3. Integral ring extensions form adistinguished class. Proof. Let ACBee be atower ofrings. IfCisintegral over A,then it isclear that Bisintegralover Aand Cisintegralover B.Conversely, assume that each step inthetower isintegral. Let aEC.Then asatisfies anintegral equation nbn- 1+ +b 0 a+ n-la,..0= with biEB.Let B1=A[b o,...,bn-1]. Then B1isafinitely generated A- module byProposition 1.2,and isobviously faithful. Then B1[a] isfinite over Bbhence over A,and hence aisintegral over A.Hence Cisintegral over A. Finally letB,Cbeextension rings ofAand assume Bintegral over A.Assume that B,Care subrings ofsome ring. Then C[B] isgenerated byelements of Bover C,and each element ofBisintegralover C.That C[B] isintegral over Cwillfollow immediately from our next proposition. Proposition 1.4. LetAbeasubring ofC.Then theelements ofCwhich are integral over Aformasubring ofc. Proof. Let a,pECbeintegralover A.Let M =A[a] and N =A[P]. Then MNcontains 1,and istherefore faithful asanA-module. Furthermore, aM cMandpNcN. Hence MNismapped into itself bymultiplication with a+pand a{3. Furthermore MNisfinitely generated over A(if{Wi} are generators ofMand{Vj}aregenerators ofNthen {WiVj}aregenerators of MN). This proves ourproposition. InProposition 1.4,the setofelements ofCwhich areintegral over Ais called theintegral closure ofAinC Example. Consider theintegers Z.LetKbe afinite extension ofQ.We call Kanumber field. The integral closure ofZinKiscalled thering of algebraic integers ofK.This isthe most classical example. Inalgebraic geometry,one considers afinitely generated entire ring Rover Zorover afield k.Let Fbethequotient field ofR.One then considers the integral closure ofRinF,which isproved tobefinite over R.IfKisafinite extension ofF,one also considers theintegral closure ofRinK. Proposition 1.5. Let AcBbe anextension ring, and letBbeintegral over A.Let ubeahomomorphism ofB.Then u(B) isintegral over u(A). Proof. Let aEB,and let n n-l0 a+an-1a +.. .+ao= VII,1 INTEGRAL RING EXTENSIONS 337 beanintegral equation for rxover A.Applying(Jyields (J(rx)n +(J(a n_1)(J(rx)n-l +...+(J(ao)=0, there byprovingour assertion. Corollary 1.6. Let Abe anentire ring, kitsquotient field, and Eafinite extension ofk.Let rxEEbeintegral over A.Then the norm and trace ofrx (from Etok)areintegral over A,and soarethecoefficients oftheirreducible polynomial satisfied byrxover k. Proof. For each embedding(JofEover k,(Jrxisintegral over A.Since the norm istheproduct of(Jrxover allsuch (J(raised toapower ofthecharacteristic), itfollows that the norm isintegral over A.Similarly forthetrace, andsimilarly forthecoefficients ofIrr(rx, k,X),which areelementary symmetric functions of the roots. Let Abeanentire ring and kitsquotient field. We saythat Aisintegrally closed ifitisequal toitsintegral closure ink. Proposition 1.7. Let Abeentire andfactorial. Then Aisintegrally closed. Proof. Suppose that there exists aquotient a/bwith a,bEAwhich is integralover A,and aprime element pinAwhich divides bbutnot a.Wehave, for some integern>1,and aiEA, (a/b)n +an_l(a/b)n-l +...+ao=0 whence an+an_1ban-1+.. .+aobn=o. Since pdivides b,itmust divide an,and hence must divide a,contradiction. Letf:A-.Bbe aring-homomorphism (A,Bbeing commutative rings). We recall that such ahomomorphism isalso called anA-algebra. We may view BasanA-module. We saythat Bisintegral over A(for thisring-homo- morphism f)ifBisintegral overf(A). This extension ofour definition of integrality isuseful because there areapplications when certain collapsings take place, and westill wish tospeak ofintegrality. Strictly speakingweshould not saythat Bisintegral over A,butthatfisanintegral ring-homomorphism, orsimply thatfisintegral. We shall usethisterminology frequently. Some ofour preceding propositions have immediate consequences for integral ring-homomorphisms; forinstance, iff:A-.Band g:B-.Care integral, then g0f:A-.Cisintegral. However, itisnotnecessarily true that ifg0fisintegral,soisf. Letf:A-.Bbeintegral, and letSbe amultiplicativesubset ofA.Then wegetahomomorphism S-1f:S-1A-.S-1B, where strictly speaking, S-1B=(f(S))-1 B,andS-lf isdefined by (S-If)(x/s)=f(x)/f(s). 338 EXTENSION OFRINGS VII,1 Itistrivially verified that this isahomomorphism. We have acommutative diagram B)s-1B fl IS-If A s-1A thehorizontal maps being thecanonical ones: x-.x/l. Proposition1.8. Letf:A-.Bbeintegral, and letSbe amultiplicative subset of'A.Then S-If':S-1A-.s-1Bisintegral. Proof. IfrxEBisintegral overf(A), then writing rxpinstead off(a){Jr for aEAandpEBwehave rxn+an_1rxn-1+...+ao=0 with aiEA.Taking thecanonical image inS-1Aand S-1Brespectively,we seethat this relation proves theintegrality ofrx/I over S-1A,thecoefficients beingnow ai/I. Proposition 1.9. Let Abeentire andintegrally closed. Let Sbeamultipli- cative subset ofA,0ftS.Then S-1Aisintegrally closed. Proof. Let rxbeanelement ofthequotient field, integralover S-1A.We have anequation n+an- 1 n- 1+ +ao-0 rx-rx...--, Sn-1 So aiEAand SiES.Let sbetheproduct sn- 1...so. Then itisclear that srxis integral over A,whence inA.Hence rxlies inS-1A,and S-1Aisintegrally closed. Let pbeaprime ideal ofaring Aand letSbethecomplement ofpinA. Wewrite S=A-p.Iff: A-.BisanA-algebra (i.e.aring-homomorphism), weshall writeB"instead ofS-1 B.We can viewB"asanA"=S-1A-module. Let Abeasubring ofB.Let pbeaprime ideal ofAand let beaprime ideal ofB.We saythat liesabove pif nA=p.Ifthat isthe case, then theinjection A-.Binduces aninjection ofthefactor rings A/p-.B/, and infact wehave acommutative diagram: B B/ I I A )A/p VII,1 INTEGRAL RING EXTENSIONS 339 thehorizontal arrows being thecanonical homomorphisms, and thevertical arrows being injections. IfBisintegralover A,thenB/ isintegral over AlpbyProposition1.5. Proposition 1.10. Let Abeasubring ofB,letpbeaprime ideal ofA,and assume Bintegral over A.Then pB =FBand there exists aprime ideal of Blying above p. Proof. We know thatB"isintegral over A"and thatA"isalocal ring with maximal idealm"=S-Ip,where S=A-p.Since weobviously have pB"=pA"B"=m"B", itwill suffice toprove ourfirst assertion when Aisalocal ring. (Note that the existence ofaprime ideal pimplies that 1=F0,andpB=Bifandonly if1EpB.) Inthat case, ifpB=B,then 1has anexpressionasafinite linear combination ofelements ofBwith coefficients inp, 1=a1b1+...+anbn with aiEpand biEB.We shall now usenotation asifA"cB".We leave it tothe reader asanexercise toverify that our argumentsarevalid when we deal only with acanonical homomorphism A"-.B".LetBo=A[b h...,bnJ. Then pBo=Boand Boisafinite A-module byProposition 1.2. Hence Bo=0 byNakayama's lemma, contradiction. (See Lemma 4.1ofChapter X.) Toprove our second assertion, note thefollowing commutative diagram: ' lB I A)A" Wehave justproved m"B"=FB".Hencem"B"iscontained inamaximal ideal 9JlofB".Taking inverse images,we seethat theinverse image of9JlinA"isan ideal containing m"(inthe case ofaninclusion A"cB"theinverse image is 9JlnA,,).Sincem"ismaximal, wehave 9J1nA"=m".Let betheinverse image of9J1inB(inthe case ofinclusion,=9JlnB). Then isaprime ideal ofB.The inverse image ofm"inAissimply p.Taking theinverse image of9Jlgoing around both ways inthediagram,wefind that nA=p, aswas tobeshown. Proposition1.11. Let Abe asubring ofB,and assume that Bisintegral over A.Let beaprime ideal ofBlying over aprime ideal pofA,Then ismaximal ifandonlyifpismaximal. 340 EXTENSION OFRINGS VII,2 Proof. Assume pmaximal inA.Then A/p isafield, andB/ isanentire ring, integral over A/p. IfrxEB/, then rxisalgebraic over A/p, and weknow that A/p[rx] isafield. Hence every non-zero element ofB/ isinvertible in B/, which istherefore afield. Conversely,assume that ismaximal inB. ThenB/ isafield, which isintegralover theentire ring A/p. IfA/p isnot a field, ithas anon-zero maximal ideal m.ByProposition 1.10, there exists a prime ideal 9JlofB/ lying above m,9Jl =F0,contradiction. 2. INTEGRAL GALOIS EXTENSIONS We shall now investigate therelationship between theGalois theory ofa polynomial, and theGalois theory ofthis same polynomial reduced modulo a prime ideal. Proposition 2.1. Let Abeanentire ring, integrally closed initsquotient field K.LetLbe afinite Galois extension ofKwith group G.Let pbea maximal ideal ofA,and let,.Qbeprime ideals oftheintegral closure Bof AinLlying above p.Then there exists (JEGsuch that(J=.Q. Proof. Suppose that .Q=F(Jforany(JEG.Then t.Q =F(Jforanypair ofelements (J,!EG.There exists anelement xEBsuch that x=0(mod (J), x=1(mod (J.Q),all (JEG all (JEG (use theChinese remainder theorem). The norm N(x)=n(Jx tiEG lies inB(\K =A(because Aisintegrally closed), and lies in nA=p. But xrt(J.Qforall (JEG,sothat (JXrt.Qforall (JEG.This contradicts thefact that the norm ofxliesinp=.QnA. Ifone localizes, one can eliminate thehypothesis that pismaximal; just assume that pisprime. Corollary 2.2 LetAbeintegrally closed initsquotient field K.Let Ebea finite separable extension ofK,and Btheintegral closure ofAinE.Let pbe amaximal ideal ofA.Then there exists only afinite number ofprime ideals of Blying above p. Proof. LetLbethesmallest Galois extension ofKcontaining E.If.Q t, .Q2aretwo distinct prime ideals ofBlying above p,andh2 aretwoprime ideals oftheintegral closure ofAinLlying above .Qtand .Q2respectively, thent =F2. This argument reduces our assertion tothe case that EisGalois over K,and itthen becomes animmediate consequence oftheproposition. VII,2 INTEGRAL GALOIS EXTENSIONS 341 Let Abeintegrally closed initsquotient field K,and letBbeitsintegral closure inafinite Galois extension L,with group G.Then (JB =Bforevery aEG.Let pbeamaximal ideal ofA,and amaximal ideal ofBlying above p. We denote byG'J\thesubgroup ofGconsisting ofthose automorphismssuch that(J=. ThenG'J\operates inanatural way ontheresidue class field B/, and leaves Alp fixed. Toeach (JEG'J\wecan associate anautomorphism iiofB/ over Alp, and themap given by (JI-+(J induces ahomomorphism ofG'J\into the group ofautomorphisms ofB/ over Alp. The group G'J\will becalled thedecomposition groupof. Itsfixed field will bedenoted byLdec,and will becalled thedecomposition fieldof. Let Bdecbetheintegral closure ofAinLdec,and .Q = nBdec. ByProposition 2.1, weknow that istheonly prime ofBlying above .Q. Let G=U(JjG'J\be acoset decomposition ofGinG.Then theprime ideals (Jjareprecisely thedistinct primes ofBlying above p.Indeed, fortwo elements (J,!EGwehavea=tifandonly ift-l(J=,i.e.t-1(Jliesin G'J\.Thus t,(Jlieinthe same coset modG. Itisthen immediately clear that thedecomposition group ofaprime (J isaG'J\(J- 1. Proposition 2.3. Thefield Ldecisthesmallest subfield EofLcontaining Ksuch that istheonly prime ofBlying above nE(which isprime in BnE). Proof. Let Ebeasabove, and letHbetheGalois group ofLover E.Let q= nE.ByProposition 2.1, allprimes ofBlying above qareconjugate by elements ofH.Since there isonly oneprime, namely,itmeans that Hleaves invariant. Hence GcG'J\and E::JLdec. We have already observed that Ldechas therequired property. Proposition 2.4. Notation beingasabove, wehave Alp=Bdec/.Q (under thecanonical injection Alp-+BdecI.Q). Proof. Ifaisanelement ofG,not inG'J\'then(J =F and(J-l =F. Let .Qu=(J-l nBdec. Then .Qa=F.Q. Let xbeanelement ofBdec. There exists anelement yofBdec such that y=x(mod.Q) y= 1(mod .Qa) 342 EXTENSION OFRINGS VII,2 foreach uinG,but notinG'J\.Hence inparticular, y=x(mod) y=1(mod u-1) foreach unotinG'J\.This second congruence yields uy= 1(mod) forallurtG'J\.The norm ofyfrom LdectoKisaproduct ofyand other factors uywith urtG'J\.Thus weobtain NLdeC{)_ K Y=X(mod). But the norm liesinK,and even inA,since itisaproduct ofelements integral over A.This last congruence holds mod .Q,since both xand the norm liein Bdec .This isprecisely themeaning oftheassertion inourproposition. Ifxisanelement ofB,weshall denote byxitsimage under thehomo- morphism B-+B/. Then uistheautomorphism ofB/ satisfying therelation ux =(ux). Iff{X) isapolynomial with coefficients inB,wedenote by!{X) itsnatural image under theabove homomorphism. Thus, if f{X)=bnXn+...+bo, then !(X)=Dnxn+...+Do. Proposition 2.5. Let Abeintegrally closed initsquotient field K,and let Bbeitsintegral closure inafinite Galois extension LofK,with group G. Let pbeamaximal ideal ofA,and amaximal ideal ofBlying above p. ThenB/ isanormal extension ofA/p, and themapu1-+Uinduces ahomo- morphism ofG'J\onto theGalois group ofB/ over A/p. Proof. LetB=B/ and A=A/p. Any element ofBcan bewritten as xfor some xEB.Let xgenerateaseparable subextension ofBover A,and let fbetheirreducible polynomial for xover K. The coefficients offlieinA because xisintegral over A,and allthe roots offareintegral over A.Thus m f{X)=n(X-Xi) i=1 VII,2 INTEGRAL GALOIS EXTENSIONS 343 splits into linear factors inB.Since m J(X)=L(X-Xi) i=1 and alltheXilieinB,itfollows thatJsplits into linear factors inB.Weobserve thatf(x)=0implies J(x)=o.Hence Bisnormal over A,and [A(x):A]<[K(x): K]<[L:K]. ThIs implies that the maximal separable subextension ofAinBisoffinite degreeover A(using theprimitive element theorem ofelementary field theory). This degree isinfact bounded by[L:K]. There remains toprove that the map(J1---+ifgives asurjective homo- morphism ofG'J\onto theGalois group ofBover A.Todothis, weshall give anargument which reduces ourproblem tothe case when istheonly prime ideal ofBlying above p.Indeed, byProposition 2.4,theresidue class fields of theground ring and thering Bdecinthedecomposition field are the same. This means that toprove oursurjectivity, wemay take Ldec asground field. This isthedesired reduction, and we can assume K =Ldec, G=G'J\. This being the case, take agenerator ofthemaximal separable subextension of13over A,and letitbex,for some element xinB.Letfbetheirreducible polynomial ofxover K.Any automorphism ofBisdetermined byitseffect onX,and maps xon some root ofJ.Suppose that x=x1.Given any root Xi off,there exists anelement (JofG=G'J\such that (JX =Xi. Hence ifx =Xi. Hence theautomorphisms ofBover Ainduced byelements ofGoperate transitively onthe roots ofJ.Hence they give usallautomorphisms ofthe resid ueclass field, aswas tobeshown. Corollary 2.6. LetAbeintegrally closed initsquotient field K.LetLbea finite Galois extension ofK,and Btheintegral closure ofAinL.Let pbe a maximal ideal ofA.Let cp:A Alp bethecanonical homomorphism, and let t/1I' t/12betwohomomorphisms ofBextending cpinagiven algebraic closure ofAlp.Then there exists anautomorphism(]"ofLover Ksuch that t/J1=t/J20(J. Proof. The kernels of1/11,1/12areprime ideals ofBwhich areconjugate byProposition 2.1. Hence there exists anelement! oftheGalois group G such that t/J1,1/120!have the same kernel. Without loss ofgenerality,wemay therefore assume that1/1l'1/12have the same kernel. Hence there exists an automorphism wof1/1 1(B) onto1/12(B) such that W01/11=1/12.There exists an element (JofG'J\such that w01/11=1/1 10(J,bythepreceding proposition. This proves what wewanted. 344 EXTENSION OFRINGS VII,2 Remark. Inalltheabove propositions, wecould assume pprime instead ofmaximal. Inthat case, one has tolocalize atptobeable toapply ourproofs. Intheabove discussions, thekernel ofthemap G'J\-+G'J\ iscalled the inertia groupof. Itconsists ofthose automorphisms ofG'J\ which induce thetrivial automorphism ontheresidue class field. Itsfixed field iscalled theinertia field, and isdenoted byLin . Corollary 2.7. Let theassumptions beasinCorollary 2.6and assume that istheonly prime ofBlying above p.Letf(X) be apolynomial inA[X] with leading coefficient1.Assume thatfisirreducible inK[X], and has a root rxinB.Then thereduced polynomial fisapower ofanirreducible poly- nomial inA[X]. Proof. ByCorollary 2.6, weknow that any two roots offareconjugate under some isomorphism ofBover A,and hence thatfcannot split into relative prime polynomials. Therefore, fisapower ofanirreducible polynomial. Proposition 2.8. Let Abe anentire ring, integrally closed initsquotient field K.LetLbeafinite Galois extension ofK.LetL=K(rx), where rxis integral over A,and let f(X)=xn+an_1Xn-1+...+ao betheirreducible polynomial ofrxover k,with aiEA.Let pbeamaximal ideal inA,let beaprime ideal oftheintegral closure BofAinL, lying above p.Letf(X) bethereduced polynomial with coefficients inAlp. Let G'J\bethedecomposition group. Iffhas nomultiple roots, then the map (J1-+Uhastrivial kernel, and isanisomorphism ofG'J\ontheGalois group offover Alp. Proof. Let f(X)=n(X-Xi) bethefactorization offinL.We know that allXiEB.If(JEG",then we denote byathehomomorphic image of (Jinthegroup G'J\'asbefore. We have f(x)=n(X-Xi)' Suppose that ax;=Xiforalli.Since ((Jx;)=axi,and since fhas nomultiple roots, itfollows that (Jisalso theidentity. Hence our map isinjective, thein- ertia group istrivial. The fieldA[x l'.. .,xn] isasubfield ofBand any auto- VII,2 INTEGRAL GALOIS EXTENSIONS 345 morphism ofBover Awhich restricts totheidentity onthis subfield must be theidentity, because themap G'J\-.G'J\isonto theGalois group ofBover A. Hence Bispurely inseparable over A[x I'. ..,xn] and therefore G'J\isiso- morphic totheGalois group ofJover A. Proposition 2.8isonlyaspecialcase ofthemore-general situation when theroot ofapolynomial does notnecessarily generate aGalois extension. We state aversion useful tocompute Galois groups. Theorem 2.9. Let Abeanentire ring, integrally closed initsquotient field K. Letf(X)EA[X] have leading coefficient1and beirreducible over K (orA,it'sthe same thing). Let pbeamaximal ideal ofAand letJ=fmod p. Suppose thatJhas nomultiple roots inanalgebraic closure ofA/p. Let Lbe asplitting fieldforfover K,and letBbetheintegral closure ofAin L.Let beanyprime ofBabove pand let abar denote reduction mod p. Then themap G'J\-.G'J\ isanisomorphism ofG'J\with theGalois group ofJover A. Proof. Let (rxl,...,rxn)bethe roots off inBand let(ai'...,an)betheir reductions mod. Since n f(X)=n(X-rx;), i=1 itfollows that n J(X)=n(X-a;). i=I Any element ofGisdetermined byitseffect asapermutation oftheroots, and for (JEG'J\'wehave (j'ii=(Jrx;. Hence ifu=idthen (J=id,sothemap G'J\-.G'J\isinjective. Itissurjective byProposition 2.5, sothetheorem isproved. This theorem justifies thestatement used tocompute Galois groups inChapter VI,2. Theorem 2.9givesavery efficient tool foranalyzing polynomialsover a rlng. Example. Consider the"generic" polynomial fw(X)=xn+wn_Ixn-1 +... +Wo 346 EXTENSION OFRINGS VII,3 where wo,. . .,Wn-Iarealgebraically independentover afield k.Weknow that theGalois group ofthispolynomialover thefield K=k(wo,. . .,wn-I) isthe symmetric group. Let tl,. . .,tnbetheroots. Let abeagenerator ofthesplitting field L;that is,L=K(a). Without loss ofgenerality,we can select atobe integral over thering k[wo,. . .,wn-I](multiply anygiven generator byasuitably chosen polynomial and useProposition 1.1). Letgw(X) betheirreducible poly- nomial ofaover k(wo,. . .,Wn-I). The coefficients of9arepolynomialsin(w). Ifwe cansubstitute values (a)for(w)with ao,. . .,an-IEksuch that garemains irreducible, then byProposition 2.8 weconclude atonce that theGalois group ofgaisthesymmetric group also. Similarly, ifafinite Galois extension of k(wo,. . .,wn-I) hasGalois group G,then we can do asimilar substitution to getaGalois extension ofkhaving Galois group G,provided thespecial polynomial garemains irreducible. Example. LetKbe anumber field; that is, afinite extension ofQ.Let 0 bethering ofalgebraic integers. LetLbe afinite Galois extension ofKand() thealgebraic integers inL.Let pbe aprime of0and aprime of()lying above p.Then ojp isafinite field, saywith qelements. Then()j isafinitextension ofojp, andbythetheory offinite fields, there isaunique element inG'J\'called theFrobenius elementFr'J\'such thatFr'J\(i)=iqforiE()j. The conditions ofTheorem 2.9 aresatisfied forallbut afinite number ofprimes p,andforsuch primes, there isaunique elementFr'J\EG'J\such thatFr'J\(x)=xqmod forall xE() .We callFr'J\theFrobenius element inG'J\.Cf.Chapter VI, 15,where some ofthesignificance oftheFrobenius element isexplained. 3. EXTENSION OF HOMOMORPHISMS When wefirst discussed the process oflocalization, weconsidered very briefly theextension ofahomomorphism toalocal ring. Inour discussion of field theory, wealso described anextension theorem forembeddings ofone field into another. Weshall now treat theextension question infullgenerality. First werecall the case ofalocal ring. Let Abe acommutative rin,g andp aprime ideal. Weknow that thelocal ring Apisthe setofallfractions x/y, with x,YEA and Yfj.p.Itsmaximal ideal consists ofthose fractions with xEp.Let Lbe afield and let cp:A Lbe ahomomorphism whose kernel isp.Then we can extendcptoahomomorphism ofA"into Lbyletting ({J(X/Y)=({J(x)/ ({J(Y) ifx/y isanelement ofA"asabove. Second, wehave integral ring extensions. Let 0bealocal ring with maximal ideal m,letBbeintegralover 0,and let({J:0-.Lbe ahomomorphism of0 VII,3 EXTENSION OFHOMOMORPHISMS 347 into analgebraically closed field L.We assume that thekernel ofqJism.By Proposition 1.10, weknow that there exists amaximal ideal 9J1ofBlying above m,i.e.such that 9Jln0=m.Then BI9Jl isafield, which isanalgebraic exten- sion ofo/m, ando/m isisomorphic tothesubfield qJ(o) ofLbecause thekernel ofqJism. We can find anisomorphism ofo/m onto qJ(o) such that thecomposite homomorphism o-+o/m-+L isequal toqJ.We now embed BI9Jl into Lsoastomake thefollowing diagram commutative: B)BI9Jl 1 1 o)o/m)L and inthis way getahomomorphism ofBinto Lwhich extendsqJ. Proposition3.1. Let Abeasubring ofBand assume that Bisintegral over A.Let qJ:A-+Lbe ahomomorphism into afield Lwhich isalgebraically closed. ThenqJhas anextension toahomomorphism ofBinto L. Proof. Let pbethekernel ofqJand letSbethecomplement ofpinA. Then wehave acommutative diagram )S-1B 1 )S-IA =A"B 1 A andqJcan befactored through thecanonical homomorphism ofAinto S-1A. Furthermore, S-1Bisintegral over S-1 A.This reduces thequestion tothe case when wedeal with alocal ring, which hasjust been discussed above. Theorem 3.2. Let Abe asubring ofafield Kand letxEK,x=FO.Let qJ:A-+Lbe ahomomorphism ofAinto analgebraically closed field L. Then qJhas anextension toahomomorphism ofA[x] orA[x-1]into L. Proof. We may first extendqJtoahomomorphism ofthelocal ringA", where pisthekernel ofqJ.Thus without loss ofgenerality,wemayassume that Aisalocal ring with maximal ideal m.Suppose that mA[x-1]=A[x-1]. 348 EXTENSION OFRINGS VII,93 Then we canwrite 1-I -n=ao+alx +...+anx with aiEm.Multiplying byxnweobtain (1-ao)xn +bn-IXn-I+...+bo=0 with suitable elements biEA. Since aoEm,itfollows that 1-aortmand hence 1-aoisaunit inAbecause Aisassumed tobealocal ring. Dividing by1-aowe seethat xisintegralover A,and hence that ourhomomorphism has anextension toA[x]byProposition 3.1. Ifontheother hand wehave mA[x-l]=FA[x-I] then mA[x-l]iscontained insome maximal ideal ofA[x- I]and nA contains m. Since mismaximal, wemust have nA=m.SinceqJand the canonical map A-+Aim have the same kernel, namely m, we can find an embedding t/JofAim into Lsuch that thecomposite map A-+Aim L isequal toqJ.We note that Aim iscanonically embedded inB/ where B=A[x-I],and extend t/Jtoahomomorphism ofB/ into L,which we can dowhether theimage ofx-1inB/ istranscendental oralgebraic over Aim. Thecomposite BB/-+Lgivesuswhat wewant. Corollary 3.3. Let Abeasubring ofafield Kand letLbeanalgebraically closed field. LetqJ:A-+Lbeahomomorphism. Let Bbeamaximal subring ofKtowhichqJhas anextension homomorphism into L.Then Bisalocal ring andifxEK,x=F0,then xEBorX-I EB. Proof. Let Sbethe setofpairs (C,t/J)where Cisasubring ofKand t/J:C-+Lisahomomorphism extending qJ.Then Sisnotempty (containing (A, qJ)],and ispartially ordered byascending inclusion and restriction. In other words, (C,t/J)<(C', t/J')ifCcC'and therestriction oft/J'toCisequal tot/J.Itisclear that Sisinductively ordered, andbyZorn's lemma there exists amaximal element, say(B, t/J0).Then first Bisalocal ring, otherwise t/J0extends tothelocal ringarising from thekernel, and second, Bhasthedesired property according toTheorem 3.2. Let Bbe asubring ofafield Khaving theproperty thatgiven xEK,x=t=0, then xEBorX-I EB.Then wecall Bavaluation ring inK.We shall study such rings ingreater detail inChapter XII. However, weshall also give some applications inthe next chapter,sowemake some more comments here. VII,3 EXTENSION OFHOMOMORPHISMS 349 LetFbeafield. We letthesymbol00satisfy theusual algebraicrules. If aEF,wedefine a+00 =00, a.oo=oo if a#0, 1 1 00.00 =00,-=00 and- =o.0 00 The expressions00+00,0.00, 0/0, and 00/00 arenotdefined. Aplace ({Jofafield Kinto afield Fisamapping cp:K-.{F,oo} ofKinto the setconsisting ofFand 00satisfying theusual rules for ahomo- morphism, namely lfJ(a +b)=lfJ(a) +((J(b), cp(ab)=lfJ(a)lfJ(b) whenever theexpressions ontheright-hand side ofthese formulas aredefined, and such that ({J(1)=1.We shall also say that theplace isF-valued. The elements ofKwhich arenotmapped into 00will becalled finite under theplace, and theothers will becalled infinite. The reader willverify atonce that the set 0ofelements ofKwhich are finite under aplace isavaluation ring ofK.The maximal ideal consists ofthose elements xsuch that ({J(x)=O.Conversely, if0isavaluation ring ofKwith maximal ideal m, weletcp: 0-.o/m bethecanonical homomorphism, and define ({J(x)=00for xEK,xrto.Then itistrivially verified thatlfJisaplace. If({Jl:K-.{FbOO} and({J2:K-.{F2,oo}areplaces ofK, wetake their restrictions totheir images. We may therefore assume that theyaresurjective. Weshall saythat they areequivalent ifthere exists anisomorphismA.:F1-.F2 such that({J2=({JlOA.. (We put A.(00)=00.) One sees that two placesare equivalent ifandonly ifthey have the same valuation ring. Itisclear that there isabijection between equivalence classes ofplaces ofK,and valuation rings of K.Aplace iscalled trivial ifitisinjective. The valuation ring ofthetrivial place issimply Kitself. Aswith homomorphisms,weobserve that thecomposite oftwoplaces isalso aplace (trivial verification). Itisoften convenient todeal with places instead ofvaluation rings, justasitis convenient todeal with homomorphisms and notalways with canonical homo- morphisms oraring modulo anideal. The general theory ofvaluations and valuation rings isdue toKrull, All- gemeine Bewertungstheorie, J.reine angew. Math. 167(1932), pp. 169-196. However, theextension theory ofhomomorphismsasabove was realized only around 1945 byChevalley and Zariski. 350 EXTENSION OFRINGS VII,3 We shall now give some examples ofplaces and valuation rings. Example 1. Letpbe aprime number. LetZ(p)bethering ofallrational numbers whose denominator isnotdivisible byp.ThenZ(p)isavaluation ring. The maximal ideal consists ofthose rational numbers whose numerator isdivisible byp. Example 2. Let kbe afield and R=k[X] thepolynomial ring inone variable. Letp=p(X) beanirreducible polynomial. Let 0bethering ofrational functions whose denominator isnotdivisible byp.Then 0isavaluation ring, similar tothat ofExample1. Example 3. Let Rbethering ofpower series k[[X]] inone variable. Then Risavaluation ring, whose maximal ideal consists ofthose power series divisible byX.The residue class field iskitself. Example 4. Let R=k[[X l'. . .,Xn]] bethering ofpower series inseveral variables. Then Risnot avaluation ring, butR isimbedded inthefield ofrepeated power series k«X 1))«X 2))· · ·«X n))=Kn.ByExample 3,there isaplace of Knwhich isKn-l-valued. By induction and composition,we can define a k-valued place ofKn. Since thefield ofrational functions k(Xl'. . .,Xn)is contained inKn, therestriction ofthisplace tok(Xl'. . .,Xn)givesak-valued place ofthefield ofrational functions innvariables. Example 5. InChapter XI weshall consider thenotion ofordered field. Letkbeanordered subfield ofanordered field K.Let 0bethesubset ofelements ofKwhich are notinfinitely large with respect tok.Let mbethe subset of elements of 0which areinfinitely small with respect tok.Then 0isavaluation ring inKand misitsmaximal ideal. Thefollowing property ofplaces will beused inconnection with projective space inthe next chapter. Proposition 3.4. Let cp:K {L,oo}beanL-valued place ofK.Given a finite number ofnon-zero elements Xl,. . .,XnEKthere exists anindexjsuch thatcpisfinite onxiiXjfori=1,..., n. Proof. LetBbethevaluation ring oftheplace. DefineXi<Xjtomean that xilxjEB.Then therelation<istransitive, that isifXi<XjandXj<Xrthen Xi<Xr-Furthermore, bytheproperty of avaluation ring,wealways have Xi< XjorXj<Xiforallpairs ofindices i,j.Hence wemay order our ele- ments, and we select the indexjsuch that Xi<Xjfor alli.This index j satisfies therequirement oftheproposition. We can obtain acharacterization ofintegral elements bymeans ofval- uation rings.We shall use thefollowing terminology. If0,.0 are local rings with maximal ideals m,9Jlrespectively,weshall saythat .0lies above 0 if0c.0and 9Jln0=m.Wethen have acanonical injection o/m .o/9Jl. VII,3 EXTENSION OFHOMOMORPHISMS 351 Proposition 3.5. Let 0bealocal ring contained inafield L.Anelement xo.f Lisintegral over 0if'andonly ifxlies inevery valuation ring .0oj'L lying above o. Proof Assume that xisnotintegral over o.Let mbethemaximal ideal ofo. Then theideal (m,1/x) ofo[l/x] cannot betheentire ring, otherwise we can write -1 =an(1/x)n +...+at(l/x) +y with yEmand aiEo.From this weget (I+y)xn +...+an=O. But 1+Yisnotinm,hence isaunit ofo.Wedivide theequation by1+Yto conclude that xisintegralover 0,contrary toourhypothesis. Thus (m,l/x) is not theentire ring, and iscontained inamaximal ideal,whose intersection with 0contains mand hence must beequal tom.Extending thecanonical homo- morphism 0[1/x]-.o[l/x]/ toahomomorphism ofavaluation ring .0ofL, we seethat theimage ofI/xis0and hence that xcannot beinthisvaluation ring. Conversely,assume that xisintegralover 0,and let xn+an-lxn-l+·..+ao=0 beanintegral equation for xwith coefficients ino.Let0beany valuation ring ofLlying above o.Supposexfj..0.Letcpbetheplace given bythecanonical homomorphism of.0modulo itsmaximal ideal. Then cp(x)=00socp(1/x)=0.. Divide theabove equation byxn ,andapply cpoThen each term except thefirst maps to0undercp,so weget cp(1)=0, acontradiction which proves the proposition... Proposition 3.6. Let Abearing contained inafield L.Anelement xofL isintegral over Aifandonlyifxlies inevery valuation ring .0ofLcontaining A.Interms ofplaces, xisintegral over Aifandonlyifevery place ofLfinite onAisfinite onx. Proof. Assume that every place finite onAisfinite onx .We mayassume x=t=o.If1/xisaunit inA[1/x]then we can write x=Co+cI(I/x)+... +cn_I(I/x)n-l withCiEAand some n.Multiplying byxn-lweconclude that xisintegral over A.If1/xisnot aunit inA[1/x],then 1/xgeneratesaproper principal ideal. ByZorn's lemma thisideal iscontained inamaximal ideal IDl.Thehomomorphism A[1/x] A[1/x]/m can beextended toaplace which isafinite onAbutmaps 352 EXTENSION OFRINGS VII, Ex 1/xon0,soxon 00,which contradicts thepossibility that 1/xisnot aunit in A[1/x]and proves that xisintegralover A.The converse implication isproved justasinthesecond part ofProposition 3.5. Remark. LetKbe asubfield ofLand letxEL.Then xisintegral over Kifandonly ifxisalgebraicover K.Soifaplace cpofLisfinite onK,and x isalgebraic over K,thencpisfinite onK(x). Of course this isatrivial case of theintegrality criterion which can be seen directly. Let xn+an_Ixn-1+...+ao=0 betheirreducible equation for xover K.Supposex=t=o.Then ao=t=O.Hence cp(x)=t=0immediately from theequation,socpisanisomorphism ofK(x) onits Image. The next result isageneralizationwhose technique ofproofcan also beused inExercise 1ofChapter IX(the Hilbert-Zariski theorem). Theorem 3.7. General Integrality Criterion. Let Abeanentire ring. Let ZI'. ..,Zmbeelements ofsome extension field ofitsquotient field K.Assume that eachZs(s=1,..., m)satisfiesapolynomial relation Zs+gs(zl, ..., zm)=0 where gs(ZI,.. .,Zm)EA[ZI'...,Zm] isapolynomial oftotal degree <ds, and that any pure power ofZsoccuring with non-zero coefficient ingsoccurs with apower strictly less than ds.Then ZI,. . .,Zmareintegral over A. Proof. Weapply Proposition 3.6. SupposesomeZsisnotintegral over A. There exists aplace cpofK,finite onA,such that cp(zs)= 00for some s.By Proposition 3.4 we canpickanindex ssuch thatcp(Zj/Zs)=t=00forallj.We divide thepolynomial relation ofthehypothesis inthelemma byz'jsandapply theplace. Bythehypothesisongs,itfollows thatcp(gs(z)/z'fs)=0,whence we get1=0,acontradiction which proves thetheorem. EXERCISES 1.Let Kbe aGaloIs extensIon oftherationals Q,wIth group G.Let BbetheIntegral closure ofZinK,and let aEBbesuch that K =Q(a). Letf(X)=Irr(a, Q,X), Let pbe aprIme number, and assume thatfremains irreducIble mod pover Z/pZ, What canyou sayabout theGalois group G?(ArtIn asked thisquestion toTate onhisqualify- ingexam,) 2,Let AbeanentIre ring and KitsquotIent field. Let tbetranscendental over K,IfA isintegrally closed, show thatA[t] isintegrally closed, VII, Ex EXERCISES 353 For thefollowing exercises, youcan use 91ofChapter X. 3,LetAbeanentire nng, Integrally closed initsquotient field K,LetLbeafinite separable extension ofK,and letBbetheintegral closure ofAinL.IfAisNoetherian, show that BISafinite A-module. [Hint: Let{Wt,...,WII} be abasis ofLover K.Multiplying allelements ofthis basis byasuitable element ofA,wemay assume without loss of generality that allWiareintegral over A.Let{w't,...,w} bethedual basis relative to thetrace, sothatTr(w;wj)=b;j'Write anelement (1ofLintegralover Aintheform (1=b1w't+' ..+bllw with hjEK,Taking the trace Tr(aw;), fori=1". .,n,conclude that Biscontained inthefinite module Aw;+... +Aw.] Hence BisNoetherian. 4,The preceding exercise applies tothe case when A=Zand k=Q.LetLbe afinite extension ofQand let 0Lbethering ofalgebraic integers inL.Letai'. . .,anbe thedistinct embeddings ofLinto thecomplex numbers. Embedded0Linto aEuclidean space bythemap a (ala,. . .,ana), Show that inany bounded region ofspace, there isonlyafinite number ofelements ofOLe[Hint: The coefficients inanintegral equation for aareelementary symmetric functions oftheconjugates ofaand thus arebounded integers,] Use Exercise 5of Chapter IIItoconclude that 0Lisafree Z-module ofdimension<n.Infact, show that thedimension isn,abasis of0Lover Zalso beingabasis ofLover Q, 5.Let Ebe afinite extension ofQ,and let 0Ebethering ofalgebraic integers ofE.Let Ubethegroup ofunits of0E.Letai'. . .,anbethedistinct embeddings ofEinto C,Map Uinto aEuclidean space, bythemap I:a(log 100tal,..., log 10"11(11). Show that l(U)ISafree abelian group, finitely generated, byshowing that inany finIte region ofspace, there isonlyafinite number ofelements ofl(U), Show that thekernel oflis afinite group, and istherefore thegroup ofroots ofunity inE.Thus Uitself isa finitely generated abelian group. 6.Generalize the results of92toinfinite Galois extensions, especially Propositions 2.1 and 2.5,using Zorn's lemma, 7.Dedekind rings. Let 0beanentire ring which isNoetherian, integrally closed, and ,such that every non-zero prime ideal ismaximal. Define afractional ideal atobean o-submodule =1=0ofthequotient field Ksuch that there exists cEO, c=1=0forwhich caCO. Prove that the fractional ideals form agroup under multiplication. Hint followingvan derWaerden: Prove thefollowingstatements inorder: (a)Given anideal a=1=0in0,there exists aproduct ofprime ideals PI...PrCa. (b)Every maximal idealPisinvertible, i.e.ifweletp-Ibethe setofelements xEKsuch that xpC 0,then p-IP=0, (c)Everynon-zero ideal isinvertible, byafractional ideal. (Use theNoetherian property thatifthis isnottrue, there exists amaximal non-invertible ideal a,and getacontradiction.) 354 EXTENSION OFRINGS VII, Ex 8.Using prime ideals instead ofprime numbers for aDedekind ringA,define thenotion ofcontent asintheGauss lemma, andprove thatiff(X), g(X) EA[X] arepolynomials ofdegree>0with coefficients inA,then cont(fg)=cont(j)cont(g). Also ifKis thequotient field ofA,prove the same statement forf,9EK[X]. 9.Let Abeanentire ring, integrally closed. Let Bbeentire, integral over A.LetQ., Q2beprime ideals ofBwith QI:JQ2butQI=1=Q2. LetP;=Q;nA,Show that PI=1=P2. 10. Let nbe apositive integer and let(,('beprimitive n-th roots ofunity. (a)Show that (1-0/(1-(')isanalgebraic integer. (b)Ifn>6isdivisible byatleast twoprimes, show that 1-(isaunit inthe ringZ[. 11. Letpbe aprime and(aprimitive p-th root ofunity. Show that there isaprincipal ideal JinZ[(] such thatJP-I=(p)(theprincipal ideal generated byp), Symmetric Polynomials 12.LetFbeafield ofcharacteristic o.Let t) ,.,.,tnbealgebraically independentover F. Let 5),,..,Snbetheelementary symmetric functions. Then R=F[t),,. .,tn]isan integral extension ofS=F[s),,,,,sn],and actually isitsintegral closure inthe rational fieldF(t)".., tn). Let Wbethe group ofpermutation ofthe variables t),...,t n. (a)Show that S=RWisthefixed subring ofRunder W. (b)Show that theelements t1...tnwith 0<';<n-iform abasis ofRover S,soinparticular,Risfree over S. Iamtold that theabove basis isdue toKronecker. There isamuch more interesting basis, which can bedefined asfollows. Leta),.,.,anbethepartial derivatives with respect tot)",.,tn, soa;=a/at;, Let PEF[t]=F[t) ,,,,,tn].Substituting a;for ti(i=1,.,,,n)givesapartial differential operator P(a)=P(a),...,an)onR.Anelement ofScan also beviewed asanelement of R.LetQER.We saythat QisW-harmonic ifP(a)Q=0forallsymmetric polynomials PES with 0constant term. Itcan beshown that the W-harmonic polynomials form a finite dimensional space. Furthermore, if{H) ,.,.,HN}isabasis forthis space over F, then itisalso abasis forRover S.This isaspecialcase ofageneral theorem ofChe- valley. See[La99b], where thespecial case isworked outindetail. CHAPTER VIII Transcendental Extensions Both fortheir own sake and forapplications tothe case offinite exten- sions oftherational numbers, one isledtodeal with ground fields which are function fields, i.e.finitely generated over some field k,possibly byelements which are notalgebraic. This chapter gives some basic properties ofsuch fields. 1. TRANSCENDENCE BASES Let Kbe anextension field ofafield k.Let Sbe asubset ofK. We recall that S(orthe elements ofS)issaid tobealgebraically independent over k,ifwhenever wehave arelation o=La(v)M(v)(S)=La(v)nxvex) xeS with coefficientsa(v)Ek,almost alla(v)=0,then wemust necessarily have all a(v)=o. We can introduce anordering among algebraically independent subsets of K,byascending inclusion. These subsets areobviously inductively ordered, and thus there exist maximal elements. IfSisasubset ofKwhich is algebraically independent over k,andifthecardinality ofSisgreatest among allsuch subsets, then wecall this cardinality the transcendence degree or dimension ofKover k.Actually, weshall need todistinguish only between finite transcendence degree orinfinite transcendence degree. We observe that 355 356 TRANSCENDENTAL EXTENSIONS VIII,1 thenotion oftranscendence degree bears tothenotion ofalgebraic indepen- dence the same relation asthenotion ofdimension bears tothenotion of linear independence. Wefrequently deal with families ofelements ofK,sayafamily {Xi}iel' and saythat such afamily isalgebraically independent over kifitselements aredistinct (inother words, Xi=Fxjifi=Fj)and ifthe setconsisting ofthe elements inthisfamily isalgebraically independent over k. Asubset SofKwhich isalgebraically independent over kand ismaximal with respect totheinclusion ordering will becalled atranscendence base of Kover k.From themaximality, itisclear that ifSisatranscendence base ofKover k,then Kisalgebraic over k(S). Theorem 1.1. LetKbe anextension ofafield k.Any two transcendence bases ofKover khave the same cardinality. Ifrisasubset ofKsuch that Kisalgebraic over k(r), and Sisasubset ofrwhich isalgebraically indepen- dent over k,then there exists atranscendence base ofKover ksuch that SC(BCr. Proof. Weshall prove thatifthere exists one finite transcendence base, say {xI'. . .,xm},m>1,mminimal, then any other transcendence base must also have melements. For this itwill suffice toprove: IfWb. . .,Wnareelements ofKwhich arealgebraically independentover kthen n<m(for we can then usesymmetry). Byassumption, there exists anon-zero irreducible polynomial flinm+ 1variables with coefficients inksuch that fl(wI'XI'. . .,xm)=o. After renumbering xI'. . .,Xmwemay writefl=gj(WI' X2,. . .,xm)x1with some gN=t=0with some N>1.Noirreducible factor ofgNvanishes on (WI' X2,. . .,xn),otherwise WIwould bearoot oftwodistinct irreducible polyno- mials over k(XI'...,xm).Hence XIisalgebraicover k(wI' x2'...,xm)and Wb X2,...,Xmarealgebraically independentover k,otherwise theminimal- ityofmwould becontradicted. Suppose inductively that after asuitable re- numbering ofX2,. . .,Xmwehave found WI'...,Wr(r<n)such that Kis algebraic over k(w l,...,WnXr+I'...,xm).Then there exists anon-zero polynomial finm+1variables with coefficients inksuch that f(wr+l, WI'...,Wr,Xr+l,...,xm)=o. Since thew'sarealgebraically independent over k,itfollows bythe same argument asinthefirst step that someXj'say xr+I'isalgebraic over k(WI'...,wr+I' xr+2,...,xm).Since atower ofalgebraic extensions isalgebraic, itfollows that Kisalgebraicover k(wI'...,Wr+I'Xr+2,...,xm).We can repeat the procedure, andifn>mwe canreplace allthex'sbyw's, tosee that Kis algebraic over k(wI'. . .,wm).This shows that n>mimpliesn=m, asdesired. VIII,2 NOETHER NORMALIZATION THEOREM 357 We have now proved: Either the transcendence degree isfinite, and is equal tothecardinality ofany transcendence base, oritisinfinite, and every transcendence base isinfinite. The cardinality statement intheinfinite case will beleft asanexercise. We shall also leave asanexercise the statement that asetofalgebraically independent elements can becompletedto a transcendence base, selected from agiven setIsuch that Kisalgebraicover k(f). (The reader will note thecomplete analogy ofour statements with those concerning linear bases.) Note. The preceding section istheonly one used inthe next chapter. The remaining sections are more technical, especially 3and which will not be used inthe restofthe book. Even 2and5will only bementioned a couple oftimes, and sothereader may omit them until they arereferred to again. 2. NOETHER NORMALIZATION THEOREM Theorem 2.1. Letk[x l,...,xn]=k[x] beafinitely generated entire ring over afield k,and assume that k(x) has transcendence degree r.Then there exist elements YI, ..., Yrink[x] such thatk[x] isintegral over kEy]=k[Yl' ...,Yr]. Proof If(xI'...,xn)arealready algebraically independent over k,we aredone. Ifnot, there isanon-trivial relation La(j)x{l...xn=0 with each coefficienta(j)Ekanda(j)=Fo.The sum istaken over afinite number ofdistinct n-tuples ofintegers (jI'...,jn)'jv>O.Let m2, ..., mnbe positive integers, and put m2 mn Y2=X2-XI ,..., Yn=Xn-XI. Substitute Xi=Yi+Xi(i=2,...,n)inthe above equation. Using vector notation, weput (m)=(1,m2, ..., mn)and use the dot product (j)'(m) to denote jl+m2j2+...+mnjn' Ifweexpand the relation after making the above substitution, weget " (j)'(m)+f( )-0 c(j)XIXI'Y2'...,Yn- wherefisapolynomial inwhich nopure power of XIappears. We now select dtobe alarge integer [say greater than any component ofavector (j) such thatc(j)=F0]and take (m)=(1,d,d2 ,...,dn). 358 TRANSCENDENTAL EXTENSIONS VIII,2 Then all(j).(m) aredistinct forthose (j)such that cU)=Fo.Inthis way we obtain anintegral equation forXl over k[Y2'.'" Yn]. Since each Xi(i>1) isintegral over k[x l,Y2, ...,Yn]' itfollows that k[x] isintegral over k[Y2' ...,Yn]. We can now proceed inductively, using thetransitivity of integral extensions toshrink the number ofy'suntil we reach analge- braically independent setofy's. The advantage oftheproof ofTheorem 2.1isthat itisapplicable when k isafinite field. The disadvantage isthat itisnot linear inXl' ..., Xn. We now deal with another technique which leads into certain aspects ofalgebraic geometry onwhich weshall comment after the next theorem. We start again with k[x l,..., xn]finitely generated over kand entire. Let(Uij) (i,j=1,...,n)bealgebraically independent elements over k(x), and letku=k(u)=k(Uij)all i,j.Put n Y.="U..X.I UJ. j=l This amounts to ageneric linear change ofcoordinates inn-space, to use geometric terminology. Again welet rbethe transcendence degree ofk(x) over k. Theorem 2.2. With the above notation, ku[x] isintegral over ku[yl,.. .,Yr]. Proof Suppose some Xiisnotintegral over ku[yl,...,Yr]. Then there exists aplace qJofku(Y) finite onku[yl,...,Yr]buttaking the value 00on some Xi.Using Proposition 3.4ofChapter VII, andrenumbering theindices ifnecessary, sayqJ(xj/x n)isfinite foralli.Letzj=qJ(Xj/x n)forj=1,...,n. Then dividing theequations Yi=LuijxjbyXn(for i=1,...,r)andapplying theplace, weget 0=UllZ; +Ul2Z +... +Uln, o=Ur1Z+Ur2Z+...+Urn. The transcendence degree ofk(z') over kcannot ber,forotherwise, theplace qJwould be anisomorphism ofk(x) onitsimage. [Indeed, if,say,z;,...,z; arealgebraically independent and Zi=xi/x n,then zl,..., Zrare also alge- braically independent, and soform atranscendence base fork(x) over k. Then theplace isanisomorphism from k(zl,...,zr) tok(z;,..., z;), and hence isanisomorphism from k(x) toitsimage.] We then conclude that Uln,..., UrnEk(uij, z') with i=1,...,r;j=1,..., n-1. Hence thetranscendence degree ofk(u) over kwould be<rn-1,which isa contradiction, proving thetheorem. VIII,2 NOETHER NORMALIZATION THEOREM 359 Corollary 2.3. Let kbe afield, and letk(x) be afinitely generated extension oftranscendence degreer.There exists apolynomial P(u)= P(Uij)Ek[u] such thatif(c)=(cij)isafamily ofelements cijEksatisfying P(c) =F0,and weletY;=LCijXj,thenk[x] isintegral overk[y, ...,Y;]. Proof ByTheorem 2.2, each Xiisintegral over ku[yI'...,Yr].The coefficients ofanintegral equation arerational functions inku. We letP(u) be acommon denominator for these rational functions. IfP(c) =F0,then there isahomomorphism qJ:k(x)[u,p(U)-I]-.k(x) such that qJ(u)=(c),and such that lfJistheidentity onk(x). We canapply lfJ toanintegral equation for Xiover ku[y] toget anintegral equation for Xi over kEy'], thus concluding theproof. Remark. After Corollary 2.3, there remains theproblem offinding ex- plicitly integral equations forXl' ..., Xn(or Yr+l, ...,Yn) over ku[YI' ...,Yr]. This isanelimination problem, and Ihave decided torefrain from further involvement inalgebraic geometry atthis point. But itmay beuseful to describe thegeometric language used tointerpret Theorem 2.2and further results inthat line. After thegeneric change ofcoordinates, themap (YI' ...,Yn)I---+(YI' ...,Yr) isthegeneric projection ofthevariety whose coordinate ring isk[x] on affine r-space. This projection isfinite, and inparticular, theinverse image of apointonaffine r-space isfinite. Furthermore, ifk(x) isseparable over k(a notion which will bedefined in4),then theextension ku(Y) isfinite separable over ku(Yt,..., Yr)(inthe sense ofChapter V). Todetermine thedegree of this finite extension isessentially Bezout's theorem. Cf.[La 58], Chapter VIII, 6. The above techniques were created by van der Waerden and Zariski, cf., forinstance, also Exercises 5and 6.These techniques have unfortunately not been completely absorbed in some more recent expositions ofalgebraic geometry. Togiveaconcrete example: When Hartshorne considers the intersection ofavariety and asufficiently general hyperplane, hedoes not discuss the"generic" hyperplane (that is,with algebraically independent coefficients over agiven ground field), and he assumes that thevariety is non-singular from the start (see hisTheorem 8.18 ofChapter 8,[Ha 77]). But thedescription ofthe intersection can bedone without simplicityas- sumptions,asinTheorem 7of[La 58], Chapter VII,6,and the corre- sponding lemma. Something was lost indiscarding thetechnique ofthe algebraically independent (uij). After two decades when themethods illustrated inChapter Xhave been prevalent, there isareturn tothe more explicit methods ofgeneric construc- tions using thealgebraically independent (uij)and similar ones for some 360 TRANSCENDENTAL EXTENSIONS VIII,3 applications because part ofalgebraic geometry and number theory are returning tosome problems asking forexplicitoreffective constructions, with bounds onthedegrees ofsolutions ofalgebraic equations. See, forinstance, [Ph91-95], [So90], and thebibliographyattheendofChapter X,6.Return- ing to some techniques, however, does not mean abandoning others; it means only expanding available tools. Bibliography [So90]R.HARTSHORNE, Algebraic Geometry, Springer-Verlag, New York, 1977 S.LANG, lntroduction toAlgebraic Geometry, Wiley-Interscience, New York, 1958 P.PHILIPPON, Sur deshauteurs alternatives, IMath, Ann. 289(1991) pp.255-283; IIAnn. lnst. Fourier 44(1994) pp.1043-1065; IIIJ.Math, Pures Appl. 74(1995) pp.345-365 C.SOULE, Gcometrie d'Arakelov etthcorie desnombres transcendants, Asterisque 198-200 (1991)pp. 355-371[Ha 77] [La58] [Ph91- 95] 3. LINEARLY DISJOINT EXTENSIONS Inthis section wediscuss theway inwhich two extensions Kand Lofa field kbehave with respect toeach other. We assume that allthe fields involved arecontained inone field Q,assumed algebraically closed. Kissaid tobelinearly disjoint from Lover kifevery finite setof elements ofKthat islinearly independent over kisstill such over L. The definition isunsymmetric, but weprove right away that theproperty ofbeing linearly disjoint isactually symmetric forKand L. Assume K linearly disjoint from Lover k.Let Y1,...,Ynbeelements ofLlinearly independent over k.Suppose there isanon-trivial relation oflinear depen- dence over K, (1) XIYl +X2Y2 +...+XnYn=o. SayXl' ..., Xrarelinearly independent over k,and xr+l'..., Xnarelinear r combinations Xi=Lai/lx/l'i=r+1,..., n.We can write therelation (1)as /l=1 follows: ,.t.x,.Y,.+i=t.Ct.ai,.x,.)Yi=0 andcollecting terms, after inverting thesecond sum, weget J.(y,.+i=t.(ai,.Yi»)x,.=o. VIII,3 LINEARLY DISJOINT EXTENSIONS 361 The y's arelinearly independent over k,sothe coefficients ofx/lare =FO. This contradicts thelinear disjointness ofKand Lover k. We now give two criteria forlinear disjointness. Criterion 1.Suppose that Kisthequotient field ofaring Rand Lthe quotient field ofaring S.To test whether Land Karelinearly disjoint, it suffices toshow that ifelements Yl, ..., YnofSarelinearly independentover k,then there isnolinear relation among they'swith coefficients inR. Indeed, ifelements Yl,...,YnofLarelinearly independent over k,and if there isarelation XlYl+...+XnYn=0with XiEK,then we can select Yin Sand XinRsuch that xy =F0,YYiESfor alli,andXXi ERforall i. Multiplying therelation byxygivesalinear dependence between elements of Rand S.However, the YYi areobviously linearly independent over k,and this proves our criterion. Criterion 2.Again letRbe asubring ofKsuch that Kisitsquotient field and Risavector space over k.Let{Uti} be abasis ofRconsidered asa vector space over k.Toprove Kand Llinearly disjoint over k,itsuffices to show that theelements {Uti}ofthis basis remain linearly independent over L. Indeed, suppose this isthe case. LetXl'...' xmbeelements ofRlinearly independent over"k. They lieinafinite dimension vector space generated by some ofthe Uti'sayUl'...,Un.Theycan becompleted to abasis forthis space over k.Lifting this vector space ofdimension nover L,itmust conserve itsdimension because the u'sremain linearly independent byhy- pothesis, and hence thex'smust also remain linearly independent. Proposition 3.1. Let Kbe afield containing another field k,and let L::JEbetwo other extensions ofk.Then Kand Larelinearly disjoint over kifandonlyifKand Earelinearly disjointover kand KE, Lare linearly disjointover E. KL/\ KE L/\/ \/E k 362 TRANSCENDENTAL EXTENSIONS VIII,3 Proof Assume first that K,Earelinearly disjoint over k,and KE, Lare linearly disjoint over E.Let{K}be abasis ofKasvector space over k(we use the elements ofthis basis astheir own indexing set), and let{rx}be a basis ofEover k.Let{l}be abasis ofLover E.Then {rxl}isabasis ofL over k.IfKand Lare notlinearly disjoint over k,then there exists a relation L(LCJCA.CZK)lrx =0 A,cz JCwith some CJCACZ =F0, CJCA.CZEk. Changing theorder ofsummation gives L(LCJCAczKrx)A.=0 A JC,A contradicting thelinear disjointness ofLand KE over E. Conversely, assume that Kand Larelinearly disjoint over k.Then a fortiori, Kand Eare also linearly disjoint over k,and the field KEisthe quotient field oftheringE[K] generated over Ebyallelements ofK.This ring isavector space over E,and abasis forKover kisalso abasis forthis ringE[K] over E.With this remark, and thecriteria forlinear disjointness, we seethat itsuffices toprove that the elements ofsuch abasis remain linearly independent over L.Atthispointwe seethat thearguments given inthefirst part oftheproof are reversible. We leave theformalism tothe reader. We introduce another notion concerning two extensions Kand Lofa field k.We shall say that Kisfree from Lover kifevery finite setof elements ofKalgebraically independent over kremains such over L.If(x) and (y) are two sets ofelements inQ,wesay that they are free over k(or independent over k)ifk(x) andk(y) arefree over k. Just aswith linear disjointness, our definition isunsymmetric, and we prove that therelationship expressed therein isactually symmetric. Assume therefore that Kisfree from Lover k.Let Yt,..., Ynbeelements ofL, algebraically independent over k.Suppose they become dependent over K. They become soin asubfield FofKfinitely generated over k,say of transcendence degree rover k.Computing the transcendence degree ofF(y) over kintwo ways givesacontradiction (cf.Exercise 5). F(y)7"'" F'"k(y) r'"/" k VIII,4 SEPARABLE AND REGULAR EXTENSIONS 363 Proposition 3.2.IfKandLarelinearly disjoint over k,then theyarefree over k. Proof LetXl' ..., Xnbeelements ofKalgebraically independentover k. Suppose they become algebraically dependent over L.We get arelation LYaM«(x)=0 between monomials M«(x) with coefficients y«inL.This givesalinear relation among theM«(x). But these arelinearly independent over kbecause thex's areassumed algebraically independent over k.This isacontradiction. Proposition 3.3. LetLbeanextension ofk,and let(u)=(u1,...,ur)bea setofquantities algebraically independent over L. Then thefield k(u) is linearly disjoint from Lover k. Proof. According tothe criteria for linear disjointness, itsuffices to prove that theelements ofabasis fortheringk[u] that arelinearly indepen- dent over kremain soover L.Infact themonomials M(u) give abasis of k[u] over k.They must remain linearly independent over L,because as wehave seen, alinear relation gives analgebraic relation. This proves our proposition. Note finally that theproperty that two extensions Kand Lofafield k arelinearly disjoint orfree isoffinite type. Toprove that they have either property, itsuffices todoitfor allsubfields Ko and LoofKand L respectively which arefinitely generated over k.This comes from the fact that thedefinitions involve onlyafinite number ofquantities atatime. 4. SEPARABLE AND REGULAR EXTENSIONS LetKbe afinitely generated extension ofk,K=k(x). We shall saythat itisseparably generated ifwe can find atranscendence basis (tI'..., tr)of K/k such that Kisseparably algebraic over k(t). Such atranscendence base issaid tobeaseparating transcendence base forKover k. Wealways denote bypthecharacteristic ifitisnot O.The field obtained from kbyadjoining allpm-th roots ofallelements ofkwill bedenoted by kl/pm .Thecompositum ofallsuch fields for m=1,2,..., isdenoted bykl/poo . Proposition 4.1. Thefollowing conditions concerning anextension field K ofkareequivalent: (i)Kislinearly disjoint from kl/pOO. (ii)Kislinearly disjoint from kl/pmfor some m. 364 TRANSCENDENTAL EXTENSIONS VIII,4 (iii)Every subfield ofKcontaining kandfinitely generated over kis separably generated. Proof Itisobvious that (i)implies (ii). Inorder toprove that (ii) implies (iii), wemay clearly assume that Kisfinitely generated over k,say K =k(x)=k(xI'.. .,xn). Let the transcendence degree ofthis extension be r.Ifr=n,theproof is complete. Otherwise, say xI'.. .,Xrisatranscendence base. Then Xr+l is algebraic over k(xI'.. .,xr).Letf(X I'...,Xr+l)be apolynomial oflowest degree such that f(XI"'"xr+l)=O. Thenfisirreducible. We contend that not allXi(i=1,..., r+1)appear to thep-th power throughout. Ifthey did, wecould write f(X)=Lc(JM(J(X)P where M(J(X)aremonomials inXI'...,Xr+l andC(JEk.This would imply that theM(J(x)arelinearly dependent over kl/p(taking thep-th root ofthe equation Lc(JM(J(x)P=0).However, theM(J(x) arelinearly independentover k(otherwise wewould get anequation forXl' ..., Xr+l oflower degree) and wethus getacontradiction tothelinear disjointness ofk(x) and kl/p .Say X1does not appear tothep-th power throughout, butactually appears in f(X). We know thatf(X) isirreducible ink[X I,...,Xr+IJ and hence f(x)=O isanirreducible equation for Xl over k(x 2,..., xr+l).Since Xl does not appear tothep-th power throughout, thisequation isaseparable equation for Xl over k(x 2,...., xr+l),inother words, Xlisseparable algebraic over k(x 2,...,xr+l).From this itfollows that itisseparable algebraic over k(x 2,..., xn).If(X2'...,xn)isatranscendence base, theproof iscomplete. If not, saythat X2isseparableover k(X3' ..., xn).Then k(x) isseparableover k(x 3,..., xn).Proceeding inductively, we see that the procedurecan be continued until wegetdown toatranscendence base. This proves that (ii) implies (iii). Italso proves that aseparating transcendence base fork(x) over kcan beselected from thegiven setofgenerators (x). Toprove that (iii)implies (i)wemay assume that Kisfinitely generated over k.Let(u)be atranscendence base forKover k.Then Kisseparably algebraicover k(u). ByProposition 3.3, k(u) and k1/pooarelinearly disjoint. Let L=kl/pOO. Then k(u)L ispurely inseparable over k(u), and hence is linearly disjoint from Kover k(u)bytheelementary theory offinite algebraic extensions. Using Proposition 3.1, weconclude that Kislinearly disjoint from Lover k,thereby proving our theorem. Anextension Kofksatisfying theconditions ofProposition 4.1 iscalled separable. This definition iscompatible with the use ofthe word foralge- braic extensions. The first condition ofour theorem isknown asMacLane's criterion. It has thefollowing immediate corollaries. VIII,4 SEPARABLE AND REGULAR EXTENSIONS 365 Corollary 4.2.IfKisseparable over k,and Eisasubfield ofKcontain- ingk,then Eisseparable over k. Corollary 4.3. Let Ebe aseparable extension ofk,and Kaseparable extension ofE.Then Kisaseparable extension ofk. Proof Apply Proposition 3.1and thedefinition ofseparability. Corollary 4.4.Ifkisperfect, every extension ofkisseparable. Corollary 4.5. Let Kbe aseparable extension ofk,andfree from an extension Lofk.Then KL isaseparable extension ofL. Proof Anelement ofKL has anexpression interms ofafinite number ofelements ofKand L. Hence any finitely generated subfield ofKL containing Liscontained inacomposite field FL,where Fisasubfield ofK finitely generated over k.ByCorollary 4.2, wemay assume that Kisfinitely generated over k.Let (t)be atranscendence base ofKover k,soKis separable algebraic over k(t). Byhypothesis, (t)isatranscendence base of KL over L,and since every element ofKisseparable algebraic over k(t), it isalso separable over L(t). Hence KL isseparably generated over L.This proves thecorollary. Corollary 4.6. Let Kand Lbetwo separable extensions ofk,free from each other over k.Then KL isseparable over k. Proof Use Corollaries 4.5and 4.3. Corollary 4.7. LetK,Lbetwo extensions ofk,linearly disjointover k. Then Kisseparable over kifandonlyifKL isseparable over L. Proof IfKisnotseparable over k,itisnotlinearly disjoint from k1/p over k,and hence afortiori itisnotlinearly disjoint from Lk1/pover k.By Proposition 4.1, thisimplies that KL isnotlinearly disjoint from Lk1/pover L,and hence that KL isnotseparable over L.The converse isaspecialcase ofCorollary 4.5,taking into account that linearly disjoint fields are free. We conclude our discussion ofseparability with two results. The first one hasalready been proved inthefirst part ofProposition 4.1, but we state it here explicitly. Proposition 4.8.IfKisaseparable extension ofk,and isfinitely gener- ated, then aseparating transcendence base can beselected fromagiven set ofgenerators. To state the second result wedenote byKpm thefield obtained from K byraising allelements ofKtothepm-th power. 366 TRANSCENDENTAL EXTENSIONS VIII,4 Proposition 4.9. Let Kbe afinitely generated extension ofafield k.If Kpmk =Kfor some m,then Kisseparably algebraic over k.Conversely, if Kisseparably algebraic over k,then Kpmk =Kforall m. Proof. IfK/k isseparably algebraic, then the conclusion follows from theelementary theory offinite algebraic extensions. Conversely, ifK/k is finite algebraic but notseparable, then themaximal separable extension ofk inKcannot beallofK,and hence KPkcannot beequal toK.Finally, if there exists anelement tofKtranscendental over k,then k(t1/pm)hasdegree pmover k(t), and hence there exists atsuch that t1/pmdoes notlieinK.This proves ourproposition. There isaclass ofextensions which behave particularly well from the point ofview ofchanging theground field, and areespecially useful in algebraic geometry. Weput some results together todeal with such exten- sions. Let Kbe anextension ofafield k,with algebraic closure K8 .We claim that thefollowing two conditions areequivalent: REG 1.kisalgebraically closed inK(i.e. every element ofKalgebraic over kliesink),and Kisseparable over k. REG 2.Kislinearly disjoint from k8over k. We show theequivalence. Assume REG 2.ByProposition 4.1, weknow that Kisseparably generated over k.Itisobvious that kmust bealgebraically closed inK.Hence REG 2implies REG 1.Toprove the converse weneed alemma. Lemma 4.10. Let kbealgebraically closed inextension K.Let xbe some element ofanextension ofK,butalgebraic over k.Then k(x) and K arelinearly disjoint over k,and[k(x):k]=[K(x):K]. Proof Letf(X) bethe irreducible polynomial for xover k.Thenf remains irreducible over K;otherwise, itsfactors would have coefficients algebraic over k,hence ink.Powers ofxform abasis ofk(x) over k,hence the same powers form abasis ofK(x) over K.This proves thelemma. Toprove REG 2from REG 1,wemay assume without loss ofgenerality that Kisfinitely generated over k,and itsuffices toprove that Kislinearly disjoint from anarbitrary finite algebraic extension Lofk.IfLisseparable algebraic over k,then itcan begenerated by oneprimitive element, and we canapply Lemma 4.10. More generally, letEbethemaximal separable subfield ofLcontaining k.ByProposition 3.1, we seethat itsuffices toprove that KE and Lare linearly disjoint over E.Let (t)be aseparating transcendence base forK over k.Then Kisseparably algebraic over k(t). Furthermore, (t)isalso a separating transcendence base forKE over E,and KE isseparable algebraic VIII,4 SEPARABLE AND REGULAR EXTENSIONS 367 over E(t). Thus KE isseparable over E,and bydefinition KE islinearly disjoint from Lover Kbecause Lispurely inseparable over E.This proves that REG 1implies REG 2. Thuswe can define anextension Kofktoberegular ifitsatisfies either one oftheequivalent conditions REG 1orREG 2. Proposition 4.11. (a)LetKbearegular extension ofk,and letEbeasubfield ofKcontaining k.Then Eisregularover k. (b)Let Ebe aregular extension ofk,and Karegular extension ofE. Then Kisaregular extension ofk. (c)Ifkisalgebraically closed, then every extension ofkisregular. Proof. Each assertion isimmediate from thedefinition conditions REG 1and REG 2. Theorem 4.12. Let Kbe aregular extension ofk,letLbe anarbitrary extension ofk,both contained insome larger field, and assume that K,L arefree over k.Then K,Larelinearly disjoint over k. Proof (Artin). Without loss ofgenerality,wemay assume that Kis finitely generated over k.Let xt,...,Xnbeelements ofKlinearly indepen- dent over k.Suppose wehave arelation oflinear dependence XtYt +...+XnYn=0 with YiEL.LetlfJbe aka-valued place ofLover k.Let(t)be atranscen- dence base ofKover k.Byhypothesis, the elements of(t)remain alge- braically independent over L,and hence lfJcan beextended toaplace ofKL which isidentity onk(t). This place must then beanisomorphism ofKon itsimage, because Kisafinite algebraic extension ofk(t)(remark atthe end ofChapter VII,3). After asuitable isomorphism, wemay take aplace equivalent tolfJwhich istheidentity onK.Say ({J(Yi/Yn) isfinite foralli(use Proposition 3.4ofChapter VII). Wedivide therelation oflinear dependence by Ynand apply ({JtogetLXi({J(Yi/Yn)=0,which givesalinear relation among the Xiwith coefficients inka ,contradicting the linear disjointness. This proves thetheorem. Theorem 4.13. LetKbe aregular extension ofk,free fromanextension Lofkover k.Then KL isaregular extension ofL. Proof. From thehypothesis,wededuce that Kisfree from thealgebraic closure LaofLover k.ByTheorem 4.12, Kislinearly disjoint from Laover k.ByProposition 3.1,KL islinearly disjoint from Laover L,and hence KL isregular over L. 368 TRANSCENDENTAL EXTENSIONS VIII,4 Corollary 4.14. LetK,Lberegular extensions ofk,free from each other over k.Then KL isaregular extension ofk. Proof. UseCorollary 4.13 andProposition 4.11(b). Theorem 4.13 isone ofthe main reasons foremphasizing the class of regular extensions: they remain regular under arbitrary base change ofthe ground field k.Furthermore, Theorem 4.12 inthebackground isimportant inthe study ofpolynomial ideals asinthe next section, and we add some remarks here onitsimplications. We now assume that the reader is acquainted with the most basic properties ofthe tensor product (Chapter XVI, 1and2). Corollary 4.15. Let K =k(x) be afinitely generated regular extension, free from anextension Lofk,and both contained insome larger field. Then thenatural k-algebra homomorphism L@kk[x]-+L[x] isanisomorphism. Proof ByTheorem 4.12 thehomomorphism isinjective, and itisobvi- ously surjective, whence thecorollary follows. Corollary 4.16. Letk(x) beafinitely generated regular extension, and let Pbetheprime ideal ink[X] vanishing on(x), that is,consisting ofall polynomials f(X)Ek[X] such thatf(x)=O.Let Lbe anextension ofk, free from k(x) over k.Let PLbetheprime ideal inL[X] vanishing on(x). Then PL=pL[X], that isPListheideal generated byPinL[X], and in particular, this ideal isprime. Proof Consider the exact sequence o-+P-+k[X]-+k[x]-+o. Since we aredealing with vector spaces over afield, the sequene remains exact when tensored with any k-space,soweget anexact sequence o-+L@kP-+L[X]-+L0kk[x]-+O. ByCorollary 4.15, weknow that Lkk[x]:::::::L[x], and theimage ofL0k P inL[X] ispL[X], sothelemma isproved. Corollary 4.16 shows another aspect whereby regular extensions behave well under extension ofthe base field, namely the way theprime ideal P remains prime under such extensions. VIII,5 DERIVATIONS 369 5. DERIVATIONS Aderivation Dofaring Risamapping D:R-+RofRinto itself which is linear and satisfies theordinary rule forderivatives, i.e., D(x +y)=Dx+Dy and D(xy)=xDy +yDx. As anexample ofderivations, consider thepolynomial ringk[X] over afield k.For each variable Xi' thepartial derivative a/aXitaken inthe usual manner isaderivation ofk[X]. Let Rbeanentire ring and letKbeitsquotient field. Let D:R-+Rbea derivation. Then Dextends uniquely toaderivation ofK,bydefining D(u/v)=vDu uDv . v Itisimmediately verified that the expression ontheright-hand side is independent ofthe way werepresentanelement ofKasu/v(u,vER),and satisfies theconditions definingaderivation. Note. Inthis section, weshall discuss derivations offields. For deriva- tions inthe context ofrings and modules, seeChapter XIX, 3. Aderivation ofafield Kistrivial ifDx =0forallxEK.Itistrivial over asubfield kofKifDx =0forallxEk.Aderivation isalways trivial over theprime field: One sees that D(I)=D(1.1)=2D(I), whence D(I)=o. We now consider theproblem ofextending derivations. Let L=K(x)=K(x 1,..., xn) be afinitely generated extension. IffEK[X], we denote byaf/ax ithe polynomials af/aXievaluated at(x). Given aderivation DonK,does there exist aderivation D* onLcoinciding with DonK? Iff(X) EK[X] isa polynomial vanishing on(x),then any such D*must satisfy (1) o=D*f(x)=fD(X) +L(af/ax;)D*Xi' where fDdenotes thepolynomial obtained byapplying Dtoallcoefficients off.Note that ifrelation (1)issatisfied forevery element inafinite setof generators oftheideal inK[X] vanishing on(x),then (1)issatisfied byevery polynomial ofthis ideal. This isanimmediate consequence ofthe rules for derivations. The preceding ideal will also becalled the ideal determined by (x)inK[X]. 370 TRANSCENDENTAL EXTENSIONS VIII,5 The above necessary condition forthe existence ofaD* turns out tobe sufficien 1. Theorem 5.1. Let Dbeaderivation ofafield K.Let (x)=(Xl' ..., xn) beafinite family ofelements inanextension ofK.Let{h(X)}beasetof generators for theideal determined by(x)inK[X]. Then, if(u)isany set ofelements ofK(x) satisfying theequations o=hD(x) +L(Oh/OXi)U i, there isone andonlyone derivation D*ofK(x) coinciding with DonK, and such that D*Xi=Uifor every i. Proof The necessity has been shown above. Conversely, ifg(x), h(x) are inK[x], and h(x) #0,one verifies immediately that themapping D*defined bytheformulas ogD*g(x)=gD(X) +L Ui,uX'I *(/h)=hD*g-gD*hD gh2' iswell defined and isaderivation ofK(x). Consider thespecialcase where (x)consists ofone element x.Let Dbe a given derivation onK. Case 1. xisseparable algebraic over K. Letf(X) bethe irreducible polynomial satisfied byxover K.Then f'(x) #o.We have o=fD(x) +f'(x)u, whence u= -fD(x)/f'(x). Hence Dextends toK(x) uniquely. IfDistrivial onK,then Distrivial onK(x). Case 2. xistranscendental over K. Then Dextends, and ucan be selected arbitrarily inK(x). Case 3. xispurely inseparable over K, soxP-a=0,with aEK.Then Dextends toK(x) ifand only ifDa =o.Inparticular ifDistrivial onK, then ucan beselected arbitrarily. Proposition 5.2. Afinitely generated extension K(x) over Kisseparable algebraic ifandonlyifevery derivation DofK(x) which istrivial onKis trivial onK(x). Proof IfK(x) isseparable algebraic over K,this isCase 1.Conversely, ifitisnot, we can make atower ofextensions between Kand K(x), such VIII,5 DERIVATIONS 371 that each step iscovered by one ofthethree above cases. Atleast one step will becovered byCase 2or3.Taking the uppermost step ofthis latter type, one sees immediately how toconstruct aderivation trivial onthe bottom and nontrivial ontopofthe tower. Proposition 5.3. Given Kand elements (x)=(Xl' ..., xn)insome extension field,assume that there exist npolynomials /;,EK[X] such that: (i)/;,(x)=0,and (ii)det(o/;,jox j)=FO. Then (x)isseparably algebraic over K. Proof. Let Dbe aderivation onK(x), trivial onK.Having h(x)=0we must have D/;,(x)=0,whence the DXisatisfynlinear equations such that the coefficient matrix has non-zero determinant. Hence DXi=0,soDistrivial onK(x). Hence K(x) isseparable algebraic over KbyProposition 5.2. Thefollowing proposition willfollow directly from Cases 1and 2. Proposition 5.4. Let K =k(x) be afinitely generated extension ofk.An element zofKisinKPkifandonlyifevery derivation DofKover kis such that Dz=O. Proof. IfzisinKPk, then itisobvious that every derivation DofK over kvanishes on z.Conversely, ifzr$KPk,then zispurely inseparable over KPk, and byCase 3oftheextension theorem, we can find aderivation Dtrivial onKPksuch that Dz =1.This derivation isatfirst defined onthe field KPk(z). One can extend ittoKasfollows. Suppose there isanelement WEKsuch that wr$KPk(z). Then wPEKPk, and Dvanishes on wp .We can then again apply Case 3toextend Dfrom KPk(z) toKPk(z, w).Proceeding stepwise, wefinally reach K,thus proving ourproposition. The derivations Dofafield Kform avector space over Kifwedefine zD for zEKby(zD)(x)=zDx. Let Kbe afinitely generated extension ofk,ofdimension rover k.We denote by 1>theK-vector space ofderivations DofKover k(derivations of Kwhich aretrivial onk).For each zEK,wehave apairing (D,z)1---+Dz of(1),K)into K. Each element zofKgives therefore aK-linear functional of1>.This functional isdenoted bydz. We have d(yz)=ydz+zdy, d(y+z)=dy+dz. These linear functionals form asubspace tFofthedual space ofD,ifwe define ydzby(D,ydz)1---+yDz. 372 TRANSCENDENTAL EXTENSIONS VIII,5 Proposition 5.5. Assume that Kisaseparably generated andfinitely generated extension ofkoftranscendence degree r.Then the vector space (over K)ofderivations ofKover khas dimension r.Elements tl'...,tr ofKfromaseparating transcendence base ofKover kifand only if dtl'...,dtrform abasis ofthedual space of over K. Proof Iftl'..., trisaseparating transcendence base forKover k,then we can find derivations D1,...,DrofKover ksuch that Ditj=ij'byCases 1and 2ofthe extension theorem. Given DE, let Wi=Dti.Then clearly D=LwiD i,and sothe Diform abasis for over K,and the dtiform the dual basis. Conversely, ifdtl'...,dtrisabasis forftover K,and ifKisnot separably generated over k(t), then byCases 2and 3we can find aderivation Dwhich istrivial onk(t) butnontrivial onK.IfD1,...,Dristhedual basis ofdtl'...,dtr(soDitj=ij)then D,D1,...,Drwould belinearly independent over K,contradicting thefirst part ofthetheorem. Corollary 5.6. Let Kbe afinitely generated and separably generated extension ofk.Let zbeanelement ofKtranscendental over k.Then Kis separable over k(z)ifandonlyifthere exists aderivation DofKover k such that Dz =FO. Proof IfKisseparable over k(z), then zcan becompleted toaseparat- ing base ofKover kand we can apply theproposition. IfDz =F0,then dz=F0,and we can complete dzto abasis of over K.Again from the proposition, itfollows that Kwill beseparableover k(z). Note. Here wehave discussed derivations offields. For derivations in the context ofrings and modules, seeChapter XVI. As anapplication, weprove: Theorem 5.7. (Zariski-Matsusaka). Let Kbe afinitely generated sepa- rable extension ofafield k.Let y,zEKand zrtKPkifthecharacteristic isp>O.Let ubetranscendental over K,and putku=k(u), Ku=K(u). (a)For allexcept possiblyone value ofcEk,Kisaseparable extension of k(y+cz). Furthermore, Kuisseparable over ku(Y +uz). (b) Assume that Kisregular over k,and that itstranscendence degree isat least 2.Then for allbut afinite number ofelements cEk,Kis aregular extension ofk(y+cz). Furthermore, Ku isregular over ku(Y +uz). Proof. We shall usethroughout the fact that asubfield of afinitely generated extension isalso finitely generated (see Exercise 4). If Wisanelement ofK,and ifthere exists aderivation DofKover ksuch that Dw =F0,then Kisseparable over k(w), byCorollary 5.6. Also byCorollary 5.6, there exists Dsuch that Dz =FO.Then for allelements cEk,except possibly one, we have D(y +cz)=Dy+cDz =FO.Also we may extend DtoKu over kubyputting Du =0,and then one sees that VIII,5 DERIVATIONS 373 D(y+uz)= Dy+uDz =F0,soKisseparable over k(y+cz)except possibly for one value ofc,and Ku isseparable over ku(Y +uz). Inwhat follows, we assume that the constants Cl'C2'... are different from theexceptional constant, and hence that Kisseparable over k(y+ciz)fori=1,2. Assume next that Kisregular over kand that the transcendence degree isatleast 2.Let Ei=k(y+ciz)(i=1,2)and letE;bethealgebraic closure ofEiinK. We must show that E;=Eifor allbut afinite number of constants. Note that k(y, z)=E1E2isthecompositum ofEland E2,and that k(y, z)has transcendence degree 2over k.Hence EandE;are free over k.Being subfields ofaregular extension ofk,they areregular over k, and aretherefore linearly disjoint byTheorem 4.12. K I/L E'l(y,z) E(y, z)/k(Y.Z( E'l/ /E2 k(y+C1z) k(y+c2z) k Byconstruction, Eand E;' arefinite separable algebraic extensions ofE1 and E2respectively. Let Lbetheseparable algebraic closure ofk(y, z)inK. There isonly afinite number ofintermediate fields between k(y, z)and L. Furthermore, byProposition 3.1the fields E(y,z)and E;'(y, z)arelinearly disjoint over k(y, z).Let c1range over thefinite number ofconstants which will exhaust theintermediate extensions between Land k(y, z)obtainable by lifting over k(y, z)afield oftype E;.IfC2isnow chosen different from any one ofthese constants Cl'then theonly way inwhich thecondition oflinear disjointness mentioned above can becompatible with our choice of C2isthat E;(y, z)=k(y, z),i.e.that E;=k(y+c2z). This means that k(y+c2z)is algebraically closed inK,and hence that Kisregular over k(y+C2Z). AsforKu, let u1,u2,...beinfinitely many elements algebraically indepen- dent over K. Let k'=k(Ul'u2,...) and K' =K(u 1,u2,...) bethe fields obtained byadjoining these elements tokand Krespectively. Bywhat has already been proved,weknow that K' isregular over k'(u +UiZ) for all but afinite number ofintegers i,say for i=1.Our assertion (a)isthen aconsequence ofCorollary 4.14. This concludes theproof ofTheorem 5.7. 374 TRANSCENDENTAL EXTENSIONS VIII, Ex Theorem 5.8. LetK=k(xl'...,Xn)=k(x) be afinitely generated regular extension ofafield k.Let ul'...,Unbealgebraically independent over k(x). Let Un+1=U1X1+...+UnX n' and letku=k(ul'...,Un'un+1).Then ku(x) isseparable over ku,andifthe transcendence degree ofk(x) over kis>2,then ku(x) isregular over ku. Proof Bytheseparability ofk(x) over k,some Xidoes not lieinKPk, sayXnrtKPk.Then wetake y=U1X1+...+Un-1Xn-l and Z=Xn' sothat Un+1=Y+UnZ, and weapply Theorem 5.7toconclude theproof. Remark. Inthegeometric language ofthe next chapter, Theorem 5.8 asserts that theintersection ofak-variety with ageneric hyperplane UIXl+...+unX n-Un+1=0 isaku-variety, ifthe dimension ofthek-variety is>2.Inany case, the extension ku(x) isseparable over ku. EXERCISES 1.Prove that thecomplex numbers have infinitely many automorphisms. [Hint: Use transcendence bases.] Describe allautomorphisms and their cardinality. 2.Asubfield kofafield Kissaid tobealgebraically closed inKifevery element of Kwhich isalgebraicover kiscontained ink.Prove: Ifkisalgebraically closed inK,and K,Lare free over k,and Lisseparable over kor1).isseparableover k,then Lisalgebraically closed inKL. 3.Let kcEcKbeextension fields. Show that tr.deg.(Kjk)=tr.deg.(KjE)+tr.deg.(Ejk). If{Xi}isatranscendence base ofEjk, and{Yj}isatranscendence base ofKjE, then {Xi'Yj}isatranscendence base ofKjk. 4.LetKjk be afinitely generated extension, and letK =>E=>kbe asubextension. Show thatEjk isfinitely generated. 5.Let kbe afield and k(x 1,...,X,.)=k(x) afinite separable extension. Let ul'...,U,.bealgebraically independentover k.Let W=U1Xl +...+U,.X,.. Let ku=k(u 1,...,U,.). Show that ku(w)=ku(x). VIII, Ex EXERCISES 375 6.Letk(x)=k(xl'...,XII) be aseparable extension oftranscendence degreer>1. Letuij(i=1,...,r;j=1,...,n)bealgebraically independent over k(x). Let II Y.= U..X.1 I)J. j=1 Let ku=k(Uij)all i,j. (a)Show that ku(x) isseparable algebraic over k(Yl' ...,Yr). (b)Show that there exists apolynomial P(u) Ek[u] having thefollowing prop- erty. Let(c)=(ci)beelements ofksuch that P(c) :FO.Let II Y= C..X.I IJJ. j=1 Then k(x) separable algebraic over k(y'). 7.Let kbe afield andk[x 1,...,XII]=Rafinitely generated entire ringover kwith quotient field k(x). Let Lbe afinite extension ofk(x). LetIbetheintegral closure ofRinL.Show that Iisafinite R-module. [Use Noether normalization, and deal with theinseparability problem and theseparable case intwosteps.] 8.Let Dbe aderivation of afield K. Then D":K Kisalinear map. Let p,.=Ker D", sop,.isanadditive subgroup ofK.Anelement XEKiscalled a logarithmic derivative (inK)ifthere exists YEKsuch that X=Dyjy. Prove: (a)An element XEKisthelogarithmic derivative of anelement YEP" but y1;.-1 (n>0)ifandonly if (D+x)"(I)=0 and (D+X)"-1 (1):FO. (b)Assume that K =UP,.,i.e.givenXEKthen xEp,.for some n>O.Let Fbe asubfield ofKsuch that DF cF.Prove that xisalogarithmic derivative in Fifand only ifxisalogarithmic derivative inK.[Hint: Ifx=Dyjy then (D+x)=y-lD 0yandconversely.] 9.Let kbe afield ofcharacteristic 0,and let zl'..., Zrbealgebraically independent over k.Let(eij),i=1,..., mandj=1,..., rbe amatrix ofintegers with r>m, and assume that this matrix hasrank m.Let W.=ze1il...zeir I r for i=1,..., m. Show that WI' ..., wmarealgebraically independent over k.[Hint: Consider the K-homomorphism mapping theK-space ofderivations ofKjk into K(r)given by D......(Dz1/Zb.. .,Dzr/Zr), and derive alinear condition forthose Dvanishing onk(w 1,...,wm).] 10.Letk,(z)be asinExercise 9.Show that ifPisarational function then d(P(z))=grad P(z). dz, using vector notation, i.e.dz=(dzl'...,dzr)andgrad P=(D 1P,...,DrP). Define dlogPand express itinterms ofcoordinates. IfP,Qarerational functions in k(z) show that dlog(PQ)=dlogP+dlogQ. CHAPTER IX Algebraic Spaces This chapter gives thebasic results concerning solutions ofpolynomial equa- tions inseveral variables over afield k.First itwill beproved that ifsuch equations have acommon zero insome field, then they have acommon zero in thealgebraic closure ofk,and such azero can beobtained bytheprocess known asspecialization. However, itisuseful todeal with transcendental extensions ofkaswell. Indeed, ifpisaprime ideal ink[X]=k[X b. . .,Xn],then k[X]/p isafinitely generated ringover k,and theimages XiofXiinthisring may betranscendental over k,sowe areled toconsider such rings. Even ifwewant todeal only with polynomial equations over afield, we are ledinanatural way todeal with equationsover theintegers Z.Indeed, ifthe equationsarehomogeneous inthevariables, then weshall prove in 3and4 that there areuniversal polynomialsintheir coefficients which determine whether these equations have acommon zero ornot. "Universal" means that thecoef- ficients areintegers, and anygiven specialcase comes from specializing these universal polynomialstothespecialcase. Being led toconsider polynomial equations over Z,wethen consider ideals ainZ[X]. The zeros ofsuch anideal form what iscalled analgebraic space. If pisaprime ideal, the zeros ofpform what iscalled anarithmetic variety. We shall meet thefirst example inthediscussion ofelimination theory, forwhich Ifollow van der Waerden' streatment inthefirst two editions ofhisModerne Algebra, Chapter XI. However, when taking thepolynomial ringZ[X]/a for some ideal a,itusually happens that such afactor ring hasdivisors ofzero, oreven nilpotent elements. Thus itisalso natural toconsider arbitrary commutative rings, and tolaythe foundations ofalgebraic geometryover arbitrary commutative ringsasdidGroth- endieck. Wegivesome basic definitions forthis purpose in5. Whereas the present chapter gives the flavor ofalgebraic geometry dealing with specific polynomial ideals, thenext chapter gives theflavor ofgeometry developing from commutative algebra, and itssystematic application tothe more generalcases just mentioned. 377 378 ALGEBRAIC SPACES IX,91 The present chapter and the next will also serve thepurpose ofgiving the reader anintroduction tobooks onalgebraic geometry, notably Hartshorne's systematic basic account. Forinstance, Ihave included those results which are needed forHartshorne's Chapter Iand II. 1. HILBERT'S NULLSTELLENSATZ The Nullstellensatz has todowith aspecialcase oftheextension theorem forhomomorphisms, applied tofinitely generated rings over fields. Theorem 1.1. Let kbe afield, and letk[x]=k[x b...,xn] be afinitely generated ring over k.Let ({J:k-.Lbe anembedding ofkinto analge- braically closed field L.Then there exists anextension of ({Jtoahomo- morphism ofk[x] into L. Proof. Let 9Jlbeamaximal ideal ofk[x]. Let (Jbethecanonical homo- morphism (J:k[x]-.k[x]/9Jl. Then (Jk[(Jxb...,(Jxn]isafield, and isinfact anextension field of(Jk.Ifwe canproveourtheorem when thefinitely generated ring isinfact afield, then weapply ({J0(J-1on(Jkand extend this toahomo- morphism of(Jk[(Jx 1,...,(Jxn]into Ltogetwhat wewant. Without loss ofgenerality,wetherefore assume thatk[x] isafield. Ifitis algebraicover k,we aredone (by theknown result foralgebraic extensions). Otherwise, let tb...,trbe atranscendence basis, r>1.Without loss of generality,wemayassume that({Jistheidentityonk.Each element xl'...,Xn isalgebraic over k(tt,...,tr).Ifwemultiply the irreducible polynomial Irr(xi' k(t),X)byasuitable non-zero element ofk[t], then wegetapolynomial allofwhose coefficients lieink[t]. Letal(t),...,an(t) bethe setoftheleading coefficients ofthese polynomials, and leta(t) betheir product, a(t)=a1(t).. .an(t). Since a(t) =t=0,there exist elements t;,. . .,t;Ekasuch thata(t')=t=0,and hence ai(t')=t=0forany i.Each Xiisintegral over thering k[tl'...,t" ()'. ..,)]. alt ar(t Consider thehomomorphism ({J:k[tt,..., tr]-.ka such thatlfJistheidentityonk,and ((J(tj)=tj.Let pbeitskernel. Then a(t) p. IX,91 HILBERT'S NULLSTELLENSATZ 379 Our homomorphism ({Jextends uniquely tothe local ringk[t]"and bythe preceding remarks, itextends toahomomorphism of k[t],,[x l'...,xn] into ka ,using Proposition 3.1ofChapter VII. This proves what wewanted. Corollary 1.2. Let kbe afield andk[x 1,...,Xn]afinitely generatedex- tension ringofk.Ifk[x] isafield, thenk[x] isalgebraic over k. Proof. Allhomomorphisms ofafield areisomorphisms (onto theimage), and there exists ahomomorphism ofk[x] over kinto thealgebraic closure ofk. Corollary 1.3. Letk[x l'...,xn]be afinitely generated entire ring over a field k,and letY1,...,Ymbenon-zero elements ofthisring. Then there exists ahomomorphism t/1:k[x] k8 over ksuch thatt/1(Yj)=F0forallj=1,...,m. Proof. Consider the ring k[x b...,xn,Y11 ,...,Y';1]and apply the theorem tothisring. Let Sbe asetofpolynomials inthepolynomial ringk[X 1,...,Xn]inn variables. LetLbeanextension field ofk.By azero ofSinLone means an n-tuple ofelements (cl'. ..,cn)inLsuch that f(c 1,...,Cn)=0 foralIIES.IfSconsists ofonepolynomial.!: then wealso saythat (c)isazero off The setofallzeros ofSiscalled analgebraic setinL(or more accurately inL(n». Let Qbetheideal generated byallelements ofS.Since SC Qitisclear that everyzero of Qisalso azero ofS.However, the converse obviously holds, namely every zero ofSisalso azero of Qbecause every element of Qisoftype g1(X)fl(X) +...+gm(X)fm(X) withjjESand giEk[X]. Thus when considering zeros ofasetS,wemay just consider zeros ofanideal. We note parenthetically that every ideal is finitely generated, and soevery algebraicsetisthe setofzeros ofafinite number ofpolynomials. Asanother corollary ofTheorem 1.1,weget: Theorem 1.4. Let Qbe anideal ink[X]=k[X b...,Xn]. Then either Q=k[X] or Qhas azero ink3 . 380 ALGEBRAIC SPACES IX, 1 Proof. Suppose0=Fk[X]. Then 0iscontained insome maximal ideal m,andk[X]/m isafield, which isafinitely generated extension ofk,because itisgenerated bytheimages ofXl'. . .,Xnmod m.ByCorollary 2.2, this field isalgebraic over k,and cantherefore beembedded inthealgebraic closure ka .Thehomomorphismonk[X] obtained bythecomposition ofthecanonical map mod m,followed bythis embedded gives thedesired zero of0,and con- cludes theproof ofthetheorem. In3 weshall consider conditions on afamily ofpolynomials tohave a common zero. Theorem 1.4implies thatifthey have acommon zero insome field, then they have acommon zero inthealgebraic closure ofthefield generated bytheir coefficients over theprime field. Theorem 1.5. (Hilbert's Nullstellensatz). Let abeanideal ink[X]. Let fbeapolynomial ink[X] such thatf(c)=0for every zero (c)=(c1,...,Cn) of0inka .Then there exists anintegerm>0such tha/I'mEo. Proof. We may assume thatf=FO.We use the Rabinowitsch trick of introducinganew variable Y,and ofconsidering the ideal 0'generated by oand 1-Yfink[X, Y]. ByTheorem 1.4, and the current assumption, the ideal 0'must bethewhole polynomial ringk[X, Y], sothere exist polynomials giEk[X, Y]and hiE0such that 1=go(1-Yf)+glh 1+...+grhr. We substitute f-1for Yandmultiply byanappropriate power fmoffto clear denominators ontheright-hand side. This concludes theproof. Forquestions involving how effective theNullstellensatz can bemade, see thefollowing references also related tothe discussion ofelimination theory discussed later inthischapter. Bibliography [BeY 91] C,BERENSTEIN and A.YGER, Effective Bezout identities inQ[ZI'...,zn], Acta Math. 166(1991), pp.69-120 [Br87] D.BROWNAWELL, Bounds forthedegree inNullstellensatz, Ann. ofMath, 126(1987), pp.577-592 [Br88] D,BROWN AWELL, Local diophantine nullstellen inequalities, J,Amer .Math. Soc, 1(1988), pp.311-322 [Br89] D.BROWNAWELL, Applications ofCayley-Chow forms, Springer Lecture Notes 1380: Number Theory, Vim 1987, H.P,Schlickewei and E.Wirsing (eds.), pp. 1-18 [Ko 88] J,KOLLAR, Sharp effective nullstellensatz, J.Amer. Math, Soc, 1No.4 (1988), pp.963-975 IX,2 ALGEBRAIC SETS, SPACES AND VARIETIES 381 2. ALGEBRAIC SETS, SPACES AND VARIETIES Weshall make some very elementary remarks onalgebraicsets. Letkbea field, and letAbeanalgebraicsetofzeros insome fixed algebraically closed extension field ofk.The setofallpolynomials fEk[X b...,Xn]such that f(x)=0forall(x)EAisobviously anideal 0ink[X], and isdetermined by A.We shall call ittheideal belonging toA,orsaythat itisassociated with A. IfAisthe setofzeros ofasetSofpolynomials, then Sc0,but 0may bebigger than S.Ontheother hand, weobserve that Aisalso the setofzeros ofo. LetA,Bbealgebraic sets, and 0,btheir associated ideals. Then itisclear that AcBifandonly ifQ=>b.Hence A=Bifandonly if0=b.This has an important consequence. Since thepolynomial ring k[X] isNoetherian, it follows that algebraicsetssatisfy thedual property, namely every descending sequence ofalgebraicsets Al=>A2=>... must besuch that Am=Am+ 1=...for some integer m,i.e.allAvareequal for v>m.Furthermore, dually toanother property characterizing theNoetherian condition, weconclude that every non-emptysetofalgebraicsets contains a minimal element. Theorem 2.1. Thefinite union and thefinite intersection ofalgebraic sets arealgebraic sets.IfA,Barethealgebraic setsofzeros ofideals 0,b,respec- tively, then AuBisthe setofzeros of0nband AnBisthe setofzeros of (Q,b). Proof'. We first consider AuB.Let(x) EAuB.Then (x) isazero of0nb.Conversely, let(x) be azero of0nb,and suppose (x)rtA.There exists apolynomial fE0such thatf(x) ;/=O.But obcon band hence (fg)(x)=0forallgEb,whence g(x)=0forallgEb.Hence (x)liesinB,and AuBisanalgebraic setofzeros of0nb. Toprove that AnBisanalgebraic set,let(x) EAnB.Then (x)isazero of(0,b).Conversely, let(x)beazero of(0,b).Then obviously (x)EAnB,as desired. This proves our theorem. Analgebraic setViscalled k-irreducible ifitcannot beexpressedasaunion V=AuBofalgebraicsets A,Bwith A,Bdistinct from V.We also sayir- reducible instead ofk-irreducible. Theorem 2.2. Let Abeanalgebraic set. (i)Then Acan beexpressedasafinite union ofirreducible algebraic sets A=VIu . . .U. (ii)Ifthere isnoinclusion relation among the"1,i.e.if"1ctfori=t=j,then therepresentation isunique. 382 ALGEBRAIC SPACES IX,2 (iii) Let W,Vi,. .., beirreducible algebraic sets such that WCVIu . . .U. Then WC"1forsome i. Proof. Wefirst show existence. Suppose the setofalgebraic sets which cannot berepresentedasafinite union ofirreducible ones isnotempty. Let Vbeaminimal element inits.Then Vcannot beirreducible, and we canwrite V=AuBwhere A,Barealgebraic sets, but A=FVand B=FV.Since each one ofA,Bisstrictly smaller than V,we can express A,Basfinite unions of irreducible algebraic sets, and thus getanexpression forV,contradiction. The uniqueness will follow from (iii), which weprove next. Let Wbecon- tained intheunion VIU . . .U.Then W=(WnVI)U . . .U(Wn). Since each Wn"1isanalgebraic set,bytheirreducibility ofWwemust have W=Wn"1for some i.Hence WC"1for some i,thus proving (iii). Now toprove (ii),apply (iii) toeach"}.Then foreachjthere issome isuch that"}C"1.Similarly foreach ithere exists vsuch that "1CWV.Since there isnoinclusion relation among the"}'s,wemusthave"}=\.';=Wv-This proves thateach"} appears among the\.';'s and each \.';appears among the"}'s,and proves theuniqueness oftherepresentation. Italso concludes theproof ofTheo- rem 2.2. Theorem 2.3 Analgebraic setisirreducible ifandonlyifitsassociated ideal isprime. Proof. Let Vbeirreducible and letpbeitsassociated ideal. Ifpisnot prime,we can find twopolynomials f,9Ek[X] such thatffj.p ,9fj.p,but fgEp.Let Q=(p,f)and b=(p,g).Let Abethealgebraic setofzeros of Q, and Bthealgebraic setofzeros ofb.Then ACV,A =t=Vand BCV,B =t=V. Furthermore AUB=V.Indeed, AUBCVtrivially. Conversely, let(x) EV. Then (fg)(x)=0implies f(x) org(x)=O.Hence (x) EAor(x) EB,proving V=AUB,and Visnotirreducible. Conversely, letVbethealgebraic set ofzeros of aprime ideal p.Suppose V=AUBwith A =t=Vand B =t=V. Let Q,bbetheideals associated with Aand Brespectively. There exist poly- nomials fEQ,ffj.pand 9Eb,9fj.p.Butfg vanishes onAUBand hence lies inp,contradiction which proves thetheorem. Warning. Given afield kand aprime ideal pink[X], itmay bethat the ideal generated bypinka[X] isnotprime, and thealgebraic setdefined over ka bypka[X] has more than one component, and soisnotirreducible. Hence the prefix referring tokisreally necessary. Itisalso useful toextend theterminology ofalgebraic sets asfollows. Given anideal aCk[X], toeach field Kcontaining kwe can associate to Qthe set IX,2 ALGEBRAIC SETS, SPACES AND VARIETIES 383 a(K)consisting ofthe zeros of ainK.Thusaisanassociation a:K a(K)C[«n). We shall speak ofaitself asanalgebraic space,sothataisnot aset, but toeach field Kassociates the seta(K). Thusaisafunctor from extensions Kofktosets(functorial with respect tofield isomorphisms). Byak-variety we mean thealgebraic space associated with aprime ideal p. The notion ofassociated ideal applies also tosucha'and theassociated ideal ofaisalso rad(a). We shall omit thesubscriptaand write simply for thisgeneralized notion ofalgebraic space. Of course wehave a= rad(a). We saythata(K)isthe setofpoints ofainK.BytheHilbert Nullstellensatz, Theorem 1.1, itfollows that ifKCK' are two algebraically closed fields containing k,then theideals associated witha(K) and 9la(K') areequal toeach other, and also equal torad(a). Thus thesmallest algebraically closed field ka containing kalready determines these ideals. However, itisalso useful toconsider larger fields which contain transcendental elements, asweshall see. Asanother example, consider thepolynomial ring k[Xl'. . .,Xn]=k[X]. Let Andenote thealgebraic space associated with the zero ideal. Then An iscalled affine n-space. Let Kbe afield containing k.For each n-tuple (CI,. . .,cn)EK(n) wegetahomomorphism cp:k[Xl'. . .,Xn] K such that cp(X i)=Ciforalli.Thus points inAn(K) correspond bijectively to homomorphisms ofk(X) into K. More generally, letVbe ak-variety with associated prime ideal p.Then k[X]/p isentire. Denote byitheimage ofXiunder thecanonical homomorphism k[X] k[X]/p.We call(fJthegeneric point ofVover k.Ontheother hand, let(x)be apoint ofVinsome field K.Then pvanishes on(x), sothehomomor- phism cp:k[X] k[x] sending Xi xifactors through k[X]/p=k[gj,whence weobtain anatural homomorphism k[gj k[x]. Ifthishomomorphism isan isomorphism, then wecall (x) ageneric point ofVinK. Given two points (x) EAn(K) and(x') EAn(K'), we say that (x') isa specialization of(x)(over k)ifthemap Xi xiisinduced byahomomorphism k[x] k[x']. From thedefinition of ageneric point of avariety, itisthen immediate that: Avariety Visthe setofspecializations ofitsgeneric point, orofageneric point. Inother words, V(K) isthe setofspecializations of(fJinKforevery field K containing k. Let uslook atthe converse construction ofalgebraicsets. Let (x)= (xl'. . .,xn)be ann-tuple with coordinates XiEKfor some extension field Kofk.Let pbetheideal ink[X] consisting ofallpolynomials f(X) such that 384 ALGEBRAIC SPACES IX,2 f(x)=o.We call pthe ideal vanishing on(x). Then pisprime, because if fgEPsof(x)g(x)=0,thenfEPorgEPsince Khas nodivisors ofo.Hence p isak-variety V,and(x)isageneric point ofVover kbecause k[X]j p=k[x]. Forfuture use, westate thenext result forthepolynomial ring over afactorial ring rather than over afield. Theorem 2.4. LetRbeafactorial ring, andlet"'},. . .,Wmbemindependent variables over itsquotient field k.Letk(wI'. . .,wm)beanextension oftran- scendence degree m-1.Then theideal inR[W] vanishing on(w) isprincipal. Proof. Byhypothesis there issome polynomial P(W) ER[W] ofdegree ::>1vanishing on(w), and after takinganirreducible factor wemay assume that thispolynomial isirreducible, and soisaprime element inthefactorial ring R[W]. LetG(W) ER[W] vanish on(w). Toprove that Pdivides G,after selecting some irreducible factor ofGvanishingon(w)ifnecessary,wemay assume without loss ofgenerality that Gisaprime element inR[W]. One ofthevariables ""ioccurs inP(W), say Wm,sothat Wmisalgebraic over k(WI'. . .,wm-I). Then (wI'. ..,wm-l)arealgebraically independent, and hence Wmalso occurs in G.Furthermore, P(w},..., wm-I, Wm)isirreducible as apolynomial in k(wI'. ..,wm-l)[W m]bythe Gauss lemma asinChapter IV, Theorem 2.3. Hence there exists apolynomial H(W m)Ek(WI,. . .,Wm-I )[WmJ such that G(W)=H(Wm)P(W). Let R'=R[WI'. .., Wm-d.Then P,Ghave content 1aspolynomials in R'[W m].ByChapter IVCorollary 2.2 weconclude that HER'[W m]=R[W], which proves Theorem 2.4. Next weconsider homogeneous ideals andprojective space. Apolynomial f(X) Ek[X] can bewritten asalinear combination f(X)=2:c(II)M(II)(X) with monomialsM(II)(X)=XI...xn andC(II)Ek.We denote thedegree of M(II)by Ivi=degM(II)=2:Vi' Ifinthis expression forfthedegrees ofthemonomials X<II) areallthe same (whenever thecoefficientC(II)is =1=0),then wesaythatf isaform, oralso that fisahomogeneous (ofthatdegree). Anarbitrary polynomial f(X) inK[X] can also bewritten f(X)=2:f(d)(X), where each f(d) isaform ofdegree d(which may be0). We callf(d) the homogeneous part offofdegree d. An ideal Qofk[X] iscalled homogeneous ifwhenever fEQthen each homogeneous partfd) also lies ina. IX,2 ALGEBRAIC SETS, SPACES AND VARIETIES 385 Proposition 2.5. Anideal Qishomogeneous ifandonlyifQhas asetof generators over k[X] consisting offorms. Proof. Supposeaishomogeneous and thatfb. . .,fraregenerators. By hypothesis, foreach integer d::>0thehomogeneous components f/d) also liein Q,and the setofsuch fi(d)(for alli,d)form asetofhomogeneous generators. Conversely, letfbe ahomogeneous element in Qand letgEK[X] bearbitrary. For each d,g(d)f lies inQ,andg(d)f ishomogeneous, soallthehomogeneous components ofgfalso lieinQ.Applying this remark tothe case whenfranges over asetofhomogeneous generators for Qshows that Qishomogeneous, and concludes theproof oftheproposition. Analgebraic space ?Iiscalled homogeneous ifforevery point (x) ECfland ttranscencental over k(x), thepoint (tx) also lies in.Ift,uaretranscendental over k(x), then there isanisomorphism k[x, t] k[x, u] which sends ton uand restricts totheidentityonk[x], sotoverify theabove condition, itsuffices toverify itfor some transcendental tover k(x). Proposition 2.6. Analgebraic space Cflishomogeneous ifand onlyifits associated ideal Qishomogeneous. Proof. SupposeCflishomogeneous. Letf(X)Ek[X] vanish on .For each (x)ECfland ttranscendental over k(x) wehave o=f(x)=f(tx)=Ltdf(d)(x). d Thereforef(d)(x)=0foralld,whencefd)EQforalld.Hence Qishomogeneous. Conversely, supposeQhomogeneous. BytheHilbert Nullstellensatz, weknow that consists ofthe zeros ofQ,and hence consists ofthe zeros of asetof homogeneous generators for Q.Butiffisoneofthose homogeneous generators ofdegree d,and(x)isapoint ofC'fl,then forttranscendental over k(x) wehave o=f(x)=tdf(x)=f(tx), so(tx) isalso azero ofQ.Hence Cflishomogeneous, thusproving theproposition. Proposition 2.7. Let beahomogeneous algebraic space. Then each irre- ducible component Vof isalso homogeneous. Proof. Let V=VI'..., betheirreducible components ofCfl,without inclusion relation. ByRemark 3.3 weknow that VIctV2U . . .U,sothere isapoint (x) EVIsuch that(x)fj.\t}fori=2,. . .,r.Byhypothesis, forttranscen- dental over k(x) itfollows that(tx) ECflso(tx) E\'ifor some i.Specializing to t=1,weconclude that (x) E\';,soi=1,which proves that VIishomoge- neous, aswas tobeshown. Let Vbe avariety defined over kbyaprime ideal pink[X]. Let(x)be a generic point ofVover k.We say that (x)ishomogeneous (over k)iffor t 386 ALGEBRAIC SPACES IX,2 transcendental over k(x), thepoint (tx) isalso apoint ofV,orinother words, (tx) isaspecialization of(x).Ifthis isthe case, then wehave anisomorphism k[xl'. . .,Xn]=k[txl'. . .,txn] , which istheidentityonkand sends XiontXi.Itthen follows from thepreceding propositions that thefollowing conditions areequivalent for avariety Vover k: Vishomogeneous. Theprime ideal ofVink[X] ishomogeneous. Ageneric point ofVover kishomogeneous. Ahomogeneous ideal always has azero, namely theorigin (0), which will becalled thetrivial zero. Weshall want toknow when ahomogeneous algebraic sethas anon-trivial zero (insome algebraically closed field). Forthis weintroduce theterminology ofprojective spaceasfollows. Let(x)besome point inAnand Aanelement ofsome field containing k(x). Then wedenote by(Ax) thepoint (Ax},. . .,Axn).Two points (x),(y)EAn(K) for some field Karecalled equivalent ifnotalltheir coordinates are0,and there exists some element AEK,A=t=0, such that (Ax)=(y). Theequivalence classes ofsuch points inAn(K) arecalled thepoints ofprojective space inK.We denote thisprojective space bypn-l, and the setofpoints ofprojective space inKbypn-l (K).Wedefine analgebraic space inprojective space tobethenon-trivial zeros of ahomogeneous ideal, with two zeros identified ifthey differ byacommon non-zero factor. Algebraic spaces over rings As weshall seeinthe next section, itisnotsufficient tolook only atideals ink[X] for some field k.Sometimes, even often, one wants todeal with polynomial equations over theintegers Z,forseveral reasons. Intheexample ofthe next sections, weshall find universal conditions over Zonthecoefficients ofasystem offorms sothat these forms have anon-trivial common zero. Furthermore, in number theory-diophantine questions-one wants toconsider systems ofequa- tions with integer coefficients, and todetermine solutions ofthese equations in theintegers orintherational numbers, orsolutions obtained byreducing mod pfor aprime p.Thus one isled toextend thenotions ofalgebraic space and varietyasfollows. Even though theapplications ofthe next section will beover Z,weshall now give general definitions over anarbitrary commutative ring R. Letf(X)ER[X]=R[X l'. . .,Xn] be apolynomial with coefficients inR. Let R Abe anR-algebra, bywhich forthe rest ofthischapterwe mean a homomorphism ofcommutative rings. Weobtain acorresponding homomorphism R[X] A[X] onthepolynomial rings, denoted byf fAwhereby thecoefficients offAare theimages ofthecoefficients off under thehomomorphismR A.Byazero offinAwe mean azero offAinA.Similarly, letSbe asetofpolynomials in R[X]. Byazero ofSinAwe mean acommon zero inAofallpolynomials fES.Let abetheideal generated bySinR[X]. Then azero ofSinAisalso IX,2 ALGEBRAIC SETS, SPACES AND VARIETIES 387 azero of QinA.We denote the setofzeros ofSinAby?1s(A), sothat wehave ?1s(A)=a(A). We calla(A)analgebraic set over R.Thus wehave anassociation ?1a:A ?1a(A) which toeach R-algebra associates the setofzeros of Qinthatalgebra. We note that R-algebras form acategory, wherebyamorphism isaring homomorphism cp:A A'making thefollowing diagram commutative: A R/jA' Then itisimmediately verified that ?1aisafunctor from thecategory ofR- algebras tothecategory ofsets. Again wecall ?1aanalgebraic space over R. IfRisNoetherian, then R[X] isalso Noetherian (Chapter IV,Theorem 4.1), and soifQisanideal, then there isalwayssome finite setofpolynomialsS generatingtheideal, sos=a. The notion ofradical of Qisagain defined asthe setofpolynomials hER[X] such that hNEQfor some positive integer N.Then thefollowing state- ment isimmediate: Suppose that Risentire. Then for every R-algebra R Kwith afield K, we have ?1a(K)=?1rad(a)(K). We can define affine space An over R.Itspoints consist ofalln-tuples (xl'. . .,Xn)=(x)withXiinsome R-algebra A.Thus Anisagainanassociation A An(A) from R-algebras to sets ofpoints. Such pointsare Inbijection with homormorphisms R[X] A from thepolynomial ringover Rinto A.Inthenext section weshall limit ourselves tothe case when A=Kisafield, and weshall consider only the functor K An(K) forfields K.Furthermore, weshall deal especially with the case when R=Z, soZhas aunique homomorphism into afield K.Thus afield K canalways beviewed asaZ-algebra. Suppose finally that Risentire (forsimplicity).We canalso consider projective space over R.Let Qbeanideal inR[X]. We define atobehomogeneous justas before. Then ahomogeneous ideal inR[X] can beviewed asdefininganalgebraic subset inprojective space pn(K) foreach field K(as anR-algebra). IfR=Z, 388 ALGEBRAIC SPACES IX,3 then adefines analgebraic subset inpn(K) forevery field K.Similarly,one can define thenotion of ahomogeneous algebraic spaceCflover R,and over the integers Zafortiori. Propositions 2.6 and 2.7 and their proofsarealso valid in this more general case, viewingCfl=Cflaasafunctor from fields Ktosetspn(K). IfQisaprime ideal P,then wecall CflpanR-variety V.IfRisNoetherian, soR[X] isNoetherian, itfollows asbefore that analgebraic spaceCflover Ris afinite union ofR-varieties without inclusion relations. We shall carry this out in5, inthevery general context ofcommutative rings. Just aswedid over a field, wemay form thefactor ringZ[X]/p and theimage (x)of(X)inthis factor ring iscalled ageneric point ofV. 3. PROJECTIONS AND ELIMINATION Let(W)=(WI'. . .,Wm)and(X)=(XI'. . .,Xn)betwo setsofindependent variables. Then ideals ink[W, X]define algebraic spaces intheproduct space Am+n.Let Qbeanideal ink[W,X]. Let QI=Qnk[W]. Let Cflbethealgebraic space ofzeros of Qand let Cfllbethealgebraic space ofzeros of QI.We have theprojection pr:Cflm+nCflm or pr:Am+n Am which mapsapoint (w,x)toitsfirst setofcoordinates (w). Itisclear that prCflCCfll.Ingeneral itisnot true that prCfl=Cfll.Forexample, theideal pgen- erated bythesingle polynomial WI-W2XI=0isprime. Itsintersection with k[W I,W2]isthe zero ideal. But itisnot true that every point inthe affine (WI' W2)-space istheprojection ofapoint inthevariety Cflp.Forinstance, the point (1,0)isnot theprojection ofanyzero ofp.One says insuch acase that theprojection isincomplete.We shall now consider asituation when such a phenomenon does not occur. Inthefirst place, letPbe aprime ideal ink[W, X]and letVbeitsvariety ofzeros. Let(w,x)be ageneric point ofV.Let PI=Pnk[W]. Then (w) isa generic point ofthevariety VIwhich isthealgebraic spacezeros ofPI.This is immediate from thecanonical injective homomorphism k[W]/PI k[W,X]/p. Thus thegeneric point (w)ofVIistheprojection ofthegeneric point (w,x)of V.The question iswhether aspecial point (w') ofVIistheprojection ofapoint ofV. Inthesubsequent applications,weshall consider ideals which arehomo- geneous only intheX-variables, andsimilarly algebraic subsets which arehomo- geneous inthe second setofvariables inAn. IX,3 PROJECTIONS AND ELIMINATION 389 Anideal Qink[W,X]which ishomogeneous in(X)defines analgebraic space inAm Xpn-I. IfVisanirreducible component ofthealgebraic setdefined by Q,then wemay view Vasasubvariety ofAm Xpn-I. Let pbetheprime ideal associated with V.Then pishomogeneous in(X). LetPI=Pnk[W]. Weshall seethat thesituation ofanincomplete projection mentioned previously iselim- inated when wedeal with projective space. We can also consider theproduct Am Xpn, defined bythe zero ideal over Z.For each field K,the setofpoints ofAmxpninKisAm(K)Xpn(K). An ideal QinZ[W, X],homogeneous in(X), defines analgebraic space ?l=ain Am Xpn. We may form itsprojection Ionthefirst factor. This applies in particularwhen Qisaprime ideal p,inwhich case wecall 9laanarithmetic subvariety ofAm Xpn. Itsprojection VIisanarithmetic subvariety ofAm, associated with theprime idealPI=PnZ[W]. Theorem 3.1. Let(W)=("),. . .,Wm)and(X)=(Xl'...,Xn)beindepen- dentfamilies ofvariables. Let Pbeaprime ideal ink[W, X](resp.Z[ X]) and assume Pishomogeneous in(X). Let Vbethecorresponding irreducible algebraic space inAm Xpn-I. Let PI=Pnk[W] (resp. pnZ[W]), and let VIbetheprojection ofVonthefirstfactor. Then VIisthealgebraic space ofzeros ofPIinAm. Proof. Let Vhave generic point (w,x).We have toprove that every zero (w') ofPIinafield istheprojection ofsome zero (w',x')ofPsuch that notall thecoordinates of(x') areequal toO.Byassumption, notallthecoordinates of (x) areequal to0,since weviewed Vas asubset ofAm Xpn-I. Fordefiniteness, saywe aredealing with the case ofafield k.ByChapter VII, Proposition 3.3, thehomomorphism k[w] k[w'] can beextended to aplace cpofk(w, x). ByProposition 3.4 ofChapter VII, there issome coordinateXjsuch that CP(Xi/Xj)=t=00foralli=1,.. .,n .Weletxi=CP(Xi/Xj)forallitoconclude the proof. Theproof issimilar when dealing with algebraic spaces over Z,replacing kbyZ. Remarks. Given thepoint (w') EAm, thepoint (w',x')inAm Xpn-l may ofcourse notlieink(w'). The coordinates (x') could even betranscendental over k(x'). Byanyone oftheforms oftheHilbert Nullstellensatz, sayCorollary 1.3ofTheorem 1.1, wedoknow that(x') could befound algebraic over k(w'), however. Inlight ofthevarious versions oftheNullstellensatz, ifasetofforms has anon-trivial common zero insome field, then ithas anon-trivial common zero inthealgebraic closure ofthefield generated bythecoefficients ofthe forms over theprime field. Inatheorem such asTheorem 1.2below, theconditions onthecoefficients fortheforms tohave anon-trivial common zero (or azero inprojective space)aretherefore also conditions fortheforms tohave such a zero inthat algebraic closure. Weshall apply Theorem 3.1toshow thatgivenafinite family ofhomogeneous polynomials, theproperty that they have anon-trivial common zero insome 390 ALGEBRAIC SPACES IX,3 algebraically closed field can beexpressed interms ofafinite number ofuniversal polynomial equations intheir coefficients .Wemake this more preciseasfollows. Consider afinite setofforms (f)=(fl,.. .,fr). Let dI'. . .,drbetheir degrees.We assume di>1for i=1,...,r.Each.lican bewritten (1) /;=Lwi,(II)M(II)(X) whereM(II)(X)isamonomial in(X)ofdegree di,andwi,(II)isacoefficient. We shall saythat(f) has anon-trivial zero (x)if(x) =t=(0)and.li (x)=0foralli. We let(w)=(w)fbethepoint obtained byarranging thecoefficients wi,(II)of theforms insome definite order, and weconsider thispointasapoint insome affine space Am, where misthenumber ofsuch coefficients. This integermis determined bythegiven degrees dl,. . .,dr-Inother words, given such degrees, the setofallforms (f)=(fl,. . .,fr)with these degrees isinbijection with thepoints ofAm. Theorem 3.2. (Fundamental theorem ofelimination theory.) Given degrees dl,...,dr'the setofallforms (fl,.. .,fr)innvariables having a non-trivial common zero isunalgebraic subspace ofAm over Z. Proof. Let(W)=(Wi,(II»)be afamily ofvariables independent of(X). Let (F)=(FI,. . .,Fr)bethefamily ofpolynomials inZ[W, X]given by (2) Fi(W, X)=LWi,(II)M(II)(X) whereM(II)(X) ranges over allmonomials in(X)ofdegree di,so(W)=(W)F. We call FI,. . .,Frgeneric forms. Let Q=ideal inZ[W, X]generated byFI,. . .,Fr- Then Qishomogeneous in(X). Thus we areinthesituation ofTheorem 3.1, with Qdefininganalgebraic spaceC1inAm Xpn-I. Note that(w)isaspecialization of(W), or, aswealso say,(f) isaspecialization of(F). AsinTheorem 3.1, let(tlbetheprojection of(tonthefirst factor. Then directly from thedefinitions, (f) has anon-trivial zero ifandonly if(w)flies inaI'soTheorem 3.2 isa specialcase ofTheorem 3.1. Corollary 3.3. Let(f) beafamily ofnforms inItvariables, and assume that(w)fisageneric point ofAm, i.e.that thecoefficients ofthese forms are algebraically independent. Then (f) does nothave anon-trivial zero. Proof. There exists aspecialization of(f)which hasonly thetrivial zero, namely fl=Xjl,. . .,f=Xn. Next wefollow vanderWaerden inshowing that C1and hence C11areirreducible. Theorem 3.4. Thealgebraic space C1.offorms having anon-trivial common zero inTheorem 3.2 isactuallyaZ-variety, i.e.itisirreducible. Theprime ideal IX,3 PROJECTIONS AND ELIMINATION 391 pinZ[W, X]associated with aconsists ofallpolynomials G(W, X) EZ[W, X] such thatfor some indexjthere isaninteger s::>0satisfying (*)j XJG(W, X)=0mod (F},. . .,Fr); that is,XJG(W, X) EQ. Ifrelation (*)holds for one index j,then itholds for everyj=1,. . .,n.(Of course, theinteger sdependsonj.) Proof. We construct ageneric point of(1.Weselect anyone ofthevariables, sayXq,and rewrite theforms Fiasfollows: F.(WX)=F+Z.Xdi " I Iq where Ffisthe sum ofallmonomials except the monomial containing Xgi. The coefficients (W) arethereby split into twofamilies, which wedenote by(Y) and (Z), where (Z)=(Z},..., Zr)are the coefficients of(Xgl,. . .,Xgr)in (F},. . .,Fr),and (Y) istheremaining family ofcoefficients ofFf,. . .,F;. We have (W)=(Y,Z), and wemay write thepolynomials Fiintheform Fi(W, X)=Fi(Y, Z,X)=Ff(Y, X)+ZiXgi. Corresponding tothevariables (Y,X) wechoose quantities (y,x)algebraically independentover Z .We let (3) Zi=-Ff(y, x)1xgi=-Ff(y, xlxq). We shall prove that(y, Z,x)isageneric point of(1. From ourconstruction, itisimmediately clear thatFi(y, Z,x)=0foralli, andconsequently ifG(W, X) EZ[W, X]satisfies (*), then G(y, z,x)=o. Conversely, letG(Y, Z,X) EZ[Y, Z,X]=Z[W, X]satisfy G(y, z,x)=o. From Taylor's formula inseveral variables weobtain G(Y, Z,X)=G(Y,..., -Fflxgi+Zi+Fflxgi,..., X) =G(Y,-Fflxgi, X)+L(Zi+Fflxgi)JLiHJLi(Y, Z,X), where the sum istaken over terms havingone factor (Zi+FiIXi)tosome power J.Li>0,and some factor HJLiinZ[Y, Z,X]. From theway (y, z,x)was constructed, and thefact thatG(y, z,x)=0,we seethat thefirst term vanishes, and hence G(Y, Z,X)=L(Zi+FfIXgi)JLiHJLi(Y, Z,X). Clearing denominators ofXq,for some integersweget XG(Y, Z,X)=0mod (Fi,. . .,Fr), orinother words, (*)qissatisfied. This concludes theproof ofthetheorem. Remark. Of course the same statement andproofasinTheorem 3.4 holds with Zreplaced byafield k.Inthat case, wedenote by Qktheideal in k[W, X]generated bythegeneric forms, andsimilarly byPktheassociated prime 392 ALGEBRAIC SPACES IX,3 ideal. Then Qk,1=Qknk[W] andPk,1=Pknk[W]. The ideal PinTheorem 3.4will becalled theprime associated with the ideal ofgeneric forms. The intersection PI=PnZ[W] will becalled theprime elimination ideal ofthese forms. If(1denotes asbefore the zeros ofP(orof Q),and (11isitsprojectiononthefirst factor, then PIistheprime associated with (11. The same terminology will beused ifinstead ofZwework over a field k.(Note: homogeneous elements ofPIhave been called inertia forms in theclassical literature, following Hurwitz. Iamavoiding thisterminology be- cause theword "inertia" isnow used inastandard way forinertia groupsasin Chapter VII,2.) The variety ofzeros ofPIwill becalled theresultant vari- ety. Itisdetermined bythegiven degrees dI'. . .,dn,sowecould denote it by(11(d b..., dn). Exercise. Show thatifPistheprime associated with theideal ofgeneric forms, then PnZ=(0)isthe zero ideal. Theorem 3.5. Assume r=n,sowedeal with nforms innvariables. Then PIisprincipal, generated byasingle polynomial,so(11iswhat one calls a hypersurface. If(w) isageneric point of(11 over afield k,then the transcen- dence degree ofk(w) over kism-1. Proof, Weprove thesecond statement first, and usethe same notation asin theproof ofTheorem 3.4. LetUj=Xj/xn.ThenUn=1and(y),(UI'...' un-I) arealgebraically independent. By(3), wehaveZi=-Ft(y, u), so k(w)=k(y, z)Ck(y, u), and sothetranscendence degree ofk(w) over kis<m-1.Weclaim that this transcendence degree ism-1.Itwill suffice toprove that UI,. . .,un-Iare algebraicover k(w)=k(y, z).Suppose this isnot the case. Then there exists a place cPofk(w, u),which istheidentityonk(w) and mapssomeUjon 00.Select anindex qsuch that CP(Ui/uq)isfinite foralli=1,.. .,n-1.Let Vi=ui/uq andv;=CP(Ui/uq).Denote by1iqthecoefficient ofX;inFiand lety*denote the variables (Y)from which Ylq,...,Ynqare deleted. By(3) we have for i=I,...,n: o=Y.ud;+z.+F*(y*U)lq qI I , =Yiq+Zi/U:'+Fi*(y*, u/uq). Applying theplace yields o=Yiq+Ft*(y*, v'). Inparticular, YiqEk(y*, v')foreach i=1,. . .,n.But thetranscendence degree ofk(v') over kisatmost n-1,while the elements(Ylq,. . .,Ynq'y*)are algebraically independent over k,which givesacontradiction proving the theorem. IX,3 PROJECTIONS AND ELIMINATION 393 Remark. There isaresult (Ilearned itfrom [Jo80]) which ismore precise than Theorem 3.5. Indeed, letaasinTheorem 3.5 bethevariety ofzeros of P,and C11itsprojection. Then thisprojection isbirational inthefollowingsense. Using thenotation oftheproof ofTheorem 3.5, theresult isnotonly thatk(w) has transcendence degreem-lover k,butactually wehave Q(y, z)=Q(w)=Q(y, u). Proof. Let PI=(R), soRistheresultant, generating theprincipal ideal PI.We shall need thefollowing lemma. Lemma 3.6. There isapositive integer swith thefollowing properties. Fix anindex iwith 1-<i-<n-1.Foreach pair ofn-tuples ofintegers>0 (a)=(aI'. . .,an) and (13)=(131'.. .,13n) withIal=1131=di,wehave s(aR _aR )=XnM(o:)(X)aw. M({3)(X)aw.-0mod (FI,. ..,Fn). 1,({3) 1,(0:) To seethis, we ust: thefact from Theorem 3.4that for some s, XR(W)=QIFI+.. ·+QnFn withQjEZ[W, X]. Differentiating with respectto"'i,({3)weget XaR=QjM(f3)(X)mod (F)o.. .,Fn), i,((3) andsimilarly XR=QjM(a)(X)mod (FI,'..,Fn)'ai,(o:) Wemultiply thefirst congruence byM(o:)(X)and thesecond byM({3)(X),and we subtract togetour lemma. From the above weconclude that aR aR M(o:)(X)aw-M({3)(X) i,((3)a"'i,(0:) vanishes onC1,i.e. onthepoint (w,u),after weputXn=1.Then weselect M(o:)(X)=Xf; andM({3)(X)=Xf;-I Xnfori=1,.. .,n-1, and we seethat wehave therational expression aR/a"'i,({3)u,= /'fori=1,...,n-1,1aRa"'f,(o:) (w)=(w) thus showing thatQ(u) CQ(w), asasserted. 394 ALGEBRAIC SPACES IX,3 We note that theargument also works over theprime field ofcharacteristic p.Theonly additional remark tobemade isthat there issome partial derivative aR/a,(a)which does notvanish on(w). This isaminor technical matter, which weleave tothereader. The above argument istaken from [Jo80],Proposition 3.3.1. Jouanolou links old-time results asinMacaulay [Ma 16]with more recent techniques ofcom- mutative algebra, including theKoszul complex (which will bediscussed in Chapter XXI). See also hismonographs [Jo90], [Jo91]. Still following van derWaerden, weshall now giveafairly explicit deter- mination ofthepolynomial generating theideal inTheorem 3.5 .We deal with thegeneric forms Fj(W, X)(i=1,. ..,n).According toTheorem 3.5, theideal PIisgenerated byasingle element. Because theunits inZ[W] consist only of +1,itfollows that this element iswell defined uptoasign. Let R(W)=R(F b. . .,Fn) beone choice ofthis element. Later weshall seehow topick inacanonical way one ofthese two possible choices .We shall prove various properties ofthis element, which will becalled theresultant ofFb. . .,Fn. For each i=1,. . .,nweletDjbetheproduct ofthedegrees with djomitted; that is, A D.=d..·d,. · ·dI I I n. We letdbethepositive integer such that d-1=L(dj-1). Lemma 3.7. Given oneoftheindices, say n,there isanelement Rn(W) lying inPI'satisfying thefollowing properties. (a)For each i,Rn(W)Xf=0mod (FI'. . .,Fn) inZ[ X]. (b)For each i,Rn(W)ishomogeneous inthe setofvariables(,(V»)'and isof degree Dnin(,(v»)'i.e. inthecoefficient ofFn. (c)As apolynomial inZ[W], Rn(W) has content 1,i.e. isprimitive. Proof. Thepolynomial Rn(W) will actually beexplicitly constructed. Let Mu(X) denote themonomials ofdegree IuI=d .Wepartition theindexing set S={u} intodisjoint subsets asfollows. Let SI={UI} bethe setofindices such thatMu)(X)isdivisible byXjl. Let S2={U2} bethe setofindices such that MU2(X) isdivisible byXq2 but notbyXjl. Let Sn={un} bethe setofindices such thatMun(X)isdivisible byXn but notbyXjl,..., X-Il. IX,93 PROJECTIONS AND ELIMINATION 395 Then Sisthedisjoint union ofSI'. . .,Sn.Write each monomial asfollows: MUl(X)=HUl(X)Xjlso degHUl=d-dl M(X)=H(X)Xdn so degH rY"=d-d .UI Un nUn n Then thenumber ofpolynomials HU'IFh. . .,HU'nFn(with (TIE Sh. . .,(TnESn) isprecisely equal tothenumber ofmonomials ofdegree d.We letRnbethe determinant ofthecoefficients ofthese polynomials, viewed asforms in(X)with coefficients inZ[W]. Then Rn=Rn(W)EZ[W].Weclaim thatRn(W) satisfies theproperties ofthelemma. First we note thatif(TnESn,thenHun(X)isdivisible byapower ofXiat most di-1,fori=1,.. .,n-1.Ontheother hand, thedegree ofHun(X)in Xnisdetermined bythecondition that thetotal degree isd-dnoHence Snhas exactly Dnelements. Itfollows atonce thatRn(W) ishomogeneous ofdegree Dn inthecoefficients ofFn,i.e. in(W n,(II»)'From theconstruction italso follows that Rnishomogeneous ineach setofvariables("",(II»)for each i=1,..., n-1. Ifwespecialize theforms Fi(i=1,.. .,n)toXfi, then Rnspecializes to1, and hence Rn=t=0and Rnisprimitive. For each(Tiwe can write HuFi= LCuuo(W)Mu(X),IuE S'I where MU'(X) «(TES)rangesover allmonomials ofdegree din(X), and Cuuo(W),I isone ofthevariables (W). Then bydefinition Rn(W)=det(CU,U'l(W)(UlESI)'. . .,Cu,un(W)(UnES n»)=det(C). where (TIESI'. . .,(TnESnindexes thecolumns, and (Tindexes the rows. Let B=Cbethematrix with components inZ[W, X]such that BC=det(C)/=Rn/. (See Chapter XIII, Corollary 4.17.) Then foreach (T,wehave Rn(W)Mu(X)=LLBiuoFi.iUiESi' I Given i,wetake for (Ttheindex such thatMu(X)=Xfinorder toobtain the first relation inLemma 3.7. ByTheorem 3.4, weconclude thatRn(W)EPI.This concludes theproof ofthelemma. Ofcourse, wepickedanindex ntofixideas. For each ione has apolynomial Risatisfying theanalogous properties, and inparticular homogeneous ofdegree Diinthevariables (Wi,(II»)which arethecoefficients oftheform Fi. 396 ALGEBRAIC SPACES IX,3 Theorem 3.8. Let Rbetheresultant ofthe ngeneric forms Fiover Z,inn variables. Then Rsatisfies thefollowing properties. (a) Risthegreatest common divisor inZ[W] ofthepolynomials RI'. . .,Rn. (b) Rishomogeneous ofdegree Diinthecoefficients ofFi. (c)LetFi= . . .+"'i,(d;)Xf;,so"'i,(d;)isthecoefficient ofXf;. Then Rcontains themonomial n +IlWD,-. (d.). .II" 1= Proof. The idea will betospecialize theforms FI,. . .,Fntoproducts of generic linear forms, where we can tell what isgoingon. For that weneed a lemma of amore general property eventually tobeproved. We shall use the following notation. Iffl,. . .,fnareforms with coefficients (w), then wewrite R(fl,. . .,fn)=R(w). Lemma 3.9. Let G,Hbegeneric independent forms with deg(GH)=dl. Then R(GH, F2,.. .,Fn) isdivisible byR(G, F2,.. .,Fn)R(H, F2,.. .,Fn). Proof. ByTheorem 3.5, there isanexpression XR(FI'. . .,Fn)=QIF I+... +QnFn with QiEZ[W, X]. Let WG,WH,WF2,. . .,WFnbethecoefficients ofG,H,F2,. . .,Fnrespectively, and let(w) bethecoefficients ofGH, F2,. . .,Fn.Then R(w)=R(GH, F2,. . .,Fn), and weobtain XR(w)=QI(W, X)GH +Q2(w, X)F 2+Qn(w, X)Fno Hence R(GH, F2,. . .,Fn)belongs totheelimination ideal ofG,F2,. . .,Fnin thering Z[W G,WH,WF2,.. .,WFn]'andsimilarly with Hinstead ofG.Since WHisafamily ofindependent variables over Z[W G,WF2,. . .,WFn]'itfollows thatR(G, F2,. . .,Fn)divides R(GH, F2,. . .,Fn)inthatring, andsimilarly for R(H, F2,. . .,Fn). But(WG)and(WH)areindependent sets ofvariables, and so R(G, F2,. . .,Fn),R(H, F2,.. .,Fn)aredistinct prime elements inthatring,so their product divides R(GH, F2,. . .,Fn)asstated, thus proving thelemma. Lemma 3.9applies toanyspecialized family ofpolynomials g,h,fl,. . ., fnwith coefficients inafield k.Observe that for asystem ofnlinear forms in nvariables, theresultant issimply thedeterminant ofthecoefficients. Thus if LI'. . .,Lnaregenerically independent linear forms inthevariables XI'. . .,Xn, then their resultant R(L I'. . .,Ln) ishomogeneous ofdegree1inthecoefficients ofLiforeach i.Weapply Lemma 3.9 tothe case offorms fl,. . .,fn-I, which areproducts ofgenerically independent linear forms. ByLemma 3.9 weconclude that forthisspecialized family ofform, their resultant hasdegree atleast Dnin IX,93 PROJECTIONS AND ELIMINATION 397 thecoefficients ofFn,soforthegeneric forms FI,. . .,Fntheir resultant has degreeatleast Dninthecoefficients ofFnoSimilarly R(F I,. . .,Fn) hasdegree atleast Diinthecoefficients ofFifor each i.But Rdivides the nelements RI(W),. . .,Rn(W) constructed inLemma 3.7. Therefore weconclude that Rhas degree exactly Diinthecoefficients ofFi.ByTheorem 3.5, weknow that R divides each Ri.Let Gbethegreatestcommon divisor ofRI'. . .,RninZ[W]. Then Rdivides Gand has the same degree ineach setofvariables (,(v»)for i=1,..., n.Hence there exists cEZsuch that G=cR. We must have c=+1,because, say, Rnisprimitive inZ[W]. This proves (a)and(b)ofthe theorem. Astothethird part,wespecialize theforms to/;=Xf;, i=1,.. .,n.Then Rnspecializesto1,and since Rdivides Rnitfollows that Ritself specializes to +1.Since allcoefficients oftheforms specialize to0except those which we denoted by,(d;)'itfollows thatR(W) contains themonomial which istheproduct ofthese variables tothe power Di,uptothesign+1.This proves (c), and concludes theproof ofTheorem 3.8. We can now normalize theresultant bychoosing thesign such that Rcontains themonomial n M-ITWD; - i(d.), i=1" with coefficient +1.This condition determines Runiquely, and wethen denote Ralso by R=Res(F I,. . .,Fn). Given forms II,. . .,Inwith coefficients (w) inafield K(actually any commu- tative ring),we can then define their resultant Res(fl,. . .,fn)=R(w) with thenormalized polynomialR 0With this normalization, wethen have a stronger result than Lemma 3.9. Theorem 3.10. Letfl=ghbeaproduct oflorms such thatdeg(gh)=dl. Let12,. . .,Inbearbitrary lorms ofdegrees d2,. . .,dn.Then Res(gh, 12,. . .,In)=Res(g, 12,. . .,In)Res(h, 12'. . .,fn). Proof. From thefact that thedegrees have toaddinaproduct ofpolynomials, together with Theorem 3.8(a) and(b), we now seeinLemma 3.9that wemust have theprecise equality inwhat was onlyadivisibility before weknew the precise degree ofRineach setofvariables. Theorem 3.10isvery useful inproving further properties ofthedeterminant, because itallows areduction tosimplecases under factorization ofpolynomials. 398 ALGEBRAIC SPACES IX,3 For instance one has: Theorem 3.11. LetFI'. . .,Fnbethegeneric forms innvariables, and let FI,. . .,Fnbetheforms obtained bysubstituting Xn=0,sothat FI,.. .,Fn-I are thegeneric forms inn-1variables. Let n>2.Then Res(F I,. . .,Fn-I,Xn)=Res( FI,. . .,Fn_l)dn . Proof. ByTheorem 3.10 itsuffices toprove theassertion when dn=1.By Theorem 3.4, foreach i=1,.. .,n-1wehave anexpression (*) XfRes(F 1,..., Fn-I,Xn)=QIF I+... +Qn-IFn-1+QnXn withQjEZ[W, X](dependingonthechoice ofi).The left-hand side can be written asapolynomialinthecoefficients ofF1,. . .,Fn-I with thenotation XfR(WFI'. . .,WFn_I'Ixn)=XfP("FI'. . .,WFn_I)=XtP(W(n-I)),say; thus inthegeneric linear form inXI'. . .,Xnwehave specialized allthecoef- ficients to0except thecoefficient ofXn,which wehave specialized to1.Sub- stitute Xn=0intheright side of(*). ByTheorem 3.4, weconclude that p(w(n-I))lieinthe resultant ideal ofFI'...' Fn-I,and therefore Res(F 1,. ..,Fn-I)divides p(w(n-I)). By Theorem 3.8 we know that p(w(n-I)) has the same homogeneity degree inWp.(i=1,..., n-1)_ _ I asRes(F I,. . .,Fn-I).Hence there iscEZsuch that cRes( FI,..., Fn-I)=Res(F I,..., Fn-I,Xn). One finds c=1byspecializing FI,. . .,Fn-ItoXii,. . .,X"--II respectively, thus concluding theproof. The next basic lemma isstated forthegeneric case, forinstance inMacaulay [Ma 16], and istaken upagain in[Jo90], Lemma 5.6. Lemma 3.12. Let Abeacommutative ring. Letfl'. . .,fn'gl'. . .,gnbe homogeneous polynomials inA[X I'. . .,Xn].Assume that (gI'. . .,gn)Cifl,. . .,fn) asideals inA[X].Then Res(fl'... ,fn) divides Res(gl'...' gn) inA. Proof. Express each gi=2:hi}hwithhijhomogeneous inA[X]. Byspe- cialization, wemay then assume that gi=2:HijFjwhereHi}andFjhave alge- braically independent coefficients over Z.ByTheorem 3.4, foreach iwehave arelation XfRes(gl'.. .,gn)=QIgI+... +Qngn with some QiEZ[W H,WF], IX,3 PROJECTIONS AND ELIMINATION 399 where WH,WFdenote theindependent variable coefficients ofthepolynomials Hi}andFjrespectively. Inparticular, (*) XfRes(gl'.. .,gn)=0mod (FI'.. .,Fn)Z[W H,WF,X]. Note that Res(gl,. . .,gn)=P(W H,WF)EZ[W H,WF]isapolynomial with integer coefficients. If(wF) isageneric point oftheresultant varietyC11over Z,then P(W H,wF)=0by(*). Hence Res(F I,. . .,Fn)divides P(W H,WF),thus proving thelemma. Theorem 3.13. LetAbeacommutative ring and letdl,...,dnbeintegers >1asusual. Let./; behomogeneous ofdegree diinA[X]=A[X I,..., Xn]. Let dbeaninteger>1,and letgi,. ..,gnbehomogeneous ofdegree din A[X]. Then ./;09=./;(gI'. . .,gn) ishomogeneous ofdegree ddi,and Res(fl0g,... ,fn0g)=Res(gl'...' gn)dJ."dnRes(fl'... ,fn)dn-1inA. Proof. We start with thestandard relation ofTheorem 3.4: (*) XfRes(F I,. . .,Fn)=0mod (FI,. . .,Fn)Z[W F,X]. We letGI,. . .,Gnbeindependent generic polynomials ofdegree d,and letWG denote their independent variable coefficients. Substituting GiforXiin(*), we find GfRes(F I,. . .,Fn)=0mod (FloG,. . .,Fn0G)Z[W F,WG,X]. Abbreviate Res(F I,. . .,Fn)byR(F), and letgi=GfR(F). ByLemma 3.12, it follows that Res(fl0G,. . .,Fn0G)divides Res(GR(F),. . .,GR(F)) inZ[W F,WG]. ByTheorem 3.10 and thehomogeneity ofTheorem 3.8(b) wefind that Res(GR(F),. . .,Gs"R(F))=Res(G.,. . .,Gn)M Res(FI,. . .,Fn)N with integers M,N>o.Since Res( GI'. . .,Gn)andRes(F I,. . .,Fn)aredistinct prime elements inZ[W G,WF](distinct because they involve independent vari- abies), itfollows that (**) Res(F I0G,..., Fn0G)=ERes(G I,..., Gn)a Res(F I,..., Fn)b with integers a,b>0and E=1or-1.Finally,wespecializeFito"'fXfi and wespecialize GitoViXf, with independent variables (WI'. . .,Wn,VI'. . .,Vn). 400 ALGEBRAIC SPACES IX,3 Substituting in(**), weobtain Res(W IUjlXjd1 ,. . .,WnUnXd n) =eRes(UIXj,.. .,UnX)a Res(WIXjl,. . .,WnXn)b. Bythehomogeneity ofTheorem 3.8(b) weget I1("'iUdi)d1didndn-1=eI1u1n-laI1W11Jidnb i i i From this wegetatonce e=1and a,barewhat theyarestated tobeinthe theorem. Corollary 3.14. LetC=(cij)beasquare matrix with coefficients inA.Let fi(X)=Fi(CX) (where CX ismultiplication ofmatrices, viewing Xasacolumn vector). Then Res(fl,. . .,fn)=det(C)d1...dnRes(F I,. . .,Fn). Proof. This isthe case when d=1and giisalinear form foreach i. Theorem 3.15. Letfl,...,inbehomogeneous inA[X], and suppose dn>difor alli.Lethibehomogeneous ofdegree dn-diinA[X]. Then n-I Res(fl,. . .,fn- bfn+hjh)=Res(fb.. .,fn)inA. }=I Proof. We may assumefi=Fiarethegeneric forms, Hiareforms generic independent from FI'...,Fn, and A=Z[W F,WH],where (WF)and (WH) are the coefficients oftherespective polynomials. We note that the ideals (FI,. . .,Fn) and (FI,. . .,Fn+.LHjFj)areequal. From Lemma 3.12 we j=Fn concl ude that thetwo resultants inthestatement ofthetheorem differ byafactor of 1or-1.We maynow specialize Hijto0todetermine that thefactor is+1, thusconcluding theproof. Theorem 3.16. Let 7Tbeapermutation of{I,. . .,n},and lete(7T)beits sign. Then Res(F 1T(l)'...'F1T(n»)=e(7T)dl"dnRes(F I,..., Fn). Proof. Again using Lemma 3.12with the ideals (FI'...' Fn) and (F 1T(1)'. . .,F1T(n»),which areequal, weconclude thedesired equality uptoa factor+1,inZ[WF].Wedetermine thissign byspecializing FitoXfi, andusing themultiplicativity ofTheorem 3.10. We arethen reduced tothe case when Fi=Xi'soalinear form; and we canapply Corollary 3.14 toconclude theproof. The next theorem was anexercise invan derWaerden's Moderne Algebra. IX,3 PROJECTIONS AND ELIMINATION 401 Theorem 3.17. LetLI'...,Ln- I,Fbegeneric forms innvariables, such that LI,. . .,Ln-Iareofdegree 1,and Fhasdegree d=dn.Let djU=1,..., n) be(-1)n-jtimes thej-th minor determinant ofthecoefficient matrix ofthe forms (LI,. . .,Ln-I).Then Res(L I,..., Ln-I,F)=F(d l,..., dn). Proof. We first claim that forallj=1,.. .,nwehave thecongruence (*) Xndj-Xjd n=0mod (LI'. . .,Ln-I)Z[ X], where asusual, (W) are thecoefficients oftheforms LI'. . .,Ln-I,F.To see this, weconsider thesystem oflinear equations WIIXI+...+WI,n-IX n-I=LI(X)-WI,nXn -l,IXI+... +-l,n-IXn-l=Ln-I(X)-Wn-l,nXn. IfC=(Cl ,. . .,Cn-I)isasquare matrix with columns Cj, then asolution of asystem oflinear equations CX=Cnsatisfies Cramer's rule Xjdet(CI,..., cn-I)=det(CI,..., Cn,..., cn-I). Using thefact that thedeterminant islinear ineach column, (*)falls out. Then from thecongruence (*)itfollows that XF(dl'...' dn)=dF(XI'...' Xn)mod(LI'...' Ln-I)Z[ X], whence XF(db.. .,dn)=0mod(LI'.. .,Ln-I,F). Hence byTheorem 3.4 and thefact that Res(L I,. . .,Ln-I,F)=R(W) generates theelimination ideal, itfollows that there exists cEZ[W] such that F(d l,..., dn)=cRes(L I,..., Ln-I,F). Since theleft side ishomogeneous ofdegree1inthecoefficients WFand homo- geneous ofdegree dinthecoefficients WL;foreach i=1,. . .,n-1,itfollows from Theorem 3.8that cEZ.Specializing LitoXiandFtoXmakes djspecialize to0ifj=t=nand dnspecializes to1.Hence theleft side specializesto1,and sodoes theright side, whence c=1.This concludes theproof. 402 ALGEBRAIC SPACES IX,4 Bibliography [Jo80] J.P.JOUANOLOU, Ideaux resultants, Advances inMathematics 37No.3 (1980), pp,212-238 [Jo90] J,P,JOUANOLOU, Leformalisme duresultant, Advances inMathematics 90 No.2 (1991) pp, 117-263 [Jo91] J.P.JOUANOLOU, Aspects invariants del'elimination, Department deMath- ematiques, Universite Louis Pasteur, Strasbourg, France (1991) [Ma 16] F.MACAULAY, Thealgebraic theory ofmodular systems, Cambridge University Press, 1916 4. RESULTANT SYSTEMS The projection argument used toprove Theorem 3.4 has theadvantage of constructingageneric point inaveryexplicit way. Ontheother hand, noexplicit, oreven effective, formula was given toconstruct asystem offorms defining at.We shall now reformulate aversion ofTheorem 3.4 over Zand weshall prove itusingacompletely different technique which constructs effectivelya system ofgenerators for anideal ofdefinition ofthearithmetic variety Cliin Theorem 3.2. Theorem 4.1. Given degrees dl,. . .,dr>1,andpositive integers m, n.Let (W)=(,(JI»)bethevariables asin3,(2)viewed asalgebraically independent elements over theintegers Z.There exists aneffectively determinable finite number ofpolynomials Rp(W)EZ[W] having thefollowing property. Let(f) be asin(1), asystem offorms ofthegiven degrees with coefficients (w) in some field k.Then(f)has anon-trivial common zeroifandonlyifRp(w)=0 forallp. Afinite family {Rp}having theproperty stated inTheorem 4.1will becalled aresultant system for thegiven degrees. According tovan der Waerden (Moderne Algebra, first and second edition, 80), thefollowing technique of proof using resultants goes back toKronecker elimination, and to apaper of Kapferer (Uber Resultanten undResultantensysteme, Sitzungsber. Bayer. Akad. Munchen 1929, pp.179-200). Thefamily ofpolynomials {Rp(W)}iscalled a resultant system, because oftheway theyareconstructed. They form asetof generators for anideal blsuch that thearithmetic variety Cliisthe setofzeros ofbl.Idon't know how close thesystem constructed below istobeingasetof generators fortheprime ideal PIinZ[W] associated with Cli.Actuallyweshall not need thewhole theory ofChapter IV, 10; weneed onlyone ofthechar- acterizing properties ofresultants. IX,4 RESULTANT SYSTEMS 403 Letp,qbepositive integers. Let I'=v-yp+VIXP-IX2+... +vXPJV (Y1 I I P 2 9=Woxq+WIXq-IX2+.. .+wxqw I I q 2 betwogeneric homogeneous polynomials inZ[v,w,XI'X2]=Z[v,w][X]. In Chapter IV, 10wedefined their resultant Res(fv, gw) incase X2=I,but we find itnow more appropriatetowork with homogeneous polynomials. For our purposes here, weneed only thefact that theresultant R(v, w)ischaracterized bythefollowing property. Ifwehave aspecialization (a ,b)of(v,w)inafield K,andiffa'fbhave afactorization P fa=aoIT(XI-a;X 2) ;=I q gb=boIT(XI-f3jX2) j=1 then wehave thesymmetric expressions interms ofthe roots: R(a, b)=Res(!a, fb)=agbgD(a;-f3j) I,} =agIJgb(a;, 1)=(-I)pqbb IJfa(f3j, I). I } From thegeneral theory ofsymmetric polynomials, itisapriori clear that R(v,w)lies inZ[v,w], andChapter IV, 10givesanexplicit representation 'Pv,wfv+t/lv,wgw=X+q-IR(v, w) where'Pv,wandt/lv,wEZ[v, W,X].This representation will not beneeded. The next property willprovide thebasic inductive step forelimination. Proposition 4.2. Letfa'gbbehomogeneous polynomials with coefficients in afield K.Then R(a, b)=0ifandonlyifthesystem ofequations fa(X)=0,gb(X)=0 has anon-trivial zero insome extension ofK(which can betaken tobefinite). Ifao=0then azero ofgbisalso azero offa;andifbo=0then azero offa isalso azero ofgb.Ifaob o=t=0then from theexpression oftheresultant asa product ofthedifference ofroots (ai-f3j)theproposition follows atonce. Weshall now prove Theorem 4.1byusing resultants. Wedothisbyinduction on n. 404 ALGEBRAIC SPACES IX,4 Ifn=1,thetheorem isobvious. Ifn=2,r=1,thetheorem isagain obvious, taking theemptysetfor(Rp). Ifn=2,r=2,then thetheorem amounts toProposition 4.2. Assume now n=2and r>2,sowehave asystem ofhomogeneous equations o=fl(X)=f2(X)= . . .=fr(X) with (X)=(XI,X2).Let dibethedegree of.f; and letd=max di.Wereplace thefamily {Jj(X)} bythefamily ofallpolynomials /;(X)X1-d;and /;(X)X1-d;,i=1,..., r. These two families have the same sets ofnon-trivial zeros, sotoprove Theorem 4.1 wemayassume without loss ofgenerality that allthepolynomials fl,. . ., frhave the same degree d. With n=2,consider thegeneric system offorms ofdegree din(X): (4)fj( X)=0with i=1,..., r,intwo variables (X)=(XI'X2), where thecoefficients ofFiare"'i,D,. . .,"'i,dsothat (W)=("),0'...' WI,d'...' ,o,..., ,d). The next proposition isaspecialcase ofTheorem 4.1, butgives thefirst step ofaninduction showing how togettheanalogue ofProposition 4.2forsuch a larger system. Let TI,. . .,Trand UI'. . .,Urbeindependent variables over Z[W, X]. LetFI'...' Frbethegeneric forms of3,(2). Let f=FI( X)T I+·..+Fr( X)T r 9=FI( X)U I+·..+Fr( X)U r sof, 9EZ[W, T,U][X]. Thenf, 9arepolynomials in(X)with coefficients in Z[W, T,U]. We may form their resultant Res(f, g)EZ[W, T,U]. Thus Res(f, g)isapolynomial inthevariables (T,U)with coefficients inZ[W]. We let(QJL(W))bethefamily ofcoefficients ofthispolynomial. Proposition 4.3. The system {QJL(W)} just constructed satisfies theproperty ofTheorem 4.1, i.e.itisaresultant system for rforms ofthe same degree d. Proof. Suppose that there isanon-trivial solution of aspecial system fj(W, X)=0with (w) insome field k.Then (w, T,U)isacommon non-trivial zero off,g,soRes(f, g)=0and thereforeQJL(w)=0forallJ..L.Conversely, suppose thatQJL(w)=0forallJ..L.Let.f;(X)=Fi(w, X). We want toshow thatfi(X) fori=1,.. .,rhave acommon non-trivial zero insome extension of IX,4 RESULTANT SYSTEMS 405 k.Ifall.liare0ink[X I,X2]then they have acommon non-trivial zero. If,say, fl=t=0ink[X], then specializingT2,. . .,Trto0and TIto 1intheresultant Res(f, g), we seethat Res(fl,/2 U2+... +frUr)=0 asapolynomial ink[U2'. . .,Ur].After makingafinite extension ofkifneces- sary,wemayassume thatfl(X)splits into linear factors. Let{ai} bethe roots offl(X I,1).Then some (ai' 1)must also be azero of12U2+... +IrUr, which implies that(ai' 1)isacommon zero ofII'. . .,Irsince U2'. ..,Ur arealgebraically independentover k.This proves Proposition 4.3. We are now readytodotheinductive step with n>2.Again, let .Ii(X)=Fi(w, X)forj=1,..., r bepolynomials with coefficients (w) insome fields k. Remark 4.4. There exists anon-trivial zero ofthesystem fi=0(i=1,. . .,r) insome extension ofkifandonlyifthere exist (XI'.. .,Xn-I)=t=(0,.. .,0) and (xn,t)=t=(0,0) insome extension ofksuch that .Ii(txb. . .,txn-I'Xn)=0fori=1,..., r. So wemay now construct thesystem (Rp)inductivelyasfollows. Let Tbe anew variable, and letx(n-I)=(Xb. . .,Xn-I).Let 9i(W,X(n-I),Sn'T)=Fi(W, TXI'. . .,TXn-I'Xn)EZ[W,X(n-I)][Xn'T]. Then giishomogeneous inthe two variables (Xn,T).Bythetheorem for two variables, there isasystem ofpolynomials (QJL)inZ[W, x(n-I)] having the property: if(w, .in-I) isapoint inafield K,then gi(W, x(n-I), Xn'T)have anon-trivial common zerofori=1,..., r. QJL(w,x(n-I)=0forallJ..L. Viewing eachQJLasapolynomialinthevariables (x(n-I», wedecompose each QJLas asum ofitshomogeneous terms, and welet(HA( x(n-I)) bethefam- ilyofthese polynomials, homogeneous in(x<n-I». From thehomogeneity property ofthe formsFjin(X), itfollows that iftistranscendental over K and gi(w, x(n-I), Xn,T)have anon-trivial common zero forj=1,..., r then gi(w, tx(n-I), Xn,T)also have anon-trivial common zero. Therefore 406 ALGEBRAIC SPACES IX,4 QJL(W, tX(n-l))=0forall J.L,and soHA(w,x(n-l))=O.Therefore wemayuse the family ofpolynomials (HA)instead ofthefamily (QJL)'and weobtain theproperty: if(w,x(n-l)) isapoint inafield K,then gi(W, x(n-l), Xn'T)have anon-trivial common zerofori=1,..., r HA(w, x(n-l))=oforallA. Byinduction onn,there exists afamily (Rp(W))ofpolynomialsinZ[W] (actually homogeneous), having theproperty: if(w) isapoint inafield K,then HA(w,x(n-l)) have anon-trivial common zeroforallA Rp(w)=0forallp. Inlight ofRemark 4.4, this concludes theproof ofTheorem 4.1bytheresultant method. 5. SPEC OF ARING We shall extend thenotions of2toarbitrary commutative rings. LetAbe acommutative ring. Byspec(A)we mean the setofallprime ideals ofA.Anelement ofspec(A) isalso called apoint ofspec(A). IffEA,weview the setofprime ideals pofspec(A) containing fasthe set ofzeros offIndeed, itisthe setofpsuch that theimage offinthecanonical homomorphism A Alp isO.Let abe anideal, and let (a)(the setofzeros ofa)bethe setofall primes ofAcontaininga.Let a,bbeideals. Then wehave: Proposition 5.1. (i)(ab)=(a)U (b). (ii)If{ai}isafamily ofideals, then (La;)=n (a;). (iii) Wehave (a)C(b) ifandonlyifrad(a) :Jrad(b), where rad(a), the radical ofa,isthe setofallelements xEAsuch that xnEaforsome positive integer n. Proof. Exercise. SeeCorollary 2.3ofChapter X. Asubset Cofspec(A) issaid tobeclosed ifthere exists anideal aofAsuch that Cconsists ofthose prime ideals psuch that acp.Thecomplement ofa closed subset ofspec(A) iscalled anopen subset ofspec(A). The following statements arethen very easy toverify, andwill belefttothereader. IX,5 SPEC OFARING 407 Proposition 5.2. The union ofafinite number ofclosed sets isclosed. The intersection ofanarbitrary family ofclosed sets isclosed. The intersection ofafinite number ofopen sets isopen. The union ofan arbitrary family ofopen sets isopen. The empty setandspec(A) itself areboth open and closed. IfSisasubset ofA,then the setofprime ideals pEspec(A) such that Scp coincides with the setofprime ideals pcontaining theideal generated byS. The collection ofopensets asinProposition 5.2 issaid tobe atopologyon spec( A), called theZariski topology. Remark. Inanalysis,one considers acompact Hausdorff space S."Haus- dorff" means thatgiven twopoints P,Qthere exists disjoint opensets Up, UQ containing Pand Qrespectively. Inthepresent algebraic context, thetopology isnotHausdorff. Intheanalytic context, letRbethering ofcomplex valued continuous functions onS.Then themaximal ideals ofRareinbijection with thepoints ofS(Gelfand-Naimark theorem). Toeach point PES, weassociate the ideal Mp offunctions fsuch thatf(P)=o.The association P Mp gives thebijection. There areanalogous results inthecomplex analyticcase. For anon-trivial example,seeExercise 19ofChapter XII. LetA,Bbecommutative rings and cp:A Bahomomorphism. Thencp induces amap qJ*=spec(qJ)=qJ-1 :spec(B)-.spec(A) by p qJ-1(p). Indeed, itisimmediately verified that({J-l(p) isaprime ideal ofA.Note however that theinverse image ofamaximal ideal ofBisnotnecessarilyamaximal ideal ofA.Example? The reader willverify atonce that spec«({J) iscontinuous, inthe sense that ifUisopen inspec(B), thenqJ- 1(U)isopen inspec(A). We can then view specas acontravariant functor from thecategory of commutative ringstothecategory oftopological spaces. Byapoint ofspec(A) inafield Lone means amapping spec«({J): spec(L) spec(A) induced byahomomorphism ({J:A LofAinto L. Forexample, foreach prime number p,wegetapoint ofspec(Z), namely thepoint arising from thereduction map Z-.Z/pZ. 408 ALGEBRAIC SPACES IX,5 Thecorresponding point isgiven bythereversed arrow, spec(Z) spec(ZjpZ). Asanother example, consider thepolynomial ringk[Xl'...,Xn]over a field k.For each n-tuple (Cl,...,cn)ink8(n)wegetahomomorphism qJ:k[X 1,...,Xn]-.k8 such thatqJistheidentity onk,and qJ(X i)=Ciforalli.The corresponding point isgiven bythereversed arrow speck[X] spec(k8). Thus wemay identify thepoints inn-space k8(n)with thepoints ofspeck[X] (over k)ink8 . However, one does not want totake points only inthealgebraic closure of k,and ofcourse one may deal with the case ofanarbitrary variety Vover k rather than allofaffine n-space. Thus letk[xI'. . .,xn] be afinitely generated entire ring over kwith achosen family ofgenerators. Let V=spec k[x]. Let A be acommutative k-algebra, correspondingtoahomomorphism k A.Then a point ofVinAmay bedescribed either asahomomorphism cp:k[xl'. . .,Xn] A, orasthereversed arrow spec(A) spec(k[x]) corresponding tothishomomorphism. Ifweput Ci=CP(Xi)' then one may call (c)=(cl'. . .,Cn)thecoordinates ofthepoint inA.Byageneric point ofV inafield Kwe mean apoint such that themap cp:k[x] Kisinjective, i.e. an isomorphism ofk[x] with some subring ofK. Let Abe acommutative Noetherian ring. We leave itasanexercise to verify thefollowing assertions, which translate theNoetherian condition into properties ofclosed sets intheZariski topology. Closed subsets ofspec(A) satisfy thedescending chain condition, i.e.,if C1=>C2=>C3=>... isadescending chain ofclosed sets, then wehave Cn=Cn+1forallsufficiently largen.Equivalently, let{C;} ie1beafamily ofclosed sets. Then there exists a relatively minimal element ofthisfamily, that isaclosed setCiointhefamily such that foralli,ifCicCiothen Ci=Cio.Theproof follows atonce from thecorresponding properties ofideals, and thesimple formalism relating unions and intersections ofclosed sets with products and sums ofideals. IX,5 SPEC OFARING 409 Aclosed setCissaid tobeirreducible ifitcannot beexpressedastheunion oftwo closed sets C;/=C 1UC2 with C1;/=Cand C2;/=C. Theorem 5.3. Let AbeaNoetherian commutative ring. Then every closed setCcan beexpressedasafinite union ofirreducible closed sets, and this expression isunique ifintheunion C=C1U. ..UCr ofirreducible closed sets, wehave CicFCjifi;/=j. Proof. Wegive theproofasanexample toshow how theversion ofTheorem 2.2 has animmediate translation inthe more general context ofspec(A). Suppose thefamily ofclosed setswhich cannot berepresentedasafinite union ofirreducible ones isnotempty. Translating theNoetherian hypothesis inthis case shows that there exists aminimal such setC.Then Ccannot beirreducible, and we can write Casaunion ofclosed sets C=C'UC", with C';/=Cand C" ;/=C.Since C'and C"arestrictly smaller than C,then we can express C'and C" asfinite unions ofirreducible closed sets, thus gettinga similar expression forC,and acontradiction which proves existence. Astouniqueness, let C=C1U...UCr=Z1U...UZs be anexpression ofCasunion ofirreducible closed sets, without inclusion relations. For each Zjwe canwrite Zj=(ZjnC1)u...u(ZjnCr). Since eachZjnCiisaclosed set, wemust haveZj=ZjnCifor some i.Hence Zj=Cifor some i.Similarly, Ciiscontained insome Zk. Since there isno inclusion relation among theZ/s,wemust haveZj=Ci=Zk. This argument can becarried outforeachZjand each Ci.This proves that eachZjappears among theC;'s and each Ciappears among theZ/s,and proves theuniqueness ofourrepresentation. This proves thetheorem. Proposition 5.4. LetCbeaclosed subset ofspec(A).Then Cisirreducible ifandonlyifC=Cfl(p)for some prime ideal p. Proof. Exercise. More propertiesatthe same basic level will begiven inExercises 14-19. 410 ALGEBRAIC SPACES IX,Ex EXERCISES Integrality 1.(Hilbert-Zariski) Let kbe afield and letVbe ahomogeneous variety with generic point (x)over k.Let bethealgebraic setofzeros inkaofahomogeneous ideal in k[X] generated byforms fl,. , .,frink[X]. Prove that Vn hasonly thetrivial zero ifandonly ifeach x;isintegral over theringk[f(x)]=k[fl(X),.. .,fr(x)]. (Compare with Theorem 3,7ofChapter VII.) 2.Letfl'. . .,frbeforms innvariables and supposen>r.Prove that these forms have anon-trivial common zero. 3.Let Rbeanentire ring. Prove that Risintegrally closed ifandonly ifthelocal ring Rpisintegrally closed foreach prime ideal p, 4.Let Rbe anentire ring with quotient field K.Let tbetranscendental over K.Let f(t)=La;t; EK[t], Prove: (a)Iff(t)isintegralover R[t],then alla;areintegralover R, (b)IfRisintegrally closed, then R[t] isintegrally closed, For the next exercises, weletR=k[x]=k[X]/p, where pisahomogeneous prime ideal. Then (x)isahomogeneous generic point for ak-variety V,WeletIbetheintegral closure ofRink(x), We assume forsimplicity thatk(x) isaregular extension ofk, 5,Let z=LC;X; with c;Ek,and z=1=O.Ifk[x] isintegrally closed, prove thatk[x/z] isintegrally closed. 6,Define anelement fEk(x) tobehomogeneous iff(tx)=tdf(x) forttranscendental over k(x) and some integer d.LetfEI.Show thatfcan bewritten intheform f=L/; where each/; ishomogeneous ofdegree i>0,and where also/;EI,(Some /;may be0,ofcourse.) We letRm denote the setofelements ofRwhich arehomogeneous ofdegreem. Similarly for1m. We note that Rmand1mare vector spacesover k,and that R(resp. I) isthedirect sum ofallspaces Rm(resp. 1m)for m=0,1,.. .This isobvious forR,and itistrue forIbecause ofExercise 6, 7.Prove thatIcan bewritten asasum I=RZI+...+Rzs, where each z;ishomoge- neous ofsome degree d;. 8.Define anintegerm>1tobewell behaved if1m=Iqmforallintegers q>1.If R=I,then all marewell behaved. InExercise 7,supposem>max d;,Show that miswell behaved. 9.(a) Prove that 1misafinite dimensional vector space over k.Letwo,. . .,WM be a basis for1mover k.Then k[Im]=k[w]. (b)Ifmiswell behaved, show thatk[Im]isintegrally closed. (c) Denote byk«x» thefield generated over kbyallquotients x;/Xjwith xj=1=0, andsimilarly fork«w», Show that k«x»=k«w». (Ifyou want tosee Exercises 4-9 worked out, see myIntroduction toAlgebraic Geometry, Interscience 1958, Chapter V.) IX,Ex EXERCISES 411 Resultants 10. Prove that theresultant defined for nforms innvariables in3actually coincides with theresultant ofChapter IV, or4when n=2. 11. Let a=(II'. . ,,Ir) be ahomogeneous ideal ink[X I,. . .,Xn)(with kalgebraically closed). Assume that theonlyzeros of aconsist of afinite number ofpoints (x(l», .,., (x(d» inprojective space pn-I, sothecoordinates ofeach x(j) can be taken ink.Let uI', . .,unbeindependent variables and let Lu(X)=ulX) +...+unXn. Let R)(u),, , ,,RS<u)Ek[u] be aresultant system forII'. . ,,Ir,Lu. (a) Show that the common non-trivial zeros ofthesystem R;(u) (i=1,..., s) inkarethe zeros ofthepolynomial nLu(x(j» Ek[u].j (b)LetD(u) bethegreatestcommon divisor ofRI(u),, . ,,RS<u) ink[u]. Show that there exist integersmj>1such that (up toafactor ink) d D(u)=nLu(x(j»mJ. j=1 [See van derWaerden, Moderne Algebra, Second Edition, Volume II,79.] 12, For forms in2variables, prove directly from thedefinition used in4that one has Res(lg, h)=Res(f, h)Res(g, h) Res(f, g)=(-I)(degf)(degg)Res(g, I). 13. Let kbe afield and letZ kbethecanonical homomorphism. IfFEZ[W, X], we denote byFtheimage ofFink["W: X]under thishomomorphism. Thus wegetR, theimage oftheresultant R, (a) Show that Risagenerator oftheprime idealPk,1ofTheorem 3.5 over the field k,Thus wemay denote RbyRk, (b) Show that Risabsolutely irreducible, and soisRk,Inother words, Rkis irreducible over thealgebraic closure ofk, Spec ofaring 14. Let Abe acommutative ring. Define spec(A) tobeconnected ifspec(A) isnot the union oftwodisjoint non-empty closed sets(orequivalently, spec(A)isnottheunion oftwodisjoint, non-empty open sets), (a)Suppose that there areidempotents el, e2inA(that isey=elande=e2), =1=0,1,such that ele2=0and el+e2=1.Show that spec(A) isnot connected. (b)Conversely, ifspec(A) isnotconnected, show that there exist idempotents asinpart (a). Ineither case, the existence oftheidempotents isequivalent with thefact that the ring Aisaproduct oftwo non-zero rings, A=A1XA2. 412 ALGEBRAIC SPACES IX,Ex 15. Prove that theZariski topology iscompact, inother words: let{Vi}iel beafamily of opensets such that UVi=spec(A), i Show that there ISafinite number ofopensetsVii',..,Vi"whose union isspec(A). [Hint: Use closed sets, and usethefact that ifasum ofideals istheunit ideal, then 1 can bewritten asafinite sum ofelements.] 16, Let1be anelement ofA.Let Sbethemultiplicative subset {I,I,12 ,13 ,. .,}con- sisting ofthe powers ofI.We denote byAfthering S-IAasinChapter II, 3. From the natural homomorphism AAfone gets the corresponding map spec(A f) spec(A). (a) Show that spec(A f)mapsontheopensetofpoints inspec(A) which are not zeros ofI. (b)Given apoint pEspec(A), and anopensetVcontaining p,show that there exists 1such that pEspec(A f)CV. 17. LetVi=spec(A f)be afinite family ofopen subsets ofspec(A) covering spec(A). For each i,letai//;EA.t"Assume that asfunctions onVinVjwehave ai//;=aj/fj forallpairs i,j,Show that there exists aunique element aEAsuch that a=a;//; inAf,foralli. 18. Let kbe afield and letk[x.,. . .,xn]=ACKbe afinitely generated subring of some extension field K.Assume thatk(xI'. . .,xn)hastranscendence degree" Show that every maximal chain ofprime ideals A:JPI:JP2:J , . .:JPm:J{O}, with PI=1=A,Pi=1=Pi+l,Pm=1={O}, must have m=" 19. Let A=Z[XI,. . .,xn] be afinitely generated entire ring over Z.Show that every maximal chain ofprime ideals asinExercise 18must have m=,+1.Here, ,= transcendence degree ofQ(X.,, , .,xn)over Q, CHAPTER X Noetherian Rings and Modules This chapter mayserve asanintroduction tothemethods ofalgebraic geometry rooted incommutative algebra and thetheory ofmodules, mostly over aNoeth- , . erlan rIng. 1. BASIC CRITERIA Let Abe aring and Mamodule (i.e.,aleftA-module). We shall saythat MisNoetherian ifitsatisfies anyone ofthefollowing three conditions: (1)Every submodule ofMisfinitely generated. (2)Every ascending sequence ofsubmodules ofM, M1cM2cM3c..., such thatM; =FMi+1isfinite. (3)Every non-empty set Sofsubmodules ofMhas amaximal element (i.e., asubmodule M0such that forany element NofSwhich contains Mo wehave N =Mo). Weshall now prove that theabove three conditions areequivalent. (1) (2)Supposewehave anascending sequence ofsubmodules ofMas above. LetNbetheunion ofalltheMi(i=1,2,...).Then Nisfinitely gen- erated, saybyelements Xb...,Xr,and each generator isinsome Mi' Hence there exists anindexjsuch that Xb. ..,XrEMj. 413 414 NOETHERIAN RINGS AND MODULES X,1 Then <X 1,...,X r)CMjeN =<Xf,...,x r), whence equality holds and ourimplication isproved. (2) (3) LetNo beanelement ofS.IfNo isnotmaximal, itisproperly contained inasubmodule N1.IfN1isnotmaximal, itisproperly contained in asubmodule N2.Inductively, ifwehave found Niwhich isnotmaximal, itis contained properly inasubmodule Ni+1.Inthis waywecould construct an infinite chain, which isimpossible. (3) (1)Let Nbe asubmodule ofM. Let aoEN.IfN=F(ao), then there exists anelement a1ENwhich does not liein(ao). Proceeding induc- tively,we can find anascending sequence ofsubmodules ofN,namely <ao)c(ao,a1)c(ao,aba2)c... where theinclusion each time isproper. The setofthese submodules has a maximal element, sayasubmodule (ao,ab...'ar),and itisthen clear that thisfinitely generated submodule must beequal toN,aswas tobeshown. Proposition 1.1. LetMbeaNoetherian A-module. Then every submodule and every factor module ofMisNoetherian. Proof. Our assertion isclear forsubmodules (say from thefirst condi- tion). For the factor module, letNbe asubmodule andf:M-.M/N the canonical homomorphism. LetM1CM2C...beanascending chain ofsub- modules ofM/N and letMi=f-l( Mi).Then M1CM2C...isanascending chain ofsubmodules ofM,which must have amaximal element, sayMr,so that Mi=Mrfor r>i.Then f(M;)=Miand our assertion follows. Proposition 1.2. LetMbe amodule, Nasubmodule. Assume that Nand M/N areNoetherian. Then MisNoetherian. Proof. With every submodule LofMweassociate thepair ofmodules LH(L nN,(L+N)/N). We contend: IfEcFare two submodules ofMsuch that their associated pairs areequal, then E=F.To seethis, let xEF.Bythehypothesis that (E+N)/N=(F+N)/N there exist elements u,vENand yEEsuch that y+u=x+v.Then x-y=u-VEFnN =EnN. Since YEE,itfollows the xEEand our contention isproved. Ifwehave an ascending sequence E1CE2C... X,1 BASIC CRITERIA 415 then theassociated pairs form anascending sequence ofsubmodules ofNand M/N respectively, and these sequences must stop. Hence our sequence E1CE2...also stops, byourpreceding contention. Propositions 1.1and 1.2may besummarized bysaying that inanexact sequence 0-.M' -.M-.M" -+0,MisNoetherian ifandonly ifM'and M" areNoetherian. Corollary 1.3. Let Mbe amodule, and letN,N'besubmodules. If M =N+N'andifboth N,N' areNoetherian, then MisNoetherian. A finite direct sumofNoetherian modules isNoetherian. Proof. We first observe that the direct product NxN'isNoetherian since itcontains Nasasubmodule whose factor module isisomorphic toN', andProposition 1.2applies. We have asurjective homomorphism NxN'-.M such that thepair (x,x')with xENand x'EN' maps on x+x'.ByProp- osition 1.1,itfollows that MisNoetherian. Finite products (orsums) follow byinduction. Aring Aiscalled Noetherian ifitisNoetherian asaleftmodule over itself. This means that every leftideal isfinitely generated. Proposition 1.4. LetAbeaNoetherian ring and letMbeafinitely generated module. Then MisNoetherian. Proof. LetXl'...,Xnbegenerators ofM. There exists ahomomorphism f:AxAx... xA-.M oftheproduct ofAwith itself ntimes such that f(a1,...,a n)=a1x1+...+anxn. This homomorphism issurjective. Bythecorollary ofthepreceding proposition, theproduct isNoetherian, and hence MisNoetherian byProposition 1.1. Proposition 1.5. Let Abearing which isNoetherian, and letqJ:A-.Bbe asurjective ring-homomorphism. Then BisNoetherian. Proof. Let b1C.. .cbnc...beanascending chain ofleft ideals ofB and letai=qJ-l(b i).Then the aiform anascending chain ofleftideals ofA which must stop, say atar.Since qJ(a;)=biforalli,ourproposition isproved. Proposition 1.6. Let Abe acommutative Noetherian ring, and letSbea multiplicative subset ofA.Then S-1AisNoetherian. Proof. We leave theproofasanexercise. 416 NOETHERIAN RINGS AND MODULES X,2 Examples. InChapter IV, wegave thefundamental examples ofNoeth- erian rings, namely polynomial rings and rings ofpower series. The above propositions show how toconstruct other examples from these, bytaking factor rings ormodules, orsubmodules. We have already mentioned that forapplications toalgebraic geometry, itis valuable toconsider factor rings oftype k[X]/a, where aisanarbitrary ideal. For this and similar reasons, ithas been found that thefoundations should be laid interms ofmodules, notjust ideals orfactor rings. Notably,weshall first seethat theprime ideal associated with anirreducible algebraic sethas ananalogue interms ofmodules. We shall also seethat thedecomposition ofanalgebraic setinto irreducibles has anatural formulation interms ofmodules, namely by expressingasubmodule asanintersection orprimary modules. In6weshall applysome general notions togettheHilbert polynomial of amodule offinite length, and weshall make comments onhow this can be interpreted interms ofgeometric notions. Thus thepresent chapter ispartly intended toprovideabridge between basic algebra andalgebraic geometry. 2. ASSOCIATED PRIMES Throughout thissection, weletAbeacommutative ring. Modules and homo- morphisms areA-modules andA-homomorphisms unless otherwise specified. Proposition 2.1. Let Sbe amultiplicative subset ofA,and assume that S does notcontain O.Then there exists anideal ofAwhich ismaximal inthe setofideals notintersecting S,and any such ideal isprime. Proof. The existence ofsuch anideal pfollows from Zorn's lemma (the setofideals notmeeting Sisnotempty, because itcontains the zero ideal, and is clearly inductively ordered). Let pbemaximal inthe set. Let a,bEA,ab Ep, but a$pand b$p.Byhypothesis, theideals (a,p)and (b,p)generated bya and p(orband prespectively) meet S,and there exist therefore elements s,s'ES,c,c',x,x'EA,p,p'EPsuch that s=ca+xp and s'=c'b+x'p'. Multiplying these twoexpressions,weobtain ss'=cc'ab+p" with some p"Ep,whence we seethat ss'lies inp.This contradicts the fact that pdoes notintersect S,and proves that pisprime. Anelement aofAissaid tobenilpotent ifthere exists anintegern> 1such that an =o. X,2 ASSOCIATED PRIMES 417 Corollary 2.2. Anelement aofAisnilpotent ifandonlyifitlies inevery prime ideal ofA. Proof. Ifan =0,then anEpforevery prime p,and hence aEp.Ifan=F0 foranypositive integer n,weletSbethemultiplicative subset ofpowers ofa, namely {1,a,a2 ,...},and find aprime ideal asintheproposition toprove the con verse. Let abeanideal ofA.The radical ofaisthe setofallaEAsuch that anEa for some integern>1,(orequivalently, itisthe setofelements aEAwhose image inthefactor ringAla isnilpotent). Weobserve that theradical ofaisan ideal, forifan =0and bm=0then (a+b)k=0ifkissufficiently large: Inthe binomial expansion, either aorbwill appear with apower atleast equal to norm. Corollary 2.3. Anelement aofAlies intheradical ofanideal aifandonly ifitlies inevery prime ideal containinga. Proof. Corollary 2.3 isequivalent toCorollary 2.2applied totheringA/a. We shall extend Corollary 2.2 tomodules. Wefirst make some remarks on localization. Let Sbe amultiplicative subset ofA.IfMisamodule, we can define S-1Minthe same way that wedefined S-1 A.Weconsider equivalence classes ofpairs (x,s)with xEMand sES,two pairs (x,s)and (x',S')being equivalent ifthere exists SIESsuch that SI(s'x-SX')=o.We denote the equivalence class of(x,s)byxis, andverify atonce that the setofequivalence classes isanadditive group (under theobvious operations). Itisinfact an A-module, under theoperation (a,xis)1---+axis. We shall denote this module ofequivalence classes byS-1M.(We note that S-1Mcould also beviewed asanS-1A-module.) Ifpisaprime ideal ofA,and Sisthecomplement ofpinA,then S-1Mis also denoted byM". Itfollows trivially from thedefinitions that ifN-.Misaninjective homo- morphism, then wehave anatural injection S-1N-.S-1M.Inother words, if Nisasubmodule ofM,then S-1Ncan beviewed asasubmodule ofS-1M. IfxENand sES,then thefraction xis can beviewed asanelement ofS-1N orS-1M.Ifxis=0inS-1M,then there existsSIESsuch that SIX=0,and this means that xisisalso 0inS-1N.Thus ifpisaprime ideal and Nisasub- module ofM, wehave anatural inclusion ofN"inM".Weshall infactidentify N"asasubmodule ofM".Inparticular,we seethatM"isthe sum ofitssub- modules (Ax)",for xEM(but ofcourse notthedirect sum). Let xEM. The annihilator aofxistheideal consisting ofallelements aEAsuch that ax =O.We have anisomorphism (ofmodules) Ala Ax 418 NOETHERIAN RINGS AND MODULES X,2 under themap a-+ax. Lemma 2.4. Let xbeanelement ofamodule M,and let Qbeitsannihilator. Let pbeaprime ideal ofA.Then (Ax)"=F0ifandonlyifpcontains Q. Proof. The lemma isanimmediate consequence ofthedefinitions, and will belefttothereader. Let abeanelement ofA.LetMbeamodule. Thehomomorphism Xt---+ax, xEM will becalled theprincipal homomorphism associated with a,and will bede- noted byaM. We shall saythat aMislocally nilpotent ifforeach xEMthere exists aninteger n{x)> 1such that an(x)x =O.This condition implies that forevery finitely generated submodule NofM,there exists anintegern> 1 such that anN =0:We take for nthelargest power ofaannihilatingafinite setofgenerators ofN.Therefore, ifMisfinitely generated, aM islocally nilpotent ifandonlyifitisnilpotent. Proposition 2.5. Let Mbe amodule, aEA.Then aM islocally nilpotent ifandonlyifalies inevery prime ideal psuch thatM"=Fo. Proof. Assume that aMislocally nilpotent. Let pbe aprime ofAsuch thatM"=FO.Then there exists xEMsuch that(Ax)"=FO.Let nbeapositive integer such that anx =O.Let Qbetheannihilator ofx.Then anEQ,and hence wecanapply thelemma, andCorollary 4.3toconclude that aliesinevery prime psuch thatM" =FO.Conversely, suppose aMisnotlocally nilpotent,sothere exists xEMsuch that anx=0forall n>o.Let S={I, a,a2 ,...},and using Proposition 2.1 letpbe aprime notintersecting S.Then(Ax)p=t=0,so Mp=t=0and afj.p,asdesired. LetMbeamodule. Aprime ideal pofAwill besaid tobeassociated with Mifthere exists anelement xEMsuch that pistheannihilator ofx.Inpar- ticular, since p=FA,wemust have x=FO. Proposition 2.6. LetMbeamodule =FO.Let pbeamaximal element inthe setofideals which areannihilators ofelements xEM,x=Fo.Then pisprime. Proof. Let pbe'theannihilator oftheelement x=FO.Then p=FA.Let a,bEA,ab Ep,a$p.Then ax =FO.But theideal (b,p)annihilates ax,and contains p.Since pismaximal, itfollows that bEp,and hence pisprime, Corollary 2.7. IfAisNoetherian andMisamodule =t=0,then there exists aprime associated with M. Proof. The setofideals asinProposition 2.6 isnotempty since M =t=0, and has amaximal element because AisNoetherian. X,2 ASSOCIATED PRIMES 419 Corollary 2.8. Assume that both Aand MareNoetherian, M =FO.Then there exists asequence ofsubmodules M =M1::JM2::J...::JMr=0 such that each factor module Mi/M i+1isisomorphic toA/Pi forsome prime Pi. Proof. Consider the setofsubmodules having theproperty described in thecorollary. Itisnotempty, since there exists anassociated prime PofM, andifPistheannihilator ofx,then Ax A/p. LetNbeamaximal element in the set.IfN=FM,then bythepreceding argument applied toM/N, there exists asubmodule N'ofMcontaining Nsuch thatN'/N isisomorphic toA/p for some p,and this contradicts themaximality ofN. Proposition 2.9. Let AbeNoetherian, and aEA.Let Mbe amodule. Then aMisinjective ifandonlyifadoes notlieinany associated prime ofM. Proof. Assume that aMisnotinjective, sothat ax =0for some xEM, x=t=O.ByCorollary 2.7, there exists anassociated prime pofAx, and aisan element ofp.Conversely, ifaMisinjective, then acannot lieinany associated prime because adoes notannihilate any non-zero element ofM. Proposition 2.10. Let AbeNoetherian, and letMbeamodule. Let aEA. Thefollowing conditions areequivalent: (i)aMislocally nilpotent. (ii) alies inevery associated prime ofM. (iii)alies inevery prime psuch thatM"=Fo. IfPisaprime such thatMp=t=0,then pcontains anassociated prime ofM. Proof. The fact that (i)implies (ii)isobvious from thedefinitions, and does not need thehypothesis that AisNoetherian. Neither does thefact that (iii)implies (i),which has been proved inProposition 2.5. We must therefore prove that(ii)implies (iii) which isactually implied bythelast statement. The latter isprovedasfollows. Let pbe aprime such thatMp=t=o.Then there exists xEMsuch that (Ax)p=t=o.ByCorollary 2.7, there exists anassociated prime qof(Ax)"inA.Hence there exists anelement Y/sof(Ax)",with YEAx, sp,and Y/s=t=0,such that qistheannihilator ofy/s.Itfollows that qcp, forotherwise, there exists bEq,bp,and 0=by/s,whence y/s=0,contra- diction. Let bI,. . .,bnbegenerators for q.For each i,there existsSiEA, sifj.p,such that SibiY=0because biy/s=O.Let t=SI...Sn.Then itis trivially verified that qistheannihilator oftyinA.Hence qcp,asdesired. Let usdefine thesupport ofMby supp(M)=setofprimes psuch thatM"=Fo. 420 NOETHERIAN RINGS AND MODULES X,2 We also have theannihilator ofM, ann(M)=setofelements aEAsuch that aM=O. We usethenotatIon ass(M)=setofassociated primes ofM. For any ideal awehave itsradical, rad(a)=setofelements aEAsuch that anEafor some integern>1. Then forfinitely generated M, we can reformulate Proposition 2.10bythe following formula: rad(ann(M))=n p= pesupp(M)n p. peass(M) Corollary 2.11. Let AbeNoetherian, and letMbeamodule. Thefollowing conditions areequivalent: (i)There exists onlyone associated prime ofM. (ii) We have M =F0,andfor everyaEA,thehomomorphism aMisinjective, orlocally nilpotent. Ifthese conditions aresatisfied, then the setofelements aEAsuch that aM islocally nilpotent isequal totheassociated prime ofM. Proof. Immediate consequence ofPropositions 2.9 and 2.10. Proposition 2.12. Let Nbeasubmodule ofM.Every associated prime of Nisassociated with Malso. Anassociated prime ofMisassociated with N orwithMIN. Proof. The first assertion isobvious. Let pbeanassociated prime ofM, and say pistheannihilator oftheelement x=Fo.IfAx nN =0,then Axis isomorphic toasubmodule ofMIN, and hence pisassociated withMIN. Suppose AxnN =t=o.Let y=ax ENwith aEAand y=t=o.Then pannihilates y. We claim p=ann(y). Let bEAand by=o.Then ba Epbut afj.p, so bEp.Hence pistheannihilator ofyinA,and therefore isassociated with N, aswas tobeshown. X,3 PRIMARY DECOMPOSITION 421 3. PRIMARY DECOMPOSITION Wecontinue toassume that Aisacommutative ring, and that modules (resp. homomorphisms) are A-modules (resp. A-homomorphisms), unless otherwise specified. LetMbeamodule. Asubmodule QofMissaid tobeprimary ifQ=FM, and ifgivenaEA,thehomomorphism aM/Qiseither injective ornilpotent. Viewing Aasamodule over itself, we seethat anideal qisprimary ifandonly ifitsatisfies thefollowing condition: Given a,bEA,ab Eqand artq,then bnEqfor some n>1. Let Qbeprimary. Let pbetheideal ofelements aEAsuch thataM/Qis nilpotent. Then pisprime. Indeed, suppose that a,bEA,ab Epand artp. ThenaM/Qisinjective, and consequently alt/ Qisinjective forall n>1.Since (ab)M/Qisnilpotent, itfollows that bM/Qmust benilpotent, and hence that bEp, proving that pisprime. Weshall call ptheprime belonging toQ,and also say that Qisp-primary. We note thecorresponding property for aprimary module Qwith prime p: Let bEAand xEMbesuch that bxEQ.Ifxfj.Qthen bEp. Examples. Let mbe amaximal ideal ofAand letqbeanideal ofAsuch that mkCqfor some positive integer k.Then qisprimary, and mbelongs to q.We leave theproof tothereader. The above conclusion isnotalways trueifmisreplaced bysome prime ideal p.For instance, letRbe afactorial ring with aprime element t.Let Abethe subring ofpolynomials f(X)ER[X] such that f(X)=ao+alX+ . . . with a1divisiblebyt.Let p=(tX, X2).Then pisprime but p2=(t2X2 ,tX3 ,X4) isnotprimary,as one sees because X2rtp2but tkrtp2forallk>I,yet t2X2EP2. Proposition 3.1. LetMbea1nodule, andQt,. . .,Qrsubmodules which are p-primary forthe same prime p.Then Qtn...nQrisalso p-primary. Proof. LetQ=Qtn...nQr. Let aEp.Let nibesuch that(aM/Q)ni=0 foreach i=1,...,rand let nbethemaximum ofn.,...,nr.Thenalt/ Q=0, sothataM/Qisnilpotent. Conversely, supposeartp.Let xEM, xrtQjfor some j.Then anxrtQjforallpositive integers n,and consequently aM/Qis injective. This proves ourproposition. 422 NOETHERIAN RINGS AND MODULES X,3 Let Nbeasubmodule ofM. When Niswritten asafinite intersection of primary submodules, say N =Qln...nQr, weshall call this aprimary decomposition ofN.Using Proposition 3.1,we seethat bygrouping theQiaccording totheir primes, we canalways obtain from agiven primary decomposition another one such that theprimes belonging totheprimary ideals arealldistinct. Aprimary decompositionasabove such that theprime ideals Pl'...,Prbelonging toQl'...,Qrrespectively aredistinct, and such that Ncannot beexpressedasanintersection ofaproper subfamily oftheprimary ideals {Q 1,. ..,Qr}will besaid tobereduced. Bydeleting some. oftheprimary modules appearing inagiven decomposition, we seethat ifN admits some primary decomposition, then itadmits areduced one. We shall provearesult giving certain uniqueness properties of areduced primary decomposition. LetNbe asubmodule ofMand letxibethecanonical homomorphism. LetQbe asubmodule ofM=M/Nand letQbeitsinverse image inM.Then directly from thedefinition, one sees thatQisprimary ifandonly ifQisprimary; andiftheyareprimary, then theprime belonging toQisalso theprime belonging toQ.Furthermore, ifN=Q 1n . . .nQrisaprimary decomposition ofNin M,then (0)=QIn . . .nQr isaprimary decomposition of(0)inM,asthereader willverify atonce from thedefinitions. Inaddition, thedecomposition ofNisreduced ifandonly ifthe decomposition of(0)isreduced since theprimes belonging toone arethe same astheprimes belonging totheother. Let Q 1n...nQr=Nbe areduced primary decomposition, and letPi belong toQi.IfPidoes notcontainPi(j;/=i)then wesaythat Piisisolated. The isolated primes aretherefore those primes which areminimal inthe set ofprimes belonging totheprimary modules Qi. Theorem 3.2. LetNbeasubmodule ofM,and let N =Q1n...nQr=Q'ln...nQ be areduced primary decomposition ofN. Then r=s.The setofprimes belonging toQb ...,Qrand Q'l, ...,Q;isthe same. If{Ph...' Pm} isthe setofisolated primes belonging tothese decompositions, then Qi=Qfor i=1,...,m,inother words, theprimary modules corresponding toisolated primes areuniquely determined. Proof. The uniqueness ofthenumber ofterms inareduced decomposition and theuniqueness ofthefamily ofprimes belonging totheprimary components will be aconsequence ofTheorem 3.5below. X,3 PRIMARY DECOMPOSITION 423 There remains toprove theuniqueness oftheprimary module belonging toanisolated prime, say Pl.Bydefinition, foreachj=2,...,rthere exists ajEPjand ajrtPl. Let a=a2.. .arbetheproduct. Then aEPjforallj>1, but artPl. We can find anintegern>1such thatalt/ Qj=0forj=2,...,r. Let N1=setofxEMsuch that a"x EN. Wecontend thatQl=N1.This will prove thedesired uniqueness. Let xEQl. Then a"x EQln... nQr=N, so XEN1.Conversely, let xEN1,sothat a"x EN,and inparticular a"x EQ1.Since artPI' weknow bydefinition that aM/Qlisinjective. Hence xEQl,thereby proving ourtheorem. Theorem 3.3. Let MbeaNoetherian module. Let Nbe asubmodule of M. Then Nadmits aprimary decomposition. Proof. We consider the setofsubmodules ofMwhich donot admit a primary decomposition. Ifthis setisnotempty, then ithas amaximal element because MisNoetherian. Let Nbethis maximal element. Then Nisnot primary, and there exists aEAsuch thataM/Nisneither injective nornilpotent. The increasing sequence ofmodules KeraM/NcKerait/NcKera/Nc.·. stops, say ata/N.Let ({J:M/N-.M/N betheendomorphism ({J=a/N. Then Ker ({J2=Ker({J.Hence 0=Ker({Jn1m({JinM/N, and neither the kernel nor theimage oflfJiso.Taking theinverse image inM, we seethat Nis theintersection oftwo submodules ofM,unequal toN.Weconclude from the maximality ofNthat each one ofthese submodules admits aprimary de- composition, and therefore that Nadmits onealso, contradiction. We shall conclude our discussion byrelating theprimes belonging toa primary decomposition with theassociated primes discussed intheprevious section. Proposition 3.4. Let Aand MbeNoetherian. Asubmodule QofMis primary ifandonlyifM/Q hasexactly one associated prime P,and inthat case, Pbelongs toQ,i.e.Qisp-primary. Proof. Immediate consequence ofthedefinitions, andCorollary 2.11. Theorem 3.5. Let Aand MbeNoetherian. The associated primes ofM areprecisely theprimes which belong totheprimary modules inareduced primary decomposition of0inM.Inparticular, the setofassociated primes ofMisfinite. Proof. Let o=Q 1n...nQr 424 NOETHERIAN RINGS AND MODULES X,4 beareduced primary decomposition of0inM. We have aninjective homo- morphism r M -+EBM/Qi. i=1 ByProposition 2.12andProposition 3.4, weconclude that every associated prime ofMbelongs tosome Qi.Conversely, letN =Q2n...nQr. Then N=F0because ourdecomposition isreduced. Wehave N =N/(N nQ1) (N+Q1)/Q1CM/Q1. Hence Nisisomorphic toasubmodule ofM/Q1' and consequently has an associated prime which can benone other than theprime P1belonging toQ1. This provesour theorem. Theorem 3.6. Let AbeaNoetherian ring. Then the setofdivisors ofzero inAistheset-theoretic union ofallprimes belonging toprimary ideals ina reduced primary decomposition ofo. Proof. Anelement ofaEAisadivisor of0ifandonly ifaAisnotinjective. According toProposition 2.9, this isequivalent toalying insome associated prime ofA(viewed asmodule over itself). Applying Theorem 3.5concludes the proof. 4. NAKAYAMA'S LEMMA WeletAdenote acommutative ring, but notnecessarily Noetherian. When dealing with modules over aring, many propertiescan beobtained firstbylocalizing, thusreducing problems tomodules over local rings. Inpractice, asinthepresent section, such modules will befinitely generated. This section shows that some aspectscan bereduced tovector spaces over afield byreducing modulo themaximal ideal ofthelocal ring. Over afield, amodule always has abasis .We extend this propertyasfar aswe can tomodules finite over alocal ring. The first three statements which follow areknown asNakayama's lemma. Lemma 4.1. Let abeanideal ofAwhich iscontained inevery maximal ideal ofA.Let Ebe afinitely generated A-module. Suppose that aE=E.Then E={O}. X,4 NAKAYAMA'S LEMMA 425 Proof. Induction onthenumber ofgenerators ofE.Let xI'...,Xsbe generators ofE.Byhypothesis, there exist elements aI'...,asEQsuch that xs=aIxI+...+asXs , sothere isanelement a(namely as)in Qsuch that (1+a)xs liesinthemodule generated bythe first s-1generators. Furthermore 1+aisaunit inA, otherwise 1+aiscontained insome maximal ideal, and since alies inall maximal ideals, weconclude that 1liesinamaximal ideal, which isnotpossible. Hence Xsitself liesinthemodule generated bys-1generators, and theproof iscomplete byinduction. Lemma 4.1applies inparticular tothe case when Aisalocal ring, and Q=misitsmaximal ideal. Lemma 4.2. Let Abealocal ring, letEbeafinitely generated A-module, and Fasubmodule. IfE=F+mE, then E=F. Proof. Apply Lemma 4.1 toElF. Lemma 4.3. LetAbealocal ring. LetEbeafinitely generated A-module. Ifx),. . .,Xnaregenerators forEmod mE, then they aregenerators for E. Proof. Take Ftobethesubmodule generated byXI'...,Xn. Theorem 4.4. Let Abe alocal ring and Eafinite projective A-module. Then Eisfree. Infact,ifxI'...,Xnareelements ofEwhose residue classes XI'.. .,xnare abasis ofElmE over Aim, then XI'...,Xnare abasis ofE over A.IfXI'. ..,Xrare such that XI'...,xrarelinearly independent over Aim, then theycan becompleted toabasis ofEover A. Proof Iamindebted toGeorge Bergman forthefollowing proof ofthe first statement. LetFbe afree module with basis eI,. . .,en,and letf:f' E bethehomomorphism mapping eitoXi.We want toprove thatfisanisomor- phism. ByLemma 4.3,fissurjective. Since Eisprojective, itfollows thatf splits, i.e. we can write F=PoEBPhwhere Po=Kerfand PIismapped isomorphic ally onto EbyfNow thelinear independence ofXI,. . .,Xnmod mE shows that PoCmE=mP0CmP I. Hence PoCmpo- Also, asadirect summand inafinitely generated module, Po isfinitely generated.SobyLemma 4.3, Po=(0)andfisanisomorphism,as was tobeproved. Astothesecond statement, itisimmediate since we cancompleteagiven 426 NOETHERIAN RINGS AND MODULES X,5 sequence Xb. . .,Xrwith X.,. . .,xrlinearly independent over AIm, toa sequence XI,. . .,Xnwith XI,. . .,xnlineary independent over AIm, and then we canapply thefirst part oftheproof. This concludes theproof ofthetheorem. Let Ebe amodule over alocal ring Awith maximal ideal m. We let E(m)=E/mE. Iff:E-.Fisahomomorphism, thenfinduces ahomo- morphism .f(m):E(m)-.F(m). Iffissurjective, then itfollows trivially thatfem)issurjective. Proposition 4.5. Letf:E-.Fbeahomomorphism ofmodules, finite over a local ring A.Then: (i)If.f(m)issurjective,soisf. (ii)Assume! isinjective. If.f(m)issurjective, then! isanisomorphism. (iii) Assume that E,Farefree. If.f(m)isinjective (resp. anisomorphism) then fisinjective (resp. anisomorphism). Proof. Theproofs areimmediate consequences ofNakayama's lemma and will belefttothereader. Forinstance, inthefirst statement, consider the exact sequence E-.F-.F/Im!-.O andapply Nakayama tothe term ontheright. In(iii), use thelifting ofbases asinTheorem 4.4. 5. FILTERED AND GRADED MODULES LetAbeacommutative ring and Eamodule. Byafiltration ofEone means asequence ofsubmodules E=Eo::JE1::JE2::J...::JEn::J. .. Strictly speaking, this should becalled adescending filtration. We don't consider any other. Example. Let abeanideal ofaring A,and EanA-module. Let En=anE. Then thesequence ofsubmodules {En} isafiltration. More generally, let{En} beany filtration ofamodule E.We saythat itis ana-filtration ifaEncEn+1forall n.Thepreceding example isana-filtration. X,5 FILTERED AND GRADED MODULES 427 We saythat ana-filtration isa-stable, orstable ifwehave aEn=En+1forall n sufficiently large. Proposition 5.1. Let{En} and{E} bestable a-filtrations ofE.Then there exists apositive integer dsuch that En+dcE and E+dcEn forall n>o. Proof. Itsuffices toprove the proposition when E=anE. Since aEncEn+ 1forall n,wehave anE cEn. Bythestability hypothesis, there exists dsuch that En+d=anEd canE, which proves theproposition. Aring Aiscalled graded (by thenatural numbers) ifone can write Aasa direct sum (asabelian group), 00 A=EBAn, n=O such that forallintegers m, n>0wehave AnAmcAn+m. Itfollows inpar- ticular that Aoisasubring, and that each component AnisanAo-module. Let Abeagraded ring. Amodule Eiscalled agraded module ifEcan be expressedasadirect sum (asabelian group) 00 E=EBEn' n=O such that AnEmcEn+m. Inparticular, EnisanAo-module. Elements ofEnare then called homogeneous ofdegreen.Bydefinition, any element ofEcan be written uniquelyasafinite sum ofhomogeneous elements. Example. Let kbe afield, and letX0'...,Xrbeindependent variables. The polynomial ring A=k[Xo,..., Xr] isagraded algebra, with k=Ao. The homogeneous elements ofdegreenarethepolynomials generated bythe monomials inX0'...,Xrofdegree n,that is r X...X" with Ldi=n. i=O Anideal IofAiscalled homogeneous ifitisgraded,asanA-module. Ifthis isthe case, then thefactor ringAIIisalso agraded ring. Proposition 5.2. Let Abeagraded ring. Then AisNoetherian ifandonly ifAoisNoetherian, and Aisfinitely generated asAo-algebra. 428 NOETHERIAN RINGS AND MODULES X,5 Proof. Afinitely generated algebra over aNoetherian ring isNoetherian, because itisahomomorphic image ofthepolynomial ring infinitely many variables, and we canapply Hilbert's theorem. Conversely, suppose that AisNoetherian. The sum 00 A+=EBAn n=l isanideal ofA,whose residue class ring isAo, which isthus ahomomorphic image ofA,and istherefore Noetherian. Furthermore, A+has afinite number ofgenerators Xl'.. .,Xsbyhypothesis. Expressing each generator asasum of homogeneous elements, wemayassume without loss ofgenerality that these generators arehomogeneous, say ofdegrees dl'...,dsrespectively, with all d;>o.Let Bbethesubring ofAgenerated over AobyXl'...,Xs. Weclaim that AnCBforall n.This iscertainly true for n=O.Let n>o.Let Xbe homogeneous ofdegreen.Then there exist elements a;EAn-d.such that s x=La;x;. ;=1 Since d;>0byinduction, each a;isinAo[x 1,...,xs]=B,sothis shows xEB also, and concludes theproof. Weshall now seetwo ways ofconstructing graded rings from filtrations. First, letAbe aring and aanideal. We view Aasafiltered ring, bythe powers an.Wedefine thefirst associated graded ring tobe 00 Sa(A)=S=EBan. n=O Similarly, ifEisanA-module, and Eisfiltered byana-filtration, wedefine 00 Es=EBEn. 11=0 Then itisimmediately verified that Esisagraded S-module. Observe that ifAisNoetherian, and aisgenerated byelements Xl'...,Xs then SisgeneratedasanA-algebra also byXl'...,Xs,and istherefore also Noetherian. Lemma 5.3. Let AbeaNoetherian ring, and Eafinitely generated module, with ana-filtration. Then Esisfinite over Sifandonlyifthefiltration ofE isa-stable. Proof. Let n Fn=EBE;, ;=0 X,5 FILTERED AND GRADED MODULES 429 and let Gn=Eo Et>...Et>EnEt>aEnEt>a2En Et>a3En Et>... Then GnisanS-submodule ofEs,and isfinite over Ssince Fnisfinite over A. We have GncG"+1 and UG n=Es. Since SisNoetherian, weget: Esisfinite over S<=>Es=GNfor some N <=>EN+m=amEN forall m>0 <=>thefiltratIon ofEisa-stable. This proves thelemma. Theorem 5.4. (Artin-Rees). Let Abe aNoetherian ring, aanideal, Ea .finite A-module with astable a-filtration. Let Fbe asubmodule, and let Fn=F(\En. Then {Fn} isastable a-filtration ofF. Proof. We have a(F nE,.)caFnaE,.cFnEn+1, so{F,.} isana-filtration ofF.We canthen form theassociated graded S-module FS,which isasubmodule ofEs,and ISfinite over Ssince SisNoetherian. We applyLemma 5.3 toconclude theproof. Wereformulate theArtin-Rees theorem initsoriginal form asfollows. Corollary 5.5. Let Abe aNoetherian ring, Eafinite A-module, and Fa submodule. Let abe anideal. There exists anintegerssuch thatforall integersn>swehave a"E nF=a"-S(aSE nF). Proof. Specialcase ofTheorem 5.4 and thedefinitions. Theorem 5.6. (Krull). Let Abe aNoetherian ring, and let abeanideal contained inevery maxilnal ideal ofA.Let Ebeafinite A-module. Then 00 nanE =O. n=1 Proof. Let F=nanE and apply Nakayama's lemma toconclude the proof. 430 NOETHERIAN RINGS AND MODULES X,5 Corollary 5.7. Let 0bealocal Noetherian ring with maximal ideal m.Then 00 nmn=o. n=l Proof. Special case ofTheorem 5.6when E=A. The second way offormingagraded ring ormodule isdone asfollows. Let Abearing and aanideal ofA.Wedefine thesecond associated graded ring 00 gra(A)=EBan/an+1. n=O Multiplication isdefined intheobvious way. Let aEanand letadenote its residue class mod an+1.LetbEam and letDdenote itsresidue class lTIod am+1. Wedefine theproduct aDtobetheresidue class ofabmod am+n+1.Itiseasily verified that this definition isindependent ofthechoices ofrepresentatives and defines amultiplication ongra(A) which makes gra(A) into agraded ring. Let Ebeafiltered A-module. Wedefine 00 gr(E)=EBEn/En +1. n=O Ifthefiltration isana-filtration, then gr(E) isagraded gra(A)-module. Proposition 5.8. Assume that AisNoetherian, and let abeanideal ofA. Then gra(A) isNoetherian. IfEisafinite A-module with astable a-filtration, then gr(E) isafinite gra(A)-module. Proo}'. LetXl'...,Xsbegenerators ofa.LetXibetheresidue class ofXi ina/a2 .Then gra(A)=(A/a)[xl'...,xs] isNoetherian, thus proving thefirst assertion. For the second assertion, we have for some d, Ed+m=amEd forall m>o. Hence gr(E) isgenerated bythefinite direct sum gr(E)o(f)...ffigr(E)d. But each gr(E)n=En/En +1isfinitely generated over A,and annihilated bya, soisafinite A/a-module. Hence theabove finite direct sum isafinite A/a- module, sogr(E) isafinite gra(A)-module, thus concluding theproof ofthe proposition. X,6 THE HILBERT POLYNOMIAL 431 6. THE HILBERT POLYNOMIAL The main point ofthis section istostudy thelengths ofcertain filtered modules over local rings, and toshow thatthey arepolynomials inappropriate cases. However, wefirst look atgraded modules, and then relate filtered modules tograded ones byusing theconstruction attheend ofthepreceding section. We start with agraded Noetherian ringtogether with afinite graded A-module E, so x A=EBAn n=Ooc and E=EBEn. n=O We have seen inProposition 5.2thatAoisNoetherian, and that Aisafinitely generated Ao-algebra. The same type ofargument shows that Ehas afinite number ofhomogeneous generators, andEnisafinite Ao-module forall n>o. Letcpbe anEuler-Poincare Z-valued function onthe class ofallfinite Ao-modules,asinChapter III,8.We define thePoincare series with respect tocptobethepower series 00 PqJ(E, t)=LqJ(En)tnEZ[[tJJ. n=O Wewrite P(E, t)instead ofPqJ(E, t)forsimplicity. Theorem 6.1. (Hilbert-Serre). Let sbethenumber ofgenerators ofAas Ao-algebra. Then P(E, t)isarational function oftype P(E, t)= sf(t) n(1-tdi) i= 1 with suitable positive integers di,andf(t)EZ[tJ. Proof. Induction on s.For s=0theassertion istrivially true. Let s>1. Let A=AO[XI,. . .,xs]'deg. Xi=di>1.Multiplication byXsonEgivesrise toanexact sequence o-+Kn-+En En+ds-+Ln+ds-+O. Let K =EBKnand L=EBLn. 432 NOETHERIAN RINGS AND MODULES X,6 Then K,Larefinite A-modules (being submodules and factor modules ofE), and areannihilated byXS'soareinfact graded AO[XI'. . .,Xs-d-modules. By definition ofanEuler-Poincare function, weget qJ(Kn)-qJ(En) +qJ(En+ds)-qJ(Ln+ds)=O. Multiplying bytn+dsandsummingover n,weget (1-tds)P(E, t)=P(L, t)-tdsP(K, t)+g(t), where g(t) isapolynomial inZ[tJ. The theorem follows byinduction. Remark. InTheorem 6.1, ifA=Ao[X l'...,xsJthen di=deg Xiasshown intheproof. The next result shows what happens when allthedegrees are equal to1. Theorem 6.2. Assume that Aisgenerated asanAo-algebra byhomogeneous elements ofdegree1.Let dbetheorder ofthepoleofP(E, t)att=1.Then forallsufficiently large n,qJ(En) isapolynomial innofdegree d-1.(For this statement, the zero polynomial isassumed tohave degree-1.) Proof. ByTheorem 6.1, cp(E n)isthecoefficient oftnintherational function P(E, t)=f(t)/(1-t)s. Cancelling powers of 1-t,wewrite P(E, t)=h(t)/(l-t)d,andh(l) ;/=0,with h(t) EZ[tJ. Let m h(t)=Laktk . k=O Wehave thebinomial expansion (_)-d=(d+k-1 )k1 tkO d_1t . For convenience welet(_;)=0for n>0and(_;)=1for n=-1. We then get m (d+n-k-1 )cp(En)=koak d-1forall n>m. The sum ontheright-hand side isapolynomial innwith leadingterm d- 1n (Lak)(d_I)!oFO. This proves thetheorem. X,6 THE HILBERT POLYNOMIAL 433 The polynomial ofTheorem 6.2 iscalled theHilbert polynomial ofthe gradedmodule E,with respect tocpo We now puttogetheranumber ofresults ofthischapter, andgiveanapplication ofTheorem 6.2 tocertain filtered modules. Let Abe aNoetherian local ring with maximal ideal m.Let qbeanm- primary ideal. Then A/q isalso Noetherian and local. Since some power ofm iscontained inq,itfollows that A/q hasonly one associated prime, viewed as module over itself, namely m/q itself. Similarly, ifMisafinite A/q-module, then Mhasonly one associated prime, and theonly simple A/q-moduleisin fact anA/m-module which isone-dimensional. Again since some power ofm iscontained inq,itfollows that A/q has finite length, and Malso has finite length. We now use thelength function asanEuler-Poincare function in applying Theorem 6.2. Theorem 6.3. Let Abe aNoetherian local ring with maximal ideal m. Let qbeanm-primary ideal, and letEbeafinitely generated A-module, with astable q-filtration. Then: (i)E/Enhasfinite length forn>o. (ii)Forallsufficiently large n,thislength isapolynomial g(n)ofdegree<s, where sistheleast number ofgenerators ofq. (iii) Thedegree andleading coefficient ofg(n) depend onlyonEand q,but not onthechosen filtration. Proof. Let G=grq(A)=EBqn/qn+ 1. Then gr(E)=EBEn/En +1isagraded G-module, andGo=A/q. ByProposition 5.8, GisNoetherian andgr(E) isafinite G-module. Bytheremarks preceding thetheorem, E/Enhasfinite length, andifqJdenotes thelength, then n qJ(E/En)=LqJ(Ej_1/Ej). j= 1 IfXI'...,Xsgenerate q,then theimages xl'...,Xsinq/q2generate GasA/q- algebra, and each Xihasdegree1.ByTheorem 6.2 we seethat qJ(En/En+ 1)=h(n) isapolynomial innofdegree<s-1forsufficiently largen.Since cp(E/En+l)-cp(E/En)=h(n), itfollows byLemma 6.4 below that cp(E/ En) isapolynomial g(n) ofdegree <sforalllargen.The last statement concerning theindependence ofthedegree 434 NOETHERIAN RINGS AND MODULES X,6 of9and itsleading coefficient from thechosen filtration follows immediately from Proposition 5.1, andwill beleft tothereader. This concludes theproof. From thetheorem, we seethat there isapolynomial 1..£, qsuch that 1..£,q(n)=length(Ejqn E) forallsufficiently largen.IfE=A,thenXA, qisusually called thecharacteristic polynomial ofq.Inparticular,we seethat XA,q(n)=length(Ajqn) forallsufficiently largen. For acontinuation ofthese topics into dimension theory,see[AtM 69] and [Mat 80]. We shall now studyaparticularly important specialcase having todowith polynomial ideals. Let kbe afield, and let A=k[X o,. . .,XN] bethepolynomial ring inN+ 1variable. Then Aisgraded, the elements of degreenbeing thehomogeneous polynomials ofdegreen.We let abe ahomo- geneous ideal ofA,and for anintegern>0wedefine: cp(n)=dimk An cp(n, a)=dimk an x(n, a)=dimk An/an=dimk An-dimk an=cp(n)-cp(n, a). Asearlier inthis section, Andenotes thek-space ofhomogeneous elements of degreeninA,andsimilarly foran.Then wehave (N+n )cp(n)= N. We shall consider thebinomial polynomial (1\ =T(T-1)··.(T-d+1) Td (1)dJ d!= d!+lower terms. Iffisafunction, wedefine thedifference function Ilfby Ilf(T)=f(T +1)-f(T). Then one verifies directly that (2) (:)=CT J. X,6 THE HILBERT POLYNOMIAL 435 Lemma 6.4. Let PEQ[T] be apolynomial ofdegree dwith rational coefficients. (a)IfP(n) EZforallsufficiently large integers n,then there exist integers co'. . .,cdsuch that P(T)=CO()+c,CT J+ '"+Cd' Inparticular, P(n) EZforallintegers n. (b)Iff:Z Zisanyfunction, andifthere exists apolynomial Q(T)EQ[T] such that Q(Z) CZanddf(n)=Q(n)forall nsufficiently large, then there exists apolynomial Pasin(a)such thatf(n)=P(n)for allnsufficiently large. Proof. Weprove (a)byinduction. Ifthedegree ofPis0,then theassertion isobvious. Suppose deg P>1.By(1)there exist rational numbers co'. . .,Cd such thatP(T) hastheexpression given in(a). ButdPhasdegree strictly smaller than deg P.Using (2)andinduction, weconclude that co'. ..,Cd-l must be integers. Finally Cdisaninteger because P(n) EZfor nsufficiently large. This proves (a). Asfor(b),using (a), we can write Q(T)=co(T )+ . . .+cd-Id-1 with integers co,. . .,Cd-I. LetPIbethe"integral" ofQ,that is P,(T)=CO()+ ... +Cd-'()'so AP,=Q. Thend(f-PI)(n)=0forall nsufficiently large. Hence (f-PI)(n) isequal toaconstant cdforall nsufficiently large,soweletP=PI+cdtoconclude theproof. Proposition 6.5. Let Q,bbehomogeneous ideals inA.Then cp(n,a+b)=cp(n, a)+cp(n, b)-cp(n, anb) x(n, a+b)=x(n, a)+x(n, b)-x(n,anb). Proof. The first isimmediate, and the second follows from thedefinition ofx. 436 NOETHERIAN RINGS AND MODULES X,6 Theorem 6.6. LetFbeahomogeneous polynomial ofdegree d.Assume that Fisnot adivisor ofzero mod a,that is:ifGEA,FG Ea,then GEa. Then x(n, +(F))=X(n,a)-X(n-d,a). Proof. First observe thattrivially cp(n, (F))=cp(n-d), because thedegree of aproduct isthe sum ofthedegrees. Next, using the hypothesis that Fisnotdivisor of0mod a,weconclude immediately cp(n,an(F))=cp(n-d,a). Finally, byProposition 6.5(the formula forX), weobtain: x(n,a+(F))=x(n, a)+x(n, (F))-x(n,an(F)) =x(n, a)+cp(n)-cp(n, (F))-cp(n)+cp(n,an(F)) =x(n, a)-cp(n-d)+cp(n-d,a) =x(n, a)-x(n-d,a) thus proving thetheorem. We denote bymthemaximal ideal m=(Xo,. . .,XN)inA.We call mthe irrelevant prime ideal. Anideal iscalled irrelevant ifsome positive power of miscontained intheideal. Inparticular,aprimary ideal qisirrelevant ifand only ifmbelongs toq.Note thatbytheHilbert nullstellensatz, thecondition that some power ofmiscontained inaisequivalent with thecondition that the onlyzero ofa(in some algebraically closed field containing k)isthetrivial zero. Proposition 6.7. Let abeahomogeneous ideal. (a)Ifaisirrelevant, then x(n, a)=0fornsufficiently large. (b)Ingeneral, there isanexpressiona=q1n ...nqsasareduced primary decomposition such that allqiarehomogeneous. (c)Ifanirrelevant primary ideal occurs inthedecomposition, letbbethe intersection ofallother primary ideals. Then x(n, a)=x(n, b) forall nsufficiently large. Proof. For(a),byassumptionwehave An=anfor nsufficiently large,so theassertion (a)isobvious. We leave (b) asanexercise. As to(c), say qsis irrelevant, and letb=q1n . . .nqs-l' ByProposition 6.5, wehave x(n, b+qs)=x(n, b)+x(n, qs)-x(n, a). But b+qsisirrelevant, so(c)follows from (a), thus concluding theproof. X,6 THE HILBERT POLYNOMIAL 437 We now want toseethat foranyhomogeneous ideal athefunction fsuch that f(n)=x(n, a) satisfies theconditions ofLemma 6.4(b). First, weobserve thatifwechange theground field from ktoanalgebraically closed field Kcontaining k,and we letAK=K[X o,. . .,XN],OK=Ka, then dimk An=dimK AK,nand dimk an=dimK aK,n. Hence we can assume that kisalgebraically closed. Second, weshall need ageometric notion, that ofdimension. Let Vbe a variety over k,sayaffine, with generic point (x)=(Xb. . .,XN). We define its dimension tobethetranscendence degree ofk(x) over k.For aprojective variety, defined byahomogeneous prime ideal p,wedefine itsdimension tobethe dimension ofthehomogeneous variety defined bypminus 1. We now need thefollowing lemma. Lemma 6.8. Let V,Wbevarieties over afield k. ffV:J Wand dim V=dim W,then V=W. Proof. Say V,Ware inaffine space AN. LetPvand Pwbetherespective prime ideals ofVand Wink[X]. Then wehave acanonical homomorphism k[X]/pv=k[x] k[y]=k[X]/pw from theaffine coordinate ring ofVonto theaffine coordinate ring ofW.Ifthe transcendence degree ofk(x) isthe same asthatofk(y), and sayYl'. . .,Yrform atranscendence basis ofk(y) over k,then Xl'. . .,Xrisatranscendence basis ofk(x) over k,thehomomorphism k[x] k[y] induces anisomorphism k[XI'...' xr] k[Yb. .".,Yr]' and hence anisomorphismonthefinite extension k[x] tok[y],asdesired. Theorem 6.9. Let °be ahomogeneous ideal inA.Let rbethemaximum dimension oftheirreducible components ofthealgebraic space inprojective space defined bya.Then there exists apolynomial PEQ[T] ofdegree<r, such thatP(Z) CZ,and such that P(n)=x(n, a) forall nsufficiently large. 438 NOETHERIAN RINGS AND MODULES X,6 Proof. ByProposition 6.7(c), wemayassume that noprimary component intheprimary decomposition ofaisirrelevant. LetZbethealgebraic space of zeros of ainprojective space. We mayassume kalgebraically closed asnoted previously. Then there exists ahomogeneous polynomial LEk[X] ofdegree1 (alinear form) which does notlieinanyoftheprime ideals belonging tothe primary ideals inthegiven decomposition. Inparticular, Lisnot adivisor of zero mod a.Then thecomponents ofthealgebraic space ofzeros of a+(L) must have dimension<r-1.Byinduction and Theorem 6.6, weconclude that thedifference x(n, a)-x(n-1,a) satisfies theconditions ofLemma 6.4(b), which concludes theproof. The polynomial inTheorem 6.9 iscalled theHilbert polynomial ofthe ideal a. Remark. The above results giveanintroduction forHartshorne's [Ha77], Chapter I,especially 7.IfZisnotempty, andifwewrite nr x(n, a)=c,+lower terms,r. then c>0and ccan beinterpretedasthedegree ofZ,oringeometric terms, thenumber ofpoints ofintersection ofZwith asufficiently general linear variety ofcomplementary dimension (counting thepoints with certain multiplicities). Forexplanations anddetails, see[Ha77],Chapter I,Proposition 7 .6andTheorem 7.7; van derWaerden [vdW 29]which does the same thing formultihomogeneous polynomial ideals; [La58], referred toattheend ofChapter VIII, 2; and the papers [MaW 85], [Ph86], making thelink with van der Waerden some six decades before. [AtM 69] [Ha77] [MaW 85] [Mat 80] [Ph86] [vdW 29]Bibliography M. ATIYAH and I.MACDoNALD, Introduction tocommutative algebra, Addison-Wesley, 1969 R.HARTSHORNE, Algebraic Geometry, Springer Verlag, 1977 D,MASSER and G,WOSTHOLZ, Zero estimates ongroup varieties II,Invent, Math. 80(1985), pp,233-267 H, MATSUMURA, Commutative algebra, Second Edition, Benjamin- Cummings, 1980 P,PHILIPPON, Lemmes dezeros dans lesgroupes algebriques commutatifs, Bull. Soc, Math. France 114(1986), pp.355-383 B.L.VAN DER WAERDEN, OnHilbert's function, series ofcomposition of ideals and ageneralization ofthetheorem ofBezout, Proc. R.Soc. Amster- dam 31(1929), pp,749-770 X,7 INDECOMPOSABLE MODULES 439 7. INDECOMPOSABLE MODULES Let Abe aring, notnecessarily commutative, and EanA-module. We say that EisArtinian ifEsatisfies thedescending chain condition onsub- modules, that isasequence E1::JE2::JE3... must stabilize: there exists aninteger Nsuch that ifn>Nthen En=En +1. Example1.Ifkisafield, Aisak-algebra, and Eisafinite-dimensional vector space over kwhich isalso anA-module, then EisArtinian aswell as Noetherian. Example 2. Let Abeacommutative Noetherian local ring with maximal ideal m,and letqbeanm-primary ideal. Then forevery positive integer n, Alqn isArtinian. Indeed, Alqn has aJordan-Holder filtration inwhich each factor isafinite dimensional vector space over thefield Aim, and isamodule offinite length. SeeProposition 7.2. Conversely, suppose that Aisalocal ring which isboth Noetherian and Artinian. Let mbethemaximal ideal. Then there exists some positive integer nsuch that mn=O.Indeed, thedescending sequence mnstabilizes, and Nakayama's lemma implies our assertion. Itthen also follows that every primary ideal isnilpotent. Aswith Noetherian rings and modules, itiseasy toverify thefollowing statements: Proposition 7.1. Let Abearing, and let o-+E' -+E-+E" -+0 beanexact sequence ofA-modules. Then EisArtinian ifandonlyifE'and E"areArtinian. Weleave theproof tothereader. Theproof isthe same asintheNoetherian case, reversing theinclusion relations between modules. Proposition 7.2. Amodule Ehas afinite simple filtration ifandonlyifE isboth Noetherian and Artinian. Proof. Asimple module isgenerated byoneelement, and soisNoetherian. Since itcontains noproper submodule =t=0,itisalso Artinian. Proposition 7.2 isthen immediate from Proposition7.1. Amodule Eiscalled decomposable ifEcan bewritten asadirect sum E=E1(f)E 2 440 NOETHERIAN RINGS AND MODULES X,7 with E1=FEand E2=FE.Otherwise, Eiscalled indecomposable. IfEis decomposable asabove, let e1betheprojection onthe first factor, and e2=1-e1theprojection onthesecond factor. Then ebe2areidempotents such that e1=F1, e2=F1, e1+e2=1and e1e2=e2el=O. Conversely, ifsuch idempotents exist inEnd(E) for some module E,then Eis decomposable, and eiistheprojectiononthesubmodule eiE. Let u:E-+Ebeanendomorphism ofsome module E.We can form the descending sequence 1m u::>1m u2::J1m u3::J. . . IfEisArtinian, this sequence stabilizes, and wehave 1m un =1mun+1forallsufficiently largen. Wecall this submodule uOO(E), or1m UOO . Similarly, wehave anascending sequence Ker ucKer u2cKer u3c... which stabilizes ifEisNoetherian, and inthis case wewrite Ker UOO=Ker un for nsufficiently large. Proposition 7.3. (Fitting's Lemma). Assume that EisNoetherian and Artinian. Let uEEnd(E). Then Ehas adirect sumdecomposition E=1m UOO Et>Ker UOO . Furthermore, therestriction ofu to1m UOOisanautomorphism, and therestric- tionofutoKer UOOisnilpotent. Proof. Choose nsuch that 1m UOO=1munand Ker UOO=Ker un. We have 1m UOOnKer UOO={O}, forifxlies intheintersection, then x=un(y) for some yEE,and then o=un(x)=u2n(y). SoyEKer u2n=Ker un,whence x=un(y)=O. Secondly, letxEE.Then for some yEun(E) wehave un(x)=un(y). X,7 INDECOMPOSABLE MODULES 441 Then we can write x=x-Un(y) +Un(y), which shows that E=1m UOO+Ker UOO .Combined with thefirst step ofthe proof, this shows that Eisadirect sum asstated. The final assertion isimmediate, since therestriction ofuto1m UOOissur- jective, and itskernel is0bythefirst part oftheproof. The restriction ofuto Ker UOOisnilpotent because Ker UOO=Ker un.This concludes theproof ofthe proposition. We now generalize thenotion ofalocal ring toanon-commutative ring. Aring Aiscalled local ifthe setofnon-units isatwo-sided ideal. Proposition 7.4. LetEbeanindecomposable module over thering A.Assume ENoetherian and Artinian. Any endomorphism ofEiseither nilpotent oran automorphism. Furthermore End(E) islocal. Proof. ByFitting's lemma, weknow that foranyendomorphism u,we have E=1m UOOorE=Ker uoo .So wehave toprove that End(E) islocal. Let ube anendomorphism which isnot aunit, so uisnilpotent. For any endomorphismvitfollows that uvand vuarenotsurjective orinjective respec- tively, soarenotautomorphisms. Let U1, U2beendomorphisms which arenot units. We have toshowU1+U2isnot aunit. Ifitisaunit inEnd(E), let Vi=Ui(U1 +U2)-1. Then V1+V2=1.Furthermore, V1=1-V2isinvertible bythegeometric series sinceV2isnilpotent. But v1isnot aunit bythefirstpart oftheproof, contradiction. This concludes theproof. Theorem 7.5. (Krull-Remak-Schmidt). Let E=F0be amodule which is both Noetherian andArtinian. Then Eisafinite direct sumofindecomposable modules. Up toapermutation, theindecomposable components insuch a direct sum areuniquely determined uptoisomorphism. Proof. The existence ofadirect sum decomposition into indecomposable modules follows from theArtinian condition. Iffirst E=E1Et>E2,then either E1,E2areindecomposable, and we aredone; or, say, E1isdecomposable. Repeating theargument,we seethat wecannot continue thisdecomposition indefinitely without contradicting theArtinian assumption. There remains toprove uniqueness. Suppose E=E1Et>...Et>Er=F1Et>...Et>Fs where EbF..jareindecomposable. We have toshow that r=sand after some permutation, EiFi.Let eibetheprojection ofEonEi,and letUjbethe projection ofEonFj,relative totheabove direct sum decompositions. Let: v.=e1u, and w.=u.e1 J J J J. 442 NOETHERIAN RINGS AND MODULES X,7 ThenLuj=idEimplies that s LvjwjlEI=idE.. j==1 ByProposition 7.4,End(E 1)islocal, and therefore someVjWjisanautomor- phism ofE1.After renumbering,wemay assume that V1W1isanautomorphism ofE1.We claim that VIand WIinduce isomorphisms between EIandF., This follows from alemma. Lemma 7.6. Let M,Nbemodules, and assume Nindecomposable. Let u:M Nand v:N Mbesuch that vuisanautomorphism. Then u,v areisomorphisms. Proof. Let e=U(VU)-1V.Then e2=eisanidempotent, lying inEnd(N), and therefore equal to0or 1since Nisassumed indecomposable. But e=F0 because idM=F0and o=FidM=id=(vu)-IVU(VU)-1VU . So e=idN.Then uisinjective because vuisanautomorphism;visinjective because e=idNisinjective; uissurjective because e=idN;and vissurjective because vuisanautomorphism. This concludes theproof ofthelemma. Returning tothetheorem, wenow seethat E=F1ffi(E2ffi... (f)Er). Indeed, e1induces anisomorphism from F1toE1,and since thekernel ofe1 isE2ffi...ffiEritfollows that F1n(E2ffi...ffiEr)=O. Butalso, F1=E1(mod E2ffi...ffiEr),soEisthe sum ofF1and E2ffi...ffiEr, whence Eisthedirect sum, asclaimed. But then ElF 1 F2ffi... ffiFs E2E9." ffiEr' Theproof isthen completed byinduction. Weapply thepreceding results toacommutative ring A.We note that an idempotent inAasaring isthe same thingasanidempotent asanelement of End(A), viewing Aasmodule over itself. Furthermore End(A) A.Therefore, we.find thespecialcases: Theorem 7.7. Let AbeaNoetherian and Artinian commutative ring. X,Ex EXERCISES 443 (i)IfAisindecomposable asaring, then Aislocal. (ii)Ingeneral, Aisadirect product oflocal rings, which areArtinian and Noetherian. Another way ofderiving this theorem will begiven intheexercises. EXERCISES 1,Let Abe acommutative ring. Let Mbe amodule, and N asubmodule. Let N =QIn."nQrbe aprimary decomposition ofN. LetQi=QJN. Show that o=<21n',.nQrisaprimary decomposition of0inMIN. State and prove the converse, 2,Let pbeaprime ideal, and a,bideals ofA,Ifabcp,show that acporbcp. 3.Let qbe aprimary ideal. Let a,bbeideals, and assume abcq.Assume that bis finitely generated, Show that acqorthere exists some positive integernsuch that b"cq, 4,LetAbeNoetherian, and letqbeap-primary ideal. Show that there exists some n> 1 such that p"cq, 5,Let Abeanarbitrary commutative ring and letSbe amultiplicative subset. Let p be aprime ideal and letqbe ap-primary ideal. Then pintersects Sifandonly ifq intersects S.Furthermore, ifqdoes not intersect S,then S-lqisS-l p-primary in S-I A, 6.Ifaisanideal ofA,letas=S-la,IfqJs:A S-I Aisthecanonical map, abbreviate lfJs1(as)byasnA,even though qJsisnotinjective. Show that there isabijection between theprime ideals ofAwhich donotintersect Sand theprime ideals ofS-1A, givenby p ps and ps PsnA=p. Prove asimilar statement forprimary ideals instead ofprime ideals. 7.Let a=q1n..,nqrbe areduced primary decomposition ofanideal. Assume that qI'.,,,q;donot intersect S,butthat qjintersects Sforj>i.Show that as=qls n.. ,nqiS isareduced primary decomposition ofas, 8.Let Abe alocal ring. Show that anyidempotent #-0inAisnecessarily theunit element. (Anidempotent isanelement eEAsuch that e2=e.) 9,Let AbeanArtinian commutative ring. Prove: (a)Allprime ideals aremaximal. [Hint: Given aprime ideal p,letxEA,x(p)=o. Consider thedescending chain (x)::>(x2)::>(x3)::>'.'.] 444 NOETHERIAN RINGS AND MODULES X,Ex (b)There ISonlyafinite number ofpnme, ormaximal, Ideals. [Hint: Among all finite Intersections ofmaximal Ideals, pickaminimal one,] (c)The Ideal Nofnilpotent elements InAisnilpotent, that ISthere eXists apositive Integer kuch that N" =(0).[Hillt: Letkbesuch that N"=N" J.Let a=N". Let bbeaminimal ideal i=0such that bai=o.Then bIpnnclpal and ba=b.] (d) AISNoetherian. (e)There eXists anIntegerrsuch that A =nA/n{ where theproductIStaken over allmaximal ideals. (f)We have A=nAp' where again theproductIStaken over allprime ideals p. 10,LetA,Bbelocal nngs with maximal ideals mA,mB,respectively. Letf:A Bbea homomorphism. We say thatfISlocal Iff-l(m B)=mA,Suppose this isthe case, Assume A,BNoetherian, and assume that: 1.A/nt A B/Ut HISanIsomorphism, 2.mAmH/ntISsurjective: 3.BISafinite A-module, viaf. Prove thatrISsurjective, [Hint: Apply Nakayama tWice.] For anideal a,recall from Chapter IX,5that ?1(a)isthe setofprimes containing a. 11,Let Abeacommutative ring and ManA-module. Define thesupport ofMby supp(M)={pEspec(A):Mpi=O}. IfMisfinite over A,show thatsupp(M)=?1(ann(M», where ann(M) istheannihilator ofMinA,that isthe setofelements aEAsuch that aM=o. 12. Let Abe aNoetherian ring and Mafinite A-module. Let/beanideal ofAsuch that supp(M) C?1(/). Then /nM=0for some n>0, 13, Let Abeany commutative ring, andM,Nmodules over A.IfMisfinitely presented, and Sisamultiplicative subset ofA,show that S-1HomA(M, N) Homs-lA(S-l M,S-IN), This isusually applied when AisNoetherian and Mfinitely generated, inwhich case Misalsofinitely presented since themodule ofrelations isasubmodule ofafinitely generated free module. 14.(a)Prove Proposition 6.7(b). (b) Prove that thedegree ofthepolynomial PinTheorem 6.9 isexactlyr. Locally constant dimensions 15.LetAbe aNoetherian local ring. LetEbe afinite A-module. Assume that Ahas no nilpotent elements, For each pnme Ideal pofA,letk(p) betheresidue class field, If dlm,,(p) Ep/pEpisconstant forallp,show that EISfree, [Hint: LetXl',,.,XrEAbe X,Ex EXERCISES 445 such that theresidue classes mod themaximal ideal form abasis forE/mEover k(m). Wegetasurjective homomorphism ArE-+O. LetJbethekernel. Show that JpCmpA;forallpsoJcpforallpand J=0.] 16. Let AbeaNoetherian local ring without nilpotent elements. Letf: E-+Fbeahomo- morphism ofA-modules, and suppose E,Farefinite free. For each prime pofAlet I(p):Ep/pE p-+Fp/pF p bethecorresponding k(p)-homomorphism, where k(p)=Ap/pApistheresidue class field atp,Assume that dimk(p)1mhp) isconstant. (a) Prove thatFilm fand 1mfarefree, and that there isanisomorphism F 1mfe>(Film f). [Hint: Use Exercise 15.] (b)Prove that Kerjisfree and E (Kerf)e>(1mf).[Hint: Use that finite projective isfree.] The next exercises dependonthenotion ofacomplex, which wehave notyetformally defined. A(finite) complex Eisasequence ofhomomorphisms ofmodules dO d1dn o EO EI...En 0 andhomorphisms di:Ei Ei+1such that di+10di=0foralli.Thus Im(di)CKer(di+1). Thehomology Hiofthecomplex isdefined tobe Hi=Ker(di+1)/Im(di). Bydefinition, HO=EOand Hn=En/lm(dn),You maywant tolook atthefirst section ofChapter XX, because all we use here isthebasic notion, and thefollowing property, which youcaneasily prove. LetE,Fbetwocomplexes. Byahomomorphismf: E F we mean asequence ofhomomorphisms fi:Ei Fi making thediagram commutative foralli: dk )Ei+1 [!i+IEi /;[ d}Fi+l Fi Show that such ahomomorphismf induces ahomomorphism H(f): H(E) H(F) onthe homology; that is,foreach iwehave aninduced homomorphism Hi(f): Hi(E) Hi(F). 446 NOETHERIAN RINGS AND MODULES X,Ex The following exercises areinspired from applications toalgebraic geometry,asfor instance inHartshorne, Aigebraic Geometry, Chapter III,Theorem 12.8. See alsoChapter XXI, 1tosee how one can construct complexes such asthose considered inthe next exercises inorder tocompute thehomology with respect toless tractable complexes. Reduction ofacomplex mod p 17.Let 0 KO K1 ...Kn0be acomplex offinite free modules over alocal Noetherian ring Awithout nilpotentelements. For each prime pofAand module E, letE(p)=Ep/pEp,andsimilarly letK(p) bethecomplex localized and reduced mod p. For agiven integer i,assume that dimk(p) Hi(K(p» isconstant, where Hiisthei-thhomology ofthereduced complex. Show thatHi(K) isfree and that wehave anatural isomorphism Hi(K)(p) Hi(K(p». [Hint: First write d:p)forthemap induced bydionKi(p). Write dimk(p)Kerd:p)=dimk(p) Ki(p)-dimk(p)1md:p). Then show that thedimensionsdimk(p)1md:p)anddimk(p)1md:;>1must beconstant. Then apply Exercise 12.] Comparison ofhomology atthespecial point 18.Let AbeaNoetherian local ring. LetKbeafinite complex, asfollows: o KO -+...-+Kn 0, such that Kiisfinite free forall;, For some index iassume that Hi(K)(m) Hi(K(m)) issurjective. Prove: (a)This map isanisomorphism. (b)Thefollowing exact sequences split: o Ker diKi-+1mdi0 o 1mdiKi+1 (c)Every term inthese sequences isfree. 19. Let AbeaNoetherian local ring. LetKbeacomplexasintheprevious exercise. For some iassume that Hi(K)(m) Hi(K(m» issurjective (orequivalently isanisomorphism bytheprevious exercise). Prove that X,Ex EXERCISES 447 thefollowing conditions areequivalent: (a)Hi-1(K)(m)-...Hi-1(K(m» issurjective. (b)Hi-l(K)(m) Hi-1(K(m» isanisomorphism. (c)Hi(K)isfree. [Hint: Lift bases until youareblue intheface.] (d)Ifthese conditions hold, then each oneofthetwo inclusions 1mdi-1CKer dicKi splits, and each one ofthese modules isfree, Reducing mod myields the corresponding inclusions 1md:;'/cKerd:m)cKi(m), and induce theisomorphismoncohomologyasstated in(b).[Hint: Apply thepreceding exercise,] CHAPTER XI Real Fields 1. ORDERED FIELDS LetKbeafield. Anordering ofKisasubset PofKhaving thefollowing properties: ORD 1.Given xEK,wehave either xEP,orx=0,or-xEP,and these three possibilitiesaremutually exclusive. Inother words, Kisthe disjoint union ofP,{O}, and-P. ORD 2.Ifx,YEP, then x+Yand xyEP. We shall also saythat Kisordered byP,and wecall Pthe setofpositive elements. Let usassume that Kisordered byP.Since 1=F0and 1=12=(_1)2 we seethat 1EP.ByORD 2,itfollows that 1+...+1EP,whence Khas characteristic o.IfxEP,and x=F0,then xx-1= 1EPimplies that x-1EP. Let x,YEK.Wedefine x<Y(or Y>x)tomean that y-xEP.Ifx<0 wesaythat xisnegative. This means that-xispositive. One verifies trivially theusual relations forinequalities, forinstance: x<y and y<z impliesx<z, x<y and z>O impliesxz<yz, implies1 1 x<y and x,y>0-<-. yx Wedefine x<ytomean x<yorx=y.Then x<yand y<ximplyx=y. IfKisordered and xEK,x=F0,then x2ispositivebecause x2=(-X)2 and either xEPor-xEP.Thus asum ofsquares ispositive,orO. Let Ebeafield. Then aproduct ofsums ofsquares inEisasumofsquares. Ifa,bEE are sums ofsquares and b=F0then a/b isasumofsquares. 449 450 REAL FIELDS XI,1 The first assertion isobvious, and the second also, from theexpression a/b=ab{b-1)2. IfEhas characteristic =F2,and-1isasum ofsquares inE,then every element aEEisasum ofsquares, because 4a={1+a)2-(1-a)2. IfKisafield with anordering P,and Fisasubfield, then obviously, PnF defines anordering ofF,which iscalled the induced ordering. We observe that our two axioms ORD 1and ORD 2apply toaring. If Aisanordered ring, with 1=F0,then clearly Acannot have divisors of0,and one can extend theordering ofAtothequotient field intheobvious way: A faction iscalled positive ifitcan bewritten intheform a/bwith a,bEAand a,b>O.One verifies trivially that this defines anorderingonthequotient field. Example. We define anordering onthepolynomial ringR[t] over the real numbers. Apolynomial f{t)=antn+...+ao with an=F0isdefined tobepositive ifan>o.The two axioms arethen trivially verified. We note that t>aforall aER.Thus tisinfinitely large with respect toR.The existence ofinfinitely large (orinfinitely small) elements inanordered field isthemain aspect inwhich such afield differs from asubfield ofthereal numbers. We shall now make some comment onthisbehavior, i.e.theexistence of infinitely large elements. LetKbeanordered field and letFbeasubfield with theinduced ordering. Asusual, weput IxI=xifx>0andIxI= -xifx<O.We saythat anelement rxinKisinfinitely large over FifIrxI>xforallxEF.We saythat itisinfinitely small over FifO <Irxl<IxlforallxEF,x=FO.We seethat rxisinfinitely large ifandonly ifrx-1isinfinitely small. We saythat Kisarchimedean over FifK has noelements which areinfinitely large over F.Anintermediate field Fh K::JFl::JF,ismaximal archimedean over FinKifitisarchimedean over F, and noother intermediate field containing Flisarchimedean over F.IfFlis archimedean over Fand F2isarchimedean over Flthen F2isarchimedean over F.Hence byZorn's lemma there always exists amaximal archimedean subfield FlofKover F.We say that Fismaximal archimedean inKifitismaximal archimedean over itself inK. LetKbeanordered field and Fasubfield. Let 0bethe setofelements ofK which arenotinfinitely large over F.Then itisclear that 0isaring, and that for anyrxEK, wehave rxorrx-1Eo.Hence 0iswhat iscalled avaluation ring, containing F.Let mbetheideal ofall rxEKwhich areinfinitely small over F. Then mistheunique maximal ideal of0,because any element in0which isnot inmhas aninverse ino.Wecall 0thevaluation ring determined bytheordering ofKIF. XI,2 REAL FIELDS 451 Pfoposition1.1. Let Kbeanordered field and Fasubfield.Let 0bethe valuation ring determined bytheordering ofKIF, and letmbeitsmaximal ideal. Then o/m isarealfield. Proof Otherwise, wecould write -1=Lrxf+a with rxiE0and aEm. SinceLrxfispositive and aisinfinitely small, such a relation isclearly impossible. 2. REAL FIELDS Afield Kissaid tobefealif-1isnot asum ofsquares inK.Afield Kis said tobefeal closed ifitisreal, andifanyalgebraic extension ofKwhich isreal must beequal toK.Inother words, Kismaximal with respect totheproperty ofreality inanalgebraic closure. Proposition 2.1. LetKbearealfield. (i)IfaEK,thenK(fi)orK( )isreal.Ifaisasumofsquares inK, thenKCv') isreal.IfK() isnotreal, then -a isasumofsquares inK. (ii)Iff isanirreducible polynomial ofodddegreeninK[X] andifrxisaroot off,then K(rx) isreal. Proof Let aEK.Ifaisasquare inK,thenK(fi)=Kand hence isrealby assumption. Assume that aisnot asquare inK.IfK(fi)isnotreal, then there exist bi,CiEKsuch that -1 =L(bi+cifi)2 =L(bf+2cibifi+cfa). Sincefiisofdegree2over K,itfollows that -1=Lbf+aLcf. Ifaisasum ofsquares inK,thisyieldsacontradiction. Inany case, we con- clude that 1+Lbf-a= L2C.I isaquotient ofsums ofsquares, andbyaprevious remark, that-aisasum of squares. HenceK() isreal, thereby provingourfirst assertion. 452 REAL FIELDS XI,2 Astothesecond, suppose K(lL) isnot real. Then we canwrite -1=Lgi(lL)2 with polynomials giinK[X] ofdegree<n-1.There exists apolynomial h inK[X] such that -1=Lgi(X)2 +h(X)f(X). The sum ofgi(X)2 has even degree, and thisdegree must be>0,otherwise -1 isasum ofsquares inK.This degree is<2n-2.Sincefhasodd degree n,it follows that hhasodd degree<n-2.IfPisaroot ofhthen we seethat-1 isasum ofsquares inK(P). Since deg h<degf,ourproofisfinished by induction. LetKbeareal field. Byareal closure weshall mean areal closed field L which isalgebraic over K. Theorem 2.2. LetKbearealfield. Then there exists areal closure ofK. IfRisreal closed, then Rhas aunique ordering. The positive elements are thesquares ofR.Every positive element isasquare, and every polynomial of odddegree inR[X] has aroot inR.Wehave Ra=R(v=I ). Proof ByZorn's lemma, our field Kiscontained insome real closed field algebraic over K.Now letRbeareal closed field. LetPbethe setofnon-zero elements ofRwhich are sums ofsquares. Then Pisclosed under addition and multiplication. ByProposition 2.1,every element ofPisasquare inR,andgiven aER,a=F0,wemust have aEPor-aEP.Thus Pdefines anordering. Again byProposition 2.1, every polynomial ofodddegreeover Rhas aroot inR.Our assertion follows byExample 5ofChapter VI,2. Corollary 2.3. LetKbearealfield and aanelement ofKwhich isnot a sumofsquares. Then there exists anordering ofKinwhich aisnegative. Proof The field K( )isreal byProposition1.1and hence has an orderingasasubfield ofareal closure. Inthisordering,-a>0and hence ais negative. Proposition 2.4. Let Rbeafield such that R=FRabut Ra =R(J=l).Then Risreal and hence real closed. Proof Let Pbethe setofelements ofRwhich are squares and =Fo.We contend that Pisanordering ofR.Let aER,a=FO.Suppose that aisnot a square inR.Let lLbearoot ofX2-a=O.Then R(lL)=R(J=l ),and hence there exist c,dERsuch that lL=C+dJ=l.Then lL2=c2+2cdJ=l-d2 . XI,2 REAL FIELDS 453 Since 1,J=1arelinearly independent over R,itfollows that c=0(because aftR2),and hence-aisasquare. Weshall now prove that asum ofsquares isasquare. Forsimplicity,write i=J=1.Since R(i) isalgebraically closed, given a,bERwe canfind c,dER such that (c+di)2=a+bi.Then a=c2-d2and b=2cd. Hence a2+b2=(c2+d2)2, aswas tobeshown. IfaER,a=F0,then notboth aand -a can besquares inR.Hence Pisan ordering and ourproposition isproved. Theorem 2.5. Let Rbeareal closed field, andf(X) apolynomial inR[X]. Let a,bERand assume thatf(a) <0andf(b) >O.Then there exists c between aand bsuch thatf(c)=o. Proof Since R(Fi )isalgebraically closed, itfollows thatfsplits into a product ofirreducible factors ofdegree1or2.IfX2+rxX+pisirreducible (rx,PER) then itisasum ofsquares, namely (X+r+(p-). and wemust have 4p> rx2since our factor isassumed irreducible. Hence the change ofsign offmust bedue tothechange ofsign ofalinear factor, which is trivially verified tobearoot lying between aand b. Lemma 2.6. LetKbeasubfield ofanorderedfield E.Let rxEEbealgebraic over K,and arootofthepolynomial f(X)=xn+an-lxn-1+...+aO with coefficients inK.ThenIrxI<1+ Ian-1I+...+ IaoI. Proof IfIrxI<1,theassertion isobvious. IfIrxI>1,weexpress IrxInin terms ofthe terms oflower degree, divide by IrxIn- 1,and getaproof for our lemma. Note that thelemma implies that anelement which isalgebraicover an ordered field cannot beinfinitely large with respect tothat field. Letf(X) be apolynomial with coefficients inareal closed field R,and assume thatfhas nomultiple roots. Let u<vbeelements ofR.ByaSturm sequence forfover theinterval [u,v]weshall mean asequence ofpolynomials S={f=fo,f'=fh...,fm} having thefollowing properties: 454 REAL FIELDS XI,92 ST 1.The lastpolynomial fmisanon-zero constant. ST2. There isnopointxE[u,v]such thatfj(x)=fi+I(X)=0forany value 0<j<m-1. ST3.IfxE[u,v]andf;{x)=0for somej=1,...,m-1,thenfj-l(X) and.fj+ I(x)have opposite signs. ST4. We have.fj(u);/=0andf;{v)=F0forallj=0,...,m. For any xE[u,v]which isnot aroot ofanypolynomial/;wedenote by J.iiS(x) thenumber ofsign changes inthesequence {f(x), fl(x),...,fm(x)}, and call Ws(x) thevariation ofsigns inthesequence. Theorem 2.7. (Sturm's Theorem). The number ofrootsoff between uand v isequal toWs(u)-Ws(v)for any Sturm sequence S. Proof. We observe that if(XI<(X2<...< exristheordered sequence of roots ofthepolynomials fjin[u,v]U=0,...,m-1),then Ws(x) isconstant ontheopen intervals between these roots, byTheorem 2.5. Hence itwill suffice toprove that ifthere ispreciselyone element exsuch that u< (X<vand exisa root ofsome fj,then Ws(u)-Ws(v)=1ifexisaroot off,and 0otherwise. Suppose that (Xisaroot ofsomejj,for 1<j<m-1.Then.fj_ 1(C(),jj+ 1(ex) have opposite signs byST3,and these signs donotchange when wereplaceex byuorv.Hence thevariation ofsigns in {fj-1(u),h{u),jj+I(u)}and {.fj-1(v),jj(v),fj+I(v)} isthe same, namely equal to2.Ifexisnot aroot off,weconclude that WS(u)=Ws(v). If exisaroot off,thenf(u) andf(v) have opposite signs, butf'(u) andf'(v) have the same sign, namely, thesign off'(ex). Hence inthis case, WS(u)=Ws(v) +1. This proves our theorem. Itiseasy toconstruct aSturm sequence for apolynomial without multiple roots. We usetheEuclidean algorithm, writing f=glf'-f2, 12=g2fl-f3, fm-2=gm-Ifm-I-fm' XI,2 REAL FIELDS 455 usingf'=fl.Since f,f'have nocommon factor, thelast term ofthis sequence isnon-zero constant. The other properties ofaSturm sequencearetrivially verified, because iftwo successive polynomials ofthe sequence have acom- mon zero, then they must allbe0,contradicting thefact that thelast one isnot. Corollary2.8. LetKbe anordered field,fanirreducible polynomial of degree>lover K.The number ofroots offintworeal closures ofKinducing thegiven ordering onKisthe same. Proof We can take vsufficiently large positive and usufficiently large negative inKsothat allroots offand allroots ofthepolynomials intheSturm sequence liebetween uand v,using Lemma 2.6. Then J.iiS(u)-Ws(v)isthe total number ofroots offinany real closure ofKinducing thegiven ordering. Theorem 2.9. LetKbeanordered field, and letR,R'bereal closures ofK, whose orderings induce thegiven ordering onK. Then there exists aunique isomorphismU:R-+R'over K,and thisisomorphism isorder-preserving. Proof Wefirst show that givenafinite subextension EofRover K,there exists anembedding ofEinto R'over K.Let E=K(ll), and let f(X)=Irr(ll, K,X). Then f(ll)=0and thecorollary ofSturm's Theorem (Corollary 2.8) shows that fhas aroot PinR'.Thus there exists anisomorphism ofK(ll) onK(P) over K, mappingIIonp. Let 1l1,...,llnbethedistinct roots offinR,and letPI'...,Pmbethedistinct roots offinR'.Say 111<...<llnintheordering ofR, PI<...<Pm intheordering ofR'. Wecontend that m =nand that wecan select anembeddingUofK(llf,...,lln) into R'such that Ulli=Pifori=1,...,n.Indeed, letYibeanelement ofR such that Y?=lli+1-llifor i=1,...,n-1 and letE1=K(1l 1,..., lln,Yf, ...,Yn-l)' Bywhat wehave seen, there exists anembeddingUofE1into R',and then Ulli+1-Ulliisasquare inR'. Hence U1l1<...<Ulln. This proves that m>n.Bysymmetry, itfollows that m=n.Furthermore, thecondition that (Jlli=Pifor i=1,. . .,ndetermines the effect of Uon 456 REAL FIELDS XI,2 K(rx1,...,rx n).We contend that (Jisorder-preserving. 'Let YEK(rxb'..'rxn) and 0<y.LetYERbesuch that y2=y.There exists anembedding of K(rx1,...,rxn,Yl'...,Yn-1,y) into R'over Kwhich must induce aonK(rxb...,rxn)and issuch that ayisa square, hence >0,ascontended. Using Zorn's lemma, itisnow clear that wegetanisomorphism ofRonto R' over K. This isomorphism isorder-preserving because itmaps squareson squares, thereby provingour theorem. Proposition 2.10. LetKbeanorderedfield, K'anextension such that there is norelation n -1 ='a.rxI I i=1 with aiEK,ai>0,and rxiEK'.LetLbethefield obtained from K'byadjoining thesquare roots ofallpositive elements ofK.Then Lisreal. Proof Ifnot, there exists arelation oftype n -1 ='a.rxI I i=1 with aiEK,ai>0,and rxiEL.(We can take ai=1.)Let rbethe smallest integer schthat we canwrite such arelation with rxiinasubfield ofL,oftype K'(A,...,A) with bjEK,bj>O.Write rx,=x.+Y'fbI I I'VUr with Xi'YiEK'(A,...,).Then -1 =Lai(x i+Yifir)2 =Lai(xf +2XiYifir +yfb r). Byhypothesis, firisnotinK'(bb...,).Hence -1 =Laixf +Laibryf, contradicting theminimality ofr. Theorem 2.11. LetKbeanorderedfield. There exists areal closure RofK inducing thegiven ordering onK. XI,3 REAL ZEROS AND HOMOMORPHISMS 457 Proof Take K' =KinProposition 2.10. Then Lisreal, and iscontained inareal closure. Our assertion isclear. Corollary 2.12. LetKbeanorderedfield, andK'anextension field. Inorder that there exist anordering onK'inducing thegiven ordering ofK,itis necessary andsufficient that there isnorelation oftype n -1 = a.rx?-I I i= 1 with aiEK,ai>0,andrxiEK'. Proof. Ifthere isnosuch relation, then Proposition 2.10 states that Lis contained inarealclosure, whose ordering induces anordering onK',and the given orderingonK, asdesired. The converse isclear. Example. LetQ8bethefield ofalgebraic numbers. One sees atonce that Qadmits onlyoneordering, theordinary one. Hence any two real closures ofQ inQ8areisomorphic, bymeans ofaunique isomorphism. The realclosures ofQ inQ8 areprecisely those subfields ofQ8which areoffinite degree under Q8. LetKbeafinite real extension ofQ,contained inQ8. Anelement rxofKisa sum ofsquares inKifandonly ifevery conjugate of rxinthereal numbers is positive,orequivalently, ifandonly ifevery conjugate of rxinone ofthereal closures ofQinQ8ispositive. Note. Thetheory developed inthis andthepreceding section isdue toArtin- Schreier. See thebibliography attheendofthechapter. 3. REAL ZEROS AND HOMOMORPHISMS Just aswedevelopedatheory ofextension ofhomomorphisms into an algebraically closed field, and Hilbert's Nullstellensatz for zeros inanalge- braically closed field, wewish todevelop thetheory forvalues inareal closed field. One ofthemain theorems isthefollowing: Theorem 3.1. Let kbe afield, K =k(x b...,xn)afinitely generated extension. Assume that Kisordered. LetRkbeareal closure ofkinducing the same ordering onkasK.Then there exists ahomomorphism qJ:k[x 1,...,Xn]-+Rk over k. 458 REAL FIELDS XI,3 Asapplications ofTheorem 3.1, onegets: Corollary 3.2. Notation being asinthetheorem, letY1,...,YmEk[x] and assume Y1<Y2<...<Ym isthegiven ordering ofK.Then one can choose qJsuch that qJY 1<...<qJYm. Proof Let YiEK8besuch that yf=Yi+ 1-Yi. Then K(Y1'...' Yn-1) has anordering inducing thegiven orderingonK.Weapply thetheorem tothe rIng k[-1 -1] Xl'...,xn,Y1,..., Ym-bYh...,Ym-1. Corollary 3.3. (Artin). Let kbearealfield admitting only oneordering. Letf(X 1,...,Xn)Ek(X) bearational function having theproperty thatfor all(a)=(ab...,an)ERin) such thatf(a) isdefined, wehavef(a)>O.Then j(X) isasumofsquares ink(X). Proof Assume that our conclusion isfalse. ByCorollary 2.3, there exists anordering ofk(X) inwhichfisnegative. Apply Corollary 3.2tothering k[X1,...,Xn'h(X)-1] where h(X) isapolynomial denominator forf(X). We can find ahomo- morphism qJofthisring into Rk(inducing theidentity onk)such thatqJ(f) <O. But qJ(f)=f(qJXl'...,qJXn). contradiction. Weletai=qJ(X i)toconclude theproof. Corollary 3.3 was aHilbert problem. Theproof which weshall describe for Theorem 3.1differs from Artin' sproof ofthecorollary inseveral technical aspects. Weshall first seehow one can reduce Theorem 3.1tothe case when Khas transcendence degree1over k,and kisreal closed. Lemma 3.4. Let Rbeareal closed field and letRobeasubfield which is algebraically closed inR(i.e. such that every element ofRnot inRoistran- scendental over Ro). Then Roisreal closed. Proof Letf{X) beanirreducible polynomial over Ro. Itsplits inRinto linear andquadratic factors. Itscoefficients inRarealgebraic over Ro, and hence must lieinRo. Hence f(X) islinear itself, orquadratic irreducible already over Ro. Bytheintermediate value theorem, wemay assume thatfispositive XI,3 REAL ZEROS AND HOMOMORPHISMS 459 definite, i.e.f(a) >0forallaERo. Without loss ofgenerality,wemayassume thatf(X)=X2+b2for some bERo.Any root ofthispolynomialwillbring J=1with itand therefore theonly algebraic extension ofRoisRo(J=1 ). This proves that Roisreal closed. Let RKbeareal closure ofKinducing thegiven orderingonK.LetRobe thealgebraic closure ofkinRK.Bythelemma, Roisreal closed. Weconsider thefield Ro(x 1,...,xn).Ifwe can prove our theorem forthe ringRo[x 1,...,xn],and find ahomomorphism t/J:Ro[x b...,xn] Ro, then welet (J:Ro RKbeanisomorphism over k(itexists byTheorem 2.9), and weletqJ=(J0t/Jtosolve ourproblem over k.This reduces our theorem tothe case when kisreal closed. Next, letFbeanintermediate field, K =>F =>k,such that Kisoftran- scendence degree lover F.Again letRKbeareal closure ofKpreserving the ordering, and letRFbethereal closure ofFcontained inRK.Ifweknow our theorem forextensions ofdimension 1,then we can find ahomomorphism t/J:RF[x b...,xn] RF. We note that the field k(t/JXb...' t/Jxn) has transcendence degree<n-1, and isreal, because itiscontained inRF.Thus we arereduced inductively to the case when Khasdimension 1,and aswe saw above, when kisreal closed. One caninterpretour statement geometricallyasfollows. We can write K =R(x, y)with xtranscendental over R,and(x,y)satisfying some irreducible polynomial f(X, Y)=0inR[X, Y]. What weessentially want toprove isthat there areinfinitely many points onthe curve f(X, Y)=0,with coordinates lying inR,i.e.infinitely many realpoints. The main idea isthat wefind some point (a,b)ER(2) such thatf(a, b)=0 but D2f(a, b)=Fo.We can then use theintermediate value theorem. We see thatf(a, b+h)changes signashchanges from asmall positive toasmall negative element ofR.Ifwe take a'ERclose toa,thenf(a', b+h)also changes sign forsmall h,and hence f(a', Y)has azero inRforalla'sufficiently close toa. Inthis way wegetinfinitely manyzeros. Tofind ourpoint, weconsider thepolynomial f(x, Y)asapolynomial inone variable Ywith coefficients inR(x). Without loss ofgenerality wemay assume that thispolynomial hasleading coefficient 1.We construct aSturm sequence forthispolynomial, say {f(x, Y),fl(x,Y),...,fm(x, Y)}. Let d=degf.Ifwedenote byA(x)=(ad- 1(x),...,ao(x)) thecoefficients of f(x, Y),then from theEuclidean alogrithm, we seethat thecoefficients ofthe 460 REAL FIELDS XI,3 polynomials intheSturm sequence can beexpressedasrational functions {Gv(A(x))} interms ofad-1(x),..., ao(x). Let v(x)=1+ad- 1(x)+. ..+ao(x) +s, where sisapositive integer, and thesignsareselected sothat each term inthis sum givesapositive contribution. We letu(x)= -v(x), and select ssothat neither unor visaroot ofanypolynomial intheSturm sequence forf.Now weneed alemma. Lemma 3.5. Let Rbeareal closed field, and{hi(x)}afinite setofrational functions inone variable with coefficients inR.Suppose therational field R(x) ordered insome way,sothat each hi(x) has asign attached toit.Then there exist infinitely many special values cofxinRsuch thathi(c) isdefined and hasthe same signashi(x),for alli. Proof. Considering the numerators and denominators ofthe rational functions, wemay assume without loss ofgenerality that thehiarepolynomials. We then write hi(x)=an(x-A)np(x), where thefirstproduct isextended over allroots AofhiinR,and thesecond product isover positive definite quadratic factors over R.For anyER,p() is positive. Itsuffices therefore toshow that thesigns of(x-A.)can bepreserved forallAbysubstituting infinitely many values afor x.Weorder allvalues ofA. and ofxand obtain ...<,1.1<X<A.2<... where possibly ,1.1or,1.2isomitted ifxislarger orsmaller than any A.Any value aofxinRselected between A.1and ,1.2will then satisfy therequirements ofour lemma. Toapply thelemma totheexistence ofourpoint,welettherational functions {h 1(x)} consist ofallcoefficients ad-l (x),. ..,ao(x), allrational functions Gv(A(x)), and allvalues jj(x, u(x)), Jj(x, v(x)) whose variation insigns satisfied Sturm's theorem. Wethen findinfinitely many special values aofxinRwhich preserve thesigns ofthese rational functions. Then thepolynomialsf(a, Y)have roots inR,and forallbut afinite number ofa,these roots have multiplicity1. Itisthen amatter ofsimple technique toseethat forallbut afinite number of points onthe curve, theelements xl'...,Xnlieinthelocal ring ofthehomo- morphism R[x, y] Rmapping (x,y)on(a,b)such thatf(a, b)=0but XI,Ex EXERCISES 461 D2f(a, b)#o.(Cf. forinstance theexample attheendof4,Chapter XII, and Exercise 18ofthat chapter.) One could also give direct proofshere. Inthis way, weobtain homomorphisms R[xl'...,Xn]-+R, thereby proving Theorem 3.1. Theorem 3.6. Let kbe areal field, K =k(x 1,...,xn,y)=k(x,y) a finitely generated extension such that Xl'.. .,Xnarealgebraically independent over k,and yisalgebraic over k(x). Letf(X, Y)betheirreducible polynomial ink[X, Y]such thatf(x, y)=O.Let Rbeareal closed field containing k, and assume that there exists (a,b)ER(n+1)such thatf(a, b)=0but Dn+lf(a, b)#o. Then Kisreal. Proof Let tI'...,tnbealgebraically independent over R.Inductively,we can putanorderingonR(t b. ..,tn)such that each tiisinfinitely small with respect toR,(cf.theexample in1).Let R'be areal closure ofR(t1,...,tn) preserving theordering. Let Ui=ai+tiforeach i=1,...,n.Thenf(u, b+h) changes sign forsmall hpositive and negative inR,and hence f(u, Y)has a root inR',sayv.Sincefisirreducible, theisomorphism ofk(x) onk(u)sending Xion Uiextends toanembedding ofk(x,y)into R',and hence Kisreal, aswas to beshown. Inthelanguage ofalgebraic geometry, Theorems 3.1and 3.6state that the function field ofavariety over areal field kisrealifandonly ifthevariety has a simple point insome real closure ofk. EXERCISES I.Let rxbealgebraic over Qand assume that Q(rx) isareal field, Prove that exisasum of squaresInQ(rx) ifandonly ifforevery embedding(JofQ(ex) inRwehave Gex>O. 2.Let Fbe afinite extension ofQ. Let qJ:F Qbe aQ-linear functional such that cp(x2)>0forallxEF,x#O.Let rxEF,ex#O.Ifcp(exx2)>0forallxEF,show that exis asum ofsquares inF,and that Fistotally real, i,e.every embedding ofFinthecomplex numbers iscontained inthereal numbers. [Hint: Use thefact that the trace givesan identification ofFwith itsdual space over Q,and usetheapproximation theorem of Chapter XII, 91.] 462 REAL FIELDS XI,Ex 3.Let (1<t<(Jbearealinterval, and letf(t) bearealpolynomial which ispositiveonthis Interval. Show thatf(t)can bewritten intheform C(LQ:+L(t-cx)Q; +L({J-t)Qi) where Q2denotes asquare, and c>0,Hint: Split thepolynomial, and usetheIdentity: (t-CX)2({J-t)+(t-cx)({J-t)2(t-cx)({J-t)= {J. -(1 Remark. The above seemingly innocuous result isakey step indeveloping the spectral theorem forbounded hermitian operatorsonHilbert space, See theappendix of[La72] and also [La85], 4,Show that thefield ofreal numbers hasonly theidentity automorphism. [Hint: Show that anautomorphism preserves theordering.] Real places For the next exercises, cf,Krull [Kr32] andLang [La53], These exercises form a connected sequence, and solutions will befound in[La53]. 5.Let Kbe afield and suppose that there exists areal place ofK;that is, aplace cp with values inareal field L.Show that Kisreal. 6,LetKbeanordered real field and letFbe asubfield which ismaximal archimedean inK,Show that thecanonical place ofKwith respecttoFisalgebraicover F(i.e. if0isthevaluation ring ofelements ofKwhich are notinfinitely largeover F,and misitsmaximal ideal, then o/m isalgebraic over F). 7,LetKbe anordered field and letFbe asubfield which ismaximal archimedean in K,LetK'bethereal closure ofK(preserving theordering), and letF'bethereal closure ofFcontained inK'.Let cpbethecanonical place ofK'with respecttoF'. Show that cp(K') isF'-valued, and that therestriction ofcptoKisequivalenttothe canonical place ofKover F, 8.Define areal field Ktobequadratically closed ifforall aEKeither or lies inK.Theordering ofaquadratically closed real field Kisthen uniquely determined, and soisthereal closure ofsuch afield, uptoanisomorphismover K. Suppose that Kisquadratically closed. Let Fbe asubfield ofKand suppose that Fismaximal archimedean inK.Letcpbe aplace ofKover F,with values ina field which isalgebraic over F.Show thatcpisequivalent tothecanonical place of Kover F, 9.LetKbeaquadratically closed real field. Letcpbearealplace ofK,taking itsvalues inareal closed field R.LetFbeamaximal subfield ofKsuch thatcpisanisomorphism onF,andidentify Fwith cp(F). Show that such Fexists and ismaximal archimedean inK,Show that theimage ofcpisalgebraic over F,and thatcpisinduced bythe canonical place ofKover F. 10. LetKbe areal field and let cpbe arealplace ofK,taking itsvalues inareal closed field R.Show that there isanextension ofcptoanR-valued place ofareal closure ofK,[Hint: first extend cptoaquadratic closure ofK,Then use Exercise 5.] XI,Ex EXERCISES 463 11. LetKCKICK2bereal closed fields. Suppose that Kismaximal archimedean in KIand KIismaximal archimedean inK2.Show that Kismaximal archimedean in K2. 12. Let Kbe areal closed field. Show that there exists areal closed field Rcontaining Kandhaving arbitrarily large transcendence degree over K,and such thatKismaximal archimedean inR. 13, Let Rbe areal closed field. LetII'. , .,Irbehomogeneous polynomialsofodd degrees innvariables over R.Ifn>r,show that these polynomialshave anon- trivial common zero inR.(Comments: Iftheforms aregeneric (inthe sense ofChapter IX), and n=r+1,itisatheorem ofBezout that inthealgebraic closure Rathe forms have exactly dI...dmcommon zeros, where diisthedegree offi.You may assume this toprove theresult asstated. Ifyou want toseethis worked out, see [La53], Theorem 15.Compare with Exercise 3ofChapter IX.) Bibliography [Ar24] E.ARTIN, Kennzeichnung desKorpers derreellen algebraischen Zahlen, Abh. Math. Sem. Hansischen Univ. 3(1924), pp.319-323 [Ar27] E.ARTIN, Uber dieZerlegung definiter Funktionen inQuadrate, Abh. Math. Sem, Hansischen Univ. 5(1927), pp. 100-115 [ArS 27] E.ARTIN and E.SCHREIER, Algebraische Konstruktion reeller Korper,Abh. Math. Sem. Hansischen Univ, 5(1927), pp.85-99 [Kr32] W.KRULL, Allgemeine Bewertungstheorie, J.reine angew. Math. (1932), pp. 169-196 [La53] S.LANG, The theory ofrealplaces, Ann. Math. 57No.2 (1953), pp.378- 391 [La72] S.LANG, Differential manifolds, Addison- Wesley, 1972; reprinted bySpringer Verlag, 1985; superceded by[La99a]. [La85] S.LANG, Real andfunctional analysis. Third edition, Springer Verlag, 1993 [La99a] S.LANG, Fundamentals ofDifferential Geometry, Springer Verlag, 1999 CHAPTER XII Absolute Values 1. DEFINITIONS, DEPENDENCE, AND INDEPENDENCE LetKbeafield. Anabsolute value vonKisareal-valued function x1-+IxIv onKsatisfying thefollowing three properties: AV1. We haveIxIv>0forallxEK,andIxIv=0ifandonly ifx=o. AV 2. For allx,YEK,wehave Ixylv=Ixlvlylv. AV3. For allx,yEK,wehaveIx+yIv<IxIv+ IyIv. Ifinstead ofAV3theabsolute value satisfies thestronger condition AV 4.Ix+ylv<max(lxl v,Iylv} then weshall saythat itisavaluation, orthat itisnon-archimedean. The absolute value which issuch thatIxIv= 1forallx=F0iscalled trivial. Weshall writeIxIinstead ofIxIvifwedeal withjustone fixed absolute value. Wealso refer tovastheabsolute value. Anabsolute value ofKdefines ametric. The distance between two elements x,yofKinthis metric isIx-yI.Thus anabsolute value defines atopologyon K.Two absolute values arecalled dependent ifthey define the same topology. Ifthey donot,theyarecalled independent. We observe that 111=112 1=1(-1)21=1112whence 111=1-11=1. Also, I-xl=IxlforallxEK,andlx-II=Ixl-1for xi=o. 465 466 ABSOLUTE VALUES XII,1 Proposition 1.1. LetIIIandIbbenon-trivial absolute values onafield K. They aredependent ifandonlyiftherelation Ixll < 1 implies Ixb<1.Ifthey aredependent, then there exists anumber A.>0 such thatIxl l=IxlforallxEK. Proof Ifthe two absolute values aredependent, then our condition is satisfied, because the setofxEKsuch that/X/I < 1isthe same asthe setsuch that lim xn=0for n--+00.Conversely,assume thecondition satisfied. Then IxII>1implies Ixb> 1sinceIx-I 11<1.Byhypothesis, there exists an element xoEK such that Ixoll >1.Let a=Ixoll and b=Ixol2. Let A.=log b . loga Let xEK,x;/=O.ThenIxII=IXoIfor some number (1.Ifm,nareintegers such that m/n > (1and n>0,wehave Ix11>IXo17/n whence Ixn/xII<1, and thus Ixn/x 12<1. This implies thatIxb<IXoI/n. Hence Ixb<Ixol2. Similarly, one proves the reverse inequality, and thus one gets Ixb=IXol2 forall xEK, x;/=O.The assertion oftheproposition isnow obvious, i.e. Ixb=Ixl. We shall givesome examples ofabsolute values. Consider first therational numbers. We have theordinary absolute value such thatImI=mforanypositive integerm. For each prime number p,wehave thep-adic absolute valuevp,defined bythe formula Iprm/n Ip=l/pr XII,1 DEFINITIONS, DEPENDENCE, AND INDEPENDENCE 467 where risaninteger, and m,nareintegers ;/=0,notdivisible byp.One sees at once that thep-adic absolute value isnon-archimedean. One cangiveasimilar definition ofavaluation forany field Kwhich isthe quotient field ofaprincipal ring. For instance, letK =k(t) where kisafield and tisavariable over k.Wehave avaluation vpforeach irreducible polynomial p(t)ink[t], defined asfortherational numbers, butthere isnoway ofnormalizing itinanatural way. Thus weselect anumber cwith 0<c< 1and forany rational function PJ/gwhere f,garepolynomials notdivisible byp,wedefine Ipl/gl p=cr. The various choices ofthe constant cgive rise todependent valuations. Any subfield ofthecomplex numbers (orreal numbers) has anabsolute value, induced bytheordinary absolute value onthecomplex numbers. Weshall seelater how toobtain absolute values oncertain fields byembedding them into others which arealready endowed with natural absolute values. Suppose that wehave anabsolute value on afield which isbounded onthe prime ring (i.e. theintegers Zifthecharacteristic is0,ortheintegers mod pif thecharacteristic isp).Then theabsolute value isnecessarily non-archimedean. Proof For any elements x,yand anypositive integer n,wehave I(x+yrl<L(:)xvyn-v<nCmax(lxl, Iyl)n, Taking n-th roots andlettingngotoinfinity provesourassertion. We note that this isalways the case incharacteristic> 0because theprime ring isfinite! Iftheabsolute value isarchimedean, then werefer thereader toany other book inwhich there isadiscussion ofabsolute values for aproof ofthefactthat itisdependent ontheordinary absolute value. This fact isessentially useless (and isnever used inthesequel), because wealways start with aconcretely given setofabsolute values onfields which interest us. InProposition 1.1 wederived astrong condition ondependent absolute values. Weshall now derive acondition onindependentones. Theorem 1.2. (Approximation Theorem). (Artin-Whaples). Let Kbe afield andI11'...' IIsnon-trivial pairwise independent absolute values onK. LetXI'...,Xsbeelements ofK,andl>O.Then there exists xEKsuch that IX-Xili<l foralli. 468 ABSOLUTE VALUES XII,92 Proof Consider first two ofour absolute values, say VIand V2.Byhypo- thesis we canfind rxEKsuch thatIrxII<1andIrxIs>1.Similarly,we canfind PEKsuchthat/PII>land/Pis <1.Puty=P/rx.Thenlyll>landlyls <1. We shall now prove that there exists zEKsuch thatIzII> 1andIzIj<1 forj=2,...,s.We prove thisbyinduction, the case s=2having just been proved. Supposewehave found zEKsatisfying Izil >1and Izlj<1 forj=2,...,s-1. IfIzIs<1then theelement znyforlargenwillsatisfy ourrequirements. IfIzIs>1,then thesequence zn tn= 1+zn tends to1atVIandvs,and tendstoOatvjU=2,..., s-1).Forlarge n,itisthen clear that tnYsatisfies ourrequirements. Using theelement zthat wehave justconstructed, we seethat thesequence zn/(1+zn)tends to 1at VIand to0atvjforj=2,...,s.For each i=1,...,s we cantherefore construct anelement Ziwhich isvery close to 1atViand very close to0atVj(j=1=i).The element x=ZIX I+...+ZsXs then satisfies therequirement ofthetheorem. 2. COMPLETIONS LetKbeafield with anon-trivial absolute value v,which will remain fixed throughout this section. One canthen define intheusual manner thenotion ofa Cauchy sequence. Itisasequence {xn} ofelements inKsuch that, given l>0, there exists aninteger Nsuch that foralln,m>Nwehave IXn-XmI<l. We saythat Kiscomplete ifevery Cauchy sequence converges. Proposition 2.1. There exists apair(Kv, i)consisting ofafield Kv,complete under anabsolute value, and anembedding i:K-+Kvsuch that theabsolute value onKisinduced bythatofKv(i.e. IxIv=Iix/for xEK),and such thatiK isdense inKv.If(K, i')isanother such pair, then there exists aunique XII,2 COMPLETIONS 469 isomorphism qJ:Kv-+Kpreserving the absolute values, and making the following diagram commutative: KvqJK' ) v\1 K Proof The uniqueness isobvious. One proves theexistence inthewell- known manner, which weshall now recall briefly, leaving thedetails tothereader. TheCauchy sequences form aring, addition andmultiplication being taken componentwise. One defines anull sequence tobeasequence {xn} such that lim Xn=O.The n-CX) null sequences form anideal inthering ofCauchy sequences, and infactform a maximal ideal. (IfaCauchy sequence isnot anull sequence, then itstays away from 0forall nsufficiently large, and one canthen take theinverse ofalmost all itsterms. Uptoafinite number ofterms, onethen getsagainaCauchy sequence.) The residue class field ofCauchy sequences modulo null sequences isthe field Kv.We embedKinKv"on thediagonal", i.e.send xEKonthesequence (x,x,x,...). Weextend theabsolute value ofKtoKvbycontinuity. If{xn} isaCauchy sequence, representinganelement inKv,wedefineII=limIXnI.Itiseasily proved that thisyields anabsolute value (independent ofthechoice ofrepre- sentative sequence {xn} for),and thisabsolute value induces thegiven one onK. Finally, one proves that Kviscomplete. Let{n} be a,Cauchy sequence in Kv. For each n,we canfind anelement XnEKsuch thatIn-XnI<1/n. Then one verifies immediately that {xn} isaCauchy sequence inK .We let beits limit inKv.Byathree-l argument, one sees that{n} converges to,thus proving thecompleteness. Apair (Kv, i)asinProposition 2.1may becalled acompletion ofK.The standard pair obtained bythepreceding construction could becalled the completion ofK. LetKhave anon-trivial archimedean absolute value v.Ifoneknows that the restriction ofvtotherationals isdependent ontheordinary absolute value, then thecompletion Kvisacomplete field, containing thecompletion ofQasa closed subfield, i.e.containing thereal numbers Rasaclosed subfield. Itwill be worthwhile tostate thetheorem ofGelfand-Mazur concerning thestructure of such fields. First wedefine thenotion ofnormed vector space. LetKbeafield with anon-trivial absolute value, and letEbeavector space over K.Byanorm onE(compatible with theabsolute value ofK) weshall mean afunction -+IIofEinto thereal numbers such that: NO 1.II>0forall EE,and =0ifandonly if =o. 470 ABSOLUTE VALUES XII,2 NO 2. For allxEKandgEE wehave Ixgl-Ixllgl. NO 3.If,'EEthenI+'I<II+I'I. Two normsIIIandI12arecalled equivalent ifthere exist numbers CI'C2>0 such that forall EEwehave CIIII<Ib<C2111. Suppose that Eisfinite dimensional, and letWb...,Wnbe abasis ofE over K.Ifwewrite anelement =XIWI +...+XnW n interms ofthis basis, with XiEK,then we can define anorm byputting II=maxlxd. i The three properties defininganorm aretrivially satisfied. Proposition 2.2. LetKbeacomplete field under anon-trivial absolute value, and letEbe afinite-dimensional space over K. Then any two norms onE (compatible with thegiven absolute value onK) areequivalent. Proof Weshall first prove that thetopologyonEisthat ofaproduct space, i.e.ifWI'...,Wnisabasis ofEover K,then asequence (v) =x(v)w +...+x(v)wIInn' Xv) EKI , isaCauchy sequence inEonly ifeach one ofthe nsequences xv) isaCauchy sequence inK.Wedothisbyinduction on n.Itisobvious for n=1.Assume n>2.Weconsider asequenceasabove, and without lossofgenerality,wemay assume that itconverges toO.(Ifnecessary, consider (v)-(Jl)for v,J.l-+00.) We must then show that the sequences ofthecoefficients converge to0also. Ifthis isnot the case, then there exists anumber a>0such' that wehave for some j,sayj=1, Ix<y) I>a forarbitrarily largev.Thus for asubsequence of(v),(V)/x\v) converges to0,and we canwrite (V)xcv) xcv)2 n M-WI=MW2+...+MWn. XI XI XI We lety/(v)betheright-hand side ofthisequation. Then thesubsequence y/(V) converges (according totheleft-hand side ofourequation). Byinduction, we XII,2 COMPLETIONS 471 conclude that itscoefficients interms ofW2,...,Wnalso converge inK,say to Y2,...,Yn.Taking thelimit, weget W1=Y2W2+...+Ynwn, contradicting thelinear independence ofthe Wi. Wemust finallyseethat two norms inducing the same topologyareequivalent. LetI11andIIzbethese norms. There exists anumber C>0such that forany EEwehave I11<Cimplies IIz<1. Let aEKbesuch that 0<IaI<1.For every EEthere exists aunique integer ssuch that Cia I<las11<c. HenceIaSIz<1whence weget atonce 11z<C-1Ial-1111. The other inequality follows bysymmetry, with asimilar constant. Theorem 2.3. (Gelfand-Mazur). LetAbeacommutative algebra over the real numbers, and assume that Acontains anelement jsuch thatj2=-1. Let C=R+Rj. Assume that Aisnormed (as avector space over R), and that Ixyl<Ixllyl forallx,yEA. GivenXoEA,Xo=t=0,there exists anelement cECsuch thatXo-cisnotinvertible inA. Proof (Tornheim). Assume that Xo-zisinvertible for all ZEC. Consider themapping f:C-.Adefined by f(z)=(xo-Z)-1. Itiseasily verified (asusual) that taking inverses isacontinuous operation. Hencefiscontinuous, and for z=F0wehave j(z)=Z-1(XOZ-1_1)-1=! (1 ). zXo_1 z From this we seethatf(z) approaches 0when zgoes toinfinity (inC). Hence the mapz1---+If(z) Iisacontinuous map ofCinto thereal numbers >0,isbounded, and issmall outside some large circle. Hence ithas amaximum, sayM. LetD 472 ABSOLUTE VALUES XII,2 bethe setofelements zECsuch thatIf(z)1=M.Then Disnotempty; Dis bounded and closed. We shall prove that Disopen, hence acontradiction. LetCobeapoint ofD,which, after atranslation, wemay assume tobethe origin. Weshall seethat ifrisreal> 0andsmall, then allpointsonthecircle of radius rlieinD.Indeed, consider the sum 1n1 S(n)=-L knk=1Xo-OJr where OJisaprimitive n-th root ofunity. Taking formally thelogarithmic n derivative ofxn-rn=n(X-wkr) shows that k=l nXn-1 Xn-rnn1-LXk' k=l -OJr and hence, dividing byn,andbyXn- 1,andsubstituting XoforX,weobtain 1 S(n)= (/t-1. Xo-rrXo Ifrissmall (say Ir/xo I<1),then we seethat limIS(n)1= =M. n-+ 00 Xo Suppose that there exists acomplex number A.ofabsolute value 1such that 1 <M. Xo-ILr Then there exists aninterval ontheunit circle near A.,and there exists l>0such that forallroots ofunity' lying inthisinterval, wehave 1 y<M-c Xo-r (This istrue bycontinuity.) Let ustake nvery large. Let bnbethenumber of n-th roots ofunity lying inourinterval. Then bn/n isapproximately equal tothe length oftheinterval (times 2n): We can express S(n) asasum 1 [1 1 ]S(n)= nLIxk+Ln k' o-wrXo-wr XII,2 COMPLETIONS 473 thefirst sumLIbeing taken over those roots ofunity wklying inourinterval, and thesecond sum being taken over theothers. Each term inthesecond sum has norm <Mbecause Misamaximum. Hence weobtain theestimate 1 IS(n)1< -[ILl I+ILulJn 1< -(bn(M-l)+(n-bn)M)n bn<M--l. n This contradicts thefact that thelimit ofIS(n) Iisequal toM. Corollary 2.4. Let Kbe afield, which isanextension ofR,and has an absolute value extending theordinary absolute value onR.Then K =Ror K =C. Proof Assume first that Kcontains C.Then theassumption that Kisa field and Theorem 2.3imply that K =C. IfKdoes notcontain C,inother words, does notcontain asquare root of -1, weletL=K(j)wherej2=-1. Wedefine anorm onL(as anR-space) by putting Ix+yjI=IxI+ IyI for x,yEK.This clearly makes Linto anormed R-space. Furthermore, if z=x+yjand z'=x'+y'jareinL,then Izz'l=lxx'-yy'l +Ixy' +x'yl <lxx' I+Iyy'l +Ixy'l +Ix'yl <IxIIx'I+ IyIIy'I+ IxIIy'I+ Ix'IIyI <(IxI+ IyI)(Ix'I+ Iy'I) <IzIIz'I, and we can therefore apply Theorem 2.3again toconclude theproof. As animportant application ofProposition 2.2, wehave: Proposition 2.5. Let Kbecomplete with respect toanontrivial absolute value v.IfEisanyalgebraic extension ofK,then vhas aunique extension to E.IfEisfinite over K,then Eiscomplete. Proof Inthe archimedean case, the existence isobvious since wedeal with thereal andcomplex numbers. Inthenon-archimedean case, wepostpone 474 ABSOLUTE VALUES XII,2 theexistence proof toalater section. Ituses entirely different ideas from the present ones. Astouniqueness, wemayassume that Eisfinite over K.By Proposition 2.2, anextension ofvtoEdefines the same topologyasthe max norm obtained interms ofabasis asabove. Given aCauchy sequence (v)inE, (v)=Xv 1WI+.. .+XvnWn, the nsequences {Xvi} (i=1,..., n)must beCauchy sequences inKbythe definition ofthe max norm. If{xv;} converges toanelement ZiinK,then it isclear that the sequence (v) convergesto Z1W1+...+ZnWn.Hence Eis complete. Furthermore, since any two extensions of vtoEareequivalent, we canapply Proposition 1.1,and we seethat wemust have A.=1,since the extensions induce the same absolute value vonK.This proves what wewant. From theuniquenesswe can getanexplicit determination oftheabsolute value onanalgebraic extension ofK.Observe first thatifEisanormal extension ofK,and (Jisanautomorphism ofEover K,then thefunction X1---+I(JXI isanabsolute value onEextending that ofK.Hence wemust have l(Jxl=IxI forallxEE.IfEisalgebraic over K,and (Jisanembedding ofEover KinK8 , then the same conclusion remains valid, asone seesimmediately byembedding Einanormal extension ofK.Inparticular, ifrxisalgebraicover K,ofdegree n, andifrxb...,rxnareitsconjugates (counting multiplicities, equal tothedegree of inseparability), then allthe absolute valuesIrxiIareequal. Denoting byN the norm from K(rx) toK,we seethat IN(rx) I=IrxIn, andtaking then-th root, weget: Proposition 2.6. Let Kbecomplete with respect toanon-trivial absolute value. Let rxbealgebraic over K,and letNbethenormfrom K(rx) toK.Let n=[K(rx): K]. Then Irxl=IN(rx) 11/n . Inthespecial case ofthecomplex numbers over thereal numbers, we can write rx=a+biwith a,bER,and we seethat theformula ofProposition 2.6is ageneralization oftheformula fortheabsolute value ofacomplex number, rx=(a2+b2)1/2, since a2+b2isnone other than the norm of rxfrom CtoR. XII,2 COMPLETIONS 475 Comments and examples. The process ofcompletioniswidespread in mathematics. The first example occurs ingetting the real numbers from the rational numbers, with theadded property ofordering. Icarry this processout infull in[La90a], Chapter IX, 3.Inallotherexamples Iknow, theordering property does notintervene .We have seen examples ofcompletions offields in thischapter, especially with thep-adic absolute values which arefaraway from ordering thefield. But thereal numbers arenevertheless needed astherange of values ofabsolute values, ormore generally norms. Inanalysis,onecompletes various spaces with various norms. Let Vbe a vector spaceover thecomplex numbers, say. Formany applications,one must also deal with aseminorm, which satisfies the same conditions except that in NO 1werequire only that IIII::>O.We allowIIII=0even if*o. One may then form the spaceofCauchy sequences, thesubspace ofnull sequences, and thefactor space V.The seminorm can beextended toaseminorm onVbycontinuity, and this extension actually turns out tobe anorm. Itisa general fact that Visthen complete under this extension. ABanach space isa complete normed vector space. Example. Let Vbethevector space ofstep functions onR, astep function beingacomplex valued function which isafinite sum ofcharacteristic functions ofintervals (closed, open,orsemiclosed, i.e. the intervals mayor may not contain their endpoints).ForfEVwedefine theLl.seminorm by 11/11 I==JI/(x) Idx. R Thecompletion ofVwith respect tothis semi norm isdefined tobeLI(R). One then wants togetabetter idea ofwhat elements ofLI(R)look like. Itisasimple lemma thatgivenanLI-Cauchy sequence inV,andgivenE>0,there exists a subsequence which converges uniformly excepton asetofmeasure less than E. Thus elements ofLI(R) can beidentified with pointwise limits ofLI-Cauchy sequences inV.The reader will find details carried out in[La85]. Analystsuse other norms orseminorms, ofcourse, and other spaces, such asthespace ofCoofunctions onRwith compact support, and norms which may bound thederivatives. There isnoend tothepossible variations. Theorem 2.3andCorollary 2.4 arealso used inthetheory ofBanach algebras, representingacertain type ofBanach algebraasthealgebra ofcontinuous func- tions on acompact space, with theGelfand-Mazur andGelfand-Naimark theo- rems. Cf.[Ri60] and [Ru73]. Arithmetic example. Forp-adic Banach spaces inconnection with the number theoretic work ofDwork, see for instance Serre [Se62], oralso [La90b], Chapter 15. Inthis book welimit ourselves tocomplete fields andtheir finite extensions. 476 ABSOLUTE VALUES [La85] [La90a] [La90b] [Ri60] [Ru 73] [Se62]XII,3 Bibliography S.LANG, Real and Functional Analysis, Springer Verlag, 1993 S,LANG, Undergraduate Algebra, Second Edition, Springer Verlag, 1990 S,LANG, Cyclotomic Fields IandII,Springer Verlag 1990 (combined from thefirst editions, 1978 and 1980) C.RICKART, Banach Algebras, Van Nostrand (1960), Theorems 1.7.1 and 4.2.2. W,RUDIN, Functional Analysis, McGraw Hill (1973) Theorems 10.14 and 11.18, J.P.SERRE, Endomorphismes completement continus des espaces deBanach p-adiques, Pub. Math. IHES 12(1962), pp,69-85 3. FINITE EXTENSIONS Throughout this section weshall deal with afield Khavinganon-trivial absolute value v. Wewish todescribe how thisabsolute value extends tofinite extensions ofK. IfEisanextension ofKand wisanabsolute value onEextending v,then weshall write wiv. . IfweletKvbethecompletion, weknow that vcan beextended toKv, and then uniquely toitsalgebraic closure K. IfEisafinite extension ofK,oreven analgebraic one, then we can extend vtoEbyembedding EinKbyaniso- morphism over K,andtaking theinduced absolute value onE.Weshall now prove that every extension ofvcan beobtained inthis manner. Proposition 3.1. LetEbea.finite extension ofK.Let wbeanabsolute value onEextending v,and letEwbethecompletion. LetKwbetheclosure ofKin Ewandidentify EinEw. Then Ew=EKw (the composite .field). Proof We observe that Kw isacompletion ofK,and that thecomposite field EKwisalgebraic over Kwand therefore complete byProposition 2.5. Since itcontains E,itfollows that Eisdense init,and hence that Ew=EKw. Ifwestart with anembeddingu:E-+K(always assumed tobeover K), then weknow again byProposition 2.5that uE.Kviscomplete. Thus this construction and theconstruction ofthepropositionareessentially thesame, up toanisomorphism. Inthefuture, wetake theembedding point ofview. We must now determine when twoembeddings give usthe same absolute value onE. Given twoembeddings u,T:EK, weshall saythat theyareconjugate over Xvifthere exists anautomorphismAofKover Kvsuch that u=AT. We seethatactuallyAisdetermined byitseffect onTE, orTE.Kv. XII,3 FINITE EXTENSIONS 477 Proposition 3.2. Let Ebe analgebraic extension ofK. Two embeddings a,T:E-+Kgive rise tothe same absolute value onEifandonlyifthey are conjugate over Kv. Proof Suppose theyareconjugateover Kv. Then theuniqueness ofthe extension ofthe absolute value from KvtoKguarantees that theinduced absolute values onEareequal. Conversely, suppose this isthe case. Let A:TE -+aEbeanisomorphismover K. We shall prove that Aextends toan isomorphism ofTE.Kvonto aE.Kv over Kv. Since tEisdense intE.Kv, anelement xETE.Kvcan bewritten x=lim tXn with XnEE.Since theabsolute values induced by aand tonEcoincide, it follows that thesequence ATXn=aXnconverges toanelement ofaE.Kvwhich wedenote byAX. One then verifies immediately that AXisindependent ofthe particular sequence tXnused, and that themap A:tE.Kv-+aE.Kvisaniso- morphism, which clearly leaves Kvfixed. This proves ourproposition. Inview oftheprevious twopropositions, ifwisanextension ofvtoafinite extension EofK,then wemay identify Ewand acomposite extension EKv ofE and Kv.IfN =[E:K]isfinite, then weshall call Nw=[Ew: Kv] thelocal degree. Proposition 3.3. LetEbeafinite separable extension ofK,ofdegree N.Then N =LNw. wlv Proof We can write E=K{lL) for asingle element lL.Letf{X) beits irreducible polynomialover K.Then over Kv,wehave adecomposition f{X)=f1(X)...f,.{X) into irreducible factors h{X). They allappear with multiplicity1according to ourhypothesis ofseparability. Theembeddings ofEintoKcorrespondtothe maps oflLonto theroots oftheh.Two embeddingsareconjugate ifandonly if they maplLonto roots ofthe same polynomial h.Ontheother hand, itisclear that thelocal degree ineach case isprecisely thedegree ofh.This provesour proposition. Proposition 3.4. Let Ebeafinite extension ofK.Then L[Ew:Kv]<[E:K]. wlv 478 ABSOLUTE VALUES XII,3 IfEispurely inseparableover K,then there exists only oneabsolute value won Eextendingv. Proof Let usfirst prove thesecond statement. IfEispurely inseparable over K,andprisitsinseparable degree, then (:/f" EKforeveryrxinE.Hence vhas aunique extension toE.Consider now thegeneralcase ofafinite extension, and letF=EprK. Then Fisseparableover Kand Eispurely inseparableover F. Bythepreceding proposition, L[Fw: Kv]=[F:K], wlv and foreach w,wehave [Ew: Fw]<[E:F]. From this ourinequality inthe statement ofthepropositionisobvious. Whenever visanabsolute value onKsuch that foranyfinite extension EofK wehave [E:K]=L[Ew:Kv]weshall saythat viswell behaved. Supposewe wlv have atower offinite extensions, L::JE::JK.Let wrangeover theabsolute values ofEextending v,and uover those ofLextendingv.IfuIwthen Lu contains Ew. Thus wehave: L[Lu:Kv]=LL[Lu:Ew][Ew:Kv] ulv wlvulw =L[Ew:Kv]L[Lu:Ew] wlv ulw <L[Ew:Kv][L:E] wlv <[E:K][L:E]. From this weimmediatelyseethat ifviswell behaved, Efinite over K,and w extends vonE,then wiswell behaved (we must have anequality everywhere). Let Ebeafinite extension ofK.Letprbeitsinseparable degree. Werecall that the norm ofanelement rxEKisgiven bytheformula Ni(rx)=nurxP" (1 where uranges over alldistinct isomorphisms ofEover K(intoagiven algebraic closure ). Ifwisanabsolute value extendingvonE,then the norm from EwtoKvwill becalled thelocal norm. Replacing theabove product byasum, wegetthetrace, and thelocal trace. Weabbreviate the trace byTr. Proposition 3.8. Let Ebeafinite extension ofK,and assume that viswell XII,3 FINITE EXTENSIONS 479 behaved. Let rxEE.Then: N{rx)=nN:{rx) wlv Tri{rx)=LTri:{rx) wlv Proof Suppose first that E=K{rx), and letf{X) betheirreducible poly- nomial of rxover K.Ifwefactor f{X) into irreducible terms over Kv, then f{X)=fl(X)...f,.{X) where each/;(X) isirreducible, and the/;aredistinct because ofourhypothesis that viswell behaved. The norm Ni{rx) isequal to(_l)degftimes theconstant term off,andsimilarly foreach/;.Since theconstant term offisequal tothe productoftheconstant terms ofthe/;,wegetthefirst part oftheproposition. The statement forthetrace follows bylooking atthepenultimate coefficient off and each/;. IfEisnotequal toK{rx), then wesimplyusethetransitivity ofthe norm and trace. We leave thedetails tothereader. One canalso argue directly onthe embeddings. Let U1,...,Umbethedistinct embeddings ofEintoK over K,and letprbetheinseparable degree ofE over K.The inseparable degree ofuE.Kvover Kvforanyuisatmost equal toproIfweseparateub...,Urninto distinct conjugacy classes over Kv, then from ourhypothesis that viswell behaved, weconclude atonce that the inseparable degree ofuiE.Kv over Kv must beequal topralso, foreach i. Thus theformula giving the norm asaproductover conjugates with multi- plicity prbreaks upinto aproduct offactors corresponding totheconjugacy classes over Kv. Taking into account Proposition 2.6, wehave: Proposition 3.6. LetKhave awell-behaved absolute value v.Let Ebea finite extension ofK,and rxEE.Let Nw=[Ew: Kv] foreach absolute value wonEextendingv.Then nIrxlw=IN{rx)lv. wlv 480 ABSOLUTE VALUES XII,4 4. VALUATIONS Inthissection, weshall obtain, among other things, theexistence theorem concerning thepossibility ofextending non-archimedean absolute values to algebraic extensions. Weintroduce first ageneralization ofthenotion ofnon- archimedean absolute value. Letrbeamultiplicative commutative group. Weshall saythat anordering isdefined inrifwe aregivenasubset Sofrclosed under multiplication such thatristhedisjoint union ofS,theunit element 1,and the setS-Iconsisting of allinverses ofelements ofS. If,f3Erwedefine <f3tomean f3-IES.We have <1ifandonly if ES.One easily verifies thefollowing properties oftherelation <: 1.For,f3Erwehave <f3,or =f3,orf3<,and these possibilities aremutually exclusive. 2. <f3implies y<f3yforany yEr. 3. <f3and f3<yimplies <y. (Conversely,arelation satisfying thethree properties gives rise toasubset S consisting ofallelements <1.However, wedon't need this fact inthesequel.) Itisconvenient toattach toanordered group formally anextra element 0, such thatO =0,and 0< forall Er.The ordered group isthen analogous tothemultiplicative group ofpositivereaIs,except that there may benon- archimedean ordering. If Erand nisaninteger =F0,such thatn =1,then =1.This follows at once from theassumption that Sisclosed under multiplication and does not contain 1.Inparticular, themap1---+nisinjective. LetKbeafield. Byavaluation ofKweshall mean amapx1-+IxIofKinto anordered group r,together with the extra element 0,such that: VALl. Ix I=oifandonly ifx=0. VAL 2.Ixyl=Ixllylforallx,YEK. VAL 3.Ix+yl<max(lxl, Iyl). We seethat avaluation gives rise toahomomorphism ofthemultiplicative group K*intor.The valuation iscalled trivial ifitmaps K* on 1.Ifthemap giving thevaluation isnotsurjective, then itsimage isanordered subgroup ofr, andbytaking itsrestriction tothisimage,weobtain avaluation onto anordered group, called thevalue group. Weshall denote valuations also byv.IfVI' V2are two valuations ofK,we shall saythattheyareequivalent ifthere exists anorder-preserving isomorphism Aoftheimage ofVlonto theimage ofV2such that Ixb=Alxl l XII,4 VALUATIONS 481 forallxEK.(We agree that A.(O)=0.) Valuations have additional properties, like absolute values. For instance, 111= 1because 111=1112 .Furthermore, I+xl=Ixl forallxEK.Proof 0bvious. Also, ifIxI<IyIthen Ix+yl=Iyl. To seethis, note that under ourhypothesis,wehave Iyl=Iy+x-xl<max(ly +xl,IxI)=Ix+yl<max(lxl, Iyl)=Iyl. Finally, inasum XI+...+Xn=0, atleast two elements ofthe sum have the same value. This isanimmediate consequence ofthepreceding remark. LetKbe afield. Asubring0ofKiscalled avaluation ring ifithas the property that forany xEKwehave xE0orx-1EO. We shall now seethat valuation rings give rise tovaluations. Let 0be a valuation ring ofKand letVbethegroup ofunits ofo.Wecontend that 0isa local ring. Indeed suppose that x,yE0are notunits. SayxjyEO. Then 1+xjy=(x+y)jyEo. Ifx+ywere aunit thenIjyE0,contradicting theassumption that yisnot aunit. Hence x+yisnot aunit. One seestrivially that for ZE0,zxisnot aunit. Hence thenonunits form anideal, which must therefore betheunique maximal ideal ofo. Let mbethemaximal ideal of0and letm* bethemultiplicative system of nonzero elements ofm.Then -1K*=m* uVum* isthedisjoint union ofm*, V,and m*- 1.The factor group K*jV can now be given anordering. IfxEK*, wedenote the coset xVby IxI.Weput 101=o. WedefineIxI<1(i.e. IxIES)ifandonly ifxEm*. Our setSisclearly closed under multiplication, and ifweletr=K*jVthenristhedisjoint union ofS, 1,S-I. Inthis wayweobtain avaluation ofK. We note that ifx,yEKand x,y=F0,then Ixl<lyl<=>lxjyl <1<=>xjYEm*. Conversely, givenavaluation ofKinto anordered groupwelet 0bethe subset ofKconsisting ofallxsuch thatIxI<1.Itfollows atonce from the 482 ABSOLUTE VALUES XII,4 axioms ofavaluation that 0isaring. IfIxI<1thenIX-II>1sothatX-I is not ino.IfIxI=1thenIX-II=1.We seethat 0isavaluation ring, whose maximal ideal consists ofthose elements xwithIxI<1and whose units consist ofthose elements xwithIxI=1.The reader willimmediately verify that there is abijection between valuation rings ofKandequivalence classes ofvaluations. The extension theorem forplacesand valuation rings inChapter VII now givesusimmediately theextension theorem forvaluations. Theorem 4.1. LetKbeasubfield ofafield L.Then avaluation onKhas an extension toavaluation onL. Proof. Let 0bethevaluation ring onKcorresponding tothegiven valua- tion. LetqJ:0-.o/m bethecanonical homomorphismontheresid ueclass field, and extend qJtoahomomorphism ofavaluation ring .0ofLasin3ofChapter VII. Let webethemaximal ideal ofsO.Since wen0contains mbut does not contain 1,itfollows that wen0=m.LetV'bethegroup ofunits of().Then V'nK=Visthegroup ofunits ofo.Hence wehave acanonical injection K*/V-.L*/V' which isimmediately verified tobeorder-preserving. Identifying K*/Vin L*/V' wehave obtained anextension ofourvaluation ofKtoavaluation ofL. Ofcourse, when wedeal with absolute values, werequire that thevalue group beasubgroup ofthemultiplicative reals. Thus wemust still prove something about thenature ofthevalue group L*/V', whenever Lisalgebraic over K. Proposition 4.2. LetLbeafinite extension ofK,ofdegreen.Let wbe a valuation ofLwith value group r'. Letrbethevalue group ofK. Then (r':r)<n. Proof. LetYh...,Yrbeelements ofLwhose values represent distinct cosets ofrinr'.Weshall prove that theYjarelinearly independent over K.In arelation alYl +...+arYr=0withajEK, aj=F0two terms must have the same value, saylaiyd=lajyjlwith i=Fj,and hence ly;I=lai-lajIIYjl. This contradicts theassumption that thevalues ofYi,Yj(i=Fj)represent distinct cosets ofrinr',and proves ourproposition. Corollary 4.3. There exists anintegere> 1such that the map y1---+ye induces aninjective homomorphism ofr'intor. Proof. Take etobetheindex (r':r). XII,4 VALUATIONS 483 Corollary 4.4. IfKisafield with avaluation vwhose value group isan ordered subgroup oftheordered group ofpositive real numbers, andifLisan algebraic extension ofK,then there exists anextension ofvtoLwhose value group isalso anordered subgroup ofthepositive reals. Proof Weknow that wecanextend vtoavaluation wofLwith some value group r',and thevalue group rofvcan beidentified with asubgroup ofR+. ByCorollary 4.3, every element ofr'hasfinite period modulo r.Since every element ofR+has aunique e-th root forevery integere>1,we canfind inan obvious wayanorder-preserving embedding ofr'into R+which induces the identity onr.Inthis wayweget our extension ofvtoanabsolute value onL. Corollary 4.5. IfLisfinite over K,andifrisinfinite cyclic, thenr'isalso infinite cyclic. Proof UseCorollary 4.3and thefact that asubgroup ofacyclic group is cyclic. Weshall now strengthen ourpreceding proposition toaslightly strongerone. Wecall(r':r)theramification index. Proposition 4.6. LetLbeafinite extension ofdegreenofafield K,andlet.tJ beavaluation ringofL.Let 9Jlbeitsmaximal ideal, let0=.0nK,and letm bethemaximal ideal of0,i.e. m =9Jlno.Then theresidue class degree [D/9Jl: o/m] isfinite. Ifwe denote itbyf,andifeistheramification index, then ef<n. Proof LetYb...,Yeberepresentatives inL*ofdistinct cosets ofr'/r and let Zb...,Zsbeelements of.0whose residue classes mod 9Jlarelinearly inde- pendent over o/m. Consider arelation "a..z. y.=0i..J IJ J l i,j withaijEK,notallaij=O.Inaninner sum s Laijzj, j= 1 divide bythecoefficient aivhaving thebiggest valuation. We obtain alinear combination ofZl'...,Zswith coefficients in0,and atleast onecoefficient equal toaunit. Since Zb...,Zsarelinearly independent mod 9Jlover o/m, itfollows that our linear combination isaunit. Hence s Laijzj=laiv I j= 1 484 ABSOLUTE VALUES XII,4 for some index v.Inthe sum t(taijZj)Yi=0 i==1 j=1 viewed asasum oni,atleast two terms have the same value. This contradicts theindependence ofIY11,...,IYeImodrjustasintheproof ofProposition 4.2. Remark. Our proof also shows that the elements {ZjY;}arelinearly in- dependent over K.This will beused again later. Ifwisanextension ofavaluation v,then theramification index will be denoted bye(wIv)and theresidue class degree will bedenoted byf(wIv). Proposition 4.7. LetKbeafield with avaluation v,and letKcEeL be finite extensions ofK.Let wbeanextension ofvtoEand letubeanextension ofw toL.Then e(u Iw)e(w Iv)=e(u Iv), f(ulw)f(wlv)=f(ulv). Proof Obvious. We can express theabove proposition bysaying that theramification index and theresidue class degree aremultiplicative intowers. Weconclude this section byrelating valuation rings inafinite extension with theintegral closure. Proposition 4.8. Let 0be avaluation ring inafield K.Let Lbe afinite extension ofK.Let()beavaluation ringofLlying above 0,and weitsmaximal ideal. LetBbetheintegral closure of0inL,and let=wenB.Then ()is equal tothelocal ring B\!3. Proof Itisclear thatBiscontained in.o. Conversely, letxbeanelement of.0.Then xsatisfies anequation with coefficients inK,notall0,say anxn+...+ao=0, aiEK. Suppose that asisthecoefficient having thebiggest value among the aiforthe valuation associated with thevaluation ring 0,and that itisthe coefficient farthest tothelefthaving this value. Let bi=aJa s.Then allbiE0and bn,...,bs+1E9Jl. XII,4 VALUATIONS 485 Divide theequation byXS .Weget (bnxn-s+...+bs+1x +1)+(bs-1+...+boXSI)=O. Letyand zbethe twoquantities inparentheses inthepreceding equation,so that we canwrite -y=z/x and -xy=z. Toproveourproposition itwill suffice toshow that yand zlieinBand that yis notin\.p. We useProposition 3.5ofChapter VII. Ifavaluation ring ofLabove contains x,then itcontains ybecause yisapolynomial inxwith coefficients in Hence such avaluation ring also contains z= -xy.Ifontheother hand the valuation ring ofLabove contains 1/x,then itcontains zbecause zisa polynomialin1/xwith coefficients in .Hence this valuation ring also contains y.From this weconclude byChapter VII, Proposition 3.5, that y,zlieinB. Furthermore, since xED, and bn,..., bs+1are in9Jlbyconstruction, it follows that ycannot beinWl,and hence cannot bein\.p.This concludes the proof. Corollary 4.9. Let thenotation beasintheproposition. Then there isonly afinite number ofvaluation rings ofLlying above . Proof This comes from the fact that there isonlyafinite number of maximal ideals \.PofBlying above themaximal ideal of0(Corollary ofPro- position 2.1, Chapter VII). Corollary 4.10. Let thenotation beasintheproposition. Assume inaddition that LisGalois over K.If.o and.0'are twovaluation rings ofLlying above 0, with maximal ideals 9Jl,Wl'respectively, then there exists anautomorphism(J ofLover Ksuch that aD =.0'and aWl =Wl'. Proof Let=.0nBand'=.0'nB.ByProposition2.1ofChapter VII, weknow that there exists anautomorphismuofLover Ksuch that uq3=q3'.From this our assertion isobvious. Example. Let kbe afield, and letKbe afinitely generated extension of transcendence degree1.Iftisatranscendence base ofKover k,then Kisfinite algebraic over k(t). Let.o beavaluation ring ofKcontaining k,and assume that Dis=f.K.Let 0=.0nk(t). Then 0isobviouslyavaluation ring ofk(t)(the 486 ABSOLUTE VALUES XII,5 condition about inverses isafortiori satisfied), and thecorresponding valuation ofk(t) cannot betrivial. Either tort-1EO. Say tEo.Then 0nk[t] cannot be the zero ideal, otherwise thecanonical homomorphism0 o/m of0modulo its maximal ideal would induce anisomorphismonk[t] and hence anisomorphism onk(t), contrary tohypothesis. Hence mnk[t] isaprime ideal p,generated by anirreducible polynomial p(t). The local ringk[t]pisobviouslyavaluation ring, which must be0because every element ofk(t) has anexpression oftype pru where uisaunit ink[t]p.Thus wehave determined allvaluation rings ofk(t) containing k,and we seethat thevalue group iscyclic. Such valuations will be called discrete and arestudied ingreater detail below. Inview ofCorollary 4.5, itfollows that thevaluation ring DofKisalso discrete. The residue class field o/m isequal tok[t]/p and istherefore afinite exten- sion ofk.ByProposition 4.6, itfollows that()1m isfinite over k(ifmdenotes themaximal ideal of(). Finally, weobserve that there isonly afinite number ofvaluation rings D ofKcontaining ksuch that tliesinthemaximal ideal of.0.Indeed, such a valuation ring must lieabove k[t]pwhere p=(t)istheprime ideal generated by t,and we canapply Corollary 4.9. 5. COMPLETIONS AND VALUATIONS Throughout this section, wedeal with anon-archimedean absolute value vonafield K.This absolute value isthen avaluation, whose value group rKisa subgroup ofthepositive reals. Welet0beitsvaluation ring,mthemaximal ideal. Let usdenote byKthecompletion ofKatv,and let6(resp. fit)betheclosure of0(resp. m)inK.Bycontinuity, every element of0hasvalue <1,and every element ofKwhich isnot in6has value >1.IfxERthen there exists an element YEKsuch thatIx-YIisvery small, and henceIxI=IyIforsuch an element y(bythenon-archimedean property). Hence 0isavaluation ring in K,and fitisitsmaximal ideal. Furthermore, 6nK =0and fitnK =m, and wehave anisomorphism o/m 6/fit. Thus theresidue class field o/m does notchange under completion. LetEbeanextension ofK,and let0Ebeavaluation ring ofElying above o. Let mEbeitsmaximal ideal. We assume that thevaluation corresponding to0E isinfact anabsolute value, sothat wecanform thecompletion E.Wethen have XII,6 DISCRETE VALUATIONS 487 acommutative diagram: )8Elm E 1 )81m°Elm E 1 o/m thevertical arrows being injections, and thehorizontal ones being isomorphisms. Thus theresidue class field extension ofourvaluation can bestudied over the completions EofK. We have asimilar remark fortheramification index. Letrv(K) andrv(.K) denote thevalue groups ofourvaluation onKandKrespectively (i.e.theimage ofthe map xIxIfor xEK*and xEK*respectively). We saw above that rv(K)=rv(K); inother words, thevalue group isthe same under completion, because ofthenon-archimedean property. (This isofcourse false inthearchime- dean case.) IfEisagain anextension ofKand wisanabsolute value ofE extending v,then wehave acommutative diagram rw(E) 1 rv(K)=: )rw(E) L )rv(K) from which we seethat theramification index (rw(E):rv(K» also does not change under completion. 6. DISCRETE VALUATIONS Avaluation iscalled discrete ifitsvalue group iscyclic. Inthat case, the valuation isanabsolute value (ifweconsider thevalue groupasasubgroup of thepositive reals). Thep-adic valuation ontherational numbers isdiscrete for each prime number p.ByCorollary 4.5, anextension ofadiscrete valuation toa finite extension field isalso discrete. Aside from theabsolute values obtained byembeddingafield into thereals orcomplex numbers, discrete valuations are themost importantones inpractice. Weshall make some remarks concerning them. Let vbeadiscrete valuation on afield K,and let 0beitsvaluation ring. Let mbethemaximal ideal. There exists anelement nofmwhich issuch that its value InIgenerates thevalue group. (The other generator ofthevalue group is Irc-1/.) Such anelement niscalled alocal parameter for v(orform). Every 488 ABSOLUTE VALUES XII,6 element xofKcan bewritten intheform x=U1{ with some unit uof0,and some integerr.Indeed, wehaveIxI=ITtI'=ITtr I for some rEZ,whence x/Ttrisaunit ino.Wecall rtheorder ofxatv.Itis obviously independent ofthechoice ofparameter selected. Wealso saythat x has azero oforder r.(Ifrisnegative,wesaythat xhas apole oforder-r.) Inparticular, we seethat misaprincipal ideal, generated byTt.Asanexercise, weleave ittothereader toverify that every ideal of0isprincipal, and isapower ofm.Furthermore, weobserve that 0isafactorial ring with exactly oneprime element (up tounits), namelyTt. Ifx,yEK, weshall write x'"yifIxI=Iyl. LetTti(i=1,2,...) be a sequence ofelements of0such thatTti'"Tti .Let Rbeasetofrepresentatives of o/m ino.This means that thecanonical map0o/m induces abijection ofR onto o/m. Assume that Kiscomplete under ourvaluation. Then every element xofo can bewritten asaconvergent series x=ao+alTtl +a2Tt2+. .. with aiER,and theaiareuniquely determined byx. This iseasily proved byarecursive argument. Supposewehave written x=ao+...+anTtn(mod mn+1) then x-(ao+...+anTtn)=Ttn+lYfor some yEO. Byhypothesis, we can write y=an+1+TtZwith some an+1ER.From this weget x=ao+...+an+ITt n+1(mod mn+2), and itisclear that then-th term inour series tends toO.Therefore our series converges (bythenon-archimedean behavior !).The factthat Rcontains precisely onerepresentative ofeach residue class mod mimplies that the aiareuniquely determined. Examples. Consider first the case oftherational numbers with thep-adic valuation vp.Thecompletion isdenoted byQp.Itisthefield ofp-adic numbers. The closure ofZinQpisthering ofp-adic integers Zp.We note that theprime number pisaprime element inboth Zand itsclosureZp.We can select our set ofrepresentatives Rtobethe setofintegers (0,1,.. .,p-1).Thus every p- adic integer can bewritten uniquelyasaconvergentsum2:a;p; where a;isan integer, 0<a;<p-1.This sum iscalled itsp-adic expansion. Such sums areadded andmultiplied intheordinary manner forconvergent series. XII,6 DISCRETE VALUATIONS 489 Forinstance, wehave theusual formalism ofgeometric series, andifwetake p=3,then 2-1 = 1_3=2(1+3+32+...). We note that therepresentatives (0,1,...,p-1)arebynomeans theonly ones which can beused. Infact, itcan beshown thatZpcontains the(p-1)-th roots ofunity, and itisoften more convenient toselect these roots ofunityas representatives forthe non-zero elements oftheresidue class field. Next consider the case ofarational field k(t), where kisany field and tis transcendental over k.We have avaluation determined bytheprime element t intheringk[t]. This valuation isdiscrete, and thecompletion ofk[t] under this valuation isthepower series ringk[[t]]. Inthat case, wecan take theelements ofkitself asrepersentatives ofthe residue class field, which iscanonically isomorphic tok.The maximal ideal ofk[[t]] istheideal generated byt. This situation amounts toanalgebraization oftheusual situation arising in thetheory ofcomplex variables. For instance, letZobeapoint inthecomplex plane. Let 0bethering offunctions which areholomorphic insome disc around Zo. Then 0isadiscrete valuation ring, whose maximal ideal consists ofthose functions havingazero atZo.Every element of0has apower series expansion 00 f(z)=Lav(z-zo)v. v==m Therepresentatives oftheresidue class field can betaken tobecomplex numbers, avoIfam=I0,then wesaythatj(z) has azero oforder m.The order isthesame, whether viewed asorder with respect tothediscrete valuation inthealgebraic sense, ortheorder inthe sense ofthetheory ofcomplex variables. We canselect a canonical uniformizing parameter namelyz-Zo,and j(z)=(z-zo)mg(z) where g(z) isapower series beginning with anon-zero constant. Thus g(z) is invertible. LetKbeagain complete under adiscrete valuation, and letEbeafinite extension ofK.Let 0E,mEbethevaluation ring and maximal ideal inElying above 0,minK.Let mbeaprime element inE.IfrEandrKarethevalue groups ofthevaluations inEand Krespectively, and e=(rE:rK) istheramification index, then Ine I=InI, 490 ABSOLUTE VALUES XII,6 and theelements ninj , 0<.< -1.-012=I=e ,J-, ,,... have order je+iinE. LetWl,...,wfbeelements ofEsuch that their residue classes mod mEfrom abasis of0Elm E.IfRisasbefore asetofrepresentatives ofo/m in0,then the set consisting ofallelements alw l+...+afwf withajERisasetofrepresentatives of0Elm Ein0E.From this we seethat every element of0Eadmits aconvergent expansion e-l f 00 LLLav,i,jnjW vni . i==O v= 1j=O Thus theelements {W vni}form asetofgenerators of0Easamodule over o. On theother hand, wehave seen intheproof ofProposition 4.6that these elements arelinearly independent over K.Hence weobtain: Proposition 6.1. LetKbecomplete under adiscrete valuation. Let Ebea finite extension ofK,and lete,fbetheramification index and residue class degree respectively. Then ef=[E:K]. Corollary 6.2. Let rxEE,rx=IO.Let vbethevaluation onKand wits extension toE.Then ordvNi(rx)=f(w Iv)ordwrx. Proof This isimmediate from theformula INi(rx) I=Irxlef and thedefinitions. Corollary 6.3. LetKbeanyfield and vadiscrete valuation onK.Let Ebea finite extension ofK.Ifviswell behaved inE(for instance ifEisseparable over K),then Le(wlv)f(wlv)=[E:K]. wlv IfEisGalois over K,then all ewareequal tothe same number e,allfware XII,7 ZEROS OFPOLYNOMIALS INCOMPLETE FIELDS 491 equal tothe same number f,and so efr=[E:K], where risthenunlber ofextensions ofvtoE. Proof. Our first assertion comes from ourassumption, andProposition3.3. IfEisGalois over K,weknow from Corollary 4.10 that any two valuations ofE lying above vareconjugate. Hence allramification indices areequal, and similarly for the residue class degrees. Our relation efr=[E:K] isthen obvious. 7. ZEROS OF POLYNOMIALS IN COMPLETE FIELDS LetKbecomplete under anon-trivial absolute value. Let f(X)=n(X-rxi)'i beapolynomial inK[X] having leading coefficient 1,and assume the roots rxi aredistinct, with multiplicities ri.Let dbethedegree off.Let gbeanother polynomial with coefficients inKa,and assume that thedegree ofgisalso d,and that 9hasleading coefficient 1.WeletI9Ibethemaximum oftheabsolute values ofthecoefficients ofg.One seeseasily that ifIgIisbounded, then theabsolute values ofthe roots of9arealso bounded. Suppose that 9comes close tof,inthe sense thatIf-gIissmall. IfPis any root ofg,then If(P)-g(P) I=If(P)1=nIrxi-Plri issmall, and hence Pmust come close tosome root off.AsPcomes close to sayrx=lI..l,itsdistance from theother roots offapproaches thedistance oflI..l from theother roots, and istherefore bounded from below. Inthat case, wesay that Pbelongs to lI... Proposition 7.1. Ifgis'sufficiently close tof,andPl, ..., Psaretheroots ofg belonging torx(counting multiplicities), then s=rlisthemultiplicity ofrxinf. Proof Assume thecontrary. Then we can find asequence gvofpoly- nomials approaching fwith preciselysroots P\V),...,PV)belongingtorx,but with s=Ir.(We can take the same multiplicityssince there isonlyafinite number ofchoices forsuch multiplicities.) Furthermore, theother roots ofgalso 492 ABSOLUTE VALUES XII,7 belong toroots off, and wemay suppose that these roots arebunched together, according towhich rootoffthey belong to.Since limgv=f,weconclude that rx must have multiplicitysinf,contradiction. Next weinvestigate conditions under which apolynomial has aroot ina complete field. We assume thatKiscomplete under adiscrete valuation, with valuation ring 0, maximal ideal p.We letnbeafixed prime element ofp. We shall deal with n-spaceover o.We denote avector (al'...,an)with aiE0byA.Iff(X 1,...,Xn)Eo[X] isapolynomial innvariables, with integral coefficients, weshall saythat Aisazero offiff(A)=0,and wesaythat Aisa zero offmod pmiff(A)=0(mod pm). LetC=(co,...,cn)beino(n+1).Let mbeaninteger>1.Weconsider the nature ofthesolutions ofacongruence oftype (*) nm(co +CIX 1+...+CnX n)=0(mod pm+ 1). This congruence isequivalent with thelinear congruence (**) Co+c1x1+...+cnXn=0(mod p). Ifsome coefficientCi(i=1,...,n)isnot =0(mod p),then the setofsolutions is not empty, and has the usual structure ofasolution ofoneinhomogeneous linear equation over the field o/p. Inparticular, ithas dimension n-1. Acongruence (*) or(**) with someCi 0(mod p)will becalled aproper congruence. As amatter ofnotation, wewrite Diffortheformal partial derivative off with respect toXi. Wewrite gradf(X)=(Dlf(X),...,Dnf(X». Proposition 7.2. Letf(X)Eo[X]. Let rbeaninteger>1and letAEo(n)be such that f(A)=0(mod p2r-l), Dif(A)=0(mod pr-l), Dif(A) =F-0(mod pr),forall i=1,..., n, for some i=1,...,n. Let vbeaninteger>0and letBE o(n)besuch that B=A(mod pr) and f(B)=0(mod p2r-1+ V). Avector YEo(n)satisfies Y=B(mod pr+v) andf(Y)=0(mod p2r+v) XII,7 ZEROS OFPOLYNOMIALS INCOMPLETE FIELDS 493 ifandonlyifYcan bewritten intheform Y=B+nr+vc,with some CEo(n) satisfying theproper congruence f(B) +r{+v gradf(B). C=0(mod p2r+v). Proof Theproof isshorter than thestatement oftheproposition.Write Y=B+nr+vC. ByTaylor's expansion, f(B +nr+VC)=f(B) +nr+vgradf(B).C(mod p2r+ 2V). Tosolve this last congruence mod p2r+ v,weobtain aproper congruence by hypothesis, because gradf(B)=gradf(A)=0(mod pr-1). Corollary 7.3. Assumptions beingasinProposition 7.2, there exists azero offino(n)which iscongruent toAmod pro Proof We canwrite this zero asaconvergent sum A+nr+1C1+nr+2C2+... solving forCl'C2,...inductivelyasintheproposition. Corollary 7.4. Letfbeapolynomial inone variable ino[X], and let aEO besuch thatf(a)=0(mod)butf'(a)=1=0(mod ).Then there exists bE0,b=a(mod p)such thatf(b)=o. Proof Take n=1and r= 1intheproposition, andapply Corollary 7.3. Corollary 7.5. Let mbeapositive integer notdivisible bythecharacteristic ofK. There exists aninteger rsuch thatfor any aE0,a=1(mod p'), the equation xm-a=0has aroot inK. Proof Apply theproposition. Example. Inthe 2-adic field Q2, there exists asquare root of-7,Le. EQ2, because-7= 1-8. When theabsolute value isnotdiscrete, itisstillpossible toformulate a criterion for apolynomial tohave azero byNewton approximation. (Cf. my paper, "On quasi-algebraic closure," Annals ofMath. (1952) pp.373-390. Proposition 7.6. Let Kbe acomplete under anon-archimedean absolute value (nontrivial). Let 0bethevaluation ring and letf(X)Eo[X] beapoly- nomial inone variable. Let CXoE0besuch that If(cxo) I<If'(CXO)21 (heref'denotes theformal derivative off). Then thesequence f(cx i) CXi+1=CXi- f'«(1.i) 494 ABSOLUTE VALUES XII,7 converges toaroot rxoffin0,and wehave <f(rxo) loc- OCoI=f'(OCO)2<1. Proof Let c= 1f(rxo)1 f'(rx o)21<1.Weshow inductively that: 1.IrxiI<1, 2.Irxi-rxo I<c, 3.f(rx i)<2i f'(rx i)2=c . These three conditions obviously imply ourproposition. Ifi=0,theyare hypotheses. Byinduction, assume them fori.Then: 1.If(rxi)1f'(rxi)21<C2igives 1rxi+1-rxd<C2i<1,whenceIrxi+ 11<1. 2.Irxi+1-rxo I<max {Irxi+1-rxd, Irxi-rxoI}=c. 3.ByTaylor's expansion,wehave f( )f()f'()f(rxi) p(f(rx i» )2 rxi+ 1=rxi-rxif'(rxi)+ f'(rx i) for some pE0,and this islessthan orequal to f(rxi)2 f'(rx i) inabsolute value. Using Taylor's expansion onf'(rxi+ 1)weconclude that If'(rxi+l)1=If'(rxi)l. From this weget f(rxi+1) f'(rx i+1)2<2i+1=c asdesired. Thetechnique oftheproposition isalso useful when dealing with rings, saya local ring0with maximal ideal msuch that mr=0for some integerr>O. Ifone has apolynomial fino[X] and anapproximate rootrxosuch that f'(rxo) =1=0mod m, then theNewton approximation sequence shows how torefinerxotoaroot off. Example inseveral variables. LetKbecomplete under anon-archimedean absolute value. Letf(X I'. . .,Xn+I)EK[X] beapolynomial with coefficients inK.Let(aI,. . .,an'b)EKn+I.Assume thatf(a, b)=O.LetDn+Ibethe XII, Ex EXERCISES 495 partial derivative with respect tothe(n+1)-th variable, and assume that Dn+If(a, b) =/:;O.Let(a) EKnbesufficiently close to(a). Then there exists an element 5ofKclose tobsuch thatf(a, 5)=O. This statement isanimmediate corollary ofProposition 7.6. Bymultiplying allai'bbyasuitable non-zero element ofKone canchange them toelements ofo.Changing thevariables accordingly, one may assume without loss ofgen- erality that ai'bEO, and thecondition onthepartial derivative notvanishing ispreserved. Hence Proposition 7.6 may beapplied. After perturbing (a) to (a), theelement bbecomes anapproximate solution ofj(a, X).As(a)approaches (a),f(a, b)approaches0and Dn+1f(a, b)approaches Dn+1j(a, b)=/:;O. Hence for(a)sufficiently close to(a), theconditions ofProposition 7.6 are satisfied, and one may refine btoaroot off(a, X), thus proving theassertion. The result was used inakey way inmypaper "On Quasi Algebraic Closure". Itistheanalogue ofTheorem 3.6ofChapter XI, forreal fields. Inthelanguage ofalgebraic geometry (which we now assume), theresult can bereformulated asfollows. Let Vbe avariety defined over K.Let Pbe a simple point ofVinK.Then there isawhole neighborhood ofsimple points of VinK.Especially, suppose that Visdefined byafinite number ofpolynomial equationsover afinitely generated field kover theprime field. After asuitable projection,one mayassume that thevariety isaffine, and defined byone equa- tionf(X b. . .,Xn+I)=0asinthe above statement, and that thepoint is P=(aI'. . .,an,b)asabove. One can then select ai=Xiclose toaibut such that(xI'. . .,xn)arealgebraically independentover k.Let ybtherefinement ofbsuch thatf(x, y)=O.Then (x,y)isageneric point ofVover k,and the coordinates of(x,y)lieinK.Ingeometric terms, this means that thefunction field ofthevarietycan beembedded inKover k,justasTheorem 3.6ofChapter XIgave thesimilar result for anembedding inareal closed field, e.g. thereal numbers. EXERCISES 1.(a)LetKbeafield with avaluation. If l(X)=ao+atX+...+anxn isapolynomial inK[X], defineIjItobethe max onthevalues laiI(i=0,...,n). Show that this defines anextension ofthevaluation toK[X], and also that the valuation can beextended totherational field K(X). How isGauss' lemma a specialcase oftheabove statement? Generalize topolynomials inseveral variables. (b)Letfbeapolynomial with complex coefficients. DefineIfItobethemaximum oftheabsolute values ofthecoefficients. Let dbeaninteger>1.Show that 496 ABSOLUTE VALUES XII, Ex there exist constants Cl'C2(depending only ond)such that, ifI,garepolynomials inC[X] ofdegrees<d,then CIIj"g 1<Ijg I<C21f"gI. [Hint: Induction onthe number offactors ofdegree 1.Note that theright inequality istrivial.] 2.LetMQbethe setofabsolute values consisting oftheordinary absolute value and all p-adic absolute values vponthefield ofrational numbers Q.Show that foranyrational number aEQ,a=F0,wehave nlal v=1. veMQ IfKisafinite extension ofQ,andMKdenotes the setofabsolute values onKextending those ofMQ,and foreach WEMKweletNwbethelocal degree [Kw: Qv]' show that for exEK, ex=F0,wehave nlexlw=1. weMK 3.Show that thep-adic numbers Qphave noautomorphisms other than theidentity. [Hint: Show that such automorphismsarecontinuous forthep-adic topology. Use Corollary 7.5 asanalgebraic characterization ofelements close to1.] 4.LetAbeaprincipal entire ring, and letKbeitsquotient field. Let 0beavaluation ring ofKcontaining A,and assume 0=FK.Show that 0isthelocal ring A(p)for some prime element p.[This applies both tothering Zand toapolynomial ringk[X] over afieldk.] 5.Let Abeanentire ring, and letKbeitsquotient field. Assume that every finitely generated ideal ofAisprincipal. Let 0beadiscrete valuation ring ofKcontaining A. Show that 0=A(p)for some element pofA,and that pisagenerator ofthemaximal ideal ofo. 6.LetQpbe ap-adic field. Show thatQpcontains infinitely many quadratic fields of type Q( ),where misapositive integer. 7.Show that thering ofp-adic integers Zpiscompact. Show that thegroup ofunits inZp iscompact. 8.IfKisafield complete with respect toadiscrete valuation, with finite residue class field, andif0isthering ofelements ofKwhose orders are >0,show that 0iscompact. Show that thegroup ofunits of0isclosed in0and iscompact. 9.LetKbeafield complete with respect toadiscrete valuation, let 0bethering ofintegers ofK,and assume that 0iscompact. Let11'12'...beasequence ofpolynomials inn variables, with coefficients ino.Assume that allthese polynomials have degree<d, and thatthey converge toapolynomial I(i.e.that1I-/; 1-+0asi-+(0).Ifeach/;has azero in0,show thatIhas azero ino.Ifthepolynomials /;arehomogeneous ofdegree d,andifeach Iihas anon-trivial zero in0,show thatIhas anon-trivial zero ino.[Hint: Use thecompactness of0and oftheunits of0forthehomogeneous case.] (For applications ofthisexercise, and also ofProposition 7.6, cf.mypaper "On quasi-algebraic closure," Annals ofMath., SS(1952), pp.412-444.) XII, Ex EXERCISES 497 10.Show that ifp,p'are two distinct prime numbers, then thefieldsQpandQp'arenot isomorphic. 11.Prove that thefieldQpcontains all(p-1)-th roots ofunity. [Hint: UseProposition 7.6, applied tothepolynomial XP-1-1which splits into factors ofdegree1intheresidue class field.] Show that twodistinct (p-1)-th roots ofunity cannot becongruent mod p. 12. (a)Letf(X) be apolynomial ofdegree1inZ[X]. Show that thevalues f(a) for aEZaredivisible byinfinitely many primes. (b)LetFbe afinite extension ofQ.Show that there areinfinitely many primes p such that allconjugates ofF(in analgebraic closure ofQp)actuallyarecontained inQp.[Hint: Use theirreducible polynomial ofagenerator for aGalois extension ofQcontaining F.] 13. LetKbeafield ofcharacteristic 0,complete with respect toanon-archimedean absolute value. Show that theseries x2x3 exp(x)=1+x+-+-+... 2! 3! x2x3 10g(1 +x)=x- -+- - ... 2 3 converge insome neighborhood ofO. (The main problem arises when thecharacteristic oftheresidue class field isp>0,sothat pdivides thedenominators n!and n.Get an expression which determines thepower ofpoccurring inn!.) Prove that theexp and loggive mappings inverse toeach other, from aneighborhood of0toaneighborhood of1. 14. LetKbeasinthepreceding exercise, ofcharacteristic 0,complete with respect toanon- archimedean absolute value. For every integern>0,show that theusual binomial expansion for(1+X)l/" converges insome neighborhood ofO. Dothis firstassuming that thecharacteristic oftheresidue class field does notdivide n,inwhich case the asser- tion ismuch simpler toprove. 15. LetFbeacomplete field with respect toadiscrete valuation, let 0bethevaluation ring, naprime element, and assume thato/(n)=k.Prove thatifa,bE 0and a=b(mod n') with r>0then apn=bpn(mod nr+") forallintegersn>O. 16. LetFbeasabove. Show that there exists asystem ofrepresentatives Rfor0/(n)in0 such that RP =Rand that thissystem isunique (Teichmiiller). [Hint: Let exbearesidue class ink.For each v>0let avbe arepresentative in0ofaPvand show that the sequence avconverges for v--+00,and infact converges toarepresentativeaof ex, independent ofthechoices ofav.]Show that thesystem ofrepresentatives Rthus obtained isclosed under multiplication, and that ifFhascharacteristic p,then Ris closed under addition, and isisomorphic tok. 17.(a)(Witt vectors again). Let be aperfect field ofcharacteristic p.We use the Witt vectors asdescribed inthe exercises ofChapter VI. One can define an absolute value onW(k), namely IxI=p-rifXristhefirst non-zero component ofx.Show that this isanabsolute value, obviously discrete, defined onthering, and which can beextended atonce tothequotient field. Show that thisquotient field iscomplete, and note that W(k) isthevaluation ring. The maximal ideal consists ofthose xsuch thatXo=0,i.e. isequal topW(k). 498 ABSOLUTE VALUES XII,Ex (b)Assume that Fhascharacteristic O.Map each vector xEW(k) ontheelement Lf-'pi whereiisarepresentative ofXiinthespecial system ofExercise 15.Show that this map isanembedding ofW(k) Into o. 18.(Local uniformization). Letkbeafield, Kafinitely generated extension oftranscendence degree 1,and 0adiscrete valuation ring ofKover k,with maximal ideal m.Assume that o/m=k.Let xbeagenerator ofm, and assume that Kisseparable over k(x). Show that there exists anelement YEOsuch that K =k(x,y),and also having thefollowing property. Letcpbetheplace onKdetermined byo.Let a=cp(x), b=cp(y)(ofcourse a=0).Letf(X, Y)betheirreducible polynomIal ink[X, Y]such thatf(x, y)=O. Then D2f(a,b)=I:O.[Hint: Write first K=k(x,z)where zisintegral overk[x]. Let z=Zh...,zn(n>2)betheconjugates of Zover k(x), and extend 0toavaluation ring Dofk(x, Zl, ..., zn). Let Z=ao+atX +. . .+arxr+... bethepower series expansion ofzwith aiEk,and letP,(x)=ao+. ..+a,xr .For i=1,...,nlet Zj-P,(x) Yi=x' Taking rlarge enough, show that Ylhas nopole atDbut Y2,...,Ynhave poles atD. The elements Yl,...,Ynareconjugate over k(x). Letf(X, Y)betheirreducible poly- nomial of(x,y)over k.Then f(x, Y)=I/1n(x)yn +...+l/1o(x) with l/1i(x)k[x]. We may also assume 1/1;(0) =I:0(sincefisirreducible). Write f(x, Y)intheform f(x, Y)=I/1n(X)Y2...Yn(Y-Yl)(Y2"1Y-1)...(y';-1Y-1). Show that I/1n(X)Y2...Yn=udoes nothave apole atD.Ifw ED,letwdenote itsresidue class modulo themaximal ideal ofD.Then o=I:f(x, Y)=(-I)n-l u(Y-.vt). Let Y=Yh.v=b.Wefind that D2f(a, b)=(_I)n-l u=I:0.] 19. Prove the converse ofExercise 17,i.e.ifK =k(x,y),f(X, Y)istheirreducible poly- nomial of(x,y)over k,andifa,bEkare such thatf(a, b)=0,but D2f(a, b)=I:0, then there exists aunique valuation ring0ofKwith maximal ideal msuch that x=a and Y=b(mod m). Furthermore, o/m=k,and x-aisagenerator ofm.[Hint: Ifg(x,y)Ek[x,y]issuch that g(a,b)=0,show that g(x,y)=(x-a)A(x, y)IB(x, y) where A,Barepolynomials such that B(a, b)=I:O.IfA(a, b)=0repeat theprocess. Show that theprocess cannot berepeated indefinitely, and leads toaproof ofthedesired assertion. ] 20.(Iss'sa-Hironaka Ann. ofMath 83(1966), pp.34-46). This exercise requiresagood working knowledge ofcomplex variables. LetKbethefield ofmeromorphic functions onthecomplex plane C.Let,Q be adiscrete valuation ring ofK(containing the XII, Ex EXERCISES 499 constants C). Show that thefunction zisin.[Hint: LetaI'a2,.. .be adiscrete sequence ofcomplex numbers tending toinfinity, forinstance thepositive integers. LetVI' v2,. ..,be asequence ofintegers, 0<:Vi<:P-1,for some prime number p,such that Vipi isnotthep-adic expansion ofarational number. Letfbeanentire function havingazero oforder Vipiataiforeach iand noother zero. Ifzisnotin 0,consider thequotient f(z) g(z)= n n(z-ai)Vipi i= 1 From theWeierstrass factorization ofanentire function, show that g(z)=h(z)pn+1for some entire function h(z). Now analyze the zero ofgatthediscrete valuation of0in terms ofthat offandn(z-aiyipi togetacontradiction.] IfUisanon-compact Riemann surface, andListhefield ofmeromorphic functions onU,andif0isadiscrete valuation ring ofLcontaining theconstants, show that every holomorphic function lponUliesino.[Hint: Map lp:U-+C,and getadiscrete valua- tion ofKbycomposing lpwith meromorphic functions onC.Apply thefirstpart ofthe exercise.] Show that thevaluation ring isthe one associated with acomplex number. [Further hint: Ifyou don't know about Riemann surfaces, doitforthecomplex plane. For each zEU,letfz beafunction holomorphic onUandhaving onlyazero oforder 1 atz.Iffor some Zothefunction fzohasorder >1at0,then show that 0isthevaluation ring associated with zo.Otherwise, every function fzhasorder 0ato.Conclude that the valuation of0istrivial onanyholomorphic function byalimit trick analogous tothat ofthefirst part oftheexercise.] Part Three LINEAR ALGEBRA and REPRESENTATIONS We shall beconcerned with modules and vector spaces, going into their structure under various points ofview. The main theme here istostudyapair, consisting ofamodule, and anendomorphism, oraring ofendomorphisms, andtrytodecompose thispair into adirect sum ofcomponents whose structure can then bedescribed explicitly. The direct sum theme recurs inevery chapter. Sometimes, we use aduality toobtain our direct sum decomposition relative toapairing, and sometimes wegetourdecomposition directly. Ifamodule refuses todecompose into adirect sum ofsimple components, then there isno choice but toapply theGrothendieck construction and seewhat can beob- tained from it. The extension theme occurs only once, inWitt's theorem, inabrief counter- point tothedecomposition theme. 501 CHAPTER XIII Matrices and Linear Maps Presumably readers ofthischapter will have had some basic acquaintance with linear algebra inelementarycourses .Wegobeyond such courses bypointing outthat alotofresults hold forfree modules over acommutative ring. This is useful when one wants todeal with families oflinear maps, andreduction modulo anideal. Note that8and9give examples ofgroup theory inthe context oflinear groups. Throughout this chapter, weletRbe acommutative ring, and welet E,FbeR-modules. We suppress theprefix Rinfront oflinear maps and modules. 1. MATRICES By an mxnmatrix inRone means adoubly indexed family ofelements ofR,(aij), (i=1,..., mandj=1,...,n),usually written intheform (all:::aln ). am 1amn We call the elementsaijthe coefficients orcomponents ofthematrix. A 1xnmatrix iscalled arow vector (ofdimension, orsize, n)and amx1matrix iscalled acolumn vector (ofdimension, orsize, m). Ingeneral,wesaythat (m,n)isthesize ofthematrix, oralso mxn. Wedefine addition formatrices ofthe same sizebycomponents. IfA=(aij) and B=(bij)arematrices ofthe same size, wedefine A+Btobethematrix whose ij-component isaij+bij.Addition isobviously associative. Wedefine themultiplication ofamatrix Abyanelement CERtobethematrix (caij), 503 504 MATRICES AND LINEAR MAPS XIII,1 whose ij-component iscaij.Then the setofmxnmatrices inRisamodule (Le. anR-module). We define theproduct ABoftwo matrices only under certain conditions. Namely, when Ahas size (m,n)and Bhas size(n,r),i.e.only when thesizeof the rows ofAisthe same asthesizeofthecolumns ofB.Ifthat isthe case, let A=(aij)and letB=(bjk).We define AB tobethe mxrmatrix whose ik- component is n Laijbjk. j= 1 IfA,B,Carematrices such that AB isdefined and BC isdefined, then sois (AB)C andA(BC) and wehave (AB)C=A(BC). This istrivial toprove. IfC=(Ckl)' then thereader will seeatonce that the ii-component ofeither oftheabove products isequal to LLaijbjkCkl' j k An mxnmatrix issaid tobeasquare matrix ifm =n.Forexample,a 1x1matrix isasquare matrix, and will sometimes beidentified with the element ofRoccurringasitssingle component. For agiven integern>1the setofsquarenxnmatrices forms aring. This isagain trivially verified and will beleft tothereader. The unit element ofthering ofnxnmatrices isthematrix 0...00 1 0 In= 0 0 whose components areequal to0except onthediagonal, inwhich case they areequal to1.Wesometimes write Iinstead ofIn. IfA=(aij)isasquare matrix, wedefine ingeneral itsdiagonal components tobetheelements aii' Wehave anatural ring-homomorphism ofRinto thering ofnxnmatrices, given by c cIn. Thus clnisthesquarenxnmatrix having allitscomponents equal to0except thediagonal components, which areequal toc.Let usdenote thering ofnxn XIII,1 MATRICES 505 matrices inRbyMatn(R). Then Matn(R) isanalgebra over R(with respect to theabove homomorphism). LetA=(aij)beanmxnmatrix. Wedefine itstranspose tAtobethematrix (aj;)(j=1,.. .,nand i=1,.. .,m). Then tAisannXmmatrix. The reader willverify atonce thatifA,Bareofthe same size, then t(A+B)=tA+tB. IfcERthen t(cA)=c!4. IfA,Bcan bemultiplied, then tB isdefined and we have t(AB)=tBtA. We note theoperationsonmatrices commute with homomorphisms. More precisely, letlp:R R'be aring-homomorphism. IfA,Barematrices inR, wedefine lpA tobethematrix obtained byapplying lptoallthecomponents of A.Then lp(A +B)=lpA+lpB, lp(AB)=(lpA)(lpB), lpCA)=tlp(A).lp(cA)=lp(c)lpA, Asimilar remark will hold throughout our discussion ofmatrices (for instance inthe next section). LetA=(aij)beasquarenxnmatrix inacommutative ring R.Wedefine thetrace ofAtobe n tr(A)=Laii; i= 1 inother words, the trace isthe sum ofthediagonal elements. IfA,Bare nxnmatrices, then tr(AB)=tr(BA). Indeed ifA=(a..)and B=(b..)then,IJ IJ tr(AB)=LLaivbvi=tr(BA). i v As anapplication, weobserve thatifBisaninvertible nxnmatrix, then tr(B-1AB)=tr(A). Indeed, tr(B-1AB)=tr(ABB-1)=tr(A). 506 MATRICES AND LINEAR MAPS XIII,2 2. THE RANK OF AMATRIX Letkbeafield and letAbeanmxnmatrix ink.Bythe row rank ofAwe shall mean themaximum number oflinearly independent rows ofA,andbythe column rank ofAweshall mean themaximum number oflinearly independent columns ofA.Thus these ranks arethedimensions ofthe vector spaces gen- erated respectively bythe rows ofAand thecolumns ofA.We contend that these ranks areequal tothe same number, and wedefine therank ofAtobe that number. Let A1,...,Anbethecolumns ofA,and letAt,...,Ambethe rows ofA. Lettx =(Xb...,xm)have components XjEk.Wehave alinear map XxIAl +...+xmAm ofk(m) onto thespace generated bythe row vectors. Let Wbeitskernel. Then Wisasubspace ofk(m)and dim W+row rank =m. IfYisacolumn vector ofdimension m,then themap (X,y)tXY=X.Y isabilinear map into k,ifweview the 1x1matrixtXYasanelement ofk. We observe that Wistheorthogonal space tothecolumn vectors A1,...,An, i.e.itisthespace ofallXsuch that X.Aj =0forallj=1,. ..,n.Bytheduality theorem ofChapter III, weknow that k(m) isitsown dual under thepairing (X,y).-. X.Y and that k(m)IW isdual tothespace generated byAI,...,An. Hence dim k(m)IW=column rank, or dim W+column rank =m. From this weconclude that column rank =row rank, asdesired. We note that Wmay beviewed asthespace ofsolutions ofthesystem ofn linear equations x1Al+...+xmAm=0, XIII,3 MATRICES AND LINEAR MAPS 507 inmunknowns xI'...,Xm.Indeed, ifwewrite outtheprecedingvector equation interms ofallthecoordinates, wegettheusual system ofnlinear equations. Weletthereader dothisifheorshewishes. 3. MATRICES AND LINEAR MAPS Let Ebeamodule, and assume that there exists abasis CB ={l'···,n} forEover R.This means that every element ofEhas aunique expressionasa linear combination x=Xl 1+...+Xnn with XiER.Wecall(Xl'...,xn)thecomponents ofXwith respect tothebasis. Wemay view thisn-tupleasarow vector. Weshall denote byXthetranspose ofthe row vector (Xl'...'Xn).WecallXthe column vector ofxwith respect to thebasis. We observe that if{'l'...,} isanother basis ofEover R,then m=n. Indeed, letpbe amaximal ideal ofR.Then E/pE isavector space over the field R/pR, and itisimmediately clear that ifwedenote byitheresidue class ofimod pE,then{b...,n}isabasis forElpEover RlpR. Hence nisalso thedimension ofthis vector space, and weknow theinvariance ofthecardinality forbases ofvector spacesover fields. Thus m=n.Weshall call nthedimension ofthemodule Eover R. We shall view R(n) asthemodule ofcolumn vectors ofsize n.Itisafree module ofdimension nover R.Ithas abasis consisting oftheunit vectors e1,...,ensuch that tei=(0,...,0,1,0,...,0) hascomponents 0except foritsi-thcomponent, which isequalto1. An mxnmatrix Agives rise toalinear map LA:R(n) R(m) bytherule Xt---+AX. Namely, we have A(X +Y)=AX +AYand A(cX)=cAX for column vectors X,Yand cER. 508 MATRICES AND LINEAR MAPS Xiii,3 The above considerations can beextended to aslightly more general context, which can bevery useful. Let Ebeanabelian group and assume that Risacommutative subring of End z(E)=Homz(E, E). Then EisanR-module. Furthermore, ifAisanmxnmatrix inR,then weget alinear map LA:E(n) E(m) defined byarule similar totheabove, namely XHAX. However, this has to beinterpreted intheobvious way. IfA=(aij)and Xisacolumn vector of elements ofE,then (all AX= am1aln )(X t)=(Yt), amnXn Ym n where Y.="a..x.I I)). j=1 IfA,Barematrices inRwhose product isdefined, then foranycERwe have LAB=LAL Band LeA=cLA. Thus wehave associativity, namely A(BX)=(AB)X. Anarbitrary commutative ring Rmay beviewed asamodule over itself. Inthis way werecover thespecial case ofour map from R(n)into R(m). Further- more, ifEisamodule over R,then Rmay beviewed asaring ofendomorphisms ofE. Proposition 3.1. Let Ebe afree module over R,and let{xl'...,xn} be a basis. Let Yl'...,Ynbeelements ofE.Let Abethematrix inRsuch that A(:)=(;:). Then {Yb...,Yn} isabasis ofEifandonlyifAisinvertible. Proof. LetX,Ybethecolumn vectors ofour elements. Then AX =Y. Suppose Yisabasis. Then there exists amatrix CinRsuch that CY =X. XIII,3 MATRICES AND LINEAR MAPS 509 Then CAX =X,whence CA =Iand Aisinvertible. Conversely,assume that Aisinvertible. Then X =A-1Yand hence Xl'.'"Xnare inthe module generated byYl,...,Yn.Suppose that wehave arelation blYl +...+bnYn=0 with biER.Let Bbethe row vector (bl,...,bn).Then BY =0 and hence BAX =O.But{xl'...,xn} isabasis. Hence BA =0,and hence BAA-1=B=O.This proves that thecomponents ofYare linearly indepen- dent over R,and proves ourproposition. We return toour situation ofmodules over anarbitrary commutative ring R. LetE,Fbemodules. We shall seehow we can associate amatrix with a linear map whenever bases ofEandFaregiven. We assume thatE,Farefree. Welet(S={l'. . .,n}and (S'={,. ..,;,,} bebases ofEandFrespectively. Let f:E F bealinear map. There exist unique elementsaijERsuch that f(1)=a11'l+...+am 1 , j'(n)=aln'l +.. .+amn' orinother words, m f(j)=Laij i= 1 (Observe that the sum isover thefirst index.) Wedefine M,(f)=(aij). Ifx=x1 1+...+Xnnisexpressed interms ofthebasis, let usdenote the column vector Xofcomponents ofxbyM(8(x). We seethat M(8,(f(x»=M'(f)M(8(x), Inother words, ifX'isthecolumn vector off(x), andMisthematrix associated withf then X' =MX. Thus theoperation ofthelinear map isreflected bythe matrix multiplication, and wehavef=LM. 510 MATRICES AND LINEAR MAPS XIII,3 Proposition 3.2. LetE,F,Dbemodules, and letCB,CB', CB"befinite bases ofE,F,D,respectively. Let EFD belinear maps. Then <B <B' <BM<B,,(g0f)=M<B,,(g)M <B.(f). Proof'. Let Aand Bbethematrices associated with themaps f,grespec- tively, with respect toourgiven bases. IfXisthecolumn vector associated with xEE,the vector associated with g(f(x» isB(AX)=(BA)X. Hence BA isthe matrix associated with g0f.This proves what wewanted. Corollary 3.3. Let E=F.Then M,(id)M/(id)=M:(id)=I. Each matrix M/(id) isinvertible (i.e. isaunit intheringofmatrices). Proof. Obvious. Corollary 3.4. LetN =M,(id). Then M:(f)=M/(id)M(f)M'(id)=NM(f)N-l. Proof. Obvious Corollary 3.5. Let Ebeafree module ofdimension nover R.Let CBbea basis ofEover R.The map fM(f) isaring-isomorphism oftheringofendomorphisms ofEonto theringofnxn matrices inR.Infact, theisomorphism isoneofalgebras over R. We shall call thematrix M(f) the matrix associated withfwith respect to thebasis CB. Let Ebeafree module ofdimension nover R.ByGL(E) orAutR(E)one means thegroup oflinear automorphisms ofE.Itisthe group ofunits in EndR(E). ByGLn(R) one means thegroup ofinvertible nxnmatrices inR. Once abasis isselected forEover R,wehave agroup-isomorphism GL(E) GLn(R) with respect tothis basis. XIII,4 DETERMINANTS 511 LetEbeasabove. If f:E E isalinear map,weselect abasis CBand letMbethematrix associated withf relative toCB.Wedefine thetrace offtobethetrace ofM,thus tr(f)=tr(M). IfM'isthematrix offwith respect toanother basis, then there exists anin- vertible matrix Nsuch that M' =N-lMN, and hence thetrace isindependent ofthechoice ofbasis. 4. DETERMINANTS Let E1,...,En'Fbemodules. Amap f:E1X... xEn F issaid tobeR-multilinear (orsimply multilinear) ifitislinear ineach variable, i.e.ifforevery index iand elements Xl'...,Xi-I' Xi+l'...,Xn,XjEEj,themap XHf(x b...,Xi-l'X,Xi+l'...,Xn) isalinear map ofEiinto F. Amultilinear map defined onann-fold product isalso called n-multilinear. IfE1=...=En=E,wealso saythatfisamultilinear maponE,instead of saying that itismultilinear onE(n). Letfbeann-multilinear map. Ifwetake two indices i,jand i#-jthen fixing allthevariables except thei-th andj-th variable, we can viewfasa bilinear maponEixEj. Assume that E1= ... =En=E.We say that themultilinear mapfis alternating iff(x1,...,xn)=0whenever there exists anindex i,1<i<n-1, such that Xi=Xi+ 1(inother words, when twoadjacent elements areequal). Proposition 4.1. Letfbe ann-multilinear alternating map onE. Let XI'...,XnEE.Then f(...,X;,Xi+b.·.)= -f(...,Xi+bXi'...). Inother words, when weinterchange twoadjacent arguments off, thevalue offchanges byasign.IfXi=xjfori#-jthenf(x l'.··,xn)=O. 512 MATRICES AND LINEAR MAPS XIII,4 Proof. Restricting ourattention tothefactors inthei-thandj-thplace, with j=i+1,wemayassume fisbilinear forthefirst statement. Then forallx, YEEwehave o=f(x +y,x+y)=f(x, y)+f(y, x). This proves what wewant, namely f(y, x)= -f(x, y).For thesecond asser- tion, we caninterchange successively adjacent arguments offuntil weobtain ann-tuple ofelements ofEhaving twoequal adjacent arguments. This shows that when Xi=Xj,i:1=j,thenf(Xb...,xn)=O. Corollary 4.2. Letfbe ann-multilinear alternating map on E. Let Xl'...,X nEE. Let i#j and let aER. Then thevalue off on(xl,...,x n) does notchange ifwereplace XibyXi+aXjand leave allother components fixed. Proof. Obvious. Amultilinear alternating map taking itsvalue inRiscalled amultilinear alternating form. Onrepeated occasions weshall evaluate multilinear alternating maps on linear combinations ofelements ofE.Let WI=a11v 1+...+alnv n, Wn=anIV1+...+ann vn. Letfben-multilinear alternating onE.Then f(wb...'wn)=f(allv 1+...+a1nv n,..., anlvl +...+annv n). Weexpand thisbymultilinearity, and getasum ofterms oftype aI,0'(1)...an, O'(n)f(V0'(I),.·.,VO'(n», where aranges overarbitrary maps of{I,...,n}into itself. Ifaisnot abijection (i.e. apermutation), then two arguments VO'(i)andvO'(j)areequal fori:1=j,and the term isequal toO.Hence wemay restrict our sum topermutations a. Shuffling back theelements(VO'(l)'...,vO'(n»)totheir standard ordering andusing Proposition 4.1, we seethat wehave obtained thefollowing expansion: Lemma 4.3. IfWl,...,Wnare asabove, then f(w l,..., wn)=Ll(a)al,O'(1)'" an,O'(n)f(Vb.",vn) 0' where the sum istaken over allpermutations aof{I,...,n}andl(a) isthe signofthepermutation. XIII,4 DETERMINANTS 513 Fordeterminants, Ishall follow Artin's treatment inGalois Theory. Byan nxndeterminant weshall mean amapping det :Matn(R) R also written D:Mat,.(R) R which, when viewed asafunction ofthecolumn vectors A1,...,Anofamatrix A,ismultilinear alternating, and such that D(I)=1.Inthischapter,we use mostly theletter Dtodenote determinants. We shall prove later that determinants exist. For themoment, wederive properties. Theorem 4.4. (Cramer's Rule). LetA1 ,.. .,Anbecolumn vectors ofdimen- sion n.LetXl,...,XnERbesuch that xlAl+...+xnAn=B forsome column vector B.Then foreach iwehave XiD(A1,...,An)=D(A1,...,B,...,An), where Binthis last line occurs inthei-thplace. Proof. Say i=1.Weexpand n D(B, A2 ,...,An)=LxjD(Aj,A2 ,...,An), j=1 and useProposition 4.1togetwhat wewant (all terms ontheright areequal to0except the onehaving x1init). Corollary 4.5. Assume that Risafield. Then A1,...,An arelinearly dependent ifandonlyifD(A 1,...,An)=o. Proof. Assume wehave arelation XlA1+...+xnAn=0 with XiER.Then XiD(A)=0foralli.Ifsome Xi=I0then D(A)=o.Con- versely, assume that A1,...,Anarelinearly independent. Then we can express theunit vectors e1,...,enaslinear combinations e1=bllA1+...+blnAn , en=bn1A1+...+bnnAn 514 MATRICES AND LINEAR MAPS XIII,4 with bijER.But 1=D(e1,.. .,en). Usingaprevious lemma, weknow that this can beexpanded into asum of terms involving D(A 1,...,An), and hence D(A) cannot beo. Proposition 4.6. Ifdeterminants exist, they areunique. IfA1,. ..,An are thecolumn vectors ofdimension n,ofthematrix A=(aij),then D(A 1,...,An)=L£(a)aa(l), 1...aa(n),n' 0' where the sum istaken over allpermutationsaof{I,...,n},and£(a) isthe signofthepermutation. Proof. Let el ,...,enbetheunit vectors asusual. We canwrite A1=a11e1+...+an 1en, An =alnen+...+annen. Therefore D(A 1,...,An)=L£(a)aa(l),l...aa(n),n 0' bythelemma. This proves that thevalue ofthedeterminant isuniquely deter- mined and isgiven bytheexpected formula. Corollary 4.7. Let lp:R R'bearing-homomorphism into acommutative ring. IfAisasquare matrix inR,define lpA tobethematrix obtained by applying qJtoeach component ofA.Then lp(D(A»=D(lpA). Proof. Apply lptotheexpression ofProposition 4.6. Proposition 4.8.IfAisasquare matrix inRthen D(A)=D('A). Proof. Inaproduct aa(l),l...aa(n),n each integer kfrom 1tonoccurs preciselyonce among theintegers a(1),...,a(n). Hence we can rewrite thisproduct intheform al,a-t(l)...an,a-1(n). XIII,4 DETERMINANTS 515 Since £(a)=£(a- 1),wecan rewrite the sum inProposition4.6intheform L£(a-l)al,a-1(l)...a",a 1(,,). a Inthis sum, each term corresponds toapermutation a.However, asaranges over allpermutations, sodoes a-1.Hence our sum isequal to L£(a)a 1,a(l)...an,a(,,), a which isnone other than DCA), aswas tobeshown. Corollary 4.9. The determinant ismultilinear andalternating with respect tothe rows ofamatrix. We shall now prove existence, and prove simultaneously one additional important property ofdeterminants. When n=1,wedefine D(a)=aforanyaER. Assume that wehave proved theexistence ofdeterminants forallintegers <n(n>2). Let Abean nxnmatrix inR,A =(aij).We letAijbethe (n-1)x(n-1)matrix obtained from Abydeleting thei-th row andj-th column. Let ibeafixed integer,1<i<n.Wedefine inductively D(A)=(_1)i+ lailD(A il)+...+(-1)i+n ainD(A i,,). (This isknown astheexpansion ofDaccording tothe;-throw.) Weshall prove that Dsatisfies thedefinition ofadeterminant. Consider Dasafunction ofthek-th column, and consider any term _ i+j(1)aijD(A ij). Ifj=Ikthenaijdoes notdependonthek-thcolumn, andD(Aij) depends linearly onthek-th column. Ifj=k,thenaijdepends linearly onthek-th column, and D(Aij)does notdepend onthek-th column. Inanycase our term depends linearlyonthek-th column. Since D(A) isasum ofsuch terms, itdepends linearly onthek-th column, and thus Dismultilinear. Next, suppose that two adjacent columns ofAareequal, say Ak=Ak+ 1. Letj beanindex =1=kand =Ik+1.Then thematrix Aijhas twoadjacent equal columns, and hence itsdeterminant isequal toO.Thus theterm corresponding toanindexj=f.kork+1givesazero contribution toD(A). The other two terms can bewritten i+k i+k+ 1(-1) aikD(A ik)+(-1) ai,k+ 1D(Ai,k+ 1). The two matrices AikandAi,k+ 1areequal because ofourassumption that the k-th column ofAisequal tothe(k+1)-th column. Similarly, aik=ai,k+1. 516 MATRICES AND LINEAR MAPS XIII,4 Hence these two terms cancel since they occur with opposite signs. This proves that ourform isalternating, andgives: Proposition 4.10. Determinants exist and satisfy the ruleofexpansion according torows and columns. (For columns, we usethefact thatD(A)=DCA).) Example. We mention explicityoneofthe most important determinants. LetXl'. . .,Xnbeelements of acommutative ring. The Vandermonde deter- minant V=V(x I,. . .,xn)ofthese elements isdefined tobe 1 1 Xl X2 V= n-I n-IXl X21 Xn n-IXn whose value can bedetermined explicitly tobe V=n(x.-x;).'<. J I] Ifthering isentire and X;=1=Xjfori=1=j,itfollows that V =1=o.Theproof for thestated value isdone bymultiplying thenext tothelast rowbyXIandsubtracting from the last row. Then repeat this step going upthe rows, thus making the elements ofthefirst column equal to0,except for 1intheupper left-hand corner. One can then expand according tothefirst column, and use thehomogeneity property and induction toconclude theproof oftheevaluation ofV. Theorem 4.11. Let Ebeamodule over R,and let Vl,...,Vnbeelements ofE. Let A=(aij)beamatrix inR,and let A(:)=(:) Let beann-multilinear alternating map onE.Then (Wl'...,wn)=D(A) (Vl'...,vn). Proof. Weexpand (allVl +...+alnv n,..., anlv 1+...+annv n), and findprecisely what wewant, taking into account D(A)=DCA). XIII,4 DETERMINANTS 517 LetE,Fbemodules, and letL:(E, F)denote the setofn-multilinear alter- nating maps ofEinto F.IfF=R,wealsowriteL:(E, R)=L:(E). Itisclear thatL:(E, F)isamodule over R,i.e.isclosed under addition andmultiplication byelements ofR. Corollary 4.12. LetEbeafree module over R,and let{VI'...,vn}beabasis. Let Fbeany module, and let WEF.There exists aunique n-multilinear alternating map w:Ex...xE F such thatw(Vl'...,vn)=w. Proof. Without loss ofgenerality, wemay assume that E=R(n), andthen, ifA1,...,Anarecolumn vectors, wedefine w(A 1,..., An)=D(A)w. Then wobviously hastherequired properties. Corollary 4.13. IfEisfree over R,and has abasis consisting ofnelements, thenL:(E) isfree over R,and has abasis consisting of1element. Proof. Weletlbethemultilinear alternating map taking thevalue 1on a basis {v1,...,vn}.Any elementq>EL:(E) can then bewritten inaunique way asCl' with some cER,namely c=q>(Vl"..,vn).This proves what wewanted. Any two bases ofL:(E) inthepreceding corollary differ byaunit inR.In other words, if isabasis ofL:(E), then =Cl=cfor some cER,and c must beaunit. Ourldepends ofcourse onthechoice ofabasis forE.When weconsider R(n), our determinant Disprecisely l'relative tothestandard basis consisting oftheunit vectors el ,...,en. Itissometimes convenient terminology tosaythat any basis ofL:(E) isa determinant onE.Inthat case, thecorollary toCramer's rule can bestated as follows. Corollary 4.14. Let Rbeafield. Let Ebeavector space ofdimension n. Let beanydeterminant onE.Let vl,...,VnEE.Inorder that {vl,...,vn} beabasis ofEitisnecessary andsufficient that (Vl"'.'vn)=Io. Proposition 4.15. LetA,Bbenxnmatrices inR.Then D(AB)=D(A)D(B). 518 MATRICES AND LINEAR MAPS XIII,4 Proof. This isactuallyacorollary ofTheorem 4.11. We take vl'...,Vn tobetheunit vectors e1 ,...,en,and consider AB Weobtain D(w l,...,Wn)=D(AB)D(e1 ,...,en). Ontheother hand, byassociativity, applying Theorem 4.11 twice, D(Wb...,Wn)=D(A)D(B)D(e1,...,en). Since D(e1 ,...,en)=1,ourproposition follows. LetA=(aij)bean nxnmatrix inR.Welet A=(bij) bethematrix such that i+jb.. =(-1)D(A..) IJ JI. (Note thereversal ofindices!) Proposition 4.16. Let d=D(A). Then AA =AA =dIe The determinant D(A) isinvertible inRifandonlyifAisinvertible, and then 11- A- = dA. Proof For anypair ofindices i,ktheik-component ofAA is ailblk +ai2b2k +...+ainbnk=ail(-l)k+ ID(A kl)+...+ain(-l)k+nD(A kn). Ifi=k,then this sum issimply theexpansion ofthedeterminant according tothei-th row, and hence this sum isequal tod.Ifi=Ik,letAbethematrix obtained from Abyreplacing thek-th rowbythei-th row, andleaving allother rows unchanged. Ifwe delete thek-th row andthej-th column from A,weobtain the same matrix asbydeleting thek-th rowandj-th column from A.Thus Ak'=Ak' 1 l' and hence our sum above can bewritten k+ 1-k+- ail(-1) D(A kl)+...+ain(-1) nD(A kn). XIII,4 DETERMINANTS 519 This istheexpansion ofthedeterminant ofAaccording tothei-th row. Hence D(A)=0,and our sum isO.We have therefore proved that theik-component ofAA isequal todifi=k(i.e.ifitisadiagonal component), and isequalto0 otherwise. This proves thatAA =dIeOntheother hand, we seeatonce from - ......, thedefinitions that =!4.Then - -......., t(AA)=!4 = =dI, andconsequently, AA =dIalso, since t(dI)=dIeWhen disaunit inR,then A isinvertible, itsinverse being d-1A.Conversely, ifAisinvertible, and AA-1=I, then D(A)D(A -1)=1,and hence D(A) isinvertible, aswas tobeshown. Corollary 4.17. LetFbeanyR-module, and letWI'...,Wnbeelements of F.LetA=(aij)bean nxnmatrix inR.Let allwl+.. ·+alnw n=VI anlwl+.. ·+annw n=vn. Then one can solve explicitly D(A)wI WI VnVI - =D(A)=A D(A)w nWn Inparticular, ifVi=0foralli,then D(A)wi=0foralli.IfVi=0foralli and Fisgenerated byWI,...,Wn'then D(A)F=o. Proof. This isimmediate from therelation AA=D(A)I, using theremarks in3about applying matrices tocolumn vectors whose components lieinthe module. Proposition 4.18. Let E,Fbefree modules ofdimension nover R.Let f:E Fbealinear map. Let CB, CB'bebases ofE,Frespectively over R. Thenfisanisomorphism ifandonlyifthedeterminant ofitsassociated matrix M,(f) isaunit inR. Proof Let A =M,(f). Bydefinition, fisanisomorphism ifandonly ifthere exists alinear map 9:F Esuch that 90f=idandfog=ideIffis anisomorphism, and B=M'(g), then AB =BA =I.Taking thedeterminant oftheproduct,weconclude thatD(A) isinvertible inR.Conversely, ifD(A) isaunit, then we can define A-IbyProposition 4.16. This A-I istheassociated matrix ofalinear map g:F Ewhich isaninverse forf,asdesired. Finally,weshall define thedeterminant ofanendomorphism. 520 MATRICES AND LINEAR MAPS XIII,4 Let Ebeafree module over R,and let(Bbeabasis. Letf:E Ebean endomorphism ofE.Let M =M(f). If(B'isanother basis ofE,and M' =M:(f), then there exists aninvertible matrix Nsuch that M' =NMN-l . Taking thedeterminant, we seethat D(M')=D(M). Hence thedeterminant does notdependonthechoice ofbasis, andwill becalled thedeterminant ofthe linear mapfWeshall give below acharacterization ofthis determinant which does notdependonthechoice ofabasis. Let Ebeany module. Then wecanview L:(E) asafunctor inthevariable E (contravariant). Infact, we can view L:(E, F) asafunctor oftwo variables, contra variant inthefirst, and covariant inthesecond. Indeed, suppose that E'.4 E isalinear map. Toeach multilinear map q>:E(n) Fwe can asociate the composite map q>0j'(n), E' ,/(n) (()X... xE-----.E x... xE F where f(n) istheproduct offwith itself ntimes. The map L:(f):L:(E, F) L:(E', F) given by q> q>0f(n), isobviouslyalinear map, which defines ourfunctor. We shall sometimes write f*instead ofL;(f). Inparticular, consider the case when E=E'and F=R.Wegetaninduced map f*:L:(E) L:(E). Proposition 4.19. LetEbeafree module over R,ofdimension n.Let{d}bea basis ofL:(E). Letf:E Ebeanendomorphism ofE.Then f*=D(f). Proof This isanimmediate consequence ofTheorem 4.11. Namely, we let{vt,.. .,vn}beabasis ofE,and then take A(or)tobeamatrix offrelative tothis basis. Bydefinition, f*(Vl'...,vn)=(f(Vl)'...,f(v n», XIII,4 DETERMINANTS 521 andbyTheorem 4.11, this isequal to D(A) (Vb...,vn). ByCorollary 4.12, weconclude thatf*=D(A) since both ofthese forms take onthe same value on(VI'...,vn). The above considerations have dealt with thedeterminant asafunction on allendomorphisms ofafree module. One can also view itmultiplicatively,as ahomomorphism. det: GLn(R) R* from thegroup ofinvertible nxnmatrices over Rinto thegroup ofunits ofR. The kernel ofthishomomorphism, consisting ofthose matrices with deter- minant 1,iscalled thespecial linear group, and isdenoted bySLn(R). We now give anapplication ofdeterminants tothesituation ofafree module and asubmodule considered inChapter III, Theorem 7.8. Proposition 4.20. Let Rbeaprincipalentire ring. LetFbeafree module over Rand letMbeafinitely generated submodule. Let{el'...,em'. ..}be abasis ofFsuch that there exist non-zero elements ai,...,amERsuch that: (i)The elements aiel,. . .,amemformabasis ofMover R. (ii) Wehave a;Ia;+ Ifori=1,..., m-1. LetLbethe setofalls-multilinear alternating formsonF.LetJsbetheideal generated byallelements f(YI'. . .,Ys)'withfELand YI'.. .,YsEM.Then Js=(a 1...as). Proof. Wefirst show that Jsc(al...as).Indeed, anelement YEM can be written intheform Y=clalel +...+crare r. Hence ifYl'. . .,YsEM, andfis multilinear alternatingonF,thenf(YI'. . .,Ys) isequal toasum interms oftype Cit...Cisait.. .aisf(eit'...,ei s). This isnon-zero only wheneit'...,eisaredistinct, inwhich case theproduct al...asdivides this term, and hence Jsiscontained inthestated ideal. Conversely,weshow that there exists ans-multilinear alternating form which gives precisely thisproduct. Wededuce this from determinants. We canwrite Fasadirect sum F=(el'...,er) F'r 522 MATRICES AND LINEAR MAPS XIII,5 with some submodule Fr.Let}; (i=1,..., r)bethelinear map F-+Rsuch that};(ej)=ij'and such that}; hasvalue 0onFr. For Vl,...,VsEFwedefine f(vl'.. .,vs)=det(};(v j». Thenfismultilinear alternating and takes onthevalue f(e2'. ..,es)=1, aswell asthevalue f(aleb...,ase s)=al...as. This proves theproposition. The uniqueness ofChapter III,Theorem 7.8 isnow obvious, since first(al) isunique, then (ala2) isunique and thequotient (a2) isunique, and soforth by induction. Remark. Compare theabove theorem with Theorem 2.9ofChapter XIX, inthetheory ofFitting ideals, which givesafancier context fortheresult. 5. DUALITY Let Rbe acommutative ring, and letE,Fbemodules over R.An R- bilinear form onExFisamap f:ExFR having thefollowing properties: For each xEE,themap yf(x, y) isR-linear, and foreach yEF,themap xf(x, y) isR-linear. We shall omit theprefix R-inthe rest ofthis section, and write <x,y)/or<x,y)instead off(x, y).IfxEF,wewrite x-1yif<x,y)=O. Similarly, ifSisasubset ofF,wedefine x-1Sifx-1yforallYES. Wethen say that xisperpendicular toS.We let SJ..consist ofallelements ofEwhich are perpendicular toS.Itisobviously asubmodule ofE.We define perpendicu- larity ontheother side inthe same way. We define thekernel off ontheleft tobeF1.and thekernel ontheright tobeE1.. We saythatfis non-degenerate ontheleftifitskernel ontheleft isO.We saythatfisnon-degenerate onthe right ifitskernel ontheright iso.IfEoisthekernel offontheleft, then we XIII,5 DUALITY 523 getaninduced bilinear map EIEo xF R which isnon-degenerate ontheleft, asone verifies trivially from thedefinitions. Similarly, ifF0isthekernel offontheright, weget aninduced bilinear map EIEoxFIFo R which isnon-degenerateoneither side. This map arises from thefact that the value <x,y)depends onlyonthe coset ofxmodulo Eoand the coset ofy modulo Fo. Weshall denote byL2(E,F;R)the setofallbilinear maps ofExFinto R. Itisclear that this setisamodule (i.e. anR-module), addition ofmaps being the usual one, and alsomultiplication ofmaps byelements ofR. The formfgives rise toahomomorphism lpf:E HomR(F, R) such that lpf(x)(y)=f(x, y)=<x,y), forallXEEandye F.Weshall callHomR(F, R)thedual module ofF,and denote itbyFV .We have anisomorphism L2(E,F;R) HomR(E, HomR(F, R) given byft---+ lpf,itsinverse being defined intheobvious way: If lp:E HomR(F, R) isahomomorphism,weletfbesuch that f(x, y)=lp(x) (y). We shall saythatfisnon-singularontheleftiflpfisanisomorphism, in other words ifour form can beused toidentify Ewith thedual module ofF. We define non-singularontheright inasimilar way, and say thatfisnon- singular ifitisnon-singularontheleftand ontheright. Warning: Non-degeneracy does notnecessarily imply non-singularity. Weshall now obtain anisomorphism IEndR(E)f-+L2(E,F;R) depending onafixed non-singular bilinear mapf:ExF R. 524 MATRICES AND LINEAR MAPS XIII,5 Let AEEndR(E) bealinear map ofEinto itself. Then themap (x,y) <Ax, y)=<Ax, y)/ isbilinear, and inthis way, weassociate linearly with each AEEndR(E)abilinear map inL2(E,F;R). Conversely, leth:ExF Rbebilinear. Given xEE,themap hx:F R such that hx(Y)=h(x,y)islinear, and isinthedual space FV .Byassumption, there exists aunique element x'EEsuch that forallyEFwehave h(x,y)=<x',y). Itisclear that theassociation x x'isalinear map ofEinto itself. Thus with each bilinear map ExF Rwehave associated alinear mapE E. Itisimmediate that themappings described inthelast twoparagraphs are inverse isomorphisms between EndR(E) and L2(E, F;R). Weemphasize of course that they depend onourform f. Ofcourse, wecould also have worked ontheright, and thus wehave a similar isomorphism IL2(£,F;R)+-+EndR(F)I depending also onourfixed non-singular form f. As anapplication, letA :E Ebelinear, and let(x,Y) (Ax, y)beits associated bilinear map. There exists aunique linear map tA :F F such that <Ax, y)=<x,'Ay) forallxEEand YEF.WecalltAthetranspose ofAwith respect tof Itisimmediately clear that if,A,Barelinear maps ofEinto itself, then for CER, t(cA)=c'A, t(A+B)=tA+tB, and t(AB)=tBtA. More generally, letE,Fbemodules with non-singular bilinear forms denoted by( ,)Eand( ,)Frespectively. LetA:E Fbe alinear map. Then bythe non-singularity of( ,)Ethere exists aunique linear map tA:F Esuch that (Ax, Y)F=(x,tAY)E forallxEEand yEF. We also call tAthetranspose with respect tothese forms. Examples. For anice classical example of atranspose,seeExercise 33. For thesystematic study when alinear map isequal toitstranspose,see the XIII,5 DUALITY 525 spectraltheorems ofChapter XV. Next Igive another example of atranspose from analysisasfollows. Let Ebethe(infinite dimensional) vector space of exfunctions onR,having compact support, i.e.equal to0outside some finite interval. We define thescalar product x (f,g)=ff(x)g(x)dx. -x LetD:E Ebethederivative. Then one has theformula (Df, g)=-(f, Dg). Thus one says that tD=-D, even though thescalar product isnot"non-singular", butmuch oftheformalism ofnon-singular forms goes over. Also inanalysis, one puts various norms onthe spaces and one extends thebilinear form by continuity tothecompletions, thus leaving thedomain ofalgebra toenter the domain ofestimates (analysis). Then thespectral theorems become more com- plicatedinsuch analytic contexts. Let usassume that E=F.Letf:ExE Rbebilinear. By anauto- morphism ofthepair(E,/), orsimply off,weshall mean alinear automorphism A :E Esuch that <Ax, Ay>=<x,y> forallx,yEE.The group ofautomorphisms offisdenoted byAut(f). Proposition 5.1. Letf:ExE Rbe anon-singular bilinear form. Let A:E Ebe alinear map. Then Aisanautomorphism offifandonlyif tAA =id,and Aisinvertible. Proof From theequality <x,y>=<Ax, Ay>=<x,tAAy> holding forallx,yEE,weconclude thattAA=idifAisanautomorphism off The converse isequally clear. Note. IfEisfree and finite dimensional, then theconditionfAA =id implies that Aisinvertible. Letf:ExE Rbe abilinear form. We say thatfissymmetric if f(x, y)=f(y, x)forallx,yEE.The setofsymmetric bilinear forms onEwill bedenoted byL;(E). Let ustake afixed symmetric non-singular bilinear form fonE,denoted by(x,y) <x,y>. Anendomorphism A:E Ewill besaid tobesymmetric with respect tofiffA =A.Itisclear that the setofsym- metric endomorphisms ofEisamodule, which weshall denote bySym(E). 526 MATRICES AND LINEAR MAPS XIII,5 Depending onourfixed symmetric non-singular f,wehave anisomorphism L;(E)+-+Sym(E) which wedescribe asfollows. If9issymmetric bilinear onE,then there exists aunique linear map Asuch that g(x, y)=(Ax, y) forallx,yEE.Using thefact that bothf,9aresymmetric,weobtain <Ax, y)=<Ay, x)=<y,'Ax)=<'Ax, y). Hence A='A. The association 9 Agives usahomomorphism from L;(E) into Sym(E). Conversely, given asymmetric endomorphism AofE,we can define asymmetric form bytherule (x,y) <Ax, y),and theassociation of this form toAclearly givesahomomorphism ofSym(E) intoL;(E) which is inverse tothepreceding homomorphism. Hence Sym(E) andL;(E)are iso- morphic. We recall that abilinear form g:ExE Rissaid tobealternating if g(x,x)=0forallxEE,andconsequently g(x,y)= -g(y,x)forallx,yEE. The setofbilinear alternating forms onEisamodule, denoted byL;(E). Letfbe afixed symmetric non-singular bilinear form on E.Anendo- morphism A:E Ewill besaid tobeskew-symmetric oralternating with respect tofif'A = -A,and also <Ax, x)=0forallxEE.Ifforall aER, 2a =0implies a=0,then this second condition <Ax, x)=0isredundant, because <Ax, x)= -<Ax, x)implies <Ax, x)=O.Itisclear that the setof alternating endomorphisms ofEisamodule, denoted byAlt(E). Depending onourfixed symmetric non-singular formf,wehave anisomorphism L(E)+-+Alt(E) described asusual. If9isanalternating bilinear form onE,itscorresponding linear map Aisthe one such that g(x,y)=<Ax, y) forallx,yEE.One verifies trivially inamanner similar tothe one used inthe symmetric case that thecorrespondence 9 Agives usour desired iso- morphism. Examples. Let kbe afield and letEbe afinite-dimensional vector space over k.Letf: ExE--+Ebe abilinear map, denoted by(x,y) xy.Toeach XIII,6 MATRICES AND BILINEAR FORMS 527 XEE,weassociate thelinear map Ax:E Esuch that Ax(Y)=xy. Then themap obtained bytaking thetrace, namely (x,y) tr(A xy) isabilinear form onE.IfxY=yx,then thisbilinear form issymmetric. Next, letEbethespace ofcontinuous functions ontheinterval [0,1].Let K(s, t)be acontinuous function oftwo real variables defined onthe square o<s< 1and 0<t<1.For lp,t/JEEwedefine <cp, t/J>=IIcp(s)K(s, t)t/J(t) dsdt, thedouble integral being taken onthesquare. Then weobtain abilinear form onE.IfK(s, t)=K(t, s),then thebilinear form issymmetric. When wediscuss matrices andbilinear forms inthenext section, thereader will note thesimilarity between thepreceding formula and thebilinear form defined byamatrix. Thirdly, letUbeanopen subset ofareal Banach space E(or afinite-dimen- sional Euclidean space, ifthereader insists), and letf:U Rbeamap which istwice continuously differentiable. For each xEU, the derivative Df(x): E Risacontinuous linear map, and the second derivative D2f(x) can beviewed asacontinuous symmetric bilinear map ofExEinto R. 6. MATRICES AND BILINEAR FORMS Weshall investigate therelation between theconcepts introduced above and matrices. Letf:ExF Rbebilinear. Assume that E,Farefree over R.Let (B ={vl,...,vm}beabasis forEover R,and let(B' ={Wl,...,wn}beabasis forFover R.Letgij=<Vi'Wj).If x=X1V1+...+xmvm and Y=YlWl+...+Ynwn areelements ofEand Frespectively, with coordinates Xi'YjER,then m n <x,y)=LLgijXiYj' i= 1j=1 528 MATRICES AND LINEAR MAPS XIII,6 LetX,Ybethecolumn vectors ofcoordinates forx,yrespectively, with respect toour bases. Then <x,y)=tXGY where Gisthematrix (gij).Wecould write G=M,(f). Wecall Gthematrix associated with theform/relative tothebases CB,(1\'. Conversely, givenamatrix G(ofsize mxn),wegetabilinear form from themap (X,Y) tXG Y. Inthis way,wegetacorrespondence from bilinear forms tomatrices andback, and itisclear that thiscorrespondence induces anisomorphism (ofR-modules) L2(E,F;R)+-+Matmxn(R) given byfM,(f). The two maps between these two modules which wedescribed above areclearly inverse toeach other. Ifwe have bases CB ={Vb...'vn}and CB' ={w}, ..., wn}such that <Vi'Wj)=ij,then wesaythat these bases aredual toeach other. Inthat case, ifXisthecoordinate vector ofanelement ofE,and Ythecoordinate vector of anelement ofF',then thebilinear maponX,Yhasthevalue X.Y=XlY1+...+XnYn given bytheusual dotproduct. Itiseasy toderive ingeneral how thematrix Gchanges when wechange bases inEand F'.However, weshall write down theexplicit formula only when E=F'and CB =CB'. Thus wehave abilinear formf: ExE R.Letebe another basis ofEand write X<Band Xeforthecolumn vectors belonging to anelement xofE,relative tothe two bases. Let Cbetheinvertible matrix M(id),sothat X<B=CXe. Then ourform isgiven by <x,y)=txetCGCY e. We seethat (1) M(f)=tCM(f)C. Inother words, thematrix ofthebilinear form changes bythetranspose. XIII,6 MATRICES AND BILINEAR FORMS 529 IfF isfreeover R,with abasis {11b.", 11n}, then HomR(F, R)isalsofree, and wehave adual basis {111,. . .,11}such that 'n('no)=5..°11 O'l ')" This hasalready been mentioned inChapter III, Theorem 6.1. Proposition 6.1. Let E,Fbefree modules ofdimension nover Rand let f:ExF' Rbe abilinear form. Then thefollowing conditions areequiv- alent: fisnon-singularontheleft. fisnon-singularontheright. fisnon-singular. The determinant ofthematrix offrelative toany bases isinvertible inR. Proof Assume thatfisnon-singularontheleft. Fix bases ofEand F relati vetowhich wewrite elements ofthese modules ascolumn vectors, and givingrise tothematrix GforfThen ourform isgiven by (X,Y)'XGY where X,Yare column vectors with coefficients inR.Byassumption themap X'XG givesanisomorphism between themodule ofcolumn vectors, and themodule ofrow vectors oflengthnover R.Hence Gisinvertible, and hence itsdeter- minant isaunit inR.The converse isequally clear, and ifdet(G) isaunit, we seethat themap YGY must also beanisomorphism between themodule ofcolumn vectors and itself. This proves our assertion. We shall now investigate how thetranspose behaves interms ofmatrices. LetE,Fbefree over R,ofdimension n. Letf: ExF-+Rbeanon-singular bilinear form, and assume givenabasis (BofEand (B'ofF.Let Gbethematrix offrelative tothese bases. Let A:E-+Ebe alinear map. IfxEE,yEF',letX,Ybetheir column vectors relative to(B,(B'. LetMbethematrix ofArelative to(B.Then for xEEand yEFwehave <Ax, y)='(MX)GY='X'MGY. LetNbethematrix of'Arelative tothebasis (B'. Then NY isthecolumn vector of'Ayrelative to(B'. Hence <x,'Ay)='XGNY. 530 MATRICES AND LINEAR MAPS XIII,6 From this weconclude that tMG =GN, and since Gisinvertible, we can solve forNinterms ofM. Weget: Proposition 6.2. LetE,Fbefreeover R,ofdimension n.Letf:ExF R beanon-singular bilinear form. Let CB,CB'bebases ofEand Frespectively over R,and letGbethematrix offrelative tothese bases. Let A:E Ebea linear map, and letMbeitsmatrix relative toCB. Then thematrix oftA relative toCB'is (G-1)tMG. Corollary 6.3. IfGistheunit matrix, then thematrix ofthetranspose is equal tothetranspose ofthematrix. Interms ofmatrices and bases, weobtain thefollowing characterization for amatrix toinduce anautomorphism oftheform. Corollary 6.4. Let the notation be asinProposition 6.2, and letE=F, CB =03'. An nxnmatrix Misthematrix ofanautomorphism oftheform f(relative toourbasis) ifandonlyif tMGM=G. Ifthiscondition issatisfied, then inparticular, Misinvertible. Proof. We use the definitions, together with the formula given in Proposition 6.2. We note that Misinvertible, forinstance because itsdeter- minant isaunit inR. Amatrix Missaid tobesymmetric (resp. alternating) iftM =M(resp. tM = -Mand thediagonal elements ofMare 0). Letf:ExE Rbe abilinear form. We say thatfissymmetric if f(x, y)=f(y, x)forallx,yEE.We saythatfisalternating iff(x, x)=0for allxEE. Proposition 6.5. Let Ebeafree module ofdimension nover R,and let CB beafixed basis. The map fM(f) induces anisomorphism between themodule ofsymmetric bilinear forms on ExE(resp. themodule ofalternating forms onExE)and themodule of symmetric nxnmatrices over R(resp. the module ofalternatingnxn matrices over R). XIII,7 SESQUILINEAR DUALITY 531 Proof. Consider first thesymmetric case. Assume thatfis symmetric.In terms ofcoordinates, letG=M(f). Our form isgiven by'XGY which must beequal to'YGX bysymmetry. However, 'XGY may beviewed asa1x1 matrix, and isequal toitstranspose, namely 'Y'GX. Thus 'YGX='Y'GX forallvectors X, Y.Itfollows that G='G.Conversely, itisclear that any symmetric matrix defines asymmetric form. Asforthealternating case, replacingxbyx+yintherelation <x,x)=0 weobtain <x,y)=<y,x)=o. Interms ofthecoordinate vectors X,Yand thematrix G,thisyields 'XGY +'YGX =O. Taking thetranspose of,say, thesecond ofthe 1x1matrices entering inthis relation, yields (for allX,Y): tXGY +'X'GY =O. Hence G+'G=O.Furthermore, letting Xbeanyone oftheunit vectors '(0,...,0,1,0,.. .,0) and using the relation 'XGX =0,we seethat thediagonal elements ofG must beequal toO.Conversely, ifGisan nxnmatrix such that 'G+G=0, and such that gii=0fori=1,...,nthen one verifies immediately that the map (X,Y)'XGY defines analternating form. This proves ourproposition. Ofcourse, ifasisusually the case, 2isinvertible inR,then ourcondition tM = -Mimplies that thediagonal elements ofMmust beo.Thus inthat case, showing that G+tG=0implies that Gisalternating. 7. SESQUILINEAR DUALITY There exist forms which are notquite bilinear, and forwhich theresults described above hold almost without change, but which must behandled separately forthesake ofclarity inthenotation involved. 532 MATRICES AND LINEAR MAPS XIII,7 Let Rhave anautomorphism ofperiod 2.Wewrite thisautomorphism as a a(and think ofcomplex conjugation). Following Bourbaki, wesaythat amap f:Exf-+R isasesquilinear form ifitisZ-bilinear, and iffor xEE,YEF,and aERwe have f(ax, y)=af(x, y) and f(x, ay)=af(x, y). (Sesquilinearmeans I!times linear, sotheterminology israther good.) LetE,E'bemodules. Amap qJ:E E'issaid tobeanti-linear (orsemi- linear) ifitisZ-linear, and qJ(ax)=aqJ(x) forallxEE.Thus wemay saythat asesquilinear formislinear initsfirst variable, and anti-linear initssecond variable. We letHomR(E, E')denote the module ofanti-linear maps ofE into E'. We shall now gosystematIcally through the same remarks that wemade previously forbilinear forms. We define perpendicularityasbefore, and also thekernel ontheright and ontheleftforanysesquilinear formf.These kernels aresubmodules, sayEo andfo, and weget aninduced sesquilinear form EIEoxFIFo R, which isnon-degenerateoneither side. LetFbeanR-module. Wedefine itsanti-module Ftobethemodule whose additive group isthe same asf,and such that theoperation RxF Fis given by (a,y) aYe Then Fisamodule. We have anatural isomorphism HomR(F, R)+-+HomR(f, R), asR-modules. Thesesquilinear formf:ExF-+Rinduces alinear map qJf:E HomR(F, R). We saythatfis non-singular ontheleftifqJfisanisomorphism. Similarly,we have acorresponding linear map qJf:F HomR(E, R) XIII,7 SESQUILINEAR DUALITY 533 from Finto thedual space ofE,and wesaythatfisnon-singularontheright ifqJfisanisomorphism. We saythatfisnon-singular ifitisnon-singularon theleftand ontheright. Weobserve that oursesquilinear formf can beviewed asabilinear form f:ExF R, and that our notions ofnon-singularityarethen compatible with those defined previously forbilinear forms. Ifwehave afixed non-singular sesquilinear form onExF,then depending onthis form, weobtain anisomorphism between themodule ofsesquilinear forms onExFand themodule ofendomorph isms ofE.We also obtain an anti-isomorphism between these modules and themodule ofendomorphisms ofF.Inparticular,we can define theanalogue ofthetranspose, which inthe presentcase weshall call theadjoint. Thus, letf: ExF Rbeanon-singular sesquilinear form. Let A:E-+Ebealinear map. There exists aunique linear map A*:f"-+f" such that (Ax, y)=(x,A*y) forallxEEand yEf".Note that A*islinear, notanti-linear. Wecall A*the adjoint ofAwith respect toourform}: Wehave therules (cA)*=cA*, (A+B)*=A*+B*, (AB)*=B*A* foralllinear maps A,BofEinto itself, and CER. Let us assume that E=f".Letf:ExE-+Rbesesquilinear. By an automorphism offweshall mean alinear automorphism A:E Esuch that (Ax, Ay)=(x,y) justaswedidforbilinear forms. Proposition 7.1. Letf:ExE Rbe anon-singular sesquUinear form. Let A :E Ebealinear map. Then Aisanautomorphism o.f.fifandonly fA*A=id,and Aisinvertible. The proof, and also theproofs ofsubsequent propositions, which are completely similar tothose ofthebilinear case, will beomitted. Asesquilinear form g:ExE Rissaid tobehermitian if g(x,y)=g(y,x) forallx,yEE.The setofhermitian forms onEwill bedenoted byL;(E). Let Robethesubring ofRconsisting ofallelements fixed under ourautomorphism 534 MATRICES AND LINEAR MAPS XIII,7 a a(i.e.consisting ofallelements aERsuch that a=a).Then L(E) isan Ro-module. Let ustake afixed hermitian non-singular formfonE,denoted by (x,y) <x,y). Anendomorphism A :E Ewill besaid tobehermitian with respect tofifA*=A.Itisclear that the setofhermitian endomorphisms isanRo-module, which weshall denote byHerm(E). Depending onourfixed hermitian non-singular formf,wehave anRo-isomorphism L(E) Herm(E) described intheusual way. Ahermitian form gcorresponds toahermitian map Aifandonly if g(x,y)=<Ax, y) forallx,yEE. We can now describe therelation between ourconcepts and matrices, just aswedidwith bilinear forms. We start with asesquilinear formf:ExF R. IfE,f"arefree, and wehave selected bases asbefore, then we canagain associate amatrix Gwith theform, and interms ofcoordinate vectors X,Y oursesquilinear form isgiven by (X,y)tXG Y, where Yisobtained from Ybyapplying theautomorphism toeach component ofY. IfE=F'and we usethe same basis ontheright and ontheleft, then with the same notation asthat used informula (I),iffissesquilinear, theformula now reads (IS) M(f)=teM(f) C . Theautomorphism appears. Proposition 7.2. Let E,f"befree modules ofdimension nover R,and let f':ExF Rbe asesquilinear form. Then thefollowing conditions are equivalent. fisnon-singular ontheleft. fisnon-singularontheright. fisnon-singular. The determinant ofthematrix offrelative toany bases isinvertible inR. XIII,7 SESQUILINEAR DUALITY 535 Proposition 7.3. LetE,Fbefree over R,ofdimension n.Letf:ExF R beanon-singular sesquilinear form. Let CB,CB'bebases ofEandFrespectively over R,and letGbethematrix offrelative tothese bases. Let A:E Ebe alinear map, and letMbeitsmatrix relative to<:B. Then thematrix ofA* relative toCB'is (G-l)'MG . Corollary 7.4. IfGistheunitmatrix, then thematrix ofA*isequal to'M . Corollary 7.5. Let thenotation be asintheproposition, and let CB =CB' beabasis ofE.An nxnmatrix Misthematrix ofanautomorphism off (relative toourbasis) ifandonlyif tMG M =G. Amatrix Missaid tobehermitian iftM =M . LetRobeasbefore thesubring ofRconsisting ofallelements fixed under ourautomorphism a a(i.e.consisting ofallelements aERsuch that a=a). Proposition 7.6. Let Ebe afree module ofdimension nover R,and let CB beabasis. The map fM(f) induces anRo-isomorphism between theRo-module ofhermitian forms onE and theRo-module ofnxnhermitian matrices inR. Remark. Ifwe had assumed atthebeginning that ourautomorphism a ahasperiod 2or 1(i.e. ifweallow ittobetheidentity), then theresults onbilinear and symmetric forms become special cases ofthe results ofthis section. However, thenotational differences aresufficiently disturbing towarrant arepetition oftheresults aswehave done. Terminology For some confusing reason, thegroup ofautomorphisms ofasymmetric (resp. alternating, resp. hermitian) form onavector space iscalled theorthogonal (resp. symplectic, resp. unitary) group oftheform. The word orthogonal is especially unfortunate, because anorthogonal map preservesmore than orthogonality: Italso preserves the scalar product, i.e.length. Furthermore, theword symplectic isalso unfortunate. Itturns outthat one can carry out a discussion ofhermitian forms over certain division rings (having automorphisms oforder 2),and their group ofautomorphisms have also been called symplectic, thereby creating genuine confusion with the useoftheword relative toalter- nating forms. 536 MATRICES AND LINEAR MAPS XIII,8 Inorder tounIfy andimprove theterminology, Ihave discussed the matter with several persons, and itseems that one could adopt thefollowing con- ventions. Assaid inthetext, thegroup ofautomorphisms ofanyformf isdenoted by Aut(f). Ontheother hand, there isastandard form, described over thereal numbers interms ofcoordinates by f(x, x)=xi+. ..+x;, over thecomplex numbers by f(x, x)=XIX I+...+XnX n' and over thequaternions bythe same formula asinthecomplexcase. The group ofautomorphisms ofthis form would becalled theunitary group, and bedenoted byUn. The points ofthis group inthereals (resp. complex, resp. quaternions) would bedenoted by Un(R), Un(C), Un(K), and these three groups would becalled therealunitary group (resp. complex unitary group, resp. quaternion unitary group). Similarly, thegroup ofpoints ofUninany subfield orsubring kofthequaternions would bedenoted byUn(k). Finally, iffisthestandard alternating form, whose matrix is (-). onewould denote itsgroup ofautomorphisms byA2n'and callitthealternating form group,orsimply thealternating group, ifthere isnodanger ofconfusion with thepermutation group. The group ofpoints ofthealternating form group inafield kwould then bedenoted byA2n(k). Asusual, thesubgroup ofAut(f) consisting ofthose elements whose determinant is1would bedenoted byadding theletter Sinfront, and would still becalled thespecial group. Inthefour standard cases, thisyields SUn(R), SUn(C), SUn(K), SA2n(k). 8. THE SIMPLICITY OF SL2(F)/:I: 1 LetFbeafield. Let nbeapositive integer. ByGLn(F) we mean thegroup ofnxninvertible matrices over F.BySLn(F)we mean thesubgroup ofthose matrices whose determinant isequal to 1.ByPGLn(f) we mean thefactor group ofGLn(F') bythesubgroup ofscalar matrices (which areinthecenter). XIII,8 THE SIMPLICITY OFSL 2(F)/:t1 537 Similarly forPSLn(F). Inthissection, we areinterested ingivinganapplication ofmatrices tothegroup theoretic structure ofSL2.Theanalogous statements forSLnwith n>3will beproved inthe next section. The standard Borel subgroup BofGL2isthegroup ofallmatrices (:) with a,b,dEF'and ad =f.O.For the Borel subgroup ofSL2,werequire in addition that ad =1.By aBorel subgroup we mean asubgroup which is conjugate tothestandard Borel subgroup (whether inGL2orSL2).We let Ubethegroup ofmatrices u(b)=().with bEF. WeletAbethegroup ofdiagonal matrices ().with a,dEP. Let s(a)=( al)with aEF'* and w=(_ ). For the restofthissection, welet G=GL2(F) or SL2(F). Lemma 8.1. The matrices X(b)=()and Y(c)=() generate SL2(F). Proof. Multiplyinganarbitrary element ofSL2(F')bymatrices ofthe above typeontheright and onthe leftcorresponds toelementary row and column operations, that isadding ascalar multiple ofarow totheother, etc. Thus agiven matrix canalways bebrought into aform (a1) 538 MATRICES AND LINEAR MAPS XIII,8 bysuch multiplications.We want toexpress this matrix with a=1= 1intheform ()G)()e). Matrix multiplication will show that we can solve thisequation, byselecting x arbitrarily=1=0,then solving forb,e,and dsuccessivelysothat -x -b 1+bx=a, e= 1+bx'd= 1+be. Then one finds 1+be=(1+xb)-1and the twosymmetric conditions b+bed +d=0 e+bex +x=0, sowegetwhat wewant, andthereby prove thelemma. Let Ubethegroup oflower matrices (). Then we seethat wVw-1=U . Also note thecommutation relation (a0 )-1(dO ) wOdw =Oa' sownormalizes A.Similarly, wBw-l=B isthegroup oflower triangular matrices. We note that B=AU =VA, and also that Anormalizes V. There isadecomposition ofGintodisjoint subsets G=BuBwB. Indeed, view Gasoperatingontheleftofcolumn vectors. Theisotropy group of el=() isobviously U. The orbit Be1consists ofallcolumn vectors whose second XIII,8 THE SIMPLICITY OFSL 2(F)/::t1539 component isO.Ontheother hand, we1=(_). and therefore theorbit Bwelconsists ofallvectors whose second component is=I0,and whose first component isarbitrary. Since these two orbits ofBand BwB cover theorbit Gel, itfollows that theunion ofBand BwB isequal toG (because theisotropy group Uiscontained inB),and theyareobviously disjoint. This decomposition iscalled theBruhat decomposition. Proposition 8.2. The Borel subgroup Bisamaximal proper subgroup. Proof. BytheBruhat decomposition, any element not inBliesinBwB, sotheassertion follows since B,BwB cover G. Theorem 8.3. IfFhas atleastfour elements, then SL2(F) isequal toitsown commutator group. Proof. We have thecommutator relation (bymatrix multiplication) s(a)u(b)s(a)-lu(b)-l=u(ba2-b)=u(b(a2-1». Let G=SL2(F)forthisproof. We letG'bethe commutator subgroup, and similarly letB'bethecommutator subgroup ofB.Weprove thefirst assertion that G=G'. From thehypothesis that Fhas atleast four elements, we can find anelement a=I0inFsuch that a2=I1,whence thecommutator relation shows that B' =U.Itfollows that G' ::)U,and since G'isnormal, weget G' ::)wUw-1 . From Lemma 8.1, weconclude that G' =G. LetZdenote thecenter ofG.Itconsists of+I,that is+theidentity2x2 matrix ifG=SL2(F); and Zisthesubgroup ofscalar matrices ifG=GL2(F). Theorem 8.4. IfFhas atleastfour elements, then SL2(F)/Z issimple. Theproof will result from two lemmas. Lemma 8.5. The intersection ofallconjugates ofBinGisequal toz. Proof. We leave this tothereader, as asimple fact using conjugation with w. Lemma 8.6. Let G=SL2(F).IfHisnormal inG,then either HcZor H ::)G'. Proof. Bythemaximality ofBwemust have HB =B or HB=G. 540 MATRICES AND LINEAR MAPS XIII,9 IfHB=Bthen HcB.Since Hisnormal, weconclude that Hiscontained in every conjugate ofB,whence inthecenter byLemma 8.5. Ontheother hand, suppose that HB=G.Write w=hb with hEHand bEB.Then wUw-l=U=hbUb-lh-l=hUh-1CHU because Hisnormal. Since UcHU and U,Ugenerate SL2(F),itfollows that HU=G.Hence GIH=HUIH UI(U nH) isabelian, whence H ::)G', aswas tobeshown. Thesimplicity ofTheorem 8.4isanimmediate consequence ofLemma 8.6. 9. THE GROUP SLn(F), n>3. Inthis section welook atthe case with n>3,and follow parts ofArtin's Geometric Algebra, Chapter IV.(Artin even treats the case ofanon-commuta- tive division algebra asthegroup ring, but weomit this forsimplicity.) Fori,j=1,..., nand i=Ijand CEF,welet Eij(c)=11 c..I) o bethematrix which differs from theunit matrix byhavingCintheij-component instead ofO.Wecall suchEij(c)anelementary matrix. Note that detEij(c)=1. IfAisanynxnmatrix, then multiplication Eij(c)Aontheleftadds ctimes the j-th row tothei-th row ofA.Multiplication AEij(c)ontheright adds ctimes thei-thcolumn tothej-th column. Weshall mostly multiplyontheleft. For fixed i=Ijthemap cEi){C) XIII,9 THE GROUP SLn(F), n>3541 isahomomorphism ofFinto themultiplicative group of nxnmatrices GLn(F). Proposition 9.1. The group SLn(F) isgenerated bytheelementary matrices. IfAEGLn(F), then Acan bewritten intheform A =SD, ",'here SESLn(F) and Disadiagonal matrix oftheform D= soDhas 1onthediagonal except onthelower right corner, where the com- ponent isd=det(A). Proof. Let AEGLn(F'). Since Aisnon-singular, thefirst component of some row isnot zero, and byanelementary row operation, we can make all =IO.Addingasuitable multiple ofthefirst row tothesecond row, wemake a21 =I0,and then addingasuitable multiple ofthesecond row tothefirst we make all=1.Then wesubtract multiples ofthefirst row from theothers to make ai 1=0fori=I1. We now repeat theprocedure with the second row and column, tomake a22= 1and ai2=0ifi>2.But then we can also make a12=0bysub- tractingasuitable multiple ofthe second row from thefirst, sowe can get ai2=0fori=I2. Werepeat thisprocedure until we arestopped atann=d=I0,andanj=0 forj=f.n.Subtractingasuitable multiple ofthelast row from thepreceding ones yieldsamatrix Doftheform indicated inthe statement ofthetheorem, and concludes theproof. Theorem 9.2. For n>3,SLn(F) isequal toitsown commutator group. Proof. Itsuffices toprove thatEij(c)isacommutator. Usingn>3,let k=Ii,j.Then bydirect computation, Eij(c)=Eik(C)Ekj(I)Eik( -c)E kj(-1) expresses Eij(c)asacommutator. This proves thetheorem. We note that ifamatrix Mcommutes with every element ofSLn(F'), then itmust beascalar matrix. Indeed, just thecommutation with theelementary matrices E..(I)=1+1..I} I} 542 MATRICES AND LINEAR MAPS XIII,9 shows that Mcommutes with allmatrices 1ij(having1intheij-component, ootherwise), soMcommutes with allmatrices, and isascalar matrix. Taking thedeterminant shows that the center consists ofn(F)I, where n(F) isthe group ofn-th roots ofunity inF. We letZbethe center ofSLn(F),sowehave justseen that Zisthegroup ofscalar matrices such that thescalar isann-th root ofunity. Then wedefine PSLn(F)=SLn(F)/Z. Theorem 9.3. For n>3,PSLn(F) issimple. The restofthis section isdevoted totheproof. Weview GLn(F)asoperating onthe vector space E=Fn .IfA.isanon-zero functional onE,welet HA.=Ker A., and callH).(orsimply H)thehyperplane associated with A.Then dim H=n-1, andconversely, ifHisasubspace ofcodimension 1,then E/H hasdimension 1..and isthekernel ofafunctional. Anelement TEGLn(F) iscalled atransvection ifitkeeps every element of some hyperplane Hfixed, and forallxEE,wehave Tx =x+h for some hEH. Given any element UEH). wedefine atransvection Tuby Tux=x+A.(x)u. Every transvection isofthis type. Ifu,vEH;. ,itisimmediate that Tu+v=Tu0Tv. IfTisatransvection and AEGLn(F), then theconjugate ATA-1isob- viouslyatransvection. The elementary matricesEij(c)aretransvections, and itwill beuseful to use them with thisgeometric interpretations, rather than formallyaswedid before. Indeed, let e1,...,enbethestandard unit vectors which form abasis ofF(n). ThenEiJ{c)leaves ekfixed ifk=Ij,and theremaining vector ejismoved byamultiple ofej. We letHbethehyperplane generated by ekwith k=f.j, and thus seethatEij(c)isatransvection. Lemma 9.4. For n>3,thetransvections =IIformasingle conjugacy class inSLn(F'). Proof. First, bypicking abasis ofahyperplane H =H).and usingone more element toform abasis ofF(n), one sees from thematrix ofatransvection Tthat detT=1,i.e.transvections areinSLn(F). XIII,9 THE GROUP SLn(F),n>3543 LetT'beanother transvection relative toahyperplane H'.Say Tx =x+A(X)U and T'x =x+A'(X)U' with UEHand u'EH'. Let zand z'bevectors such that A(Z)= 1andA'(Z')=1. Since abasis forHtogether with zisabasis forF(n), andsimilarlyabasis for H'together with z'isabasis forF(n), there exists anelement AEGLn(F) such that Au =u', AH =H', Az =z'. Itisthen immediately verified that ATA-1=T', soT,T'areconjugate inGLn(F). But infact, usingn>3,thehyperplanes H, H'contain vectors which areindependent. We canchange theimage ofabasis vector inH'which isindependent ofu'by some factor inFsoastomake detA=1,soAESLn(F). This proves thelemma. We now want toshow that certain subgroups ofGLn(F) are either con- tained inthecenter, orcontain SLn(F). Let Gbe asubgroup ofGLn(F). We saythat GisSLIt-invariant if AGA-1cGforallAESLn(F). Lemma 9.5. Let n>3.Let GbeSLn-invariant, and suppose that Gcontains atransvection T=II.Then SLn(F')cG. Proof. ByLemma 9.4, alltransvections areconjugate, and the setof transvections contains theelementary matrices which generate SLn(F) by Proposition 9.1, sothelemma follows. Theorem 9.6. Let n>3.JfG isasubgroup ofGLn(F) which isSLn-invariant and which isnotcontained inthe center ofGLn(F), then SLn(F)c:G. Proof. Bythepreceding lemma, itsuffices toprove that Gcontains a transvection, and this isthekey step intheproof ofTheorem 9.3. We start with anelement AEGwhich moves some line. This ispossible since Gisnotcontained inthe center. Sothere exists avector u=I0such that Auisnot ascalar multiple ofu,sayAu=v.Then u,varecontained insome hyperplane H=Ker A.Let T=Tuand let B=ATA-IT-l . Then ATA-1=ITand B=ATA-1T-1=f.J. 544 MATRICES AND LINEAR MAPS XIII,9 This iseasilyseen byapplying say Btoanarbitrary vector x,and using the definition ofTu. Ineach case, for some xtheleft-hand side cannot equal the right-hand side. For any vector xEF(n) wehave Bx-XE(u,v), where (u,v)istheplane generated byu,v.Itfollows that BH cH,so BH =Hand Bx-xEH. We now distinguish two cases toconclude theproof. First assume that B commutes with alltransvections with respect toH.Let WEH.Then from the definitions, wefind forany vector x: BTwx=Bx+A(x)Bw TwBx=Bx+A(Bx)w=Bx+A(X)W. Since we are inthe case BTw=TwB, itfollows that Bw =w.Theretore B leaves every vector ofHfixed. Since wehave seen that Bx-xEHforallx, itfollows that Bisatransvection and isinG,thus proving thetheorem inthis case. Second, suppose there isatransvection Twwith wEHsuch that Bdoes not commute with Tw. Let C=BTwB-IT:,l. Then C=IIand CEG.Furthermore Cisaproduct ofT:,1and BTwB-1 whose hyperplanesareHand BH, which isalso Hbywhat wehave already proved. Therefore Cisatransvection, since itisaproduct oftransvections with the same hyperplane. And CEG.This concludes theproof inthesecond case, and also concludes theproof ofTheorem 9.6. We now return tothemain theorem, that PSLn(F) issimple. Let Gbe a normal subgroup ofPSLn(F), and letGbeitsinverse image inSLn(F). Then G isSLn-invariant, and ifG=I1,then Gisnotequal tothe center ofSLn(F). Therefore Gcontains SLn(F) byTheorem 9.6,and therefore G=PSL,lF), thus proving that PSLn(F) issimple. Example. ByExercise 41ofChapter I,orwhatever other means, one sees that PSL2(Fs)=As(where Fsisthefinite field with 5elements). While youare inthemood, show also that PGL2(F3)=S4 but SL2(F3)fS4; PSL2(F3)=A4. XIII, Ex EXERCISES 545 EXERCISES 1.Interpret therank ofamatrix Ainterms ofthedimensions oftheimage and kernel ofthelinear map LA. 2.(a)LetAbeaninvertible matrix inacommutative ringR. Show that('A)-I=t(A-I). (b)Letfbe anon-singular bilinear form onthe module Eover R.Let Abe an R-automorphism ofE.Show that('A)-I=t(A-I). Prove the same thing inthe hermitian case, i.e.(A*)-I=(A-I)*. 3.Let V,Wbefinite dimensional vector spaces over afield k.Suppose given non-degenerate bilinear forms onVand Wrespectively, denoted both by(, ). Let L:V Wbe asurjective linear map and lettLbeitstranspose; that IS, (Lv, w)=(v,tLw) for vEVand wEW. (a) Show that tLisinjective. (b) Assume inaddition thatifvEV,v=1=0then (v,v)=1=O.Show that V=Ker LEB1mtL, and that thetwo summands areorthogonal. (Cf. Exercise 33for anexample.) 4.LetAt...,A,berow vectors ofdimension n,over afield k.LetX =(xl'...,xn).Let bl'.. .,brEk.Byasystem oflinear equations inkone means asystem oftype A1.X =bl'.. .,Ar.X =br. Ifb1= . ..=br=0,one says thesystem ishomogeneous. Wecall nthenumber of variables, and rthenumber ofequations. Asolution Xofthehomogeneous system iscalled trivial ifXi=0,i=1,..., n. (a)Show that ahomogeneous system ofrlinear equations In'nunknowns with n>ralways has anon-trivial solution. (b) LetLbeasystem ofhomogeneous linear equationsover afield k.Letkbea subfield ofk'.IfLhas anon-trivial solution ink',show that ithas anon-trivial solution ink. 5.LetMbean nxnmatrix over afield k.Assume thattr(MX)=0forallnxnmatrices XInk.Show that M =O. 6.Let Sbeasetofnxnmatrices over afield k.Show that there exists acolumn vector Xi=0ofdimension nink,such that MX =XforallMESifandonly ifthere exists such avector insome extension field k'ofk. ,7.Let Hbethedivision ringover the reals generated byelements i,j,ksuch that i1=j1=k1=-1, and ij= -ji=k, jk=-kj=i, ki = -ik=j. Then Hhas anautomorphism oforder 2,given by ao+at;+a1j+a3kHao-a1i-a2j-a3k . Denote thisautomorphism byexHa.What isexa? Show that thetheory ofhermitian 546 MATRICES AND LINEAR MAPS XIII, Ex forms can becarrIed out over H,which iscalled thedivision rIng ofquaternions (orby abuse oflanguage, thenon-commutative field ofquaternions). 8.LetNbeastrIctly upper trIangularnxnmatrIx, that ISN =(ai)andaij=0Ifi>j. Show that Nn=O. 9.Let Ebe avector space over k,ofdimension n.Let T:E-+Ebe alInear map such that Tisnilpotent, that ISTm=0for some posItive integerm.Show that there eXists abasis ofEover ksuch that thematrix ofTwith respect tothis basis isstrictly upper triangular. 10.IfNISanilpotentnxnmatrIX, show that I+NisInvertible. 11.Let Rbethe setofallupper trIangularnxnmatrIces(aij)withaijinsome field k,so aij=0Ifi>j.LetJbethe setofallstrIctly upper triangular matrIces. Show that J isatwo-sided Ideal inR.How would you descrIbe thefactor ring R/J? 12.Let Gbethegroup ofupper triangular matrices with non-zero diagonal elements. Let Hbethesubgroup consisting ofthose matrices whose diagonal element IS1. (Actually prove that HISasubgroup). How would you descrIbe thefactor group G/H? 13. Let Rbethering ofnxnmatrices over afield k.LetLbethesubset ofmatrices which are0except onthefirst column. (a) Show that Lisaleft ideal. (b) Show that Lisaminimal leftideal; that is,ifL'CLisaleft ideal and L' =1=0,then L'=L.(For more onthissituation, seeChapter VII,5.) 14.Let Fbeany field. Let Dbethesubgroup ofdiagonal matrIces InGLn(F). Let Nbe thenormalIzer ofDInGLn(F). Show that N/Disisomorphic tothesymmetrIc group on nelements. 15.LetFbeafinite field with qelements. Show that theorder ofGLn(F)IS n (qn_1)(qn_q).. .(qn_qn- 1)=qn(n-1)/2n(qi-1). i=1 [Hint: LetXl'. ..,Xnbeabasis ofFn .Any element ofGLn(F)ISuniquely determined byitseffect onthis basis, and thus theorder ofGLn(F)ISequal tothenumber ofall possible bases. IfAEGLn(F), letAXi=Yi'For Ylwe can select any ofthe qn-1 non-zero vectors inFn .Suppose Inductively that wehave already chosen Y.,. ..,Yr with r<n.These vectors spanasubspace ofdimension rwhich contains qrelements. For Yi+ 1we can select any ofthe qn-qrelements outside ofthis subspace. The formula drops out.] 16.Again letFbeafinite field with qelements. Show that theorder ofSLn(F) is n qn(n-l)/2 n(qi-1); i=2 and that theorder ofPSLn(F)IS 1n- 1 -qn(n-1)/2n(qi-1),di=2 where disthegreatest common divisor ofnand q-1. XIII, Ex EXERCISES 547 17.Let FbeafinIte field with qelements. Show that thegroup ofallupper tnangular matnces with IonthediagonalISaSylow subgroup ofGLn(F) and ofSLn(F). 18.The reduction map Z-+ZjNZ, where Nisapositive integer defines ahomomorphism SL2(Z)-+SL2(ZjNZ). Show that thishomomorphism issurjective. [Hint: Useelementary divisors, i.e.the structure ofsubmodules ofrank 2over theprincipal ringZ.] 19.Show that theorder ofSL2(ZjNZ) isequal to N3n(1-), piN P where theproductIStaken over allprimes dividing N. 20.Show that one has anexact sequence 1-+SL2(ZjNZ)-+GL2(ZjNZ) (ZjNZ)*-+1. Infact, show that GL2(ZjNZ)=SL2(Z/NZ)G N, where GNisthegroup ofmatrices (01A d)with dE(ZjNZ)*. 21. Show that SL2(Z) isgenerated bythematrices (:)and(- ). 22. Letpbeaprime>5.LetGbe asubgroup ofSL2(Z/pnz) with n>1.Assume that theimage ofGinSL2(Z/pZ) under thenatural homomorphism isallofSL2(Z/pZ). Prove that G=SL2(Z/pnz). Note. Exercise 22isageneralization bySerre ofaresult ofShimura; seeSerre's Abelian f-adic Representations andelliptic curves, Benjamin, 1968, IV,3, Lemma 3.See also myexposition inElliptic Functions, Springer Verlag, reprinted from Addison- Wesley, 1973, Chapter 17,4. 23. Let kbe afield inwhich every quadratic polynomial has aroot. Let BbetheBorel subgroup ofGL2(k). Show that Gistheunion ofalltheconjugates ofB.(This cannot happen forfinite groups!) 24.LetA,Bbesquare matrices ofthe same size over afield k.Assume that Bisnon- singular. Iftisavariable, show that det(A +tB) isapolynomial int,whose leading coefficient isdet(B), and whose constant term isdet(A). 25. Letall'...,alnbeelements from aprincipal ideal ring, and assume thatthey generate theunit ideal. Supposen>I.Show that there exists amatrix (aij)with thisgiven first row, and whose determinant isequal toI. 548 MATRICES AND LINEAR MAPS XIII, Ex 26.Let Abeacommutative ring, andI=(x1,.. .,x,) anideal. LetcijEAand let , Y.='c..x.I'- I)). j=1 LetI'=(Yl' ...,y,). Let D=det(ci).Show that DIc1'. 27.LetLbeafree module over Zwith basis e.,. ..,en.LetMbeafree submodule ofthe same rank, with basis U1,..., Un. Let Ui=Lcijej.Show that the index (L:M) is given bythedeterminant: (L:M)=Idet(cij) I. 28.(The Dedekind determinant). Let Gbeafinite commutative group and letFbethe vector space offunctions ofGinto C.Show that thecharacters ofG(homomorphisms ofGinto the roots ofunity) form abasis forthis space. Iff:G Cisafunction, show that fora,bEG. det(f(ab- 1»=nIx(a)f(a), laeG where theproduct istaken over allcharacters. [Hint: Use both thecharacters and thecharilcteristic functions ofelements ofGasbases forF,and consider thelinear map T=Lf(a), where istranslation bya.] Also show that det(f(ab- 1»=(If(a»)det(j (ab- 1)-j(b-1», aeG where thedeterminant ontheleft istaken foralla,bEG, and thedeterminant on theright istaken only for a,b=1=1. 29.Let 9be amodule over thecommutative ring R.Abilinear map 9x9-+9,written (x,Y)....... [x,y],issaid tomake 9aLiealgebra if[x,x]=0and [[x,y],z]+[[y,z],x] +[[z,x],y]=0 forallx,y,ZE9. (a)LetMn(R) bethering ofmatrices over R.Ifx,yEMn(R), show that the product (x,y).......[x,y]=xy-yx makes Mn(R) into aLiealgebra. (b)Let 9beaLiealgebra. Let xE9,and letLx,L(x)orLie xbethelinear map given byLx(y)=[x,y].Show that Lxisaderivation of9into itself (i.e. satisfies theruleD([y, z])=[Dy, z]+[y,Dz». (c)Show that themapx LxisaLiehomomorphism of9into themodule of derivations of9into itself. 30.Given asetofpolynomials {PlX i)}inthepolynomial ringR[X ij](1<i,j<n), a zero ofthis setinRisamatrix x=(Xij)such thatxijERand P,,(xij)=0forall v. We use vector notation, and write (X)=(Xij).We letG(R) denote the setofzeros XIII, Ex EXERCISES 549 ofour setofpolynomials {P y}.Thus G(R) cMn(R), andifR'isany commutative associative R-algebra wehave G(R')cMn(R'). We saythat the set{P y}defines an algebraic group over RifG(R') isasubgroup ofthegroup GLn(R') forallR'(where GLn(R') isthemultiplicative group ofinvertible matrices inR'). Asanexample, thegroup ofmatrices satisfying theequation 'XX =InISanalge- braic group. Let R'betheR-algebra which isfree, with abasis {I,t}such that t2=O.Thus R' =R[t]. Let gbethe setofmatrices xEMn(R) such that In+txEG(R[t]). Show that gisaLiealgebra. [Hint: Note that PlIn +tX)=Plln) +grad Py(In)tX. Use thealgebra R[t,u]where t2=u2=0toshow that ifIn+txEG(R[t]) and In+uyEG(R[u]) then [x,y]Eg.] (Ihave taken theabove from thefirst four pages of[Se65]. For more information onLiealgebras and LieGroups,see[Bo 82] and [Ja79]. [Bo 82] N.BOURBAKI, LieAlgebras and LieGroups, Masson, 1982 [Ja79] N.JACOBSON, LieAlgebras, Dover, 1979 (reprinted from Interscience, 1962) [Se65] J.P.SERRE, LieAlgebras and LieGroups, Benjamin, 1965. Reprinted SpringerLecture Notes 1500. Springer/Verlag 1992 Non-commutative cocycles LetKbe afinite Galois extension ofafield k.Letr=GLn(K), and G=Gal(Kjk). Then Goperates onr.Byacocycle ofGinrwe mean afamily ofelements {A(a)} satisfying therelation A(a)aA(r)=A(u!). We saythat thecocycle splits Ifthere exists BErsuch that A(a)=B-laB forallaEG. Inthis non-commutative case, cocycles donotform agroup, but one could define an equivalence relation todefine cohomology classes. For our purposes here, we care only whether acocycle splitsornot. When every cocycle splits,wealso say that H1(G,r)=0(or1). 31.Prove that H1(G,GLn(K»=1.[Hint: Let{el'...' eN}beabasis ofMatn(k) over k, saythematrices with 1insome component and 0elsewhere. Let N X='x.e.'- I I i=1 with variables Xi.There exists apolynomial P(X) such that xisinvertible ifandonly if(x.,...,XN) i=O.Instead ofP(Xl,...,XN)wealso write P(x). Let{A(a)} be a cocycle. Let{ta} bealgebraically independent variables over k.Then P(It'lA(Y»)#0 'lEG 550 MATRICES AND LINEAR MAPS XIII, Ex because thepolynomial does not vanish when one tyisreplaced by1and theothers arereplaced by O.Bythealgebraic independence ofautomorphisms from Galois theory, there exists anelement yEKsuch thatifweput B=L(yy)A(y) y then P(B) =1=0,soBisinvertible. Itisthen immediately verified that A(O")=BO"B- 1. But when kisfinite, cf.myAlgebraic Groupsover Finite Fields, Am. J.Vol 78No. 3,1956.] 32.Invariant bases. (Kolchin-Lang, Proc. AMS Vol 11No.1, 1960). LetKbeafinite Galois extension ofk,G=Gal(K/k) asinthepreceding exercise. Let Vbe a finite-dimensional vector spaceover K,and suppose GoperatesonVinsuch a way that a(av)=a(a)a(v) for aEKand vEV.Prove that there exists abasis {WI'.'.'wn}such that UWi=Wiforalli=1,..., nand allaEG(aninvariant basis). Hint: Let{VI'.. .,vn}beany basis, and let a(V:l)=A(O")(I)Uti VtI where A(a) isamatrix inGLiK). Solve forBintheequation (O"B)A(O")=B,and let ()=B(::} The next exercises onharmonic polynomials have their source inWhittaker, Math. Ann. 1902; seealso Whittaker andWatson, Modern Analysis, Chapter XIII. 33. Harmonic polynomials. LetPolen, d)denote thevector space ofhomogeneous poly- nomials ofdegree dinnvariables XI'.. .,Xnover afield kofcharacteristic O. For ann-tuple ofintegers (VI'...,vn)withVi>0wedenote byM(J,I)asusual the monomial M(J,I)(X)=XII...Xn. Prove: (a)The number ofmonomials ofdegree dis(n-1+d ),sothis number is n-1 thedimension ofPol(n, d). (b)Let(D)=(D I,. . .,Dn) where Diisthepartial derivative with respect tothe i-th variable. Then we candefine P(D) asusual. ForP,QEPolen, d),define (P,Q)=P(D)Q(O). Prove that this defines asymmetric non-degenerate scalar producton Pol(n, d).Ifkisnotreal, itmay happen that P =1=0but(P,P)=O.However, iftheground field isreal, then (P,P)>0forP =1=O.Show also that the monomials ofdegree dform anorthogonal basis. What is(M(J,I)' M(J,I»? (c)The map P P(D) isanisomorphism ofPolen, d)onto itsdual. XIII, Ex EXERCISES 551 (d)Let=Dt+...+D. Note that:Pol(n, d) Pol(n, d-2)isalinear map. Prove that issurjective. (e)Define Har(n, d)=Ker=vectorspace ofharmonic homogeneous poly- nomials ofdegree d.Prove that dim Har(n, d)=(n+d-3)!(n+2d-2)/(n-2)!d!. Inparticular, ifn=3,then dim Har(3, d)=2d+1. (f)Let r2=Xt+.. .+X. Let Sdenote multiplication byr2 .Show that (P, Q)=(P,SQ) forPEPol(n, d)and QEPol(n, d-2), sot=S.More generally, forREPol(n, m)and QEPol(n, d-m) we have (R(D)P, Q)=(P,RQ). (g) Show that[,S]=4d+2n onPol(n, d). Here[,S]= 0S-S0. Actually, [, S]=4E +2n, where EistheEuler operator E=2:X;D;, which is,however, thedegree operatoronhomogeneous polynomials. (h) Prove thatPol(n, d)=Har(n, d)EBr2Pol(n, d-2)andthat thetwo summands areorthogonal. This isaclassical theorem used inthetheory oftheLaplace operator. (i)2Let(c.,. . .,cn)Eknbesuch that L.JC;=O.Let H(X)=(c.X.+... +cnXn)d. Show thatHisharmonic, i.e. lies inHar(n, d). (j)For any QEPol(n, d), and apositive integer m,show that Q(D)H';'(X)=m(m-1)...(m-d+l)Q(c)H,;,-d(X). 34.(Continuation ofExercise 33). Prove: Theorem. Let kbealgebraically closed ofcharacteristic O.Let n>3.Then Har(n, d)asavector spaceover kisgenerated byallpolynomials Hwith (c) Ekn such that2:ct=O. [Hint: LetQEHar(n, d)beorthogonal toallpolynomials Hwith (c) Ekn .By Exercise 33(h), itsuffices toprove that r2 1Q.Butif2:ct=0,then byExercise 33(j)weconclude thatQ(c)=O.BytheHilbert Nullstellensatz, itfollows that there exists apolynomial F(X) such that Q(X)S=r2(X)F(X) for some positive integers. But n>3implies that r2(X) isirreducible, sor2(X)divides Q(X).] 35.(Continuation ofExercise 34). Prove that therepresentation ofO(n)=Un(R)on Har( n,d)isirreducible. Readers will find aproof inthefollowing: S.HELGASON,Topics inHarmonic AnalysisonHomogeneous Spaces, Birkhauser, 1981 (see especially 3,Theorem 3.1(ii)) N.VILENKIN, Special Functions and theTheory ofGroup Representations, AMS Trans- lations ofmathematical monographs Vol. 22, 1968 (Russian original, 1965), Chapter IX,2. 552 MATRICES AND LINEAR MAPS XIII, Ex R.HOWE and E.C.TAN, Non-Abelian Harmonic Analysis, Universitext, Springer Verlag, New York, 1992. The Howe-Tan proofruns asfollows. We now use thehermitian product (P,Q)=fP(x) Q(x) da(x), S,,-I where aisthe rotation invariant measure on the(n-l)-sphere 8n-l .Let el,.. .,enbetheunit vectors inRn .We canidentify O(n-I)asthesubgroup of O(n) leaving enfixed. Observe that O(n) operatesonHar(n, d),say ontheright by composition P poA,AEO(n), and thisoperation commutes with.Let A:Har(n, d)--tC bethefunctional such that ,1(P)==P(e n).Then AisO(n-I)-Invariant, and since the hermitian product isnon-degenerate, there exists aharmonic polynomial Qnsuch that ,1(P)=<P,Qn> forallPEHar(n, d). LetMcHar(n, d)be anO(n)-submodule. Then therestriction AMofAtoMis nontrivial because O(n) acts transitively onSn-l. LetQ,';fbetheorthogonal pro- jection ofQnonM.Then QttisO(n-1)-invariant, and soisalinear combination Q(x)=L cjxI. j+2k=d Furthermore QIfisharmonic. From this youcanshow thatQisuniquely determined, byshowing theexistence ofrecursive relations among thecoefficientsCj.Thus the submodule Misuniquely determined, and must beallofHar(n, d). Irreducibility ofsln(F). 36. LetFbe afield ofcharacteristic O.Let 9==sIll(F) bethe vector space ofmatrices with trace 0,with itsLiealgebra structure [X,Y]==XY-YX. LetEijbethematrix having (i,j)-component1and allother components O.Let G==SLn(F). Let Abe themultiplicative group ofdiagonal matrices over F. (a)Let Hi==Eu-Ei+l,i+l for i=1,. . .,n-I.Show that the elementsEij (ii=j),HI,...,Hn-l form abasis of9over F. (b)For gEGletc(g) betheconjugation action ong,that isc(g)X==gXg-l . Show that eachEijisaneigenvector forthis action restricted tothegroup A. (c) Show that theconjugation representation ofGon9isirreducible, that is,if Vi=0isasubspace of9which isc(G)-stable, then V==g.Hint: Look up thesketch oftheproof in[JoL 01],Chapter VII, Theorem 1.5, and putinall thedetails. Note that forii=jthematrix Eijisnilpotent,soforvariable t, theexponential series exp(tEij)isactuallyapolynomial. The derivative with respect totcan betaken intheformal power series F[[t]],notusing limits. If Xisamatrix, andx(t)==exp(tX), show that x(t)Yx(trl=XY-YX =[X,Y]. 1=0 CHAPTER XIV Representation ofOne Endomorphism We deal here with oneendomorphism ofamodule, actuallyafree module, andespeciallyafinite dimensional vector space over afield k.We obtain the Jordan canonical form for arepresenting matrix, which has aparticularly simple shape when kisalgebraically closed. This leads toadiscussion ofeigenvalues and thecharacteristic polynomial. The main theorem can beviewed asgiving anexample forthegeneral structure theorem ofmodules over aprincipal ring. Inthepresent case, theprincipal ring isthepolynomial ringk[X] inone variable. 1. REPRESENTATIONS Letkbeacommutative ring and Eamodule over k.Asusual, wedenote by Endk(E) thering ofk-endomorphisms ofE,Le.thering ofk-linear maps ofEinto itself. Let Rbeak-algebra (given byaring-homomorphism k Rwhich allows ustoconsider Rasak-module). Byarepresentation ofRinEone means ak- algebra homomorphism R Endk(E), that isaring-homomorphism p:R Endk(E) which makes thefollowing diagram commutative: R)Endk(E)/ k 553 554 REPRESENTATION OFONE ENDOMORPHISM XIV,1 [Asusual, weview Endk(E)asak-algebra; ifIdenotes theidentity map ofE, wehave thehomomorphism ofkinto Endk(E) given bya al. Weshall also useItodenote theunit matrix ifbases have been chosen. The context will always make ourmeaning clear.] Weshall meet several examples ofrepresentations inthesequel, with various types ofrings (both commutative andnon-commutative). Inthischapter, the rings will becommutative. Weobserve that Emay beviewed asanEndk(E) module. Hence Emay be viewed asanR-module, defining theoperation ofRonEbyletting (x,v) p(x)v for xERand vEE.Weusually write xvinstead ofp(x)v. Asubgroup FofEsuch that RF cFwill besaid tobeaninvariant sub- module ofE.(Itisboth R-invariant andk-invariant.) We also say that itis invariant under therepresentation. We saythat therepresentation isirreducible, orsimple, ifE=I0,and ifthe only invariant submodules are0and Eitself. The purpose ofrepresentation theories istodetermine the structure ofall representations ofvarious interesting rings, and toclassify their irreducible representations. Inmost cases, wetake ktobeafield, which mayor may not bealgebraically closed. The difficulties inproving theorems about representa- tions may therefore lieinthecomplication ofthering R,orthecomplication of thefield k,orthecomplication ofthemodule E,orallthree. Arepresentation pasabove issaid tobecompletely reducible orsemi-simple ifEisanR-direct sum ofR-submodules Ei, E=E1(f)...(f)Em such that each Eiisirreducible. We also saythat Eiscompletely reducible. Itisnot true that allrepresentationsarecompletely reducible, and infact those considered inthischapter will not beingeneral. Certain types ofcompletely reducible representations will bestudied later. There isaspecial type ofrepresentation which will occur very frequently. Let vEEand assume that E=Rv. We shall also write E=(v). We then say that Eisprincipal (over R),and that therepresentation isprincipal. Ifthat is the case, the setofelements xERsuch that xv =0isaleftideal aofR(obvious). The map ofRonto Egiven by xxv induces anisomorphism ofR-modules, Ria E (viewing Rasaleftmodule over itself, andRiaasthefactor module). Inthis map, theunit element 1ofRcorresponds tothegeneratorvofE. XIV,1 REPRESENTATIONS 555 As amatter ofnotation, ifvl'...,VnEE,welet(VI'...,vn)denote thesub- module ofEgenerated byVb...,Vn. Assume that Ehas adecomposition into adirect sum ofR-submodules E=El...Es. Assume that each E;isfree andofdimension>lover k.Let CB1,... ,CBsbe bases forEl'. . .,Esrespectivelyover k.Then {CBI'...,<Bs}isabasis forE. Let 'PER,and let 'Pibetheendomorphism induced by 'PonE;.LetM;bethe matrix of'Piwith respecttothebasis CB;. Then thematrix Mof'Pwith respect to{CB.,. . .,CBs}looks like o 0 M2 0 o 0 o 0Ms Amatrix ofthis type issaid tobedecomposed into blocks, Ml'...Ms. When wehave such adecomposition, thestudy ofqJoritsmatrix iscompletely reduced (sotospeak) tothestudy oftheblocks. Itdoes notalways happen that wehave such areduction, butfrequently something almost asgood happens. Let E'be asubmodule ofE,invariant under R.Assume that there exists abasis of£'over k,say{Vl'...,vm},and that this basis can becompleted toabasis ofE, {vl'...,Vm,Vm+l'...,Vn}. This isalways the case ifkisafield. Let lpER.Then thematrix oflpwith respect tothis basis hastheform (Mf* )oM". Indeed, since E'ismapped into itself byqJ,itisclear that wegetM'intheupper left, and azero matrix below it.Furthermore, foreachj=m+1,..., nwecan write cpV=CjlVI+ . . .+CjmVm+Cj,m +1t),11 +I+ . . .+Cjnvn. The transpose ofthematrix(Cji)then becomes thematrix (,,) occurringontheright inthematrix representing lp. 556 REPRESENTATION OFONE ENDOMORPHISM XIV,2 Furthermore, consider anexact sequence o E' E E" o. Let vm+1,...,vnbetheimages ofVm+1,...,Vnunder thecanonical map E E". We can define alinear map ".E"E" cp. inanatural waysothat (cpr)=cp"(v) for aU vEE.Then itisclear that the matrix ofcp"with respect tothebasis {vl'...,vn}isM". 2. DECOMPOSITION OVER ONE ENDOMORPHISM Let kbeafield and Eafinite-dimensional vector spaceover k,E=f.O.Let AEEndk(E) bealinear map ofEinto itself. Let tbetranscendental over k.We shall define arepresentation ofthepolynomial ringk[t] inE.Namely,wehave ahomomorphism k[t] k[A]cEndk(E) which isobtained bysubstituting Afor tinpolynomials. The ring k[A] isthe subring ofEndk(E) generated byA,and iscommutative because powers ofA commute with each other. Thus iff(t) isapolynomial and vEE,then f(t)v=f(A)v. The kernel ofthehomomorphism f(t) f(A) isaprincipal ideal ofk[t], which is=f.0because k[A] isfinite dimensional over k.Itisgenerated bya unique polynomial ofdegree> 0,having leading coefficient 1.This polynomial will becalled theminimal polynomial ofAover k,and will bedenoted byqA(t). Itisofcourse notnecessarily irreducible. Assume that there exists anelement vEEsuch that E=k[t]v=k[A]v. This means that Eisgenerated over kbytheelements v,Av,A2V,.... Wecalled such amodule principal, andifR=k[t] wemay write E=Rv=(v). IfqA(t)=td+ad- 1td-1+...+aothen theelements A Ad- 1v,v,..., v constitute abasis forEover k.This isproved inthe same way astheanalogous statement forfinite field extensions. First wenote that they arelinearly inde pendent, because any relation oflinear dependence over kwould yield apoly- XIV,2 DECOMPOSITION OVER ONE ENDOMORPHISM 557 nomial g(t) ofdegree less than deg qAand such that g(A)=o.Second, they generateEbecause anypolynomial f(t) can bewritten f(t)=g(t)qA(t) +r(t) with degr<deg qA.Hencef(A)=r(A). With respect tothis basis, itisclear that thematrix ofAisofthefollowing type: 000 100 010o -ao o -at o -a2 .......................... . o00 0-ad-2 o00...1_ad- 1 IfE=(v)isprincipal, then Eisisomorphic tok[t]/(qA(t» under themap f(t) f(A)v. Thepolynomial qAisuniquely determined byA,and does not dependonthechoice ofgeneratorvforE.This isessentially obvious, because iffl,f2are twopol¥nomialswith leading coefficient 1,thenk[t]/(fl (t» isiso- morphic tok[t]/(f2(t» ifandonlyiffl=f2.(Decompose each polynomial into prime powers andapply thestructure theorem formodules overprincipal rings.) IfEisprincipal then weshall call thepolynomial qAabove thepolynomial invariant ofE,with respect toA,orsimply itsinvariant. Theorem 2.1. Let Ebeanon-zero finite-dimensional space over thefield k, and letAEEndk(E). Then Eadmits adirect sumdecomposition E=E1(f)...(f)Er, where each Eiisaprincipal k[A]-submodule, with invariant qi=I0such that qllq21...lqr. The sequence (qb...,qr)isuniquely determined byEand A,and qristhe minimal polynomial ofA. Proof The first statement issimplyarephrasing inthepresent language forthestructure theorem formodules over principal rings. Furthermore, itis clear that qr(A)=0since qiIqrforeach i.Nopolynomial oflower degree than qrcanannihilate E,because inparticular, such apolynomial does notannihilate Er. Thus qristheminimal polynomial. Weshall call(q1,...,qr)theinvariants ofthepair (E,A). LetE=k(n), and letAbean nxnmatrix, which weview asalinear map ofEinto itself. The invariants (ql,...,qr)will becalled theinvariants ofA(over k). Corollary 2.2. Letk'beanextension field ofkandletAbeannxnmatrix ink.The invariants ofAover kare the same asitsinvariants over k' . 558 REPRESENTATION OFONE ENDOMORPHISM XIV,2 Proof. Let{Vl,...,vn}beabasis ofk(n) over k.Then wemay view italso asabasis ofk'(n) over k'.(The unit vectors areinthek-space generated by VI'...,Vn;hence Vl,...,Vngenerate then-dimensional space k,(n) over k'.)Let E=k(n). LetLAbethelinear map ofEdetermined byA.LetLbethelinear map ofk,(n)determined byA.The matrix ofLAwith respect toourgiven basis is the same asthematrix ofL. We can select the basis corresponding tothe decomposition E=E1(f)...(f)Er determined bytheinvariants q1,...,qr.Itfollows that theinvariants don't change when weliftthebasis toone ofk,(n). Corollary 2.3. LetA,Bbe nxnmatrices over afield kand letk'be an extension field ofk.Assume that there isaninvertible matrix C'ink'such that B=C'AC,-l. Then there isaninvertible matrix CinksuchthatB=CAC-l . Proof. Exercise. The structure theorem formodules over principal rings gives ustwo kinds ofdecompositions. One isaccording totheinvariants ofthepreceding theorem. The other isaccording toprime powers. Let E=I0beafinite dimensional spaceover thefield k,and letA:E E beinEndk(E). Let q=qAbeitsminimal polynomial. Then qhas afactorization, el eq=Pl...Pss(ei>1) into prime powers (distinct). Hence Eisadirect sum ofsubmodules E=E(Pl) (f)...(f)E(ps)' such that each E(Pi) isannihilated bypfi.Furthermore, each such submodule can beexpressedasadirect sum ofsubmodules isomorphic tok[t]/(pe)for some irreducible polynomial pand some integere>1. Theorem 2.4. LetqA(t)=(t-ex)eforsome exEk,e>1.Assume that E isisomorphic tok[t]/(q). Then Ehas abasis over ksuch that thematrix ofA relative tothis basis isoftype ex0 0 1 ex 0 o 0 o 1 ex XIV,2 DECOMPOSITION OVER ONE ENDOMORPHISM 559 Proof Since Eisisomorphic tok[t]/(q), there exists anelement vEE such thatk[t]v=E.This element corresponds totheunit element ofk[t] inthe isomorphism k[t]/(q) E. Wecontend that theelements v,(t-ex)v,...,(t-ex)e- lV, orequivalently, v,(A-ex)v,...,(A-ex)e- lV, torm abasis forEover k.Theyarelinearly independent over kbecause any relation oflinear dependence would yieldarelation oflinear dependence between A Ae- 1 v,v,..., v, and hence would yieldapolynomial g(t) ofdegree less than deg qsuch that g(A)=O.Since dim E=e,itfollows that our elements form abasis forE over k.But(A-ex)e=O.Itisthen clear from thedefinitions that thematrix of Awith respect tothis basis hastheshape stated inour theorem. Corollary 2.5. Let kbealgebraically closed, and letEbeafinite-dimensional non-zero vector space over k.Let AEEndk(E). Then there exists abasis of Eover ksuch that thematrix ofAwith respect tothis basis consists ofblocks, and each block isofthetype described inthetheorem. Amatrix having theform described inthepreceding corollary issaid tobein Jordan canonical form. Remark 1. Amatrix (or anendomorphism) Nissaid tobenilpotent if there exists aninteger d>0such that Nd=0 .We seethat inthedecomposition ofTheorem 2.4 orCorollary 2.5, thematrix Miswritten intheform M=B+N where Nisnilpotent. Infact, Nisatriangular matrix (i.e.ithas zero coefficients onand above thediagonal), and Bisadiagonal matrix, whose diagonal elements aretheroots oftheminimal polynomial. Such adecompositioncanalways be achieved whenever thefield kissuch that alltheroots oftheminimal polynomial lieink.Weobserve also that theonlycase when thematrix Nis0iswhen all the roots ofthe minimal polynomial have multiplicity1.Inthis case, if n=dim E,then thematrix Misadiagonal matrix, with ndistinct elements on thediagonal. 560 REPRESENTATION OFONE ENDOMORPHISM XIV,2 Remark 2. The main theorem ofthis section can also beviewed asfalling under thegeneral pattern ofdecomposingamodule into adirect sum asfar as possible, and also giving normalized bases for vector spaces with respect to various structures, sothat one can tell inasimple way theeffect ofanendo- morphism. More formally, consider thecategory ofpairs (E,A),consisting of afinite dimensional vector space Eover afield k,and anendomorphism A:E E.Byamorphism ofsuch pairs f:(E,A) (E',A') we mean ak-homomorphism f:E E'such that thefollowing diagram is commutative: fE' ) jA'E Aj Ef)E' Itisthen immediate that such pairs form acategory,sowehave thenotion of isomorphism. One can reformulate Theorem 2.1bystating: Theorem 2.6. Twopairs (E,A)and(F,B)areisomorphic ifandonlyifthey have the same invariants. You can prove this asExercise 19.The Jordan basis givesanormalized form forthematrix associated with such apair and anappropriate basis. Inthenext chapter,weshall find conditions under which anormalized matrix isactually diagonal, forhermitian, symmetric, andunitary operatorsover the complex numbers. As anexample andapplication ofTheorem 2.6, weprove: Corollary 2.7. Let kbeafield and letKbeafinite separable extension of degree n.Let Vbeafinite dimensional vector space ofdimension nover k,and letp,p':K Endk(V) betworepresentations ofKonV,.that is,embeddings ofKinEndk(V). Then p,p'areconjugate,. that is,there exists BEAutk(V) such that p'()=Bp()B-I forall EK. Proof. Bytheprimitive element theorem offield theory, there exists an element aEKsuch that K=k[a]. Letp(t) betheirreducible polynomial ofa over k.Then (V,p(a» and (V,p'(a» have the same invariant, namely p(t). Hence these pairsareisomorphic byTheorem 2.6, which means that there exists BEAutk(V) such that p'(a)=Bp(a)B-1 . But allelements ofKarelinear combinations ofpowers ofawith coefficients ink,soitfollows immediately thatp'()=Bp()B-Iforall EK, asdesired. XIV,3 THE CHARACTERISTIC POLYNOMIAL 561 Togetarepresentation ofKasincorollary 2.7, one may ofcourse select a basis ofK,andrepresent multiplication ofelements ofKonKbymatrices with respecttothis basis. Insome sense, Corollary 2.7tells usthat this istheonly waytoget such representations.We shall return tothispoint ofview when considering Cartan subgroups ofGLn inChapter XVIII, 12. 3. THE CHARACTERISTIC POLYNOMIAL Let kbe acommutative ring and Eafree module ofdimension nover k. Weconsider thepolynomial ringk[t], andtllinear map A:E E.We have a homomorphism k[t] k[A] asbefore, mapping apolynomial f(t) onf(A), and Ebecomes amodule over thering R=k[t]. LetMbeanynxnmatrix ink(for instance thematrix ofA relative toabasis ofE). Wedefine thecharacteristic polynomial PM(t) tobethe determinant det(tl n-M) where Inistheunit nxnmatrix. Itisanelement ofk[t]. Furthermore, ifN isaninvertible matrix inR,then det(tl n-N-IMN)=det(N-l(tl n-M)N)=det(t1n-M). Hence thecharacteristic polynomial ofN-1MNisthe same asthat ofM. We may therefore define thecharacteristic polynomial ofA,and denote byPA,the characteristic polynomial ofany matrix Massociated with Awith respect to some basis. (IfE=0,wedefine thecharacteristic polynomial tobe1.) Ifcp:k k'isahomomorphism ofcommutative rings, and Misan nxn matrix ink,then itisclear that PqJM(t)=CPPM(t) where CPP Misobtained from PMbyapplying qJtothecoefficients ofPM. Theorem 3.1. (Cayley-Hamilton). We have PA(A)=o. Proof Let{vl'. ..,vn}be abasis ofEover k.Then n tv. ="a..v.J i...J IJI i= 1 where(aij)=Misthematrix ofAwith respect tothe basis. LetB(t) bethe matrix with coefficients ink[t], defined inChapter XIII, such that B(t)B(t)=PA(t)1 n. 562 REPRESENTATION OFONE ENDOMORPHISM XIV,3 Then (V1 )(pA(t)Vl )(0 )B(t)B(t) :=:=: Vn PA(t)V n ° because B(t{)=(I) - Hence PA(t)E=0,and therefore PA(A)E=0.This means that PA(A)=0, aswas tobeshown. Assume now that kisafield. LetEbeafinite-dimensional vector spaceover k,and letAEEndk(E). ByaneigenvectorwofAinEone means anelement wEE, such that there exists anelement AEkforwhich AW=AW.IfW=I0,then Aisdetermined uniquely, and iscalled aneigenvalue ofA.Ofcourse, distinct eigenvectors may have the same eigenvalue. Theorem 3.2. The eigenvalues ofAareprecisely the roots ofthecharacter- isticpolynomial ofA. Proof Let Abeaneigenvalue. Then A-AIisnotinvertible inEndk(E), and hence det(A-AI)=0.Hence Aisaroot ofPA.The argumentsare re- versible, sowealso getthe converse. Forsimplicity ofnotation, weoften write A-Ainstead ofA-AI. Theorem 3.3. Let w1,...,Wmbenon-zero eigenvectors ofA,having distinct eigenvalues. Then they arelinearly independent. Proof Suppose that wehave a1W1+...+amWm=° with aiEk,and letthis beashortest relation with notallai=°(assuming such exists). Then ai=I°foralli.Let Al,...,Ambetheeigenvalues ofour vectors. Apply A-Altotheabove relation. Weget a2(A 2-Al)W2+...+am(Am-Al)W m=0, which shortens ourrelation, contradiction. Corollary 3.4. IfAhas ndistinct eigenvalues Ab...,Anbelonging toeigen- vectors Vl,...,Vn,anddim E=n,then {vl,...,vn}isabasisfor E.Thematrix XIV,3 THE CHARACTERISTIC POLYNOMIAL 563 ofAwith respect tothis basis isthediagonal matrix: o A.2 o A.n Warning. Itisnotalways true that there exists abasis ofEconsisting of eigenvectors! Remark. Let kbeasubfield ofk'.IfMisamatrix inkwe can define its, characteristic polynomial with respect tok,and also with respect tok'.Itis clear that thecharacteristic polynomials thus obtained areequal. IfEisavector space over k,weshall seelater how toextend ittoavector space over k'.A linear map Aextends toalinear map oftheextended space, and thecharacter- isticpolynomial ofthelinear map does notchange either. Actually, ifweselect abasis forEover k,then E k(n),and k(n) ck'(n)inanatural way. Thus selecting abasis allows ustoextend the vector space, butthis seems todepend onthe choice ofbasis. Weshall give aninvariant definition later. Let E=E1(f)...(f)Erbe anexpression ofEasadirect sum ofvector spaces over k.Let AEEndk(E), and assume that AEicEiforalli=1,...,r. Then Ainduces alinear maponEi.We can select abasis forEconsisting of bases forE1,...,Er,and then thematrix forAconsists ofblocks. Hence we see that r PA(t)=nPAi(t). i=1 Thus thecharacteristic polynomial ismultiplicativeondirect sums. Our condition above that AEicEican also beformulated bysaying that Eisexpressedasak[A]-direct sum ofk[A]-submodules, oralso ak[t]-direct sum ofk[t]-submodules. Weshall apply this tothedecomposition ofEgiven inTheorem 2.1. Theorem 3.5. Let Ebe afinite-dimensional vector spaceover afield k,let AEEndk(E), and letql,...,qrbetheinvariants of(E, A). Then PA(t)=q1(t)...qr(t). Proof We assume that E=k(n)and that Aisrepresented byamatrix M. We have seen that theinvariants donotchange when weextend ktoalarger field, and neither does thecharacteristic polynomial. Hence wemayassume that kisalgebraically closed. Inview ofTheorem 2.1 wemayassume that Mhas a 564 REPRESENTATION OFONE ENDOMORPHISM XIV,3 single invariant q.Write q(t)=(t-cxl)e1...(t-rJ.s)es with distinct cxl'...,CXs.Weview Masalinear map, andsplit out vector space further into adirect sum ofsubmodules (over k[t]) having invariants (t-cxl)e1,...,(t-cxs)e. respectively (this istheprime power decomposition). For each one ofthese submodules, wecanselect abasis sothat thematrix oftheinduced linear map has theshape described inTheorem 2.4. From thisitisimmediately clear that the characteristic polynomial ofthe map having invariant (t-cx)e isprecisely (t-ex)e,and our theorem isproved. Corollary 3.6. The minimal polynomial ofAand itscharacteristic poly- nomial have the same irreducible factors. Proof. Because qristheminimal polynomial, byTheorem 2.1. We shall generalize our remark concerning themultiplicativity ofthe characteristic polynomial over direct sums. Theorem 3.7. Let kbeacommutative ring, and inthefollowing diagram, o)E' A'!)E" A"!)E A!)0 o)E')E)E")0 letthe rows beexact sequences offree modules over k,offinite dimension, and letthevertical maps bek-linear maps making thediagram commutative. Then PA(t)=PA,(t)P A,,(t). Proof. We may assume that E'isasubmodule ofE.We select abasis {Vl,...,vm}forE'. Let{vm+l'...,v}be abasis forE",and let vm+1,...,Vn beelements ofEmappingon Vm+l'...,vnrespectively. Then {V1,...,Vm,Vm+b...,Vn} isabasis forE(same proofasTheorem 5.2ofChapter III), and we areinthe situation discussed in91.The matrix forAhastheshape (',,) XIV,3 THE CHARACTERISTIC POLYNOMIAL 565 where M'isthematrix forA'and M"isthematrix forA".Taking thecharacter- isticpolynomial with respect tothis matrix obviously yields ourmultiplicative property. Theorem 3.8. Let kbeacommutative ring, and Eafree module ofdimension nover k.Let AEEndk(E). Let PA(t)=tn+cn_1tn-1+...+co. Then tr(A)=-Cn-1and det(A)=(-l)nco. Proof. For thedeterminant, weobserve that PA(O)=co.Substituting t=0inthedefinition ofthecharacteristic polynomial bythedeterminant shows that Co=(-l)n det(A). For thetrace, letMbethematrix representing Awith respect tosome basis, M=(aij).Weconsider thedeterminant det(t1n-aij).Initsexpansionasasum over permutations, itwill contain adiagonal term (t-all)...(t-ann), which willgive acontribution tothecoefficient oftn-1equal to -(a 11+...+ann)' Noother term inthisexpansion willgive acontribution tothecoefficient of tn- 1,because thepower oftoccurring inanother term will beatmost tn- 2. This proves our assertion concerning the trace. Corollary 3.9. Let thenotation beasinTheorem 3.7. Then tr(A)=tr(A') +tr(A,,) and det(A)=det(A') det(A"). Proof. Clear. Weshall now interpret our results intheEuler-Grothendieck group. Let kbeacommutative ring. Weconsider thecategory whose objects are pairs (E,A),where Eisak-module, and AEEndk(E). We define amorphism (E',A') (E,A) tobeak-linear map E' Emaking thefollowing diagram commutative: E' A-jf)E jA E')Ef 566 REPRESENTATION OFONE ENDOMORPHISM XIV,3 Then we can define thekernel ofsuch amorphism tobeagainapair. Indeed, letEbethekernel off: E' E.Then A'maps Eointo itself because fA'E=AfE=o. WeletAbetherestriction ofA'onEo. Thepair(E,A) isdefined tobethe kernel ofourmorphism. Weshall denote byfagainthemorphism ofthepair (E',A') (E,A). We canspeak ofanexact sequence (E',A') (E,A) (E",A"), meaning that theinduced sequence E' E E" isexact. Wealso write 0instead of(0,0),according toouruniversal convention tousethesymbol 0forallthings which behave like azero element. We observe that ourpairs now behave formally likemodules, andthey in fact form anabelian category. Assume that kisafield. Let(tconsist ofallpairs (E,A)where Eisfinite dimensional over k. Then Theorem 3.7 asserts that thecharacteristic polynomial isanEuler- Poincare map defined for each object inourcategory (t,with values into the multiplicative monoid ofpolynomials with leading coefficient 1. Since thevalues ofthemapareinamonoid, thisgeneralizes slightly thenotion ofChapter III,8,when wetook thevalues inagroup. Ofcourse when kisa field, which isthe most frequent application,we canview thevalues ofour map tobeinthemultiplicative group ofnon-zero rational functions, soourprevious situation applies. Asimilar remark holds now forthe trace and thedeterminant. Ifkisa field, the trace isanEuler map into theadditive group ofthefield, and thedeter- minant isanEuler map into themultiplicative group ofthefield. We note also that allthese maps (like allEuler maps) aredefined ontheisomorphism classes of pairs, and aredefined ontheEuler-Grothendieck group. Theorem 3.10. Let kbeacommutative ring, Man nxnmatrix ink,andf apolynomial ink[t]. Assume that PM(t) has afactorization, n PM(t)=n(t-exi) i= 1 into linear factors over k.Then thecharacteristic polynomial off(M) is given by n Pf(M)(t)=n(t-f(exi», i= 1 XIV, Ex EXERCISES 567 and n n det(f(M»)=nf(rxi). i= 1tr(f(M»)=Lf(rxi), i=1 Proof. Assume first that kisafield. Then using thecanonical decomposi- tion interms ofmatrices given inTheorem 2.4, wefind that our assertion is immediately obvious. When kisaring,we use asubstitution argument. Itis however necessary toknow thatifX=(x;j)isamatrix with algebraically independent coefficients over Z,then Px(t) has ndistinct roots Yl,...,Yn[in analgebraic closure ofQ(X)] and that wehave ahomomorphism z[xij,Y1,...,Yn] k mapping XonMand Yl,...,Ynonrxl,...,rxn.This isobvious tothereader who read thechapteronintegral ring extensions, and thereader who has not can forget about thispart ofthetheorem. EXERCISES 1.Let Tbeanupper triangular square matrix over acommutative ring (i.e. alltheele- ments below and onthediagonal are0).Show that Tisnilpotent. 2.Carry outexplicitly theproof that thedeterminant ofamatrix * * M2 0* 0 0...0Ms where each Miisasquare matrix, isequal totheproduct ofthedeterminants ofthe matrices M1,..., Ms. 3.Letkbeacommutative ring, and letM,M'besquarenxnmatrices ink.Show that thecharacteristic polynomials ofMM'andM'M areequal. 4.Show that theeigenvalues ofthematrix o 100 001 0 000 1 100 0 inthecomplex numbers are+1,+i. 568 REPRESENTATION OFONE ENDOMORPHISM XIV, Ex 5.LetM,M'besquare matrices over afield k.Let q,q'betheir respective minimal polynomials. Show that theminimal polynomIal of (.) istheleast common multiple ofq,q'. 6.LetAbeanilpotent endomorphism ofafinite dimensional vector spaceEover thefield k.Show thattr(A)=O. 7.LetRbeaprincipalentire ring. LetEbeafree module over R,andletEV=HomR(E, R) beitsdual module. Then Evisfree ofdimension n.LetFbe asubmodule ofE. Show that EV/F.l can beviewed asasubmodule ofFV ,and that itsinvariants are the same astheinvariants ofFinE. 8.LetEbeafinite-dImensional vector space over afield k.Let AEAutk(E). Show that thefollowing conditions areequivalent: (a) A =I+N,with Nnilpotent. (b)There exists abasis of£such that thematrix ofAwith respect tothis basis has allitsdiagonal elements equal to 1and allelements above thediagonal equal too. (c)Allroots ofthecharacteristic polynomial ofA(inthealgebraIc closure ofk) areequal to1. 9.Let kbeafield ofcharacteristic 0,and letMbean nxnmatrix ink.Show that Mis nilpotent ifandonly iftr(MV)=0for 1<v<n. 10. Generalize Theorem 3.10 torational functions (instead ofpolynomials), assuming that kisafield. 11.Let Ebeafinite-dimensional spaceover thefield k.Let exEk.Let E(1bethesubspace ofIigenerated byalleigenvectors ofagiven endomorphism Aof£,havingexasan eigenvalue. Show that every non-zero element of£(1isaneigenvector ofAhavingexas aneigenvalue. 12.Let Ebefinite dimensional over thefield k.Let AEEndk(E). Let vbeaneigenvector forA.LetBEEndk(E) besuch that AB =BA. Show that Bvisalso aneigenvector forA(ifBvi=0),with the same eigenvalue. DiagonaUzable endomorphisms Let Ebeafinite-dimensional vector space over afield k,and letSEEndk(E).We say that Sisdiagonalizable ifthere eXIsts abasis of£consistIng ofeigenvectors ofS.The matrix ofSwith respect tothis basis isthen adiagonal matrIx. 13.(a)IfSisdiagonahzable, then itsminimal polynomialover kisoftype m q(t)=n(t-Ai), i=I where AI'...,Amaredistinct elements ofk. (b)Conversely, iftheminimal polynomial ofSisofthepreceding type, then Sis diagonalizable. [Hint: The space can bedecomposedas adirect sum ofthe subspaces E).jannihilated byS-Ai.] XIV, Ex EXERCISES 569 (c)IfSisdiagonalizable, andifFisasubspace ofEsuch that SF cF,show that S isdiagonalizable asanendomorphism ofF,I.e.that Fhas abasis consisting of eigenvectors ofS. (d)Let S,Tbeendomorphisms ofE,and assume that S,Tcommute. Assume that both S,Tare diagonalizable. Show that theyaresimultaneously diagonalizable, i.e.there exists abasis ofEconsisting ofeigenvectors forboth Sand T.[Hint: IfAisaneigenvalue ofS,and E).isthesubspace ofEconsisting ofallvectors v such that Sv =AV,then TE).cE)..] t4.Let Ebe afinite-dimensional vector space over analgebraically closed field k.Let AEEndk(E). Show that Acan bewritten Inaunique wayasasum A=S+N where Sisdiagonalizable, NISnIlpotent, and SN =NS.Show that S,Ncan beex- pressedaspolynomials inA.[Hint: Let PA(t)=n(t-Ai)m,bethefactorization ofPA(t)with distinct Ai.LetEibethekernel of(A-Ai)mi .Then Eisthedirect sum of theEi.Define SonEsothat onEi,Sv =AiVforall vEEi.Let N =A-S.Show that S,Nsatisfy ourrequirements. TogetSasapolynomial inA,letgbeapolynomial such that g(t)=Aimod (t-Ai)m,for alli,and g(t)=0mod t.Then S=g(A) and N =A-g(A).] t5.After you have read thesection onthe tensor product ofvector spaces, youcaneasily dothefollowing exercise. LetE,Fbefinite-dimensional vector spaces over analge- braically closed field k,and letA :E-+Eand B:F-+Fbek-endomorphisms ofE,F, respectively. Let PA(t)=n(t-lXi)ni and PB(t)=n(t-(3)mj bethe factorizations oftheir respectively characteristic polynomials, into distinct linear factors. Then PA@B(t)=n(t-lXi(3)nimj. i.j [Hint: Decompose Einto thedirect sum ofsubspaces Ei,where Eiisthesubspace of Eannihilated bysome power ofA-lXi.Dothe same forF,gettingadecomposition into adirect sum ofsubspaces Fj.Then show that some power ofA(8)B-lXi{3j annihilates Ei(8)Fj.Use thefactthat E(8)Fisthedirect sum ofthesubspaces Ei(8)Fj, and thatdimk(E i(8)Fj)=nimj.] 16.Letrbeafree abelian group ofdimension n>1.Letr'beasubgroup ofdimension n also. Let{V.,..., vn}beabasis ofr,and let{w 1,..., wII}beabasis ofr'.Write Wi=Laijvj. Show that theindex (r:r')isequal totheabsolute value ofthedeterminant ofthe matrix(aij). 17.Prove thenormal basis theorem forfinite extensions ofafinite field. 18.LetA=(aij)beasquarenxnmatrix over acommutative ring k.LetAijbethematrix obtained bydeleting thei-th rowandj-th column from A.Letbij=(_I)i+jdet(A ji), and letBbethematrix (bij).Show thatdet(B)=det(A)n- 1,byreducing theproblem to the case when Aisamatrix with variable coefficients over theintegers. Use this same method togiveanalternative proof oftheCayley-Hamilton theorem, that PA(A)=O. 570 REPRESENTATION OFONE ENDOMORPHISM XIV, Ex 19.Let(E,A)and(E',A')bepairs consisting ofafinite-dimensional vector space over a field k,and ak-endomorphism. Show that these pairsareisomorphic ifandonly if their invariants areequal. 20. (a) How many non-conjugate elements ofGL2(C) arethere with characteristic poly- nomial (3(t +1)2(t-I)? (b)How many with characteristic polynomialt3-1001 t? 21. Let Vbe afinite dimensional vector spaceover Qand letA:V Vbe aQ-linear map such that A5=Id.Assume thatifvEVissuch that Av=v,then v=O.Prove that dim Visdivisible by4. 22. Let Vbeafinite dimensional vector space over R,and letA:V VbeanR-linear map such that A2= -Id.Show that dim Viseven, and that Visadirect sum of2- dimensional A-invariant subspaces. 23. Let Ebe afinite-dimensional vector space over analgebraically closed field k.Let A,Bbek-endomorphisms ofEwhich commute, i.e.AB =BA. Show that Aand Bhave acommon eigenvector. [Hint: Consider asubspace consisting ofallvectors having afixed element ofkaseigenvalue.] 24. Let Vbeafinite dimensional vector space over afield k.LetAbeanendomorphism ofV.LetTr(Am) bethetrace ofAm asanendomorphism ofV.Show that thefollowing power series inthevariable tareequal: (IX) )dIX) expL-Tr(Am)tm =det(I-tA) or-- dlogdet(I-tA)=LTr(Am)tm . m= I m tm= I Compare with Exercise 23ofChapter XVIII. 25. Let V,Wbefinite dimensional vector spacesover k,ofdimension n.Let(v,w) (v,w)be anon-singular bilinear form onVxW.Let cEk,and letA:V Vand V:W Wbeendomorph isms such that (Av, Bw)=c(v, w)forallvEVand wEW. Show that anddet(A)det(tl-B)=(-I)ndet(cl-tA) det(A)det(B)=cn . For anapplication ofExercises 24and 25toacontext oftopology oralgebraic geometry,seeHartshorne's Algebraic Geometry, Appendix C,4. 26. Let G==SLn(C) and letKbethecomplex unitary group. Let Abethegroup ofdi- agonal matrices with positive real components onthediagonal. (a)Show thatifgENorG(A) (normalizer ofAinG),then c(g) (conjugation by g)permutes thediagonal components ofA,thus giving rise to ahomo- morphism NorG(A)---+Wtothegroup Wofpermutations ofthediagonal coordina tes. Bydefinition, thekernel oftheabove homomorphism isthecentralizer CenG(A). (b)Show that actually allpermutations ofthecoordinates can beachieved by elements ofK,sowegetanisomorphism W NorG(A)/CenG(A) NorK(A)/CenK(A). Infact, theKontherightcan betaken tobetherealunitary group, because permutation matrices can betaken tohave real components (0or+1). CHAPTER XV Structure ofBilinear Forms There are three major types ofbilinear forms: hermitian (orsymmetric), unitary, andalternating (skew-symmetric). Inthischapter, wegive structure theorems giving normalized expressions forthese forms with respect tosuitable bases. The chapter also follows thestandard pattern ofdecomposinganobject into adirect sum ofsimple objects, insofar aspossible. 1. PRELIMINARIES, ORTHOGONAL SUMS The purpose ofthischapter istogosomewhat deeper into the structure theory forourthree types offorms. Todothis weshall assume most ofthetime that ourground ring isafield, and infact afield ofcharacteristic =f.2inthe symmetric case. We recall our three definitions. Let Ebe amodule over acommutative ring R.Letg:ExE-+Rbeamap. Ifgisbilinear, wecall gasymmetric form ifg(x,y)=g(y,x)forallx,yEE.Wecall galternating ifg(x,x)=0,and hence g(x,y)= -g(y,x)forallx,yEE.IfRhas anautomorphism oforder 2, written a a,wesaythat gisahermitian form ifitislinear initsfirstvariable, antilinear initssecond, and g(x,y)=g(y,x). We shall write g(x, y)=<x,y)ifthereference togisclear. We also oc- casionally write g(x,y)=x.yorg(x,x)=x2 .We sometimes call gascalar product. 571 572 STRUCTURE OFBILINEAR FORMS XV,1 IfVb. ..,VmEE,wedenote by(Vl,.. .,vm)thesubmodule ofEgenerated by Vb.. .,Vm. Let 9besymmetric, alternating, orhermitian. Then itisclear that theleft kernel of9isequal toitsright kernel, and itwillsimply becalled thekernel ofg. Inanyone ofthese cases, wesaythat 9isnon-degenerate ifitskernel isO. Assume that Eisfinite dimensional over thefield k.The form isnon-degenerate ifandonly ifitisnon-singular, i.e.,induces anisomorphism ofEwith itsdual space (anti-dual inthe case ofhermitian forms). Except forthe few remarks ontheanti-linearity made intheprevious chapter, wedon't usetheresults oftheduality inthat chapter. We need only theduality over fields, given inChapter III. Furthermore, wedon't essentially meet matrices again, except fortheremarks onthepfaffian in 1O. We introduce one more notation. Inthestudy offorms onvector spaces, weshall frequently decompose the vector space into direct sums oforthogonal subspaces. IfEisavector space with aform gasabove, and f"',F"aresubspaces, weshall write E=f'1-f" tomean that Eisthedirect sum off'and f",and that Fisorthogonal (or perpendicular) tof",inother words, x1-y(or<x,y)=0)forall xEf"'and yEf"". Wethen saythat Eistheorthogonal sum off'and f"".There will beno confusion with the useofthesymbol .1when wewrite f'1- f"tomean simply that f'isperpendicular tof".The context always makes ourmeaning clear. Most ofthischapter isdevoted togiving certain orthogonal decompositions ofavector space with oneofourthree types offorms,sothateachfactor inthe sum isaneasily recognizable type. Inthesymmetric and hermitian case, weshall beespecially concerned with direct sum decompositions into factors which are I-dimensional. Thus if < ,)issymmetric orhermitian, weshall saythat{VI'. . .,vn}isanorthogonal basis (with respecttotheform) if<Vi,Vj)=0whenever i=f.j.We seethat an orthogonal basis gives such adecomposition. Iftheform isnondegenerate, and if{vl'...,vn}isanorthogonal basis, then we see atonce that <Vi,Vi) i=0 foralli. Proposition 1.1. Let Ebeavector space over thefield k,and letgbeaform ofoneo.fthethree above types. Suppose that Eisexpressed asanorthogonal sum, E=E1.1...1-Em. Then 9isnon-degenerateonEifandonlyfitisnon-degenerate oneach Ei. IfE?isthekernel oftherestriction of9toEi,then thekernel of9inEisthe orthogonalsum EO =E?1-...1-E. xv, 1 PRELIMINARIES, ORTHOGONAL SUMS 573 Proof Elements v,",'of£can bewritten uniquely m V="v.I' i=1m W=LWi i=1 with Vi'"'iE£i. Then v.w=m "V.. W.I I' i=1 and V.W=0ifVi.Wi=0foreach i=1,...,m.From this our assertion is obvious. Observe that ifEl'...,Emarevector spaces over k,and g1,. ..,gmareforms onthese spaces respectively, then wecandefine aformg=gl(f)...(f)gmonthe direct sum E=E1(f). ..(f)Em;namely ifv,ware written asabove, then welet m g(v,w)=Igi(V i,Wi). i=1 Itisthen clear that, infact, wehave E=E11... .1.Em.Wecould also write g=g11.. ..1.gm. Proposition 1.2. Let Ebeafinite-dimensional space over thefield k,and let gbeaform ofthepreceding type onE.Assume that gisnon-degenerate. Let Fbe asubspace ofE.Theform isnon-degenerate onFifand onlyif F+F.1 =E,and alsoifandonlyifitisnon-degenerate onF.1. Proo.f We have (asatrivial consequence ofChapter III,95) dim F'+dim F'.1=dim £=dim(F' +F'.1) +dim(F' nf"'.1). Hence f"'+f"'.1 =£ifandonly ifdim(f"' nF'.1)=O.Our first assertion follows atonce. Since F',F'.1enter symmetrically inthedimension condition, our second assertion also follows. Instead ofsaying that aform isnon-degenerate onE,weshall sometimes say, byabuse oflanguage, that Eisnon-degenerate. Let£beafinite-dimensional space over thefield k,and letgbeaform of thepreceding type. LetEobethekernel oftheform. Then wegetaninduced form ofthe same type go:EIEox£IEo k, because g(x,y)depends only onthe coset ofxand the coset ofymodulo Eo. Furthermore, 90isnon-degenerate since itskernel onboth sides isO. Let£,£'befinite-dimensional vector spaces, with forms g,g'asabove, respectively. Alinear mapa:E E'issaid tobemetric if g'(ax, ay)=g(x,y) 574 STRUCTURE OFBILINEAR FORMS XV,2 orinthedotnotation, ax.ay=x.yforallx,yEE.If(Jisalinear isomorphism, and ismetric, then wesaythat aisanisometry. LetE,Eobeasabove. Then wehave aninduced form onthefactor space EIEo. IfWisacomplementary subspace ofEo,inother words, E=Eo(f)W, and ifwelet a:E EIEo bethecanonical map, then (Jismetric, and induces anisometry ofWonEIEo. This assertion isobvious, and shows that if E=Eo(f)W' isanother direct sumdecomposition ofE,then W'isisometric toW. Weknow that W EIEo isnondegenerate. Hence our form determines auniquenon- degenerate form, uptoisometry,oncomplementary subspaces ofthekernel. 2. QUADRATIC MAPS Let Rbeacommutative ring and letE,F'beR-modules. We suppress the prefix R- asusual. We recall that abilinear map.f: ExE F'issaid tobe symmetric iff(x, y)=f(y,x)forallx,yEE. We saythat Fiswithout 2-torsion ifforallyEf'such that 2y=0wehave y=O.(This holds if2isinvertible inR.) Letj: E F'beamapping. Weshall saythat.fis quadratic (i.e.R-quadratic) ifthere exists asymmetric bilinear map 9:ExE f'and alinear map h:E F' such that forallxEEwehave f(x)=g(x,x)+h(x). Proposition 2.1. Assume that f'iswithout 2-torsion. Let.f:E F'be quadratic, expressedasabove interms ofasymmetric bilinear map and a linear map. Then g,hareuniquely determined by.! F'orallx,yEEwehave 2g(x, y)=.f(x +y)-f(x)-f(y). Proof Ifwecompute f(x +y)-f(x)-f(y), then weobtain 2g(x, y). Ifgtissymmetric bilinear, hiislinear, and.f(x)=gl(X, x)+hl(x), then 2g(x, y)=2g I(x,y).Since Fisassumed tobewithout 2-torsion, itfollows that g(x,y)=9I(x,y)forallx,yEE,and thus that 9isuniquely determined. But then hisdetermined bytherelation h(x)=f(x)-g(x,x). Wecall g,hthebilinear and linear maps associated with! If.f: E f'isamap,wedefine f: ExE F' XV,3 SYMMETRIC FORMS, ORTHOGONAL BASES 575 by I1f(x, y)=f(x +y)-f(x)-f(y). We saythatfishomogeneous quadratic ifitisquadratic, andifitsassociated linear mapiso.We shall saythat F'isuniquely divisible by2ifforeach ZEF there exists auniqueuEfsuch that 2u=z.(Again this holds if2isinvertible inR.) Proposition 2.2. Letf:E f'be amap such that4fisbilinear. Assume thatfisuniquely divisible by2.Then the map xf(x)-tl1f(x, x)is Z-linear. Iffsatisfies thecondition f(2x)=4f(x), thenfishomogeneous quadratic. Proof Obvious. Byaquadratic form onE,one means ahomogeneous quadratic map f:E R,with values inR. Inwhat follows, we areprincipally concerned with symmetric bilinear forms. Thequadratic forms playa secondary role. 3. SYMMETRIC FORMS, ORTHOGONAL BASES Letkbeafield ofcharacteristic =1=2. LetEbeavector space over k,with thesymmetric form g.We saythat 9 isanull form orthat Eisanull space if(x,y)=0forallx,yEE.Since we assumed that thecharacteristic ofkis=I2,thecondition x2=0forallxEE implies that 9isanull form. Indeed, 4x.y=(x+y)2-(x_y)2. Theorem 3.1. LetEbe =1=0andfinite dimensional over k.Let gbeasym- metric formonE.Then there exists anorthogonal basis. Proof We assume first that 9isnon-degenerate, and prove ourassertion by induction inthat case. Ifthedimension nis1,then our assertion isobvious. Assume n>1.Let VIEEbesuch thatvI=f.0(such anelement exists since gisassumed non-degenerate). Letf=(VI) bethesubspace generated byVI. Then f'isnon-degenerate, andbyProposition 1.2, wehave E=F+F1-. Furthermore, dimf.l =n-1.Let{V2,. . .,vn}beanorthogonal basis ofF'1-. 576 STRUCTURE OFBILINEAR FORMS XV,3 Then {vl'. ..,vn}arepairwise orthogonal. Furthermore, theyarelinearly independent, forif a1VI+. . .+anVn=0 with aiEkthen wetake thescalar product withVitogetaiv;=0whence ai=0 foralli. Remark. Wehave shown infactthatif9isnon-degenerate, and vEEissuch that v2#0then wecancompletevtoanorthogonal basis ofE. Suppose that theform gisdegenerate. LetEobeitskernel. We canwrite Easadirect sum E=Eo(f)W for some subspace W. The restriction ofgtoWisnon-degenerate; otherwise there would beanelement ofWwhich isinthekernel ofE,and =f.O.Hence if {VI' ..., vr}isabasis ofEo,and{WI'...' ""n-r} isanorthogonal basIs ofW,then {VI'.. .,Vr,WI'. . .,Wn-r} isanorthogonal basis ofE,aswas tobeshown. Corollary 3.2. Let{VI'..., vn}be anorthogonal basis qfE.Assume that vl=f.0for i<rand vf=0{or i>r.Then the kernel o.fEisequal to (vr+b. ..,vn). Proof Obvious. If{VI'...,Vn}isanorthogonal basis ofEand ifwewrite x =XIVI+...+XnVn with XiEk,then X2 2 2=alx l+...+anx n where ai=<Vi'Vi). Inthisrepresentation oftheform, wesaythat itisdiagonal- ized. With respect toanorthogonal basis, we see atonce that theassociated matrix oftheform isadiagonal matrix, namely al a2 o ar o o o XV,4 SYMMETRIC FORMS OVER ORDERED FIELDS 577 Example. Note that Exercise 33ofChapter XIII gave aninteresting example ofanorthogonal decomposition involving harmonic polynomials. 4. SYMMETRIC FORMS OVER ORDERED FIELDS Theorem 4.1. (Sylvester) Let kbeanordered field and letEbe afinite dimensional vector spaceover k,with anon-degenerate symmetric form g.There exists aninteger r>0such that, if{V.,. . .,vn}isanorthogonal basis ofE, then preciselyramong the nelements vy,. . .,vare> 0,and n-ramong these elements are <o. Proof. Let ai=vf,fori=1,...,n.After renumbering thebasis elements, sayai'. . .,ar>0and ai<0fori>r.Let{wI'...,wn}beanyorthogonal basis, and letbi=wl. Say bl,...,bs>0and bj<0forj>s.We shall prove that r=s.Indeed, itwill suffice toprove that Vb...,Vr,Ws+b.. .,Wn arelinearly independent, forthen weget r+n-s<n,whence r<s,and r=sbysymmetry. Suppose that XIV l+...+XrV r+Ys+IW s+l+...+YnWn=O. Then XIVl +...+XrV r=-Ys+IW s+1- ... -YnWn. Squaring both sides yields 2 2b2+b2 a1xI +...+arX r=s+lYs+ 1+...nYn. The left-hand side is>0,and theright-hand side is<O.Hence both sides are equal to0,and itfollows that Xi=Yj=0,inother words that our vectors are linearly independent. Corollary 4.2. Assume that every positive element ofkisasquare. Then there exists anorthogonal basis {VI'...,vn}ofEsuch thatvf= 1for i<r andvf= -1for i>r,and risuniquely determined. Proof. Wedivide each vector inanorthogonal basis bythesquareroot of theabsolute value ofitssquare. Abasis having theproperty ofthecorollary iscalled orthonormal. IfXisan element ofEhaving coordinates (xl'...,xn)with respect tothis basis, then X2 2 2 2 2=XI+...+Xr-Xr+I- ...-Xn. 578 STRUCTURE OFBILINEAR FORMS XV,4 We say that asymmetric form 9ispositive definite ifX2>0forall XEE,X=/;O.This isthe case ifandonly ifr=ninTheorem 4.1. Wesay that 9isnegative definite ifX2<0forallXEE,X=/;O. Corollary 4.3. The vector space Eadmits anorthogonal decomposition E=E+ 1.E-such that gispositive definite onE+andnegative definite on E-. The dimension ofE+(orE-) isthe same inallsuch decompositions. Let usnow assume that theform gispositive definite and that every positive element ofkisasquare. Wedefine the norm ofanelement vEEby Ivl=. Then wehaveIvI>0ifv=IO.Wealso have theSchwarz inequality Iv.wl<Ivllwl forallv,WEE. This isproved intheusual way, expanding o<(av+bw)2=(av+bw). (av+bw) bybilinearity, andletting b=IvIand a=IwI.One then gets +2ab v.w<21V12 IW12 . IfIvIorIwi=0ourinequality istrivial. Ifneither is0wedivide by IvIIwItoget what wewant. From theSchwarz inequality,wededuce thetriangle inequality Iv+wi<Ivl+Iwl. We leave ittothereader asaroutine exercise. When wehave apositive definite form, there isacanonical way ofgettingan orthonormal basis, starting with anarbitrary basis {vl,...,vn}andproceeding inductively. Let ,1 VI= YvVI. Then Vlhas norm 1.Let w2=V2-(V2.V'I)V'b and then ,1 v2= lw2Iw2. XV,5 HERMITIAN FORMS 579 Inductively, welet W=v-(V.V'l)V'l- .. .-(V.v,)v' r r r r r- 1r- 1 and then ,1 Vr= IwrlWr. The {V'l'...,v}isanorthonormal basis. The inductive process just described isknown astheGram-Schmidt orthogonalization. 5. HERMITIAN FORMS Letkobeanordered field (asubfield ofthereals, ifyouwish) and letk=ko(i), where i=J=l.Then khas anautomorphism oforder 2,whose fixed field isko. LetEbeafinite-dimensional vector space over k.Weshall deal with ahermi- tian form onE,i.e. amap ExEk written (x,y) (x,y) which isk-linear initsfirstvariable, k-anti-linear initssecond variable, and such that (x,y)=(y,x) forallx,yEE. Weobserve that(x,x)EkoforallxEE.This isessentially the reason why theproofs ofstatements concerning symmetric forms hold essentially without change inthehermitian case. Weshall now make thelistoftheproperties which apply tothis case. Theorem 5.1. There exists anorthogonal basis. Iftheform isnon-degenerate, there exists aninteger rhaving thefollowing property. If{V.,. . .,vn}isan orthogonal basis, then precisely ramong the nelements (VI'VI)'. ..,(vn,vn) are> 0and n-ramong these elements are <O. 580 STRUCTURE OFBILINEAR FORMS XV,5 Anorthogonal basis {Vb...,vn}such that <Vi'Vi)= 1or-1iscalled an orthonormal basis. Corollary 5.2. Assume thattheform isnon-degenerate, andthat every positive element ofkoisasquare. Then there exists anorthonormal basis. We say that thehermitian form ispositive definite if(x,x)>0forall xEE.We saythat itisnegative definite if(x,x)<0forallxEE,x=/;O. Corollary 5.3. Assume that theform isnon-degenerate. Then Eadmits an orthogonal decomposition E=E+ .1E-such that theform ispositive definite onE+and negative definiteonE-. The dimension ofE+(orE-) isthe same inallsuch decompositions. Theproofs ofTheorem 5.1and itscorollaries areidentical with those ofthe analogous results forsymmetric forms, andwill beleft tothereader. We have thepolarization identity, forany k-linear map A :E E,namely <A(x +y),(x+y»-<A(x-y),(x-y»=2[<Ax, y)+<Ay, x)]. If<Ax, x)=0forallx,wereplace xbyixand get <Ax, y)+<Ay, x)=0, i<Ax, y)-i<Ay, x)=o. From this weconclude: If<Ax, x)=0,forallx,then A=O. This istheonly statement which has noanalogue inthe case ofsymmetric forms. The presence ofiinoneoftheabove linear equations isessential tothe conclusion. Inpractice,one uses the statement inthecomplex case, and one meets ananalogous situation inthereal case when Aissymmetric. Then the statement forsymmetric maps isobvious. Assume that thehermitian form ispositive definite, and that every positive element ofkoisasquare. Wehave theSchwarz inequality, namely I<x,y) 12<<x,x)<y, y) whose proof comes again byexpanding o« ax+py,ax +py) andsettinga=<y,y)andp= -<x,y). Wedefine the norm ofIxItobe Ixl=J<x,x). XV,6 THE SPECTRAL THEOREM (HERMITIAN CASE) 581 Then weget atonce thetriangle inequality Ix+yl<Ixl+Iyl, and for r:J.Ek, Ir:J.xI=Ir:J.11xI. Just asinthesymmetric case, givenabasis, one can find anorthonormal basis bytheinductive procedure ofsubtracting successive projections. Weleave this tothereader. 6. THE SPECTRAL THEOREM (HERMITIAN CASE) Throughout thissection, weletEbeafinite dimensional spaceover C,ofdimension >1,and weendow Ewith apositive definite hermitian form. Let A:E-+Ebealinear map (i.e. C-linear map) ofEinto itself. For fixed yEE,the map x <Ax, y)isalinear functional, and hence there exists a unique element y*EEsuch that <Ax, y>=<x,y*) forallxEE.We define themap A*:E EbyA*y=y*.Itisimmediately clear that A*islinear, and weshall call A*theadjoint ofAwith respect toour hermitian form. Thefollowing formulas aretrivially verified, forany linear maps A,BofE into itself: (A+B)*=A*+B*, (r:J.A)*=CiA* ,A** =A, (AB)*=B*A*. Alinear map Aiscalled self-adjoint (orhermitian) ifA* =A. Proposition 6.1. Aishermitian ifandonlyif(Ax, x)isrealforallxEE. Proo.f Let Abehermitian. Then <Ax, x)=<x,Ax)=<Ax, x), whence <Ax, x)isreal. Conversely,assume <Ax, x)isreal forallx.Then <Ax, x)=<Ax, x>=<x,Ax)=<A*x, x), andconsequently <(A-A*)x,x)=0forallx.Hence A=A*bypolarization. 582 STRUCTURE OFBILINEAR FORMS XV,6 Let A:E Ebealinear map. Anelement EEiscalled aneigenvector ofAifthere exists AECsuch thatA=A.If =I0,then wesaythat Aisan eigenvalue ofA,belonging to. Proposition 6.2. Let Abehermitian. Then alleigenvalues belonging to nonzero eigenvectors ofAare real.If,'are eigenvectors=1=0having eigenvalues A,Xrespectively, andifA=1=X,then .1'. Proof Let Abeaneigenvalue, belonging totheeigenvector =f.O.Then <A,>=<,A>, and these two numbers areequal respectively toA<,> andA<,>. Since =I0,itfollows that A=A,i.e.that Aisreal. Secondly, assume that,'and A,A'are asdescribed above. Then <A,'>=A<,'>=<,A'>=A'<, '>, from which itfollows that<,'>=O. Lemma 6.3. LetA:E Ebealinear map, and dim E>1.Then there exists atleast one non-zero eigenvector ofA. Proof Weconsider C[A], i.e.thering generated byAover C.As avector space over C,itiscontained inthering ofendomorphisms ofE,which isfinite dimensional, thedimension being the same asforthering ofall nxnmatrices ifn=dim E.Hence there exists anon-zero polynomial Pwith coefficients in Csuch thatP(A)=O.We can factor Pinto aproduct oflinear factors, P(X)=(X-Al)...(X-Am) withAjEC.Then (A-All)...(A-AmI)=O.Hence notallfactors A-Ajl can beisomorphisms, and there exists AECsuch that A-AIisnot aniso- morphism. Hence ithas anelement =I0initskernel, and wegetA-A=O. This shows that isanon-zero eigenvector,asdesired. Theorem 6.4. (Spectral Theorem, Hermitian Case). Let Ebe anon- zerofinite dimensional vector spaceover thecomplex numbers, with apositive definite hermitian form. LetA:E Ebeahermitian linear map. Then Ehas anorthogonal basis consisting ofeigenvectors ofA. Proof Let1beanon-zero eigenvector, with eigenvalue Al,and letE1be thesubspace generated by 1.Then Amaps Etinto itself, because <AEt, l>=<Et,Al>=<Et,All>=Al<Et, l>=0, whence AEt isperpendicular to l' Since1=I0wehave <l'1>>0and hence, since our hermitian form is non-degenerate (being positive definite), wehave E=E1(f)Et. XV,6 THE SPECTRAL THEOREM (HERMITIAN CASE) 583 The restriction ofour form toEtispositive definite (ifdim E>1).From Proposition 6.1, we see atonce that therestriction ofAtoEtis hermitian. Hence we cancomplete theproof byinduction. Corollary 6.5. Hypotheses being asinthetheorem, there exists anortho- normal basis consisting ofeigenvectors ofA. Proof. Divide each vector inanorthogonal basis byitsnorm. Corollary 6.6. LetEbeanon-zero finite dimensional vector space over the complex numbers, with apositive definite hermitian form f.Let gbeanother hermitian formonE.Then there exists abasis ofEwhich isorthogonal for bothfand g. Proof. We write f(x, y)=(x,y). Sincefisnon-singular, being positive definite, there exists aunique hermitian linear map Asuch thatg(x,y)=(Ax, y) forallx,yEE.Weapply thetheorem toA,and find abasis asinthetheorem, say{Vl,. ..,vn}.LetAibetheeigenvalue such that AVi=AiVi. Then g(vj, Vj)=(Avj, Vj)=Ai(Vi, Vj), and therefore our basis isalsoorthogonal forg,aswas tobeshown. Werecall that alinear map U:E Eisunitary ifandonly ifV*=V-I. This condition isequivalenttotheproperty that(Ux, Vy)=(x,y)forallelements x,yEE.Inother words, Visanautomorphism oftheformf. Theorem 6.7. (Spectral Theorem, Unitary Case). LetEbeanon-zero finite dimensional vector space over thecomplex numbers, with apositive definite hermitian form. LetU:E Ebeaunitary linear map. Then Ehas anorthogonal basis consisting ofeigenvectors ofV. Proof. Let1=I0beaneigenvector ofU.Itisimmediately verified that thesubspace ofEorthogonal to1ismapped into itself byU,using therelation U* =U- 1,because ifflisperpendicular to1,then (Ufl, 1)=(fl,U*1)=(fl,U-11)=(fl, A-11)=o. Thus we can finish theproof byinduction asbefore. Remark. IfAisaneigenvalue oftheunitary map U,then Ahasnecessarily absolute value 1(because Upreserves length), whence Acan bewritten inthe form ei8with ()real, and wemay view Uasarotation. Let A :E Ebe aninvertible linear map. Just asone writes anon-zero complex number z=re;() with r>0,there exists adecomposition ofAas a product called itspolar decomposition. LetP:E Ebelinear. We say that P issemipositive ifPishermitian and wehave (Px, x)>0forallxEE.Ifwe have (Px, x)>0forallx=1=0inEthen wesaythat Pispositive definite. For 584 STRUCTURE OFBILINEAR FORMS XV,7 example, ifweletP=A*Athen we seethat Pispositive definite, because (A*Ax, x)=(Ax, Ax) >0ifx=1=O. Proposition 6.8. Let Pbesemipositive. Then Phas aunique semipositive square root B:E E,i.e. asemipositive linear map such that B2=P. Proof. Forsimplicity,we assume that Pispositive definite. Bythespectral theorem, there exists abasis ofEconsisting ofeigenvectors. The eigenvalues must be>0(immediate from thecondition ofpositivity). The linear map defined bysending each eigenvector toitsmultiple bythesquare root ofthecorresponding eigenvaluesatisfies therequired conditions. Asforuniqueness, since Bcommutes with Pbecause B2=P,itfollows thatif{V.,. . .,vn}isabasis consisting of eigenvectors forP,then each Viisalso aneigenvector forB.(Cf. Chapter XIV, Exercises 12and 13(d).) Since apositive number has aunique positive square root, itfollows that Bisuniquely determined astheunique linear map whose effect onViismultiplication bythesquare root ofthecorresponding eigenvalue forP. Theorem 6.9. Let A:E Ebeaninvertible linear map. Then Acan be written inaunique wayasaproduct A=VP, where Visunitary and Pis positive definite. Proof. Let P=(A*A)1I2, and letV=AP-I .Using thedefiitions, itis immediately verified that Visunitary,sowegettheexistence ofthedecom- position. Asforuniqueness, suppose A=VIP I.Let V2=ppll=V-IVI. Then U2isunitary,soVV2=I.From thefact thatp*=PandPi=P.,we conclude that p2=pi. Since P,PIareHermitian positive definite, itfollows asinProposition 6.8 that P=PI'thus proving thetheorem. Remark. The arguments used toprove Theorem 6.9apply inthe case of Hilbert space inanalysis. Cf. myReal Analysis. However, fortheuniqueness, since there may not be"eigenvalues", one has touse another technique from analysis, described inthat book. As amatter ofterminology, theexpression A=VPinTheorem 6.9 iscalled thepolar decomposition ofA.Ofcourse, itdoes matter inwhat order wewrite thedecomposition. There isalso aunique decomposition A=PIVIwith PI positivedefinite and VIunitary (apply Theorem 6.9 toA-I, and then take inverses). 7. THE SPECTRAL THEOREM (SYMMETRIC CASE) LetEbeafinite dimensional vector space over thereal numbers, and let9be asymmetric positive definite form onE.IfA:E Eisalinear map, then weknow xv, 7 THE SPECTRAL THEOREM (SYMMETRIC CASE) 585 that itstranspose, relative tog,isdefined bythecondition <Ax, y)=<x,tAy) forallx,yEE.We saythat Aissymmetric ifA =tA. Asbefore, anelement EEiscalled aneigenvector ofAifthere exists AERsuch thatA=A,and A. iscalled aneigenvalue if =f.o. Theorem 7.1. (Spectral Theorem, Symmetric Case). Let E =1=O.Let A:E Ebe asymmetric linear map. Then Ehas anorthogonal basis consisting ofeigenvectors ofA. Proof. Ifwe select anorthogonal basis for thepositive definite form, then thematrix ofAwith respect tothis basis isarealsymmetric matrix, and wearereduced toconsidering the case when E=Rn .LetMbethematrix repre- senting A.Wemay view Masoperating onen,and then Mrepresentsahermi- tian linear map. Let z=f.0beacomplex eigenvector forM,and write z=x+iy, with x,yERn. ByProposition 6.2, weknow that aneigenvalueAforM,be- longing toz,isreal, and wehave Mz=AZ. Hence Mx=AxandMy=Ay. But wemust have x=I0ory=IO.Thus wehave found anonzero eigenvector forM,namely, A,inE.We can now proceed asbefore. Theorthogonal comple- ment ofthiseigenvector inEhasdimension (n-1),and ismapped into itself by A.We can therefore finish theproof byinduction. Remarks. The spectral theorems arevalid over areal closed field; our proofs don't need any change. Furthermore, theproofsarereasonably close tothose which would begiven inanalysis forHilbert spaces, and compact operators. The existence ofeigenvalues and eigenvectors must however be proved differently, forinstance using theGelfand-Mazur theorem which wehave actually proved inChapter XII, orusingavariational principle (Le. findinga maximum orminimum forthequadratic function dependingontheoperator). Corollary 7.2. Hypotheses being asinthetheorem, there exists anortho- normal basis consisting ofeigenvectors ofA. Proof Divide each vector inanorthogonal basis byitsnorm. Corollary 7.3. LetEbeanon-zero finite dimensional vector spaceover the reaIs ,with apositive definite symmetric form f.Let 9beanother symmetric form onE.Then there exists abasis ofEwhich isorthogonal forbothf and g. Proof We write f(x, y)=<x,y). Sincefisnon-singular, being positive definite, there exists aunique symmetric linear map Asuch that g(x,y)=<Ax, y) 586 STRUCTURE OFBILINEAR FORMS XV,8 forallx,YEE.Weapply thetheorem toA,and find abasis asinthetheorem. Itisclearlyanorthogonal basis forg(cf.the same proof inthehermitian case). The analogues ofProposition6.8 and thepolar decomposition also hold in thepresent case, with the same proofs.See Exercise 9. 8. ALTERNATING FORMS LetEbeavector spaceover thefield k,onwhich wenow make norestriction. Weletfbe analternating form onE,i.e. abilinear mapf:ExE-+ksuch that f(x, x)=x2=0forallxEE.Then x.y=-y.x forallx,YEE,asone seesbysubstituting (x+y)for xinx2=o. Wedefine ahyperbolic plane (for thealternating form) tobea2-dimensional space which isnon-degenerate. We getautomaticallyanelement wsuch that w2=0,w=/;o.IfPisahyperbolic plane, and WEP, w=t=0,then there exists anelement y=1=0inPsuch that w·y=t=O.After dividing ybysome constant, wemay assume that w·y=1.Then y·w= -1.Hence thematrix oftheform with respect tothebasis {w,y}is (-) Thepair w,yiscalled ahyperbolic pairasbefore. Given a2-dimensional vector space over kwith abilinear form, and apair ofelements {w,y}satisfying the relations w2=y2=0, y.w= -1, w.y=1, then we seethat theform ISalternating, and that (w,y)isahyperbolic plane for theform. Given analternating formfonE,wesaythat E(or.f)ishyperbolic ifEis anorthogonal sum ofhyperbolic planes. We saythat E(or1)isnullifx.y=0 forallx,YEE. Theorem 8.1. Letfbeanalternating formonthefinite dimensional vector space Eover k.Then Eisanorthogonal sumofitskernel and ahyperbolic subspace. IfEisnon-degenerate, then Eisahyperbolic space, and itsdimension ISeven. Proof. Acomplementary subspace tothekernel isnon-degenerate, and hence wemay assume that Eisnon-degenerate. LetwEE, w=f.O.There exists YEEsuch that w.y=I0and y=IO.Then (w,y)isnon-degenerate, hence isahyperbolic plane P.We have E=P(f)p.landp.lisnon-degenerate. We XV,8 ALTERNATING FORMS 587 complete theproof byinduction. Corollary 8.2. Allalternating non-degenerate forms ofagiven dimension over afield kare isometric. We seefrom Theorem 8.1that there exists abasis ofEsuch that relative to this basis, thematrix ofthealternating form is o 1 -1 0 o 1 -1 0 o 1 -1 0 o o For convenience ofwriting,wereorder thebasis elements ofourorthogonal sum ofhyperbolic planes insuch away that thematrix oftheform is (-) where]ristheunit rxrmatrix. The matrix (0Ir ) -] 0r iscalled thestandard alternating matrix. Corollary 8.3. Let Ebe afinite dimensional vector spaceover k,with a non-degenerate symmetric form denoted by < ,).Letnbe anon-de- generate alternating formon£.Then there exists adirect sum decomposition E=EIE9£2and asymmetric automorphism AofE(with respectto< ,») having thefollowing property. Ifx,yEEare written X=(XbX2) with XtEEl and X2EE2, y=(y1,Y2) with1EEl and y2EE2, 588 STRUCTURE OFBILINEAR FORMS XV,9 then Q(x,Y)=<AXl' Y2>-<Ax2,Y1>. Proof Take abasis ofEsuch that thematrix ofQwith respect tothis basis isthe standard alternating matrix. Letfbethesymmetric non-degenerate form onEgiven bythedotproduct with respect tothis basis. Then weobtain adirect sum decomposition ofEinto subspaces EbE2(corresponding tothe first n,resp. thelast ncoordinates), such that Q(x, y)=f(x t,Y2)-!(X2, Yl). Since < ,>isassumed non-degenerate, wecanfind anautomorphism Ahaving thedesired effect, and Aissymmetric because fissymmetric. 9. THE PFAFFIAN Analternating matrix isamatrix Gsuch that 'G= -Gand thediagonal elements areequal toO.As we saw inChapter XIII, 96,itisthematrix ofan alternating form. WeletGbean nxnmatrix, and assume niseven. (For odd n,cf.exercises.) We start over afield ofcharacteristic O.ByCorollary 8.2, there exists anon- singular matrix Csuch that 'CGC isthematrix (-) and hence det(C)2det(G)=1or 0 accordingasthekernel ofthealternating form istrivial ornon-trivial. Thus in any case, we seethat det(G)isasquare inthefield. Now we move over totheintegers Z.Let tij(1<i<j<n)ben(n-1)/2 algebraically independent elements over Q,lettu=0fori=1,. ..,n,and let tij= -tjifori>j.Then thematrix T=(tij)isalternating, and hence det(T) isasquare inthefield Q(t) obtained from Qbyadjoining allthevariables tij. However, det(T) isapolynomial inZ[tJ, and since wehave unique factorization inZ[t], itfollows thatdet(T) isthesquare ofapolynomial inZ[t]. We canwrite det(T)=p(t)2. Thepolynomial Pisuniquely determined uptoafactor of+1.Ifwe substitute xv, 10 WITT'S THEOREM 589 values forthetijsothat thematrix Tspecializes to (0InI2 ),-Inl2 0 then we seethat there exists aunique polynomial Pwith integer coefficients taking thevalue 1forthisspecialized setofvalues of(t). Wecall Pthegeneric Pfaffian ofsize n,and write itPf. Let Rbeacommutative ring. We have ahomomorphism Z[t]-+R[t] induced bytheunique homomorphism ofZinto R.The image ofthegeneric Pfaffian ofsize ninR[t] isapolynomial with coefficients inR,which westill denote byPf.IfGisanalternating matrix with coefficients inR,then wewrite Pf(G) forthevalue ofPf(t) when wesubstitutegijfortijinPf. Since thedeter- minant commutes with homomorphisms,wehave: Theorem 9.1. Let Rbeacommutative ring. Let(gij)=Gbeanalternating matrix withgijER.Then det(G)=(Pf(G»2. Furthermore, ifCisannxnmatrix inR,then Pf(CGtC)=det(C) Pf(G). Proof The first statement has been proved above. The second statement will follow ifwe can prove itover Z.LetUij(i,j=1,..., n)bealgebraically independent over Q,and such thatUij,tijarealgebraically independent over Q. Let Ubethematrix(uij).Then Pf(UTtU)=+det(U) Pf(T), asfollows immediately from taking thesquare ofboth sides. Substitute values forUand Tsuch that Ubecomes theunit matrix and Tbecomes thestandard alternating matrix. Weconclude that wemust have a+signontheright-hand side. Our assertion now follows asusual foranysubstitution ofUtoamatrix in R,and any substitution ofTtoanalternating matrix inR,aswas tobeshown. 10. WITT'S THEOREM We goback tosymmetric forms and weletkbeafield ofcharacteristic =/;2. 590 STRUCTURE OFBILINEAR FORMS xv, 10 Let Ebeavector space over k,with asymmetric form. We saythat Eisa hyperbolic plane iftheform isnon-degenerate, ifEhasdimension 2,and ifthere exists anelement w=I0inEsuch that ",,2 =O.We saythat Eisahyperbolic space ifitisanorthogonalsum ofhyperbolic planes. Wealso saythat theform onEishyperbolic. Suppose that Eisahyperbolic plane, with anelement w=I0such that w2=O.Let uEEbesuch that E=(w,u).Then u.w=I0;otherwise wwould beanon-zero element inthekernel. Let bEkbesuch that w.bu=bw.u=1. Then select aEksuch that (aw +bu)2=2abw.u+b2u2=O. (Thiscan bedone since wedeal with alinear equation ina.)Put v=aw+bu. Then wehave found abasis forE,namely E=(w,v)such that w2=v2=0and w.v=1. Relative tothis basis, thematrix ofourform istherefore (). We observe that, conversely,aspace Ehavingabasis {w,v}satisfying w2=v2=0and w.v= 1isnon-degenerate, and thus isahyperbolic plane. A basis {w,v}satisfying these relations will becalled ahyperbolic pair. Anorthogonal sum ofnon-degenerate spaces isnon-degenerate and hence ahyperbolic space isnon-degenerate. We note that ahyperbolic space always has even dimension. Lemma 10.1. LetEbeafinite dimensional vector spaceover k,with anon- degenerate symmetric form g.LetFbe asubspace, Fathekernel ofF,and supposewehave anorthogonal decomposition F=1-"'01.u. Let{wb...,ws}beabasis ofFo.Then there exist elements Vb.. .,VsinE perpendicular toU,such that each pair {Wi' Vi}isahyperbolic pair generating ahyperbolic plane Pi'and such that wehave anorthogonal decomposition U1.PI1....1. Ps. Proof Let U1=(w2'.. .,ws)(f)U. Then U1iscontained InF0(f)Uproperly, and consequently (1-"'0 (f)U).lIS xv, 10 WITT'S THEOREM 591 contained invtproperly. Hence there exists anelement UlEvibut Ul(Fo (f)U).L. We have Wl.Ul=I0,and hence (wl,ul)isahyperbolic plane Pl. We have seen previously that we canfind VlEPIsuch that {wl,vl}isahyperbolic pair. Furthermore, weobtain anorthogonal sumdecomposition F1=(w2'...,Ws).1.p1.1.U. Then itisclear that (W 2,. ..,ws)isthekernel ofF1,and we cancomplete the proof byinduction. Theorem 10.2 LetEbeafinite dimensional vectorspace over k,and letg beanon-degenerate symmetric formonE.LetF,F'besubspaces ofE,and leta-:F F'beanisometry. Then a-can beextended toanisometry ofEonto itself. Proof. Weshall first reduce theproof tothe case when Fisnon-degenerate. We can write F=F0.1.Vasinthelemma ofthepreceding section, and then aF =F' =(JF0.1.aV. Furthermore, aF0=Fisthekernel ofF'.Now we canenlarge both Fand F'asinthelemma toorthogonal sums V.1.P1.1.....1.Psand(JU.1.P'l.1.....1.P corresponding toachoice ofbasis inF0and itscorresponding image inF. Thus we can extend atoanisometry ofthese extended spaces, which are non- degenerate. This gives usthedesired reduction. We assume that f",f'"arenon-degenerate, andproceed stepwise. Suppose first that F' =F,i.e.that (Jisanisometry ofFonto itself. We can extend (JtoEsimply byleaving every element ofF.Lfixed. Next, assume that dim F=dim F' =1and that F=IF'.Say F=(v)and F' =(v'). Then v2=V,2.Furthermore, (v,v')hasdimension 2. If(v,v')isnon-degenerate, ithas anisometry extending (J,which mapsvon v'and v'on v.We canapply thepreceding step toconclude theproof. If(v,v')isdegenerate, itskernel hasdimension 1.Let Wbeabasis forthis kernel. There exist a,bEksuch that v'=av+bw. Then V,2 =a2v2and hence a=+1.Replacing v'by-v'ifnecessary,wemayassume a=1.Replacingw bybw, wemayassume v'=v+w.Let z=v+v'.Weapply Lemma 10.1 to the space (W,z)=(w).1.(z). We canfind anelement YEEsuch that y.z=0, y2=0,and w.y=1. 592 STRUCTURE OFBILINEAR FORMS xv, 10 The space (z,w,y)=(z)1.(w,y)isnon-degenerate, beinganorthogonal sum of(z)and thehyperbolic plane (w,y).Ithas anisometry such that zz, w -w, y-y. But v=l(z-w)ismappedon v'=l(z+w)bythis isometry. We have settled thepresentcase. Wefinish theproof byinduction. Bytheexistence ofanorthogonal basis (Theorem 3.1), every subspace Fofdimension > 1has anorthogonal de- composition into asum ofsubspaces ofsmaller dimension. LetF=F11.F2 with dim F1and dim F2>1.Then aF=aF11.af"2. Let a1=aIF 1betherestriction ofatoFl.Byinduction, we can extend a1to anisometry al:EE. Then al(Ff)=(a 1Fl).l.Since aF2isperpendiculartoaFl=alF 1,itfollows that (JF2iscontained ina1(Ft). Let a2=aIF2.Then theisometry a2:F2-+a2F2=aF2 extends byinduction toanisometry a2:Ft-+(j1(Ft). Thepair (a1,a2)givesusanisometry ofF11.Ft=Eonto itself, asdesired. Corollary 10.3. Let E,E'befinite dimensional vector spaces with non- degenerate symmetric forms, and assume that theyare isometric. LetF,F'be subspaces, and let(j:F F'beanisometry. Then (jcan beextended toan isometry ofEonto E'. Proof. Clear. Let Ebe aspace with asymmetric form g,and letFbe anull subspace. Then byLemma 10.1,we can embed Fin ahyperbolic subspace Hwhose dimension is2dim F. Asapplications ofTheorem 10.2, wegetseveral corollaries. Corollary 10.4. Let Ebe afinite dimensional vector space with anon- degenerate symmetric form. Let Wbeamaximal null subspace, and letW'be some null subspace. Then dim W'<dim W,and W'iscontained insome maximal null subspace, whose dimension isthe same asdim W. xv, 10 WITT'S THEOREM 593 Proof. That W'iscontained inamaximal null subspace follows byZorn's lemma. Suppose dim W'>dim W.Wehave anisometry ofWonto asubspace ofW'which we can extend toanisometry ofEonto itself. Then (J-I(W') isa null subspace containing W,hence isequal toW,whence dim W=dim W'. Our assertions follow bysymmetry. Let Ebeavector space with anon-degenerate symmetric form. Let Wbea null subspace. ByLemma 10.1 we can embed Winahyperbolic subspace Hof Esuch that Wisthemaximal null subspace ofH,andHisnon-degenerate. Any such Hwill becalled ahyperbolic enlargement ofW. Corollary 10.5. Let Ebe afinite dimensional vector space with anon- degenerate symmetric form. LetWand W'bemaximal null subspaces. LetH, H'behyperbolic enlargements ofW, W'respectively. Then H,H'areisometric and soareHi.andH'1-. Proof. We have obviouslyanisometry ofHonH',which can beextended toanisometry ofEonto itself. This isometry maps Hi. onH'i., asdesired. Corollary 10.6. Letgl'g2'hbesymmetric forms onfinite dimensional vector spaces over thefield ofk.IfglEBhisisometric tog2EBh,andifgl' g2are non-degenerate, then glisisometric tog2. Proof. Letglbeaform onEland g2aform onE2.Let hbeaform onF. Then wehave anisometry between F'(f)Eland F'(f)E2.Extend theidentity id :F Ftoanisometrya- ofFEBE1toFEBE2byCorollary 10.3. Since El and E2aretherespective orthogonal complements ofFintheir two spaces,we must have a-(E 1)=E2,which proves what wewanted. If9isasymmetric form onE,weshall saythat 9isdefinite ifg(x, x) =/;0 foranyxEE,x=1=0(i.e. x2=1=0ifx=1=0). Corollary 10.7. Let9beasymmetric formonE.Then 9has adecomposition asanorthogonal sum 9=go(f)ghyp (f)gdef where goisanullform, ghypishyperbolic, and gdef isdefinite. Theform ghyp (f)gdef isnon-degenerate. Theforms go, ghyp,and gdefareuniquely determined uptoisometries. Proof. The decomposition 9=go(f)glwhere goisanull form and gl isnon-degenerate isunique uptoanisometry, since gocorresponds tothe kernel ofg. We may therefore assume that 9isnon-degenerate. If 9=gh(f)gd 594 STRUCTURE OFBILINEAR FORMS xv, 11 where ghishyperbolic and gdisdefinite, then ghcorresponds tothehyperbolic enlargement ofamaximal null subspace, andbyCorollary 10.5 itfollows that ghisuniquely determined. Hence gdisuniquely determined astheorthogonal complement ofgh.(By uniquely determined, we mean ofcourse upto an isometry. ) Weshall abbreviateghypbyghand gdefbygd. 11. THE WITT GROUP Let g,cpbysymmetric forms onfinite dimensional vector spacesover k .We shall saythattheyareequivalent ifgdisisometric toCPd'The reader willverify atonce that this isanequivalence relation. Furthermore the(orthogonal)sum oftwonull forms isanullform, andthe sum oftwohyperbolic forms ishyperbolic. However, the sum oftwo definite forms need not bedefinite. We write our equivalence g-- cpoEquivalence ispreserved under orthogonal sums, and hence equivalence classes ofsymmetric forms constitute amonoid. Theorem 11.1. The monoid ofequivalence classes ofsymmetric forms (over thefield k)isagroup. Proof. We have toshow that every element has anadditive inverse. Let 9 be asymmetric form, which wemayassume definite. We let-g betheform such that(-g)(x, y)=-g(x, y).Wecontend that g(f)-g isequivalent toO. Let Ebethe space onwhich 9isdefined. Then 9(f)-9isdefined onE(f)E. Let Wbethesubspace consisting ofallpairs (x,x)with xEE.Then Wisanull space for9(f)-g.Since dim(E (f)E)=2dim W,itfollows that Wisamaximal null space, and that 9(f)-9ishyperbolic,aswas tobeshown. The group ofTheorem 11.1 will becalled theWitt group ofk,andwill be denoted byW(k). Itisofimportance inthestudy ofrepresentations ofelements ofkbythequadratic formfarising from g[i.e.f(x)=g(x, x)], forinstance when one wants toclassify thedefinite forms f. Weshall now define another group, which isofimportance inmore functorial studies ofsymmetric forms, forinstance instudying thequadratic forms arising from manifolds intopology. Weobserve that isometry classes ofnon-degenerate symmetric forms (over k)constitute amonoid M(k), thelawofcomposition being theorthogonalsum. Furthermore, thecancellation law holds (Corollary 10.6). We let cl :M(k) WG(k) XV, Ex EXERCISES 595 bethecanonical map ofM(k) into theGrothendieck group ofthis monoid, which weshall call the Witt-Grothendieckgroup over k.As weknow, the cancellation lawimplies that clisinjective. If9isasymmetric non-degenerate form over k,wedefine itsdimension dim gtobethedimension ofthespace Eonwhich itisdefined. Then itisclear that dim(g (f)g')=dim 9+dim g'. Hence dim factors throughahomomorphism dim: WG(k) z. This homomorphism splits since wehave anon-degenerate symmetric form of dimension 1. LetWGo(k) bethekernel ofourhomomorphism dim. If9isasymmetric non-degenerate form wecandefine itsdeterminant det(g) tobethedeterminant ofamatrix Grepresenting 9relative toabasis, modulo squares. This iswell defined asanelement ofk*Ik*2. Wedefine detoftheO-form tobe1.Then detis ahomomorphism det:M(k) k*lk*2, and can therefore befactored throughahomomorphism, again denoted by det, oftheWitt-Grothendieck group, det: WG(k) k*lk*2. Other properties ofthe Witt-Grothendieck group will begiven inthe exerCIses. EXERCISES 1.(a) Let Ebeafinite dImensional spaceover thecomplex numbers, and let h:ExE-+C beahermitian form. WrIte h(x,y)=g(x,y)+if(x, y) where g,fare real valued. Show that g,fare R-bilinear, gissymmetric, fis alternating. (b) Let Ebefinite dimensIonal over C.Let g:ExE-+CbeR-bilinear. Assume that forallxEE,themap y1-+g(x,y)isC-linear, and that theR-bilinear form f(x, y)=g(x,y)-g(y,x) 596 STRUCTURE OFBILINEAR FORMS XV, Ex ISreal-valued onExE.Show that there exists ahermitian form honEand a symmetrIc C-bilinear form 1/1onEsuch that 2ig=h+1/1.Show that hand1/1are uniquely determined. 2.Prove thereal case oftheunitary spectral theorem: IfEisanon-zero finite dimensional space over R,with apositive definite symmetric form, and U :E-+EISaunitary linear map, then Ehas anorthogonal decomposition into subspaces ofdimension 1or2, invariant under U.Ifdim E=2,then thematrix ofUwith respect toany ortho- normal basis ISoftheform (COS (} sin (}-sin (} )or(-1 cos (} 00 )(COS (} 1 Sln(}-sin (} )cos (}, depending onwhether det(U)=1or-1.Thus UisarotatIon, orarotation followed byareflection. 3.Let Ebe afinite-dimensional, non-zero vector space over thereals, with apositive definite scalar product. LetT:E-+Ebeaunitary automorphism ofE.Show that E isanorthogonal sum ofsubspaces E=E11....1.Em such that each EiisT-invariant, and hasdimension 1or2.IfEhasdimension 2,show that one can find abasis such that thematrix associated with Twith respect tothis basis is (cos (} sin (}-sin (} )or cos (} (-cos (} sin (}sin (} )cos (}, accordingasdetT= 1ordetT=-1. 4.Let Ebe afinite dimensional non-zero vector space over C,with apositive definite hermitian product. LetA,B:E Ebe ahermitian endomorphism. Assume that AB=BA. Prove that there exists abasis ofEconsisting ofcommon eigenvectors forAand B. 5.LetEbeafinite-dimensional space over thecomplex, with apositive definite hermitian form. Let Sbeasetof(C-linear) endomorphisms ofEhavingnoinvariant subspace except 0and E.(This means thatifFisasubspace ofEand BF cFforallBES,then F=0orF=E.) Let Abeahermitian map ofEinto Itself such that AB =BAforall BE S.Show that A=AIfor some real number A.[Hint: Show that there exists exactly oneeigenvalue ofA.Ifthere were twoeigenvalues, sayAli=Az,onecould find two polynomials fandgwith real coefficients such thatf(A) i=0,g(A) i=0but f(A)g(A)=O.LetFbethekernel ofg(A) and getacontradiction.] 6.LetEbeasinExercise 5.LetTbeaC-linear map ofEinto itself. Let A=!<T +T*). Show that Aishermitian. Show that Tcan bewritten intheform A+iBwhere A,B arehermitian, and areuniquely determined. 7.Let Sbeacommutative setofC-linear endomorphisms ofEhaving noinvariant sub- space unequal to0orE.Assume inaddition thatifBES,then B* ES.Show that each XV, Ex EXERCISES 597 element ofSisoftype rxIfor some complex number rx.[Hint: LetBoES.Let A=!(Bo +B). Show that A=),,1for some real A..] 8.Anendomorphism BofEissaid tobenormal ifBcommutes with B*. State and provea spectral theorem fornormal endomorphisms. Symmetric endomorphisms For Exercises 9,10and 11weletEbeanon-zero finite dimensional vector space over R,with asymmetric positive definite scalar product g,which gives rise toanorm lion E. LetA :E Ebe asymmetric endomorphism ofEwith respect tog.Define A;>0 tomean (Ax, x);>0forallxEE. 9.(a) Show that A;>0ifandonly ifalleigenvalues ofAbelonging tonon-zero eigenvectorsare;>O.Both inthehermitian case and thesymmetric case, one says that Aissemipositive ifA;>0,andpositive definite if(Ax, x)>0forall x=1=O. (b) Show that anautomorphism AofEcan bewritten inaunique wayasaproduct A=UPwhere Uisrealunitary (that is,tuu=I),and Pissymmetric positive definite. For two hermitian orsymmetric endomorphisms A,B,define A;>Bto mean A-B;>0,andsimilarly forA>B.Suppose A>O.Show that there are two real numbers a>0and f3>0such that al<A<f3I. 10.IfAisanendomorphism ofE,define itsnormIAItobethegreatest lower bound of allnumbers Csuch thatlAx I<clxl forallxEE. (a) Show that this norm satisfies thetriangle inequality. (b) Show that theseries A2 exp(A)=I+A+ 2!+. . . converges, andifAcommutes with B,then exp(A+B)=exp(A) exp(B). IfAissufficiently close toI,show that theseries (A-1) (A-1)210g(A)= -+. . . 1 2 converges, andifAcommutes with B,then 10g(AB)=logA+logB. (c)Using thespectral theorem, show how todefine logPforarbitrary positive definite endomorphisms P. 11.Again, letEbe non-zero finite dimensional over R,and with apositive definite symmetric form. LetA :E Ebe alinear map. Prove: (a)IfAissymmetric (resp. alternating), thenexp(A) issymmetric positive definite (resp. realunitary). (b)IfAisalinear automorphism ofEsufficiently close toI,and issymmetric 598 STRUCTURE OFBILINEAR FORMS XV, Ex positive definite (resp. real unitary), then log Aissymmetric (resp. alternating). (c)More generally, ifAispositive definite, then logAissymmetric. 12. Let Rbeacommutative ring, letE,FbeR-modules, andletf: E-+Fbeamapping. Assume thatmultiplication by2inFisaninvertible map. Show thatfis homogeneous quadratic ifandonlyiffsatisfies theparallelogram law: f(x +y)+f(x-y)=2f(x) +2f(y) forallx,yEE. 13 .(Tate) Let E,Fbecomplete normed vector spaces over the real numbers. Let f:E-+Fbeamap having thefollowing property. There exists anumber C>0such that forallx,yEEwehave If(x +y)-f(x)-f(y) I<C. Show that there exists aunique additive map g:E Fsuch thatIg-flisbounded (i.e.lg(x)-f(x) Iisbounded asafunction ofx).Generalize tothebilinear case. [Hint: Let .f(2"x)g(x)=11m".] "-CX> 2 14.(Tate) Let Sbe asetandf:S Samap ofSinto itself. Let h:S Rbe areal valued function. Assume that there exists areal number d> 1such that h0f-df isbounded. Show that there exists aunique function hfsuch that hf-hisbounded, and hf0f=dhf.[Hint: Lethf(x)=Iimh(fn(x))/dn.] 15. Define maps ofdegree> 2,from one module into another. [Hint: For degree 3, consider theexpression f(x +y+z)-f(x +y)-f(x +z)-f(y +z)+f(x) +f(y) +f(z).] GeneralIze the statement proved forquadratic maps tothese higher-degree maps, i.e. theuniqueness ofthevarious multilinear maps entering into their definitions. Alternating forms 16. Let Ebeavector space over afield kand letgbeabilinear form onE.Assume that whenever x,yEEaresuch thatg(x,y)=0,then g(y,x)=O.Show that gissymmetric oralternating. 17. Let Ebeamodule over Z.Assume that Eisfree, ofdimension n>1,and letfbea bilinear alternating form onE.Show that there exists abasis {ei}(i=1,..., n)and anintegerrsuch that 2r<n, e1.ez=a., e3.e4=a2,..., e2,-1.e2,=a, where a.,..., arEZ,aii=0,and aidivides ai+ 1for i=1,...,r-1and finally ei.ej=0forallother pairs ofindices i<j.Show that theideals Zaiareuniquely determined. [Hint: Consider theinjective homomorphism lpf:E-+EVofEinto the XV, Ex EXERCISES 599 dual space over Z,viewing cpf(E)asafree submodule ofEV.]. Generalize toprincipal nngs when you know thebasis theorem formodules over these rings. Remark. Abasis asinExercise 18iscalled asymplectic basis. For one useof such abasis, see thetheory oftheta functions, asinmyIntroduction toAlgebraic and Abelian Functions (Second Edition, Springer Verlag), Chapter VI,3. 18. Let Ebeafinite-dimensional vector space over thereals, and let< ,>beasymmetric positive definite form. Let Qbeanon-degenerate alternating form onE.Show that there exists adirect sumdecomposition E=E1EBE2 having thefollowing property. Ifx,yEEarewritten x=(Xl'X2) With XIEEl and X2EE2, y=(ybY2) withYlEE Iand Y2EE2' thenfl(x, Y)=(XI'Y2)-(X2'YI).[Hint: UseCorollary 8.3, show that Aispositive definite, and take itssquare root totransform thedirect sum decomposition obtained inthatcorollary.] 19. Show that thepfaffian ofanalternatingnxnmatnx is0when nisodd. 20. Prove alltheproperties forthepfaffian stated inArtin's Geometric Algebra (Inter- science, 1957), p.142. The Witt group 21. Show explicitly how W(k) isahomomorphic image ofWG(k). 22. Show that WG(k) can beexpressedasahomomorphic image ofZ[k*/k*2] [Hint: Use the eXistence oforthogonal bases.] 23.Witt's theorem isstill true foralternating forms. Prove itorlook itupinArtin (ref. inExercise 20). SLn(R) There isawhole area oflinear algebraic groups, giving rise toanextensive algebraic theoryaswell asthepossibility ofdoing Fourier analysis onsuch groups. The group SLn(R) (orSLn(C)can serve asaprototype, and anumber ofbasic facts can beeasily verified. Some ofthem arelisted below asexercises. Readers wanting toseesolutions can look them upin[JoL 01],Spherical Inversion onSLn(R), Chapter I. 24. Iwasawa decomposition. We start with GLn(R). Let: G =GLn(R); K =subgroup ofrealunitarynxnmatrices; U =group ofrealunipotent upper triangular matrices, that ishaving components1 onthediagonal, arbitrary above thediagonal, and 0below thediagonal; 600 STRUCTURE OFBILINEAR FORMS XV, Ex A =group ofdiagonal matrices with positive diagonal components. Prove that theproduct map UxAxK-+UAK eGis actuallyabijection. This amounts toGram-Schmidt orthogonalization. Prove thesimilar statement inthe complex case, that is,forG(C)=GLn(C), K(C)=complex unitary group.. U(C)= complex unipotent upper triangular group, and Athe same group ofpositive diag- onal matrices asinthereal case. 25. Let now G==SLn(R), and letK,Abethecorresponding subgroups having deter- minant 1.Show that theproduct UxAxK-+UAK again givesabijection with G. 26. Let abetheR-vector space ofrealdiagonal matrices with trace O.Let avbethe dual space. Let ai(i==1,...,n-1)bethefunctional defined on anelement H = diag(h l,...,hn)bylI.;(H)=h;-h;+I. (a)Show that{lI.l,"', lI.n-l} isabasis of av over R.(b)LetH;;+ Ibethediagonal matrix with h;=I,hi+I==-1, and hj==0 forj=l-i,i+l. Show that {H I,2,...,H n-l,n}isabasis of a.(c)Abbreviate Hi i+I==H;(i=I,..., n-I).LetafEavbethefunctional such thatlI.;(Hj)==Jij (==Iifi==jand 0otherwise). Thus {lI.,. ..,lI._1}isthe dual basis of {HI,' ..,Hn-l}. Show that lI.;(H)==hi+...+hi. 27. The trace form. LetMatn(R) bethe vector space ofreal nxnmatrices. Define the twisted trace form onthis space by Br(X, Y)=tr(X' Y)=(X,Y)t. Asusual,rYisthetranspose ofamatrix Y.Show that Hrisasymmetric positive definite bilinear form onMatn(R). What istheanalogous positive definite hermitian form onMatn(C)? 28.Positivity. On a(real diagonal matrices with trace 0)theform ofExercise 27can be defined bytr(XY), since elements X,YEa aresymmetric. Letd ={ai,...,an-I} denote thebasis ofExercise 26.Define anelement HEatobesemipositive (writen H>0)ifai(H)>0foralli=1,...,n-1.For each aEaV ,letHa. Earepresenta with respect toBr,that is(Ha.,H)=a(H) forallHEa.Show that H>0ifand only if n-I H==LSiHrx' I ;=Iwith S;>o. Similarly, define Htobepositive and formulate thesimilar condition with Si>O. 29. Show that theelements na;(i=I,...,n-1)can beexpressedaslinear combina- tions oflI.l ,...,an-I with positive coefficients inZ. 30. Let Wbethegroup ofpermutations ofthediagonal elements inthevector spaceaof diagonal matrices. Show thataoisafundamental domain fortheaction ofWon a (i.e., given HEa,there exists aunique H+ >0such that H+==wH for some WE W. CHAPTER XVI The Tensor Product Having considered bilinear maps,wenow come tomultilinear maps and basic theorems concerning their structure. There isauniversal module representing multilinear maps, called the tensor product. Wederive itsbasic properties, and postponetoChapter XIX thespecialcase ofalternating products. The tensor product derives itsname from the use made indifferential geometry, when this product isapplied tothetangent spaceorcotangent space ofamanifold. The tensor productcan beviewed also asprovidingamechanism for"extending the base"; thatis,passing from amodule over aring toamodule over some algebra over thering. This "extension" can also involve reduction modulo anideal, because what matters isthat we aregivenaringhomomorphismf: A B,and wepass from modules over Atomodules over B.Thehomomorphism fcan be ofboth types,aninclusion oracanonical map with B=All for some ideall, or acomposition ofthe two. Ihave tried toprovide thebasic material which isimmediately used ina variety ofapplications tomany fields (topology, algebra, differential geometry, algebraic geometry, etc.). 1. TENSOR PRODUCT Let Rbeacommutative ring. IfE1,.. .,En'}'aremodules, wedenote by Ln(Eb...,En;F) themodule ofn-multilinear maps f:E1X... xEn-+f". 601 602 THE TENSOR PRODUCT XVI,1 Werecall that amultilinear map isamap which islinear (i.e., R-linear) ineach variable. We usethewords linear andhomomorphism interchangeably. Unless otherwise specified, modules, homomorphisms, linear, multilinear refer tothering R. One may view themultilinear maps ofafixed setofmodules El,...,Enasthe objects ofacategory. Indeed, if f:E1X... xEn Fand g:E1X... xEn G aremultilinear, wedefine amorphism f gtobeahomomorphism h:F G which makes thefollowing diagram commutative: FY ElX... xEnjh G Auniversal object inthis category iscalled atensor product ofE1,...,En (over R). Weshall now prove that atensor product exists, and infact construct oneina natural way. Byabstract nonsense, weknow ofcourse that atensor product is uniquely determined, uptoaunique isomorphism. LetMbethefree module generated bythe setofalln-tuples (xl'...,xn), (XiEEi),i.e.generated bythe setE1X... xEn. Let Nbethe submodule generated byalltheelements ofthefollowing type: (xl'...,Xi+x,...,xn)-(x1,...,Xi'...,Xn)-(xl'...,X,...,Xn) (xl'...,aXi,.. .,Xn)-a(x 1,...,Xn) forallXiEEi,X;EEi,aER.We have thecanonical injection E1X... xEn M ofour setinto thefree module generated byit.Wecompose this map with the canonical map M MINonthefactor module, togetamap q>:Elx... xEn MIN. Wecontend thatq>ismultilinear and isatensor product. Itisobvious thatqJismultilinear-our definition was adjusted tothis purpose. Let f:Elx... xEn G beamultilinear map. Bythedefinition offree module generated by E1X... xEn XVI,1 TENSOR PRODUCT 603 wehave aninduced linear map M -+Gwhich makes thefollowing diagram commutative: M E1X.'. xEn(j G Sincefismultilinear, theinduced map M Gtakes onthevalue 0onN.Hence bytheuniversal property offactor modules, itcan befactored through MIN, and wehave ahomomorphism f*:MIN Gwhich makes thefollowing dia- gram commutative: MINY E1X... xEnjfOG Since theimage ofq>generates MIN, itfollows that theinduced mapf*is uniquely determined. This proves what wewanted. The module MIN will bedenoted by n El(8)...(8)En oralso (8)Ei. i=1 Wehave constructed aspecific tensor product intheisomorphism class oftensor products, and weshall callitthetensor product ofEl,...,En.IfXiEEi,wewrite q>(x l'...,Xn)=Xl(8)...(8)Xn=Xl(8)R. ..(8)RXn' We have foralli, Xl(8)...(8)aXi (8)...(8)Xn=a(x 1(8)...(8)Xn), Xl(8)...(8)(Xi+xD(8)...(8)Xn =(X 1(8)...(8)Xn)+(X 1(8)...(8)X;(8)...(8)Xn) forXi'X;EEiand aER. Ifwehave two factors, say E(8)F,then every element ofE(8)F'can be written asasum ofterms X(8)Ywith XEEand yEF,because such terms generate E(8)Fover k,anda(x (8)y)=ax(8)yfor aER. 604 THE TENSOR PRODUCT XVI, 1 Remark. Ifanelement ofthe tensor product is0,then that element can already beexpressed interms ofafinite number oftherelations defining the tensor product. Thus ifEisadirect limit ofsubmodules Eithen funF(8)Ei=F'(8)funEi=F(8)E. Inparticular, every module isadirect limit offinitely generated submodules, and one usesfrequently thetechnique oftesting whether anelement ofF(8)Eis obytesting whether theimage ofthis element inF(8)Eiis0when Eiranges over thefinitely generated submodules ofE. Warning. The tensor productcaninvolve agreat deal ofcollapsing between themodules. For instance, take the tensor product over ZofZlmZ andZlnZ where m,nareintegers>1and arerelatively prime. Then the tensor product ZlnZ (8)ZlmZ=o. Indeed, wehave n(x (8)y)=(nx) (8)y=0andm(x (8)y)=x(8)my=O.Hence x(8)y=0forallxEZlnZ and yEZlmZ. Elements oftype x(8)ygenerate the tensor product, which istherefore O.Weshall seelater conditions under which there isnocollapsing. Inmany subsequent results, weshall assert theexistence ofcertain linear maps from atensor product. This existence isproved byusing theuniversal mapping property ofbilinear maps factoring through the tensor product. The uniqueness follows byprescribing thevalue ofthelinear mapsonelements of type x(8)y(say fortwofactors) since such elements generate thetensor product. Weshall prove theassociativity ofthe tensor product. Proposition 1.1. Let El,E2,E3bemodules. Then there exists aunique isomorphism (El(8)E2)(8)E3 El(8)(E2(8)E3) such that (x(8)y)(8)z x(8)(y(8)z) for xEEl' YEE2and ZEE3. Proof. Since elements oftype (x(8)y)(8) Zgenerate thetensor product, the uniqueness ofthedesired linear map isobvious. Toprove itsexistence, let xEEl. The map Ax:E2xE3 (El(8)E2)(8)E3 XVI, 1 TENSOR PRODUCT 605 such that Ax(Y, z)=(x@y)@zisclearly bilinear, and hence factors througha linear map ofthetensor product Ax:E2@E3 (El@E2)@E3. The map ElX(E2@E3)(El@E2)@E3 such that (x,ex) Ax(ex) for xEEland exEE2@E3isthen obviously bilinear, and factors througha linear map El@(E2@E3)(E 1@E2)@E3, which has thedesired property (clear from itsconstruction). Proposition 1.2. LetE,F'bemodules. Then there isaunique isomorphism E@ff@E such that x@y y@xforxEEand yEF. Proof The map ExF-+F@Esuch that (x,y) y@xisbilinear, and factors through the tensor product E@F,sending x@[email protected] this last map has aninverse (bysymmetry)weobtain thedesired isomorphism. The tensor product has various functorial properties. First, suppose that /;:E Ei (i=1,...,n) isacollection oflinear maps. Wegetaninduced mapontheproduct, nh:nEi-+nEi. Ifwecompose nhwith thecanonical map into thetensor product, then weget aninduced linear map which wemay denote byT(fl,...,fn)which makes the following diagram commutative: E'lX... xE' )E'l@...@E n nf.J jT(J" ,f,,) ElX... xE )E1@.·.@En n 606 THE TENSOR PRODUCT XVI,1 Itisimmediately verified that Tisfunctorial, namely that ifwehave acom- posite oflinear maps h0gi(i=1,..., n)then T(fl09b...,in0gn)=T(fl'...,in)0T(g l'. ..,gn) and T(id,...,id)=ide We observe thatT(fl'...' fn)istheunique linear map whose effect on an element X'l(8)...(8)xofE'l(8)...(8)Eis X'l(8).. .(8)xfl(X'l) (8)...(8)in(x). Wemay view Tasamap n (rt n ) I\L(E;, Ei)-+L E;,iEj, and thereader will have nodifficulty inverifying that this map ismultilinear. Weshall write outwhat this means explicitly fortwofactors, sothat ourmap can bewritten (f,g) T(f, g). Given homomorphisms f:F' F'and gl,g2:E' E,then T(f, gl+g2)=T(f,gl)+T(f, g2), T(f, agl)=aT(f, gl). Inparticular, select afixed module F,and consider thefunctor t=tF(from modules tomodules) such that t(E)=F(8)E. Then tgives rise toalinear map t:L(E', E) L(t(E'), t(E» foreach pair ofmodules E',E,bytheformula t(f)=T(id, f). Remark. Byabuse ofnotation, itissometimes convenient towrite fl(8)...(8)in instead ofT(fl"." in). XVI,2 BASIC PROPERTIES 607 This should not beconfused with the tensor product ofelements taken inthe tensor product ofthemodules L(E'l' E1)(8)...(8)L(E, En). The context willalways make ourmeaning clear. 2. BASIC PROPERTIES The most basic relation relating linear maps, bilinear maps, and the tensor productisthefollowing: For three modules E,F',G, L(E, L(F, G) L2(E,F;G) L(E (8)F,G). Theisomorphisms involved aredescribed inanatural way. (i)L2(E,F;G) L(E, L(F, G». Iff:ExF Gisbilinear, and xEE,then themap fx:FG such thatfx(Y)=f(x, y)islinear. Furthermore, themapxfxislinear, and isassociated withftoget(i). (ii)L(E, L(F, G» L2(E,F;G). LetqJEL(E, L(F', G». We letj:ExF Gbethebilinear map such that fqJ(x, y)=qJ(X) (y). ThenqJ fqJdefines (ii). Itisclear that thehomomorphisms of(i)and(ii) areinverse toeach other and therefore give isomorphisms ofthefirst two objects intheenclosed box. (iji) L2(E,F;G) L(E (8)F,G). This isthemapff*which associates toeach bilinear mapftheinduced linear maponthe tensor product. The association ff*isinjective (because f*isuniquely determined byf),and itissurjective, because any linear map ofthe tensor product composed with thecanonical map ExF-+E(8)Fgives rise toabilinear map onExF. 608 THE TENSOR PRODUCT XVI,2 n Proposition 2.1. LetE=EBEibeadirect sum. Then wehave anisomor- i=1 phism n F(8)E+-+EB(F(8)Ei). i= 1 Proof. Theisomorphism isgiven byabstract nonsense. Wekeep Ffixed, and consider thefunctor! :X F(8)X.Aswe sawabove, tislinear. Wehave projections Tti:E EofEonEi.Then Tti0Tti=Tti, Tt.OTt.=OI Jifi=Ij, n LTti=ide ;=1 Weapply thefunctor!, and seethat !(Tt;) satisfies the same relations, hence gives adirect sum decomposition oft(E)=F(8)E.Note that t(Tti)=id(8)Tti. Corollary 2.2. LetIbeanindexing set, and E=EBEi.Then wehave an ieI isomorphism (fflE)@Fffl(Ei@F). Proof. Let Sbeafinite subset ofI.We have asequence ofmaps (fflEi)XF--+ffl(E;@F)--+ffl(Ei@F) thefirst ofwhich isbilinear, and thesecond islinear, induced bytheinclusion of SinI.The first istheobvious map. IfScS',then atrivial commutative diagram shows that therestriction ofthemap (.Ei)XF--+ffl(Ei@F) induces ourpreceding maponthe sum foriES.But wehave aninjection ($Ei)xF($Ei)xF. IeS IeS' Hence bycompatibility,we can define abilinear map ($Ei)xFEB(Ei(8)F), reI iel XVI,2 BASIC PROPERTIES 609 andconsequentlyalinear map (fflEi)@F--+ffl(Ei@F). Inasimilar way, one defines amap intheopposite direction, and itisclear that these maps areinverse toeach other, hence give anisomorphism. Suppose now that Eisfree, ofdimension lover R.Let{v}beabasis, and consider F(8)E.Every element ofF(8)Ecan bewritten asasum ofterms y(8)av with yEFand aER.However, y(8)av=ay(8)v.Inasum ofsuch terms, wecan then uselinearityontheleft, Jl(Yi@v)=(J/i)@v, YiEF. Hence every element isinfact oftype y(8)vwith some YEF. We have abilinear map FxEF such that (y,av) ay,inducingalinear map F(8)E F. Wealso have alinear map F F(8)Egiven byy y(8)v.Itisclear that these maps areinverse toeach other, and hence that wehave anisomorphism F(8)E F. Thus every element ofF(8)Ecan bewritten uniquely intheform y(8)v,YEF. Proposition 2.3. LetEbefree over R,with basis {viheI. Then every element ofF(8)Ehas aunique expression oftheform LYi(8) Vi' ieIYiEF with almost allYi=o. Proof. This follows atonce from thediscussion oftheI-dimensional case, and thecorollary ofProposition 2.1. Corollary 2.4. Let E,Fbefree over R,with bases {V;}ieI and{Wj}jeJre- spectively. Then E(8)Fisfree, with basis {Vi(8)Wj}.Wehave dim(E (8)F)=(dim E)(dim F). 610 THE TENSOR PRODUCT XVI,2 Proof. Immediate from theproposition. We seethat when Eisfree over R,then there isnocollapsing inthe tensor product. Every element ofF(8)Ecan beviewed asa"formal" linear combina- tion ofelements inabasis ofEwith coefficients inF. Inparticular, we seethat R(8)E(orE(8)R)isisomorphic toE,under the correspondence x x(8)1. Proposition 2.5. LetE,Fbefreeoffinite dimension over R.Then wehave an isomorphism EndR(E) (8)EndR(F) EndR(E (8)F) which istheunique linear map such that f(8)gT(f, g) forfEEndR(E) and 9EEndR(F). [We note that the tensor product onthe left ishere taken inthe tensor product ofthetwo modules EndR(E) andEndR(F).] Proof Let{vd beabasis ofEand let{Wj}beabasis ofF.Then {Vi(8)wj} isabasis ofE(8)F.For each pair ofindices (i',j')there exists aunique endo- morphism f=Ii,i'ofEand 9=9j,j'ofFsuch that f(v i)=Vi' and f(vv)=0ifv=Ii g(Wj)=wj'andg(wJl)=0ifJl=Ij. Furthermore, thefamilies {!i,i'}and {gj,j'}are bases ofEnd R(E)and EndR(F) respectively. Then T(f, g)(vv (8) W)={Vi.@Wj'f(v,Ji)=(,)Jl0 If(v,II)#-(l,)). Thus thefamily {T(!i, i',9j,j')}isabasis ofEndR(E (8)F). Since thefamily {Ii, i'(8)gj,j'}isabasis ofEndR(E) <8>EndR(F), theassertion ofourproposition is now clear. InProposition 2.5, we seethat theambiguity ofthetensor sign inf(8)9isin fact unambiguous intheimportant special case offree, finite dimensional modules. Weshall seelater animportant application ofProposition 2.5when wediscuss the tensor algebra ofamodule. Proposition 2.6. Let o-+E' E!.E" 0 XVI,2 BASIC PROPERTIES 611 beanexact sequence, and Fany module. Then thesequence F(8)E' -+F(8)E F(8)E" 0 isexact. Proof. Given x"EE"and yEF,there exists xEEsuch that x" =t/J(x), and hence y(8)x"istheimage ofy(8)xunder thelinear map F(8)E F(8)E". Since elements oftype y(8)x"generate F(8)E", weconclude that thepreceding linear map issurjective. One also verifies trivially that theimage of F(8)E' F(8)E iscontained inthekernel of F(8)E-+F(8)E". Conversely, letIbetheimage ofF(8)E' F(8)E,and let f:(F(8)E)II F(8)E" bethecanonical map. Weshall define alinear map 9:F(8)E" (F(8)E)II such that 90f=id,This obviously willimply thatfisinjective, and hence will prove thedesired converse. Let yEFand x"EE". Let xEEbesuch that t/J(x)=x". Wedefine amap FxE" (F(8)E)II byletting (y,x") y(8)x(mod I), and contend that this map iswell defined, i.e.independent ofthechoice ofx such that t/J(x)=x".Ift/J(Xl)=t/J(X2)=x",then t/J(x l-X2)=0,and by hypothesis, Xl-x2=q>(x') for some x'EE'.Then y(8)Xl-Y(8)X2=Y(8)(Xl-X2)=y(8)q>(x'). This shows that y(8)Xl=Y(8)X2(mod I),and proves that our map iswell defined. Itisobviously bilinear, and hence factors throughalinear map g,on the tensor product. Itisclear that therestriction of90fonelements oftype y(8)x"istheidentity. Since these elements generate F(8)E", weconclude thatfisinjective,aswas tobeshown. 612 THE TENSOR PRODUCT XVI,3 Itisnotalways true that thesequence o F(8)E' F(8)E F(8)E" 0 isexact. Itisexact ifthefirst sequence inProposition 2.6splits, i.e.ifEis essentially thedirect sum ofE'and E". This isatrivial consequence ofPro- position 2.1,and thereader should carry outthedetails togetaccustomed tothe formalism ofthetensor product. Proposition 2.7. Let abeanideal ofR.Let Ebeamodule. Then themap (Ria)xE ElaE induced by (a,x) ax (mod aE),aER,xEE isbilinear and induces anisomorphism (Ria) (8)E ElaE. Proof. Our map (a,x) ax(mod aE)clearly induces abilinear map of RiaxEonto ElaE, and hence alinear map ofRia (8)Eonto ElaE. We can construct aninverse, for wehave awell-defined linear map E Ria (8)E such that xI(8)x(where Iistheresidue class of1inRia). Itisclear that aE iscontained inthekernel ofthis last linear map, and thus that weobtain a homomorphism ElaE Ria (8)E, which isimmediately verified tobeinverse tothehomomorphism described in thestatement oftheproposition. The association E ElaE Ria (8)Eisoften called areduction map. In 94, weshall interpret this reduction mapasanextension ofthebase. 3. FLAT MODULES Thequestion under which conditions theleft-hand arrow inProposition 2.6 isaninjection gives rise tothetheory ofthose modules forwhich itis,and we follow Serre incalling them flat. Thus formally, thefollowing conditions are equivalent, and define aflatmodule F,which should becalled tensor exact. F1.For every exact sequence E' E E" XVI,3 FLAT MODULES 613 thesequence F(8)E' -+F(8)E F(8)E" isexact. F2. For every short exact sequence o-+E' -+E-+E" 0 thesequence o F'(8)E' -+F(8)E F(8)E" 0 isexact. F3.For every injection 0 E' Ethe sequence o-+F(8)E' F(8)E isexact. Itisimmediate that F1implies F2implies F3.Finally, we seethat F3implies F1bywriting down thekernel andimage ofthemap E' Eandapplying F3. We leave thedetails tothereader. Thefollowing proposition gives tests forflatness, and also examples. Proposition 3.1. (i)Theground ring isflat asmodule over itself. (ii)LetF=EBFibeadirect sum. Then Fisflat ifandonlyifeach Fiisflat. (iii) Aprojective module isflat. The properties expressed inthispropositionarebasically categorical, cf.the comments onabstract nonsense attheendofthesection. Inanother vein, we have thefollowing tests having todowith localization. Proposition 3.2. (i)Let Sbeamultiplicative subset ofR.Then S-1Risflat over R. (ii) Amodule Misflat over Rifandonlyifthelocalization Mpisflat overRp foreach prime ideal pofR. (iii) Let Rbeaprincipal ring. Amodule FisflatifandonlyifF istorsionfree. Theproofsaresimple, and will belefttothereader. More difficult tests for flatness will beproved below, however. Examples ofnon-flatness. IfRisanentire ring, and amodule Mover R hastorsion, then Misnotflat. (Prove this, which isimmediate.) 614 THE TENSOR PRODUCT XVI,3 There isanother type ofexample which illustrates another badphenomenon. Let Rbe some ring inafinite extension KofQ,and such that Risafinite module over Zbut notintegrally closed. LetR'beitsintegral closure. Let pbe amaximal ideal ofRand suppose thatpR' iscontained intwo distinct maximal ideals $1and$2' Then itcan beshown that R'isnotflat over R,otherwise R' would befree over thelocal ring Rp,and therank would have tobe1,thus precluding thepossibility ofthe twoprimes $1and$2. Itisgood practice for thereader actually toconstruct anumerical example ofthis situation. The same type ofexamplecan beconstructed with aring R=k[x,y], where kisan algebraically closed field, even ofcharacteristic 0,and x,yare related by an irreducible polynomial equation f(x,y)=0over k.We take Rnotintegrally closed, such that itsintegral closure exhibits the same splitting of aprime pof Rinto twoprimes. Ineach one ofthese similar cases, one says that there isa singularity atp. As athird example, letRbethepower series ring inmore than one variable over afield k.Let mbethemaximal ideal. Then misnotflat, because otherwise, byTheorem 3.8below, mwould befree, andifR=k[[x.,. . .,xn]],then x., . ..,Xnwould be abasis formover R,which isobviously not the case, since x., X2arelinearly dependentover Rwhen n>2.The same argument, ofcourse, applies toany local ring Rsuch thatmlm2has dimension>2over Rim. Next we come tofurther criteria when amodule isflat. For theproofs,we shall snake itallover theplace. Cf.theremark attheendofthesection. Lemma 3.3. LetFbeflat, and suppose that ONMFO isanexact sequence. Thenfor anyE,wehave anexact sequence o N(8)E M(8)E F(8)E O. Proof Represent Easaquotient ofaflatLbyanexact sequence o K L-+E o. XVI,3 FLAT MODULES 615 Then wehave thefollowing exact and commutative diagram: 0 j Nfg)K)M@K)F@K)0 j j j 0)N@L)M@L)F@L j j N@E)M@E j j 0 0 Thetopright 0comes byhypothesis that Fisflat. The 0ontheleft comes from thefact that Lisflat. The snake lemma yields the exact sequence ON@EM@E which proves thelemma. Proposition 3.4. Let o F' F F" 0 beanexact sequence, and assume that F"isflat. Then Fisflat ifandonlyifF' isflat. More generally, let o FO F1 ...F" 0 beanexact sequence such that Fl ,...,F"areflat. Then FOisflat. 616 THE TENSOR PRODUCT XVI,3 Proof. Let 0 E' Ebeaninjection. We have anexact and commuta- tivediagram: o0 j )F'(8)E')F'(8)E')F"(8)E')0 j j j )F'(8)E)F(8)E)F'"(8)E o The 0ontopisbyhypothesis that F"isflat, and the two zeros ontheleft are justified byLemma 3.3.IfF'isflat, then thefirst vertical map isaninjection, and thesnake lemma shows that Fisflat. IfFisflat, then themiddle column isan injection. Then thetwo zeros ontheleftand thecommutativity oftheleftsquare show that themap F'(8)E' F"(8)Eisaninjection,soF'isflat. This proves the first statement. Theproof ofthesecond statement isdone byinduction, introducing kernels and cokernels ateach stepasindimension shifting, andapply thefirst statement ateach step. This proves theproposition Togiveflexibility intesting forflatness, the next two lemmas areuseful, in relating thenotion offlatness toaspecific module. Namely,wesaythat Fis E-flat orflatforE,ifforevery monomorphism o E' E the tensored sequence o F'(8)E' F'(8)E isalso exact. Lemma 3.5. Assume that FisE-flat. Then F'isalsoflatforevery submodule and every quotient module ofE. Proof The submodule part isimmediate because ifE'lCEcEare submodules, and F(8)E'l F(8)EisamonomorphismsoisF(8)E'1 F(8)E since thecomposite map with F(8)E2 F'(8)Eisamonomorphism. Theonly question lieswith afactor module. Suppose wehave anexact sequence o N E M o. LetM'be asubmodule ofMand E'itsinverse image inE.Then wehave a XVI,3 FLAT MODULES 617 commutative diagram ofexact sequences: 0)N)E' )M' "I I 0)N)E)M)0 )o. We tensor with Ftogetthe exact and commutative diagram 0 K I I F@N)F@E')F@M')0 I I I 0)F@N)F@E..)F@M I 0 \vhere Kisthequestionable kernel which wewant toprove isO.But thesnake lemma yields the exact sequence OKO which concludes theproof. Lemma 3.6. Let{EJ beafamily ofmodules, and suppose that Fisflatfor each Ei.Then Fisflatfortheir direct sum. Proof. LetE=EBEibetheir direct sum. Wehave toprove thatgiven any submodule E'ofE,thesequence o F@E' F@E=EBF@Ei isexact. Note that ifanelement ofF@E'becomes 0when mapped into the direct sum, then itbecomes 0already inafinite subsum, sowithout loss of generalitywemay assume that the setofindices isfinite. Then byinduction, we can assume that the setofindices consists oftwo elements, sowehave two modules EIand E2,and E=EI8:)E2.LetNbe asubmodule ofE.LetN1 =NnEIand letN2bethein1age ofNunder theprojectiononE2.Then 618 THE TENSOR PRODUCT XVI,3 wehave thefollowing commutative and exact diagram: o I To I )N2 I)0 )N I o)El)E)E2 Tensoring with Fwegetthe exact and commutative diagram: 0 0 I I F'@ Nt)F'@N)F'@N 2)0 I I I 0)F'@E1)F'@ E)F@E2 The lower left exactness isdue tothefact that [email protected] thesnake lemma shows that thekernel ofthemiddle vertical map iso.This proves the lemma. The next proposition shows that totestforflatness, itsuffices todo soonly for aspecial class ofexact sequences arising from ideals. Proposition 3.7. F'isflatifandonlyiffor every ideal aofRthenatural map a@F'aF isanisomorphism. lnfact, F'isflatifandonlyforevery ideal aofRtensoring thesequence o-+a R Ria-+0 with f"yields anexact sequence. Proof IfFisflat, then tensoring with Fandusing Proposition 2.7shows that thenatural map isanisomorphism, because aM isthekernel ofM MlaM. Conversely,assume that this map isanisomorphism forallideals a.This means XVI,3 FLAT MODULES 619 that FisR-flat. ByLemma 3.6itfollows that Fisflatforanarbitrary direct sum ofRwith itself, and since any module Misaquotient ofsuch adirect sum, Lemma 3.5implies that FisM-flat, thus concluding theproof. Remark onabstract nonsense. The proofs ofProposition 3.1(i),(ii),(iii), and Propositions 3.3through 3.4 arebasically rooted inabstract nonsense, anddepend only onarrow theoretic arguments. Specifically,asinChapter XX, 6,suppose that wehave abifunctor Tontwodistinct abelian categories aand CBsuch that foreach A,thefunctor B T(A, B)isright exact and foreach B thefunctor A T(A, B)isright exact. Instead of"flat" wecall anobject A ofatrexact ifB T(A, B)isanexact functor; and wecall anobject LofCB T-exact ifA T(A, L)isexact. Then thereferences tothebase ring and free modules can bereplaced byabstract nonsense conditions asfollows. Inthe useofLinLemma 3.3, weneed toassume that forevery object EofB there isatT-exact Land anepimorphism L E O. For theanalog ofProposition 3.7, weneed toassume that there issome object RinCBforwhich FisR-exact, that isgivenanexact sequence Oa-+R then 0 T(F, a) T(F', R)isexact; and wealso need toassume that Risa generator inthe sense that every object Bisthequotient ofadirect sum ofRwith itself, then over some family ofindices, and Trespects direct sums. The snake lemma isvalid inarbitrary abelian categories, either because its proof is"functorial," orbyusing arepresentation functor toreduce ittothe category ofabelian groups. Take your pick. Inparticular, wereally don't need tohave acommutative ring asbase ring, this was done only forsimplicity oflanguage. We now pass tosomewhat different considerations. Theorem 3.8. Let Rbeacommutative local ring, and letMbeafinite flat module over R.Then Misfree. Infact,ifxI'...,XnEMareelements ofM whose residue classes are abasis ofMlmM over Rim, then Xl'...,Xnform abasis ofMover R. Proof. Let R(n) Mbethemap which sends theunit vectors ofR(n) on Xl'...' xnrespectively, and letNbeitskernel. We getanexact sequence o N R(n) M, 620 THE TENSOR PRODUCT XVI,3 whence acommutative diagram m@N II)m@R<n) gl )R(n))m@M hi o)N)M inwhich the rows are exact. Since Misassumed flat, themap hisaninjection. Bythesnake lemma one getsanexact sequence o cokerf-+coker g coker h, and the arrow ontheright ismerely R(n)ImR<n)-+MImM, which isanisomorphism bytheassumptionon xb...,Xn. Itfollows that cokerf=0,whence mN =N,whence N =0byNakayama ifRisNoetherian, soNisfinitely generated. IfRisnotassumed Noetherian, then one has toadd aslight argumentasfollows incase Misfinitely presented. Lemma 3.9. Assume that Misfinitely presented, and let O-+NEMO beexact, with Efinite free. Then Nisfinitely generated. Proof. Let Ll-+L2-+M 0 beafinite presentation ofM,that isanexact sequence with LbL2finite free. Using thefreeness, there exists acommutative diagram Ll I. 1)0)M Id .M)0 o)N)E such that L2 Eissurjective. Then thesnake lemma gives atonce the exact sequence ocoker(L lN) 0, socoker(L 1N)=0,whence Nisanimage ofLland istherefore finitely generated, thereby proving thelemma, and alsocompleting theproof ofTheorem 3.8when Misfinitely presented. XVI,3 FLAT MODULES 621 Westill have notproved Theorem 3.8inthefully generalcase. For this we useMatsumura's proof (see hisCommutative Algebra, Chapter 2),based onthe following lemma. Lemma 3.10. Assume that Misflat over R.Let a;EA,XiEMfor i=I, . . .,n,and suppose that wehave therelation n Laixi=O. i=1 Then there exists anintegersand elements b,jEAandYjEM(j=1,..., s) such that ,a.b..=0II} iforalljand Xi=LbijYjjforalli. Proof. Weconsider the exact sequence o K R(n) R where themap R(n) Risgiven by n (bl'...,bn)Laib;, i=1 and Kisitskernel. Since Misflatitfollows that K(8)M M(n) M isexact, where fMisgiven by n fM(Zl,...,Zn)= Laizi. i=1 Therefore there exist elementsPjEKandYjEMsuch that s (Xl' ..., xn)=LPjYj. j==1 WritePj=(blj,...,bnj)with bijER.This proves thelemma. Wemaynow apply thelemma toprove thetheorem inexactly the same way weproved that afinite projective module over alocal ring isfree inChapter X, Theorem 4.4, byinduction. This concludes theproof. Remark. Intheapplications Iknow of,thebase ring isNoetherian, and so onegets away with thevery simple proof given atfirst. Ididnot want toobstruct thesimplicity ofthisproof, and that isthe reason Igave theadditional tech- nicalities inincreasing order ofgenerality. 622 THE TENSOR PRODUCT XVI,3 Applications ofhomology.Weendthis section bypointingout aconnection between the tensor product and thehomological considerations ofChapter XX, 8forthose readers who want topursue thistrend ofthoughts. The tensor product isabifunctor towhich we canapply theconsiderations ofChapter XX,8.Let M,Nbemodules. Let ... Ei Ei-1Eo-+M 0 beafree orprojective resolution ofM,i.e. anexact sequence where Eiisfree or projective foralli>O.Wewrite this sequenceas EM M -+O. Then bydefinition, Tori(M, N)=i-thhomology ofthecomplex E(8)N,that isof ...Ei(8)N Ei-1(8)N-+...-+Eo(8)N O. This homology isdetermined uptoaunique isomorphism. Ileave tothereader topick whatever convention isagreeable tofixone resolution todetermine a fixed representation ofTori(M, N), towhich allothers areisomorphic by a unique isomorphism. Since wehave abifunctorial isomorphism M(8)N N(8)M, wealso get a bifunctorial isomorphism Tori(M, N) Tori(N, M) foralli.SeePropositions 8.2 and 8.2' ofChapter XX. Following general principles,we say that Mhas Tor-dimension<dif Tor;(M, N)=0foralli>dand allN.From Chapter XX,8wegetthefollow- ingresult, which merely replaces T-exact byflat. Theorem 3.11. Thefollowing three conditions areequivalent concerninga module M. (i)Misflat. (ii)Torl(M, N)=0forallN. (iii)Tori(M, N)=0foralli> 1and allN,inother words, MhasTor- dimension O. Remark. Readers willing tousethis characterization canreplacesome of thepreceding proofs from 3.3to3.6by aTor-dimension argument, which is more formal, oratleast formal inadifferent way, and may seem more rapid. The snake lemma was used adhoc ineach case toprove thedesired result. The general homology theory simply replaces this usebythecorresponding formal homological step, once thegeneral theory ofthederived functor hasbeen carried out. XVI,4 EXTENSION OFTHE BASE 623 4. EXTENSION OF THE BASE Let Rbeacommutative ring and letEbeaR-module. WespecifyRsince we aregoing towork with several rings inamoment. Let R R'beahomo- morphism ofcommutative rings,sothat R'isanR-algebra, and may beviewed as anR-module also. We have a3-multilinear map R'xR'xE-+R'(8)E defined bytherule (a,b,x) ab(8)x. This induces therefore aR-linear map R'(8)(R' (8)E) R'(8)E and hence aR-bilinear map R'x(R' (8)E) R'(8)E.Itisimmediately verified that our last map makes R'(8)Einto aR'-module, which weshall call the extension ofEover R',and denote byER,.Wealso saythat ER,isobtained by extension ofthebase ring from RtoR'. Example 1. Let abeanideal ofRand letR Ria bethecanonical homo- morphism. Then theextension ofEtoRia isalso called the reduction ofE modulo a.This happens often over theintegers, when wereduce modulo aprime p(i.e. modulo theprime ideal (p». Example 2. Let Rbeafield and R'anextension field. Then Eisavector spaceover R,and ER,isavector space over R'.Interms ofabasis, we seethat ourextension gives what was alluded tointhepreceding chapter. This example will beexpanded intheexercises. Wedraw the same diagramsasinfield theory: ER, E/R'R/ tovisualize anextension ofthebase. From Proposition 2.3, weconclude: Proposition 4.1. Let Ebe afree module over R,with basis {Vi}ieI. Let v;= 1(8)Vi.Then ER,isafree module over R',with basis {V;}ieI' Wehadalready used aspecialcase ofthisproposition when weobserved that thedimension ofafree module isdefined, i.e.that two bases have the same 624 THE TENSOR PRODUCT XVI,4 cardinality. Indeed, inthat case, wereduced modulo amaximal ideal ofRto reducethequestion toavector space over afield. When westart changing rings, itisdesirable toindicate Rinthenotation forthe tensor product. Thus wewrite ER'=R'(8)E=R'(8)R E. Then wehave transitivity oftheextension ofthebase, namely, ifR R' R"isa succession ofhomomorphisms ofcommutative rings, then wehave aniso- morphism R"(8)RE R"(8)R'(R' (8)RE) and thisisomorphism isoneofR"-modules. Theproof istrivial andwill beleft tothereader. IfEhas amultiplicative structure, we can extend the base also forthis multiplication. LetR-+Abearing-homomorphism such that every element in theimage ofRinAcommutes with every element inA(Le. anR-algebra). Let R R'beahomomorphism ofcommutative rings. We have a4-multilinear map R'xAxR'xA R'(8)A defined by (a,x,b,y) ab(8)xy. Wegetaninduced R-linear map R'(8)A(8)R'(8)A R'(8)A and hence aninduced R-bilinear map (R' (8)A)x(R' (8)A) R'(8)A. Itistrivially verified that thelawofcomposition onR'(8)Awehave just defined isassociative. There isaunit element inR'(8)A,namely,1(8)1.We have aring-homomorphism ofR'into R'(8)A,given bya a(8)1.Inthis way one sees atonce that R'(8)A =AR' isanR'-algebra. We note that themap xl(8)x isaring-homomorphism ofAinto R'(8)A,and that wegetacommutative diagram ofringhomomorphisms, R'(8)A=AR' A/R'R/ XVI,5 SOME FUNCTORIAL ISOMORPHISMS 625 For therecord, wegivesome routine tests forflatness inthecontext ofbase extension. Proposition 4.2. Let R AbeanR-algebra, and assume Acommutative. (i)Base change. IfFisaj/at R-module, then AQ9RFisaflat A-module. (ii)Transitivity. IfAisajiat commutative R-algebra andMisaflatA-module, then MisflatasR-module. Theproofsareimmediate, and will belefttothereader. 5. SOME FUNCTORIAL ISOMORPHISMS Werecall anabstract definition. Let 21, betwocategories. The functors of21into (say covariant, and inone variable) can beviewed asthe objects ofacategory, whose morphismsaredefined asfollows. IfL,Maretwo such functors, amorphism H :L Misarule which toeach object Xof21 associates amorphism Hx:L(X)-+M(X) in,such that foranymorphism f:X Yin21,thefollowing diagram iscommutative: L(X)Bx)M(X) L<JJj jM(f) L(Y) By)M(Y) We can therefore speak ofisomorphisms offunctors. Weshall seeexamples of these inthetheory oftensor products below. Inourapplications,ourcategories areadditive, that is,the setofmorphisms isanadditive group, and thecomposi- tion law isZ-bilinear. Inthat case, afunctor Liscalled additive if L(f +g)=L(f) +L(g). WeletRbeacommutative ring, and weshall consider additive functors from thecategory ofR-modules into itself. For instance wemay view the dual module asafunctor, E EV=L(E, R)=HomR(E, R). Similarly,wehave afunctor intwovariables, (E,F) L(E, F')=HomR(E, F), contravariant inthefirst, covariant inthesecond, and bi-additive. 626 THE TENSOR PRODUCT XVI,5 We shall give several examples offunctorial isomorphisms connected with thetensor product, and forthisitismost convenient tostate ageneral theorem, givingusacriterion when amorphism offunctors isinfact anisomorphism. Proposition 5.1. LetL,Mbetwofunctors (both covariant orboth contra- variant) from thecategory ofR-modules intoitself. Assume thatbothfunctors areadditive. LetH :L Mbeamorphism offunctors. IfHE:L(E) M(E) isanisomorphism forevery I-dimensional free module Eover R,then HEisan isomorphism forevery finite-dimensional free module over R. Proof. Webegin with alemma. Lemma 5.2. Let Eand Ei(i=1,...,m)bemodules over aring. Let lpi:Ei Eand t/Ji:E Eibehomomorphisms having thefollowing properties: '/1. 0(f). =id'1', 't'l , t/Ji0qJj=0ifi=Ij m LqJi0t/Ji=id, i=1 Then themap x (t/J 1X,...,t/JmX) m isanisomorphism ofEonto thedirect product nEi,and themap i= 1 (xl'...,Xm) lp1X1+...+lpmXm isanisomorphism oftheproduct onto E.Conversely, ifEisequal tothedirect sumofsubmodules Ei(i=1,..., m),ifwelett/Jibetheinclusion ofEiinE, and lpitheprojection ofEonEi,then these maps satisfy theabove-mentioned properties. Proof. Theproofisroutine, and isessentially the same asthatofProposition 3.1ofChapter III. We shall leave itasanexercise tothereader. We observe that thefamilies {lp;} and {t/J;} satisfying theproperties ofthe lemma behave functorially: IfTisanadditive contravariant functor, say, then thefamilies {T( t/Ji)}and{T(lpi)} alsosatisfy theproperties ofthelemma. Similarly ifTisacovariant functor. Toapply thelemma, wetake the modules Eitobethe I-dimensional components occurring inadecomposition ofEinterms ofabasis. Let usassume forinstance that L,Mareboth covariant. We have foreach module Eacom- XVI,5 SOME FUNCTORIAL ISOMORPHISMS 627 mutative diagram L(E) L(<p')1 L(E;)HE )M(E) 1M(<p,) HEi)M(E i) and asimilar diagram replacing qJibyt/Ji,reversing the two vertical arrows. Hence weget adirect sum decomposition ofL(E) interms ofL(t/Ji) andL(qJi)' andsimilarly forM(E), interms ofM(t/J;) and M(qJi). Byhypothesis, HEiisan isomorphism. Itthen follows trivially that HEisanisomorphism. Forinstance, toprove injectivity, wewrite anelement vEL(E) intheform v=LL(qJi)Vb with ViEL(E i).IfHEv=0,then o=LHEL(qJi)Vi=LM(qJi)HEiVi' and since the maps M(qJi) (i=1,..., m)giveadirect sum decomposition of M(E), weconclude that HEiVi=0foralli,whence Vi=0,and V=O.The surjectivity isequally trivial. When dealing with afunctor ofseveral variables, additive ineach variable, one cankeep allbut one ofthevariables fixed, and then apply theproposition. Weshall dothis inthefollowing corollaries. Corollary 5.3. LetE',E,F',Fbefree andfinite dimensional over R.Then we have afunctorial isomorphism L(E', E)(8)L(F', F) L(E' (8)F',E(8)F) such that f(8)9T(f, g). Proof. Keep E,F',Ffixed, and view L(E', E)(8)L(F', F)asafunctor inthe variable E'.Similarly, view L(E' (8)F\E(8)F) asafunctor inE'.The mapf(8)9T(f, g)isfunctorial, and thus bythelemma, itsuffices toprove that ityieldsanisomorphism when E'has dimension 1. Assume now that this isthe case; fixE'ofdimension 1,and view the two expressions inthecorollaryasfunctors ofthevariable E.Applying thelemma 628 THE TENSOR PRODUCT XVI,5 again, itsuffices toprove that our arrow isanisomorphism when Ehasdi- mension 1.Similarly, wemay assume that F,F'have dimension 1.Inthat case theverification that the arrow isanisomorphism isatriviality,asdesired. Corollary 5.4. Let E,Fbefree andfinite dimensional. Then wehave a natural isomorphism EndR(E) (8)EndR(F)-+EndR(E (8)F). Proof Special case ofCorollary 5.3. Note that Corollary 5.4 had already been proved before, and that we mention ithere only toseehow itfitswith thepresent point ofview. Corollary 5.5. LetE,Fbefreefinite dimensional over R.There isafunc- torial isomorphism EV0F--+L(E, F) given for AEEVand YEFbythemap AQ9y A).,y whereA).,yissuch thatforallxEE, wehave A).,y(x)=A(X)Y. The inverse isomorphism ofCorollary 5.5 can bedescribed asfollows. Let{VI,...,vn}be abasis ofE,and let{v(,...,vnV}bethedual basis. If AEL(E, F),then theelement n Ev/ (8)A(v;) EEv(8)F ;=1 maps toA.Inparticular, ifE=F,then theelement mapping totheidentity idE iscalled theCasimir element n Ev/(8)V;, ;=1 independent ofthechoice ofbasis. Cf.Exercise 14. Toprove Corollary 5.5,justify that there isawell-defined homomorphism ofEVQ9FtoL(E, F),bytheformula written down. Verify that this homo- morphism isboth injective andsurjective. We leave thedetails asexercises. Differential geometers arevery fond oftheisomorphism L(E, E)--+EV0E, and often use EV0Ewhen they think geometrically ofL(E, E),therebyem- phasizinganunnecessary dualization, and anirrelevant formalism, when itis easier todeal directly with L(E, E). Indifferential geometry, one applies various functors Ltothetangent space atapointon amanifold, and elements ofthespaces thus obtained arecalled tensors (oftype L). XVI,6 TENSOR PRODUCT OFALGEBRAS 629 Corollary 5.6. LetE,Fbefree andfinite dimensional over R.There isa functorial isomorphism EVFV (EF)v. given for XV EEVand yVEFVbythemap XV yVt---+A, where Aissuch that,forallxEEand yEF, A(X y)=(x,XV)(y, yV). Proof. Asbefore. Finally,weleave thefollowing results asanexercise. Proposition 5.7. Let Ebefree andfinite dimensional over R. The trace function onL(E,E)isequal tothecomposite ofthe two maps L(E, E) EVE R, where thefirst map istheinverse oftheisomorphism described inCorollary 5.5, and thesecond map isinduced bythebilinear map (XV, x)1--+(x,XV). Ofcourse, itisprecisely inasituation involving the trace that the iso- morphism ofCorollary 5.5becomes important, and that the finite dimen- sionality ofEisused. Inmany applications, this finite dimensionality plays norole, anditisbetter todeal with L(E, E)directly. 6. TENSOR PRODUCT OF ALGEBRAS Inthis section, weagain letRbe acommutative ring. ByanR-algebra we mean aring homomorphism R Ainto aring Asuch that theimage ofRis contained inthe center ofA. LetA,BbeR-algebras. We shall make A Binto anR-algebra. Given (a,b)EAxB,wehave anR-bilinear map Ma,b:AxB A0Bsuch thatMa,b(a', b')=aa'0bb'. HenceMa,binduces anR-linear map ma,b:A B A(8)Bsuch that ma,b(a', b')=aa'0bb'. Butma,b depends bilinearlyonaandb,soweobtain finallyaunique R-bilinear map A0BxABA0B 630 THE TENSOR PRODUCT XVI,6 such that (a b)(a' b')=aa' bb'. This map isobviously associative, and wehave anatural ring homomorphism R A0Bgiven byc 10c=c01. Thus A0BisanR-algebra, called theordinary tensor product. Application: commutative rings We shall now seetheimplication oftheabove forcommutative rings. Proposition 6.1. Finite coproducts exist inthecategory ofcommutative rings, and inthecategory ofcommutative algebras over acommutative ring. IfR Aand R Bare twohomomorphisms ofcommutative rings, then their coproductover RisthehomomorphismR A0Bgiven by a a(8) 1= 1(8)a. Proof. We shall limit ourproof tothe case ofthecoproduct oftworing homomorphisms R Aand R B.One can useinduction. LetA,Bbecommutative rings, and assume given ring-homomorphisms into acommutative ring C, qJ:A Cand tfJ:B c. Then we can define aZ-bilinear map AxBC by(x,y) qJ(x)tfJ(y). From this wegetaunique additive homomorphism A(8)BC such that x(8)y q>(x)tfJ(y). We have seen above that we can define aring structure onA(8)B,such that (a(8)b)(c (8)d)=ac(8)bd. Itisthen clear that ourmap A(8)B Cisaring-homomorphism. Wealso have tworing-homomorphisms A1.A(8)Band B.!4 A(8)B given by x x(8) 1and y1(8)y. The universal property ofthe tensor product shows that (A(8)B,f,g)isa coproduct ofourrings Aand B. IfA,B,Care R-algebras, and ifqJ,tfJmake thefollowing diagram com- XVI,6 TENSOR PRODUCT OFALGEBRAS 631 mutative, c A"B ""-R/ then A(8)Bisalso anR-algebra (itisinfact analgebra over R,orA,orB,de- pendingonwhat one wants touse), and themap A(8)B Cobtained above givesahomomorphism ofR-algebras. Acommutative ringcanalways beviewed asaZ-algebra (Le. asanalgebra over theintegers). Thus one sees thecoproduct ofcommutative ringsasa specialcase ofthecoproduct ofR-algebras. Graded Algebras. LetGbeacommutative monoid, written additively. By aG-graded ring,weshall mean aring A,which asanadditive groupcan be expressedasadirect sum. A=E8Ar, reG and such that theringmultiplication maps ArxAsinto Ar+s'forallr,SEG. Inparticular, we seethat Aoisasubring. The elements ofArarecalled thehomogeneous elements ofdegree r. We shall construct several examples ofgraded rings, according tothe following pattern. Suppose given foreach rEG anabelian group Ar(written additively), and foreach pair r,SEGamap ArxAs Ar+s. Assume that Ao isacommutative ring, and thatcomposition under these maps isassociative and Ao-bilinear. Then thedirect sum A=EBArisaring: We can define multiplica- reG tion intheobvious way, namely (LXr)(LYS)=L(LXrYs). reG seG reG r+s=r The above product iscalled theordinary product. However, there isanother way. Suppose thegrading isinZorZ/2Z. We define thesuper product of xEArandyEAs tobe(-l)rs xy ,where xyisthegiven product. Itiseasily veri- fied that thisproduct isassociative, and extends towhat iscalled thesuper product A0A Aassociated with thebilinear maps. IfRisacommutative ring such that Aisagraded R-algebra, i.e.RAr CArforallr(inaddition tothe condition that Aisagraded ring), then with the super product, Aisalso an R-algebra, which will bedenoted byAsu, andwill becalled thesuper algebra associated with A. 632 THE TENSOR PRODUCT XVI,7 Example. Inthenext section, weshall meet thetensor algebra T(E), which will begradedasthedirect sum ofTr(E), and soweget theassociated super tensor algebra Tsu(E) accordingtotheabove recipe. Similarly, letA,Bbegraded algebras (graded bythenatural numbers as above). We define their super tensor product A@su B tobetheordinary tensor productasgraded module, butwith thesuper product (a0b)(a' 0b')=(-I)(degb)(de ga')aa' 0bb' ifb,a'arehomogeneous elements ofBandA respectively. Itisroutinely verified that A@su Bisthen aring which isalso agraded algebra. Except forthesign, theproduct isthe same astheordinary one, butitisnecessary toverify associati vity explicitly. Suppose a'EAi'bEBj,a"EAs, and b'EBr.Then thereader will find atonce that thesign which comes outbycomputing (a@sub)(a' @sub')(a" @sub") intwo waysturns out tobethe same, namely (-l)U+js+sr.Since bilinearity is trivially satisfied, itfollows that A@su Bisindeed analgebra. The super product inmany ways ismore natural than what wecalled the ordinary product. Forinstance, itisthenatural product ofcohomology intopol- ogy. Cf.Greenberg-Harper, Algebraic Topology, Chapter 29.For asimilar con- struction with Z/2Z-grading,seeChapter XIX, 4. 7. THE TENSOR ALGEBRA OF AMODULE Let Rbe acommutative ringasbefore, and letEbe amodule (Le. an R-module). For each integerr>0,welet r Tr(E)=(8)Eand TO(E)=R. i=1 Thus Tr(E)=E(8)...(8)E(tensor product taken rtimes). Then Trisafunctor, whose effect onlinear maps isgivenasfollows. Iff: E Fisalinear map, then Tr(f)=T(f,...,f) inthe sense of91. From theassociativity ofthe tensor product, weobtain abilinear map Tr(E)xTS(E) Tr+s(E), XVI,7 THE TENSOR ALGEBRA OFAMODULE 633 which isassociative. Consequently, bymeans ofthisbilinear map,wecandefine aring structure onthedirect sum 00 T(E)=EBTr(E), r==O and infact analgebra structure (mapping RonTO(£)=R).We shall callT(E) the tensor algebra ofE,over R.Itisingeneral notcommutative. Ifx,yET(E), weshall again write x(8)yforthering operation inT(E). Letf:E-+Fbealinear map. Thenfinduces alinear map Tr(f):Tr(E)-+Tr(F) foreach r>0,and inthis way induces amap which weshall denote byT(f)on T(E). (There can benoambiguity with themap of1,which should now be written Tl(f), and isinfactequal tofsince Tl(E)=E.)Itisclear thatT(f) is theunique linear map such that forXl'...,XrEEwehave T(f)(Xl (8)...(8)Xr)=f(x 1)(8)...(8)f(xr). Indeed, theelements ofTl(E)=Earealgebra-generators ofT(E) over R.We seethatT(f) isanalgebra-homomorphism. Thus Tmay beviewed asafunctor from thecategory ofmodules tothecategory ofgraded algebras, T(f) beinga homomorphism ofdegree O. When Eisfree and finite dimensional over R,wecandetermine thestructure ofT(E) completely, using Proposition 2.3. LetPbeanalgebra over k.Weshall say that Pisanon-commutative polynomial algebra ifthere exist elements tl'...,tnEPsuch that theelements M(")(t)=t"...t"I 11 Is with 1<iv<nform abasis ofPover R.We may call these elements non- commutative monomials in(t). Asusual, byconvention, when r=0,the corresponding monomial istheunit element ofP.We seethat t1,...,tngenerate Pasanalgebra over k,and that Pisinfact agraded algebra, where Prconsists of linear combinations ofmonomials ti1.. .tirwith coefficients inR.Itisnatural to saythat tb. . .,tnareindependent non-commutative variables over R. Proposition 7.1. LetEbefree ofdimension nover R.Then T(E) isisomorphic tothenon-commutative polynomial algebra on nvariables over R.Inother words, if{vl'...,vn}isabasis ofEover R,then theelements M(i)(V)==ViI(8)...(8)V;v'1<iv<n formabasis ofTr(E), and every element ofT(E) has aunique expressionasa finitesum La(i)M(i)(v), (i)a(i)ER 634 THE TENSOR PRODUCT XVI,7 with almost alla(i>equal too. Proof. This follows atonce from Proposition 2.3. The tensor product oflinear maps will now beinterpreted inthecontext of the tensor algebra. For convenience, weshall denote themodule ofendomorph isms EndR(E) by L(E)forthe restofthis section. Weform thedirect sum 00 (LT)(E)=EBL(Tr(E», r=O which weshall also write LT(E) forsimplicity. (Of course, LT(E) isnotequal to EndR(T(E»,sowemust view LTasasingle symbol.) Weshall seethat LTisa functor from modules tograded algebras, bydefiningasuitable multiplication onLT(E). LetfEL(Tr(E», 9EL(TS(E», hEL(Tm(E». Wedefine theproduct fgEL(Tr+s(E» tobeT(f, g),inthenotation of91,inother words tobethe unique linear map whose effect on anelement x(8)ywith xETr(E) and yETS(E) is x(8)yf(x) (8)g(y). Inview oftheassociativity ofthe tensor product,weobtain atonce the as- sociativity (fg)h=f(gh), and wealso seethat ourproduct isbilinear. Hence LT(E) isak-algebra. We have analgebra-homomorphism T(L(E» LT(E) given ineach dimension rbythelinear map fl(8)...(8)f,. T(fb...,f,.)=.fl...f,.. Wespecify here that thetensor productontheleftistaken in L(E) (8)...(8)L(E). Wealso note that thehomomorphism isingeneral neither surjective norinjective. When Eisfree finite dimensional over R,thehomomorphism turns out tobe both, and thus wehave aclear picture ofLT(E)asanon-commutative poly- nomial algebra, generated byL(E). Namely, from Proposition 2.5, weobtain: Proposition 7.2. Let Ebefree, finite dimensional over R.Then wehave an algebra-isomorphism 00 T(L(E»=T(EndR(E» LT(E)=EBEndR(Tr(E» r=O XVI,8 SYMMETRIC PRODUCTS 635 given by f(8)gT(f, g). Proof. ByProposition 2.5, wehave alinear isomorphism ineach dimen- sion, and itisclear that themap preserves multiplication. Inparticular,we seethat LT(E)ISanoncommutative polynomial algebra. 8. SYMMETRIC PRODUCTS Let6ndenote thesymmetric groupon nletters, sayoperating ontheintegers (1,...,n).Anr-multilinear map f:E(r) F issaid tobesymmetric iff(xl'...,Xr)=f(X(1(l)'...,xa(r»forall (JE6r. InTr(E), weletbrbethesubmodule generated byallelements oftype Xl(8)...(8)Xr-Xa(l) (8).. .(8)Xa(r) forallXiEEand aE6r. Wedefine thefactor module sr(E)=Tr(E)/br' and let 00 S(E)=EBsr(E) r==O bethedirect sum. Itisimmediately obvious that thedirect sum 00 b=EBbr r=O isanideal inT(E), and hence thatS(E) isagraded R-algebra, which iscalled the symmetric algebra ofE. Furthermore, thecanonical map E(r) sr(E) obtained bycomposing themaps E(r) Tr(E) Tr(E)/b r=sr(E) / isuniversal forr-multilinear symmetric maps. 636 THE TENSOR PRODUCT XVI,8 Weobserve that Sisafunctor,from thecategory ofmodules tothecategory ofgraded R-algebras. The image of(x1,...,xr)under thecanonical map E(r) -+sr(E) will bedenoted simply byXl'..Xr. Proposition 8.1. Let Ebefreeofdimension nover R.Let{Vl,...,vn}bea basis ofEover k.Viewed aselements ofSl(E) inS(E), these basis elements are algebraically independent over R,and S(E) istherefore isomorphic tothe polynomial algebra innvariables over R. Proof Let tl'...,tnbealgebraically independent variables over R,and form thepolynomial algebra R[t b...,tn].LetPrbetheR-module ofhomo- geneous polynomials ofdegreer.We define amap ofE(r) Prasfollows. If wl,...,Wrareelements ofEwhich can bewritten n Wi=Laivvv, v=1i=1,..., r, then our map isgiven by (w 1,...,Wr) (a11t1+...+a1ntn)...(ar1t1+...+arntn). Itisobvious that this map ismultilinear and symmetric. Hence itfactors throughalinear map ofsr(E) into Pr: E(r))sr(E)p/r From thecommutativity ofourdiagram, itisclear that theelementVit...Visin sr(E) mapsontit...tisinPrforeach r-tuple ofintegers (i)=(il,...,ir).Since themonomials Mcn(t)ofdegreerarelinearly independent over k,itfollows that themonomialsMCi)(V)inS'(E) arealso linearly independent over R,and that ourmap sr(E)-+Prisanisomorphism. One verifies atonce that themultiplica- tion inS(E) corresponds tothemultiplication ofpolynomials inR[t], and thus that themap ofS(E) into thepolynomial algebra described asabove foreach component sr(E) induces analgebra-isomorphism ofS(E) onto R[t], asdesired. Proposition 8.2. Let E=E'(f)E"be adirect sumoffinite free modules. Then there isanatural isomorphism sn(E' (f)E") E8SPE'(8)sqE". p+q=n Infact, this isbutthen-part ofagraded isomorphism SeE' (f)E") SE' (8)SE". XVI, Ex EXERCISES 637 Proof Theisomorphism comes from thefollowing maps. The inclusions ofE'and E"into their direct sum give rise tothefunctorial maps SE' (8)SE" -+SE, and theclaim isthat this isagraded isomorphism. Note that SE'and SE" are commutative rings, and sotheir tensor product isjust the tensor product of commutative rings discussed in6.The reader caneither giveafunctorial map backward toprove the desired isomorphism, ormore concretely, SE' isthe polynomial ringon afinite family ofvariables, SE" isthepolynomial ring in another family ofvariables, and their tensor product isjust thepolynomial ring inthe two families ofvariables. The matter iseasy nomatter what, and the formal proof islefttothereader. EXERCISES 1.Letkbeafield and k(ex)afinite extension. Letf(X)=Irr(ex, k,X),and suppose thatfis separable. Letk'beany extension ofk.Show that k(ex) (8)k'isadirect sum offields. Ifk'isalgebraically closed, show that these fields correspond totheembeddings of k(ex)Ink'. 2.Let kbeafield,f(X)anIrreducible polynomialover k,and exaroot off.Show that k(ex) (8)k'isisomorphic,asak'-algebra, tok'[X]/(f(X». 3.Let Ebeafinite extension ofafield k.Show that Eisseparable over kifandonly if E(8)kLhas nonilpotent elements forallextensions Lofk,and also when L=kat 4.Letcp:A Bbe acommutative ring homomorphism. Let EbeanA-module and F aB-module. Let FAbetheA-module obtained from Fvia theoperation ofAonF through cp,that isforyEFAand aEAthisoperation isgiven by (a,y)1-+cp(a)y. Show that there ISanatural isomorphism HomB(B (8)AE,F) HomA(E, FA). 5.The norm. LetBbeacommutative algebra over thecommutative ringRand assume that Bisfree ofrank r.Let Abeany commutative R-algebra. Then A0Bisboth anA-algebra and aB-algebra.We view A(8)BasanA-algebra, which isalso free ofrank r.If{eI'. . .,er}isabasis ofBover R,then 1A(8)el'...,1A(8)er isabasis ofA(8)Bover A.Wemay then define the norm N [email protected]:A(8)B-+A astheunique map which coincides with thedeterminant oftheregular representation. 638 THE TENSOR PRODUCT XVI, Ex Inother words, ifbEBand bBdenotes multiplication byb,then NB,R(b)=det(b B); andsimilarly after extension ofthebase. Prove: (a) Let lp:A-+Cbeahomomorphism ofR-algebras. Then thefollowing diagram iscommutative: A@B Nj Atp@id )C0B jN C tp (b)Let x,YEA @B.Then N(x @By)=N(x) @N(y). [Hint: Use the com- mutativity relationseiej=eje;and theassociativity.] Alittle flatness 6.LetM,Nbeflat. Show that M@Nisflat. 7.LetFbeaflatR-module, and letaERbeanelement which isnot azero-divisor. Show that ifax =0for some xEFthen x=o. 8.Prove Proposition 3.2. Faithfully flat 9.Wecontinue toassume that rings arecommutative. LetMbeanA-module. We say that Misfaithfully flatifMisflat, andifthefunctor TM:E M@AE. isfaithful, that isE#0implies M@AE=1=O.Prove that thefollowing conditions are equivalent. (i)Misfaithfully flat. (ii)Misflat, and ifu:F-+Eisahomomorphism ofA-modules, u#0,then TM(U): M@AF-+M@A Eisalso #0. (iii) Misflat, and forallmaximal ideals mofA,wehave mM #M. (iv) Asequence ofA-modules N' -+N-+N"isexact ifandonly ifthe sequence tensored with Misexact. 10.(a)Let A-+Bbearing-homomorphism. IfMisfaithfully flat over A,then B@AM isfaithfully flat over B. (b) LetMbefaithfully flat over B.Then Mviewed asA-module viathehomomorphism A-+Bisfaithfully flat over AifBisfaithfully flat over A. 11.LetP,M,Ebemodules over thecommutative ring A.IfPisfinitely generated (resp. finitely presented) and Eisflat, show that thenatural homomorphism HomA(P, M)@A E-+HomA(P, M@AE) isamonomorphism (resp.anisomorphism). XVI, Ex EXERCISES 639 [Hint: LetF1-+F0-+P-+0beafinite presentation, say. Consider thediagram o)HomA(P, M) (8)AE)HomA(F 0,M)(8)AE)HomA(F., M)(8)A E ! ! ! o·HomA(P, M(8)AE)·HomA(F 0,M(8)AE))Horn A(F1,M(8)AE)]. Tensor products and direct limits 12.Show that the tensor product commutes with direct limits. Inother words, if{EJ isa directed family ofmodules, and Misanymodule, then there isanatural isomorphism lim(E i(8)AM) (lim Ei)(8)AM.---+ 13.(D.Lazard) Let Ebeamodule over acommutative ring A.Tensor productsareall taken over that ring. Show that thefollowing conditions areequivalent: (i)There exists adirect family {F;} offree modules offinite type such that E limFi. ----+ (ii) Eisflat. (iii) For every finitely presented module Pthenatural homomorphism HomA(P, A)(8)A E-+HomA(P, E) issurjective. (iv) For every finitely presented module Pand homomorphism f:P-+Ethere exists afree module F,finitely generated, andhomomorphisms g:P-+Fand h:F-+E such thatf=hog. Remark. The point ofLazard's theorem liesinthefirst two conditions: Eisfiat ifandonlyifEisadirect limit offree modules offinite type. [Hint: Since the tensor product commutes with direct limits, that (i)implies (ii) comes from thepreceding exercise and thedefinition offlat. Toshow that(ii)implies (iii),useExercise 11. Toshow that (Hi)implies (iv) iseasy from thehypothesis. Toshow that (iv)implies (i), use thefact that amodule isadirect limit offinitely presented modules (an exercise inChapter III), and (iv) toget the free modules instead. For complete details, seefor instance Bourbaki, Algebre, Chapter X,1, Theorem 1,p.14.] The Casimir element 14.Letkbeacommutative field and letEbeavector spaceover k,offinite dimension n.Let Bbe anondegenerate symmetric bilinear form onE,inducing aniso- 640 THE TENSOR PRODUCT XVI, Ex morphism E ---+EvofEwith itsdual space. Let{VI,...,Vn}beabasis ofE.The B- dual basis{v,...,v}consists oftheelements ofEsuch thatB(Vi,vi)=ij. (a)Show that theelement L:Vi(8)vIinE(8)Eisindependent ofthechoice of basis. Wecall this element theCasimir element (seebelow). (b)Inthesymmetric algebra S(E), letQB=L:ViVIeShow that QBisindepen- dent ofthechoice ofbasis. WecallQBtheCasimir polynomial. Itdepends on B,ofcourse. (c)More generally, letDbean(associative) algebra over k,let:E ---+Dbean injective linear map ofEinto D.Show that the element L:(Vi)(vI)= (J)B,fI)isindependent ofthechoice ofbasis. WecallittheCasimir element in D,determined by and B. Remark. The terminology oftheCasimir element isdetermined bytheclassical case, when GisaLiegroup, E=9=Lie(G)istheLiealgebra ofG(tangent space atthe origin with theLiealgebra product determined bytheLiederivative), and(v) isthe differential operator associated with V(Lie derivative inthedirection ofv).The Casimir element isthen apartial differential operator inthealgebra ofalldifferential operators onG.Cf.basic books onmanifolds and Lietheory, forinstance [JoL 01],Chapter II,1 andChapter VII,2. 15.LetE=sIn(k)=subspace ofMatn(k) consisting ofmatrices with trace O.LetBbe thebilinear form defined byB(X, Y)=tr(XY). Let G=SLn(k). Prove: (a)Bisc(G)-invariant, where c(g) isconjugation byanelement gEG. (b)Bisinvariant under thetranspose (X,Y) (tX,tY). (c)Letk=R.Then Bispositive definite onthesymmetric matrices and nega- tive definite ontheskew-symmetric matrices. (d)Suppose Gisgiven with anaction onthealgebra DofExercise 14,and that thelinear map:E ---+DisG-linear. Show that theCasimir element isG- invariant (for theconjugation action onS(E), and thegiven action onD). CHAPTER XVII Semisimplicity Inmany applications, amodule decomposesasadirect sum ofsimple sub- modules, and then one candevelopafairly precise structure theory, both under general assumptions, andparticular applications. This chapter isdevoted to those results which can beproved ingeneral. Inthe next chapter, weconsider those additional results which can beproved inaclassical andimportant special case. Ihave more orless followed Bourbaki intheproof ofJacobson's density theorem. 1. MATRICES AND LINEAR MAPS OVER NON-COMMUTATIVE RINGS InChapter XIII, weconsidered exclusively matrices over commutative rings. For our present purposes, itisnecessary toconsider amore general situation. LetKbearing. We define amatrix(lpij)with coefficients inKjust aswe didforcommutative rings. The product ofmatrices isdefined bythe same formula. Then weagain have associativity anddistributivity, whenever the size ofthematrices involved intheoperations makes theoperations defined. Inparticular, thesquarenxnmatrices over Kform aring, again denoted by Matn(K). We have aring-homomorphism K Matn(K) onthediagonal. 641 642 SEMISIMPLICITY XVII,1 Byadivision ring weshall mean aring with 1=I0,and such that every non-zero element has amultiplicative inverse. IfKisadivision ring, then every non-zero K-module has abasis, and the cardinalities oftwo bases areequal. Theproof isthe same asinthecommutative case; we never needed commutativity inthearguments. This cardinality is again called thedimension ofthemodule over K,and amodule over adivision ring iscalled avector space. We can associate amatrix with linear maps, depending onthechoice ofa finite basis, justasinthecommutative case. However, weshall consider a somewhat different situation which wewant toapply tosemisimple modules. Let Rbearing, and let E=E1...(f)En' F=Fl(f)...Fm beR-modules, expressedasdirect sums ofR-submodules. Wewish todescribe themost general R-homomorphism ofEinto F. Suppose first F=F1has onecomponent. Let lp:E1(f)...(f)En F beahomomorphism. Letlpj:EjFbetherestriction oflptothefactor Ej' Every element xEEhas aunique expression x=Xl+...+Xn,withXjEEj' Wemay therefore associate with xthecolumn vector X =t(xb...,xn),whose components areinEl'...,Enrespectively. We can associate with lpthe row vector (lpb...,lpn), lpjEHomR(E j,F),and theeffect oflpontheelement xof Eisdescribed bymatrix multiplication, ofthe row vector times thecolumn vector. More generally, consider ahomomorphism lp:E1(f).·.En F1(f)...(f)Fm' Let lri:F1(f)...(f)Fm Fibetheprojection onthei-th factor. Then we can apply ourprevious remarks tolri0lp,foreach i.Inthis way,we seethat there exist unique elementslpijEHomR(E j,Fi),such that lphas amatrix representa- tion (lpl1 M(lp)=: lpml... qJtn )lpmn whose effect onanelement xisgiven bymatrix multiplication, namely (qJ1... qJt")(X:l). lpml lpmnXn XVII, 1 MATRICES AND LINEAR MAPS OVER NON-COMMUTATIVE RINGS 643 Conversely, given amatrix(lpij)withlpijEHomR(E j,Fi),we can define an element ofHomR(E, F)bymeans ofthis matrix. We have anadditive group- isomorphism between HomR(E, F)and this group ofmatrices. Inparticular, letEbeafixed R-module, and letK =EndR(E). Then wehave aring-isomorphism EndR(E(n» Matn(K) which toeach lpEEnd R(E(n» associates thematrix (qJ1·.. qJt") lpn 1 lpnn determined asbefore, and operating ontheleft oncolumn vectors ofE(n), with components inE. Remark. Let Ebe aI-dimensional vector space over adivision ring D, and let{v} be abasis. For each aED,there exists aunique D-linear map fa:E Esuch thatfa(v)=avoThen wehave therule fafb=fba. Thus when weassociate amatrix with alinear map, depending on abasis, the multiplication gets twisted. Nevertheless, thestatement wejust made preceding this remark iscorrect!! Thepoint isthat wetook thelpijinEndR(E), and not inD,inthespecialcase that R=D.Thus Kisnotisomorphic toD(inthe non-commutative case), butanti-isomorphic. This istheonly point ofdifference oftheformal elementary theory oflinear maps inthecommutative ornon- commutative case. Werecall that anR-module Eissaid tobesimple ifitis=I0andifithas no submodule other than 0orE. Proposition 1.1. Schur's Lemma. LetE,Fbesimple R-modules. Every non-zero homomorphism ofEinto Fisanisomorphism. Thering EndR(E) is adivision ring. Proof. Letf:E Fbeanon-zero homomorphism. Itsimage and kernel aresubmodules, hence Kerf=0and 1mf=F.Hencefisanisomorphism. IfE=F,thenfhas aninverse, asdesired. The next proposition describes completely thering ofendomorphisms ofa direct sum ofsimple modules. Proposition 1.2. Let E=E\nt> (f)...(f)Enr) be adirect sum ofsimple modules, theEibeing non-isomorphic, and each Eibeing repeated nitimes in 644 SEMISIMPLICITY XVII,1 the sum. Then, uptoapermutation, El'...,Erareuniquely determined up toisomorphisms, and themultiplicitiesn1,.. .,nrareuniquely determined. The ring EndR(E) isisomorphic toaringofmatrices, oftype M2o o where Miisannixnimatrix over EndR(E i).(The isomorphism isthe one with respect toourdirect sumdecomposition.) Proof. The last statement follows from ourprevious considerations, taking into account Proposition 1.1. Supposenow that wehave twoR-modules, with direct sumdecompositions into simple submodules, and anisomorphism E\nd (f)...(f)Enr) F\md (f).. .(f)Fms), such that the Eiarenon-isomorphic, and theFjarenon-isomorphic. From Proposition 1.1, weconclude that each Eiisisomorphic tosome Fj,and con- versely. Itfollows that r=s,and that after apermutation, EiFi.Further- more, theisomorphism must induce anisomorphism End --+FmdI I foreach i.Since EiFi,wemay assume without loss ofgenerality that in fact Ei=Fi.Thus we arereduced toproving: Ifamodule isisomorphic to E(n) and toE(m), with some simple module E,then n=m.But EndR(E(n» is isomorphic tothe nxnmatrix ring over the division ring EndR(E)=K. Furthermore thisisomorphism isverified atonce tobe anisomorphismas K-vector space. The dimension ofthe space ofnxnmatrices over Kisn2 . This proves that themultiplicitynisuniquely determined, and proves our proposition. When Eadmits a(finite) direct sum decomposition ofsimple submodules, thenumber oftimes that asimple module ofagiven isomorphism class occurs inadecomposition will becalled themultiplicity ofthesimple mod ule(orof theisomorphism class ofthesimple module). Furthermore, if E=E\nd (f)...(f)Enr) isexpressedasasum ofsimple submodules, weshall call nl+...+nrthe length ofE.Inmany applications,weshall also write r E=nlEl(f).'.(f)nrEr=EBn;E;. i= 1 XVII,2 CONDITIONS DEFINING SEMISIMPLICITY 645 2. CONDITIONS DEFINING SEMISIMPLICITY Let Rbe aring. Unless otherwise specified inthis section allmodules and homomorphisms will beR-modules andR-homomorphisms. Thefollowing conditions on amodule Eareequivalent: SS1. Eisthe sum ofafamily ofsimple submodules. SS2. Eisthedirect sum ofafamily ofsimple submodules. SS3.Every submodule FofEisadirect summand, i.e.there exists a submodule F'such that E=F(f)F'. Weshall now prove that these three conditions areequivalent. Lemma 2.1. Let E=2:be asum (not necessarily direct) ofsimple sub- iEI modules. Then there exists asubset JcIsuch that Eisthedirect sum EBEj. jEJ Proof. LetJbeamaximal subset ofIsuch that the sumLEjisdirect. jeJ Wecontend that this sum isinfactequal toE.Itwill suffice toprove that each Eiiscontained inthis sum. But theintersection ofour sum with Eiisasub- module ofE;,hence equal to0orEi.Ifitisequal to0,then Jisnotmaximal, since we canadjoin itoit.Hence Eiiscontained inthesum, and our lemma is proved. The lemma shows that SS 1implies SS2.To seethat SS2implies SS3,take asubmodule F,and letJbeamaximal subset of1such that the sum F+EBEj jeJ isdirect. The same reasoning asbefore shows that this sum isequal toE. Finallyassume SS3. Toshow SS1,weshall first prove that everynon-zero submodule ofEcontains asimple submodule. Let vEE, v =1=O.Then by definition, Rvisaprincipal submodule, and thekernel ofthehomomorphism RRv isaleftideal L=f.R. Hence Liscontained inamaximal left ideal M'#R (byZorn's lemma). Then MIL Isamaximal submodule ofRIL (unequal to RIL), and hence Mv isamaximal submodule ofRv,unequal toRv,correspond- ing toMIL under theisomorphism RIL Rv. 646 SEMISIMPLICITY XVII,3 We canwrite E=Mv (f)M'with some submodule M'. Then Rv =Mv (M' nRv), because every element xERv can bewritten uniquelyasasum x=av+x' with aEMandx'EM', andx'=x-avlies inRv. Since Mv ismaximal in Rv, itfollows thatM'nRvissimple,asdesired. LetEobethesubmodule ofEwhich isthe sum ofallsimple submodules of E.IfEo =f.E,then E=Eo(f)Fwith F=1=0,and there exists asimple sub- module ofF,contradicting thedefinition ofEo. This proves that SS3implies 551. Amodule Esatisfyingour three conditions issaid tobesemisimple. Proposition 2.2. Every submodule and every factor module ofasemisimple module issemisimple. Proof. LetFbeasubmodule. LetF0bethe sum ofallsimple submodules ofF.Write E=F0(f)Fo. Every element xofFhas aunique expression x=Xo+Xowith XoEF0and XoEFo. But Xo=x-XoEF.Hence Fis thedirect sum F=F0(f)(FnFo). We must therefore have F0=F,which issemisimple. Asforthefactor module, write E=F(f)F'.Then F'isasum ofitssimple submodules, and thecanonical map E ElF induces anisomorphism ofF'onto ElF. Hence ElF issemisimple. 3. THE DENSITY THEOREM Let Ebe asemisimple R-module. LetR'=R'(E) bethering EndR(E). Then Eisalso aR'-module, theoperation ofR'onEbeing given by (cp,x) cp(x) forcpER'and xEE.Each aERinduces aR'-homomorphism fa: E Eby themapfa(x)=ax. This iswhat ismeant bythecondition cp((Xx)=(Xcp(x). We letR"=R"(E)=EndR,(E). We call R'thecommutant ofRand R"the bicommutant. Thus wegetaring-homomorphism R EndR,(E)=R"(E)=R" XVII,3 THE DENSITY THEOREM 647 byah. We now ask how big istheimage ofthisring-homomorphism. Thedensity theorem states that itisquite big. Lemma 3.1. LetEbesemisimpleover R.LetR'=EndR(E), fEEndR,(E) asabove. Let xER.There exists anelement aERsuch that ax=f(x). Proof. Since Eissemisimple, we canwrite anR-direct sum E=Rx(f)F with some submodule F.Let 7T:E Rxbetheprojection. Then 7TER',and hence f(x)=f(nx)=nf(x). This shows thatf(x)ERx, asdesired. Thedensity theorem generalizes thelemma bydealing with afinite number ofelements ofEinstead ofjust one. For theproof, we use adiagonal trick. Theorem 3.2. (Jacobson). Let Ebesemisimpleover R, and let R'=EndR(E). LetfEEndR,(E). LetXl'...,xnEE.Then there exists an element aERsuch that ax;=f(x;) for i=1,..., n. IfEisfinitely generated over R',then thenatural map R EndR,(E) issurjective. Proof. Forclarity ofnotation, weshall first carry out theproof incase E issimple. Letfen):E(n) -+E(n) betheproduct map,sothat f(n)(Yl,..., Yn)=(f(Yl)'... ,f(Yn». LetR=EndR(E(n». Then Risnone other than thering ofmatrices with coefficients inR'.Sincefcommutes with elements ofR'initsaction onE,one seesimmediately thatf(n) isinEndR(E(n». Bythelemma, there exists anelement aERsuch that (ax 1,. ..,axn)=(f(x 1)'...,f(x n», which iswhat wewanted toprove. When Eisnotsimple, suppose that Eisequal toafinite direct sum ofsimple submodules E;(non-isomorphic), with multiplicities n;: E=E\nd(f)...(f)Enr) (Ei*Ejifi=Ij), then thematrices representing thering ofendomorphisms split according to blocks corresponding tothenon-isomorphic simple components inourdirect sum decomposition. Hence here again theargument goes throughasbefore. 648 SEMISIMPLICITY XVII,3 The main point isthatf(n) lies inEnd(E(n»,and that we canapply thelemma. We add theobservation thatifEisfinitely generated over R',then anelement fEEndR,(E) isdetermined byitsvalue on afinite number ofelements ofE,so the asserted surjectivity R EndR,(E) follows atonce. Intheapplications below, Ewill be afinite dimensional vector spaceover afield k,and Rwill be ak-algebra,sothefiniteness condition isautomatically satisfied. The argument when Eisaninfinite direct sum would besimilar, but the notation isdisagreeable. However, intheapplicationsweshall never need the theorem inanycase other than the case when Eitself isafinite direct sum of simple modules, and this isthe reason whywefirst gave theproof inthat case, and letthereader write out theformal details intheother cases, ifdesired. Corollary 3.3. (Burnside's Theorem). Let Ebe afinite-dimensional vector spaceover analgebraically closed field k,and letRbeasubalgebra of Endk(E). IfEisasimple R-module, then R=EndR,(E). Proof. We contend that EndR(E)=k.Atany rate, EndR(E) isadivision ring R',containing kasasubring and every element ofkcommutes with every element ofR'.Let aER'.Then k(a)isafield. Furthermore, R'iscontained in Endk(E)asak-subspace, and istherefore finite dimensional over k.Hence k(lJ.) isfinite over k,and therefore equal toksince kisalgebraically closed. This proves that EndR(E)=k.Let now {VI'. . .,vn}be abasis ofEover k.Let AEEndk(E). According tothedensity theorem, there exists aERsuch that (xVi=AVi for i=1,...,n. Since the effect ofAisdetermined byitseffect on abasis, weconclude that R=Endk(E). Corollary 3.3 isused inthefollowing situation asinExercise 8.Let E be afinite-dimensional vector spaceover field k.Let Gbe asubmonoid of GL(E) (multiplicative). AG-invariant subspace FofEisasubspace such that a-F CFforall a-EG.We say that EisG-simple ifithas noG-invariant subspace other than 0and Eitself, and E =1=O.Let R=k[G] bethesubalgebra ofEndk(E) generated byGover k.Since weassumed that Gisamonoid, it follows that Rconsists oflinear combinations ,a.a.t- r I with aiEkand aiEG.Then we seethat asubspace FofEisG-invariant ifand only ifitisR-invariant. Thus EisG-simple ifandonly ifitissimple over Rin the sense which wehave been considering. We can then restate Burnside's theorem ashestated it: Corollary 3.4. Let Ebe afinite dimensional vector space over analge- braically closed field k,and letGbea(multiplicative) submonoid ofGL(E). XVII,3 THE DENSITY THEOREM 649 IfEisG-simple, then kEG]=Endk(E). When kisnotalgebraically closed, then westill getsome result. Quite generally, letRbearing and Easimple R-module. Wehave seen that EndR(E) isadivision ring, which wedenote byD,and Eisavector spaceover D. Let Rbe aring, and Eany R-module. We shall saythat Eisafaithful module ifthefollowing condition issatisfied. Given (XERsuch that (Xx =0 forallxEE,wehave (X=O.Intheapplications, Eisavector space over afield k,and wehave aring-homomorphism ofRinto Endk(E). Inthis way, Eisan R-module, and itisfaithful ifandonly ifthishomomorphism isinjective. Corollary 3.5. (Wedderburn's Theorem). Let Rbearing, and Easimple, faithful module over R.Let D =EndR(E), and assume that Eisfinite dimen- sional over D.Then R=EndD(E). Proof. Let{Vl"."vn}be abasis ofEover D.Given AEEndD(E), by Theorem 3.2there exists rxERsuch that rxVi=AVifor i=1,...,n. Hence themap R EndD(E) issurjective. Our assumption that Eisfaithful over Rimplies that itisinjective, and ourcorollary isproved. Example. Let Rbeafinite-dimensional algebra over afield k,and assume that Rhas aunit element, soisaring. IfRdoes not have any two-sided ideals other than 0and Ritself, then any nonzero module Eover Risfaithful, because thekernel ofthehomomorphism R Endk(E) isatwo-sided ideal =f.R.IfEissimple, then Eisfinite dimensional over k. Then Disafinite-dimensional division algebra over k.Wedderburn's theorem gives arepresentation ofRasthering ofD-endomorphisms ofE. Under theassumption that Risfinite dimensional, one can find asimple module simply bytakingaminimal left ideal =1=O.Such anideal exists merely bytakingaleft ideal ofminimal non-zero dimension over k.An even shorter proof ofWedderburn's theorem will begiven below (Rieffel's theorem) inthis case. Corollary 3.6. LetRbearing, finite dimensional algebra overafield kwhich isalgebraically closed. LetVbeafinite dimensional vector space over k,with asimple faithful representation p:R---+Endk(V). Then pisanisomorphism, inother words, R=Matn(k). Proof. Weapply Corollary 3.5, noting that Disfinite dimensional over k.Given aED, we note that k(a) isacommutative subfield ofD,whence k(a)=kbyassumption that kisalgebraically closed, and thecorollary follows. 650 SEMISIMPLICITY XVII,3 Note. Thecorollary applies tosimple rings, which will bedefined below. Suppose next thatVI'. ..,Vmarefinite dimensional vector spacesover afield k,and that Risak-algebra with representations R Endk(\';),i=1,.. .,m, so\';isanR-module. Ifwelet E=\'1EB· · ·EBVm, then Eisfinite over R'(E), sowegetthefollowing consequence ofJacobson's densi tytheorem. Theorem 3.7. Existence ofprojection operators. Let kbeafield,Ra k-algebra, and\'1,...,Vmfinite dimensional k-spaces which are also simple R-modules, and such that\';isnotR-isomorphic tofor i=1=j.Then there exist elementse;ERsuch thate;acts astheidentity on\';ande;=0 ifj=1=i. Proof. We observe that theprojection fifrom thedirect sum Etothei-th factor isinEndR,(E),because ifcpER'thencp()C\:fforallj. Wemay therefore apply thedensity theorem toconclude theproof. Corollary 3.8. (Bourbaki). Let kbeafield ofcharacteristic O.Let Rbe ak-algebra, and letE,Fbesemisimple R-modules, finite dimensional over k. Foreach aER,letaE' aFbethecorresponding k-endomorphismsonEand Frespectively. Suppose that the traces areequal; that is, tr(aE)=tr(aF) forall aER. Then Eisisomorphic toFasR-module. Proof. Each ofEand Fisisomorphic toafinite direct sum ofsimple R- modules, with certain multiplicities. Let Vbe asimple R-module, and suppose E=v<n) EBdirect summands notisomorphic toV F=v<m) EBdirect summands notisomorphic toV. Itwill suffice toprove that m=n.Let evbetheelement ofRfound inTheorem 3.7 such that evacts astheidentityonV,and is0ontheother direct summands ofEand F.Then tr(eE)=ndimk(V) and tr(eF)=mdimk(V), Since the traces areequal byassumption, itfollows that m=n,thus concluding theproof. Note that thecharacteristic 0isused here, because thevalues ofthe trace areink. Example. Inthelanguage ofrepresentations, suppose Gisamonoid, and XVII,4 SEMISIMPLE RINGS 651 wehave twosemisimple representations into finite dimensional k-spaces p:G Endk(E) and p':G Endk(F) (so pandp'map Ginto themultiplicative monoid ofEndk).Assume that trp(u)=trp'(u)foralluEG.Then pandp'areisomorphic. Indeed, welet R=k[G], sothat pandp'extend torepresentations ofR.Bylinearity, one has that trp(a)=trp'(a)' forall aER, soone canapply Corollary 3.8. 4. SEMISIMPLE RINGS Aring Riscalled semisimple if1=I0,andifRissemisimpleasaleftmodule over itself. Proposition 4.1. IfRissemisimple, then every R-module issemisimple. Proof. AnR-module isafactor module ofafreemodule, and afreemodule isadirect sum ofRwith itself acertain number oftimes. We canapply Proposi- tion 2.2toconclude theproof. Examples. 1)Let kbe afield and letR=Matn(k) bethealgebra of nxnmatrices over k.Then Rissemisimple, andactually simple,asweshall define and prove in5,Theorem 5.5. 2)Let Gbe afinite group and suppose that thecharacteristic ofkdoes not divide #(G). Then thegroup ringk[G] issemisimple,asweshall prove inChapter XVIII, Theorem 1.2. 3)The Clifford algebras enover thereal numbers aresemisimple. See Exer- cise 19ofChapter XIX. Aleftideal ofRisanR-module, and isthus called simple ifitissimpleasa module. Two ideals L,L'are called isomorphic ifthey areisomorphicas modules. We shall now decompose Rasasum ofitssimple leftideals, andthereby get astructure theorem forR. Let{LJ iEIbeafamily ofsimple leftideals, notwo ofwhich areisomorphic, and such that each simple leftideal isisomorphic tooneofthem. We saythat thisfamily isafamily ofrepresentatives fortheisomorphism classes ofsimple leftideals. Lemma 4.2. LetLbe asimple leftideal, and letEbeasimple R-module. IfLisnotisomorphic toE,then LE =o. Proof. We have RLE =LE, and LE isasubmodule ofE,hence equal to 652 SEMISIMPLICITY XVII,4 oorE.Suppose LE =E.LetyEEbesuch that Ly =f.o. Since Lyisasubmodule ofE,itfollows that Ly=E.The maplJ.lJ.yofL into Eisahomomorphism ofLinto E,which issurjective, and hence nonzero. Since Lissimple, thishomomorphism isanisomorphism. Let Ri=LL LLi bethe sum ofallsimple leftideals isomorphic toLi.From thelemma, wecon- clude that RiRj=0ifi=Ij.This will beused constantly inwhat follows. We note that Riisaleftideal, and that Risthe sum R = R.i...J p ieI because Risasum ofsimple leftideals. Hence foranyjEI, R.cR.R=R.R.cR.J J J J l' thefirst inclusion because Rcontains aunit element, and thelast because Rj isaleftideal. Weconclude that Rjisalso aright ideal, i.e.Rjisatwo-sided ideal foralljEI. We can express theunit element 1ofRasasum 1=Lei ieI with eiERi.This sum isactually finite, almost all ei=O.Say ei=I0for indices i=1,...,S,sothat wewrite r=el+...+es. For any xER,write x= x.i...J " ieIXiERi. Forj=1,...,swehaveejx=ejxjand also X. = 1.x. =elx,+...+ex.=e.x.J 1 Js1 JJ. Furthermore, x=elx +...+esx. This proves that there isnoindex i other than i=1,..., sand also that thei-thcomponent Xiofxisuniquely determined aseix=eixi. Hence the sum R=Rl+...+Rsisdirect, and furthermore, eiisaunit element for Ri,which istherefore aring. Since XVII,4 SEMISIMPLE RINGS 653 RiRj=0fori=f.j,wefind that infact s R=nRi i= 1 isadirect product oftherings Ri. Aring Rissaid tobesimple ifitissemisimple, andifithasonlyone isomorphism class ofsimple left ideals .We seethat wehave provedastructure theorem forsemisimple rings: Theorem 4.3. Let Rbesemisimple. Then there isonly afinite number of non-isomorphic simple leftideals, sayLl,...,Ls.If Ri=LL L::Li isthe sumofallsimple leftideals isomorphic toLi,then Riisatwo-sided ideal, which isalso aring (the operations being those induced byR),and Risring isomorphic tothedirect product s R=nRio i=1 Each Riisasimple ring.Ifeiisitsunit element, then 1=el+...+es,and Ri=Rei. Wehaveeiej=0ifi=f.j. Weshall now discuss modules. Theorem 4.4. Let Rbesemisimple, and letEbeanR-module =IO.Then s s E=EBRiE=EBeiE, i=l i=l and RiEisthesubmodule ofEconsisting ofthe sumofallsimple submodules isomorphic toLi. Proof. Let Eibethe sum ofallsimple submodules ofEisomorphic toLi. IfVisasimple submodule ofE,then RV=V,and hence LiV=Vfor some i. Byaprevious lemma, wehave Li V.Hence Eisthedirect sum ofE1,...,Es. Itisthen clear that RiE=Ei. Corollary 4.5. Let Rbesemisimple. Every simple module isisomorphic to oneofthesimple leftideals Li. Corollary 4.6. Asimple ring hasexactly one simple module, up to ISO- morphism. 654 SEMISIMPLICITY XVII,5 Both these corollaries areimmediate consequences ofTheorems 4.3and 4.4. Proposition 4.7. Let kbe afield and Eafinite dimensional vector space over k.Let Sbeasubset ofEndk(E). Let Rbethek-algebra generated bythe elements ofS.Then Rissemisimple ifandonlyifEisasemisimple R(orS) module. Proof. IfRissemisimple, then Eissemisimple byProposition 4.1. Con- versely, assume EsemisimpleasS-module. Then EissemisimpleasR-module, and soisadirect sum n E=EBEi i= 1 where each Eiissimple. Then foreach ithere exists anelement ViEEisuch that Ei=RVi. The map X(XV1'...' xvn) isaR-homomorphism ofRinto E,and isaninjection since Riscontained in Endk(E). Since asubmodule ofasemisimple module issemisimple byProposi- tion 2.2, thedesired result follows. 5. SIMPLE RINGS Lemma 5.1. Let Rbearing, and tfJEEndR(R) ahomomorphism ofRinto itself, viewed asR-module. Then there exists rJ.ERsuch that tfJ(x)=XrJ.for allxER. Proof. We have tfJ(x)=tfJ(x.1)=xtfJ(1).Let rJ.=tfJ(1). Theorem 5.2. Let Rbeasimple ring. Then Risafinite direct sumofsimple leftideals. There are notwo-sided ideals except 0and R.IfL,Maresimple leftideals, then there exists rJ.ERsuch that LrJ. =M .Wehave LR=R. Proof. Since Risbydefinition alsosemisimple, itisadirect sum ofsimple m leftideals, sayffiLj.We can write 1asafinite sum 1=?f3j,withf3jELj'jEJ ]=1 Thenm m R=EBRf3j=EBLj. j== 1 j==1 XVII,5 SIMPLE RINGS 655 This proves our first assertion. Astothesecond, itisaconsequence ofthe third. Lettherefore Lbeasimple leftideal. Then LR isaleftideal, because RLR=LR, hence (Rbeing semisimple) isadirect sum ofsimple leftideals, say m LR =EBLj, j=1L=Ll. LetMbeasimple leftideal. We have adirect sumdecomposition R=L L'. Let n:R Lbetheprojection. ItisanR-endomorphism. Let (J:L Mbe anisomorphism (itexists byTheorem 4.3). Then (J01t:R RisanR-endo- morphism. Bythelemma, there exists rxERsuch that (J0n(x)=xrx forall xER. Apply this toelements xEL.Wefind (J(x)=xrx forall xEL. The map x xrxisaR-homomorphism ofLinto M,isnon-zero, hence isan isomorphism. From this itfollows atonce that LR =R,thereby proving our theorem. Corollary 5.3. Let Rbeasimple ring. Let Ebeasimple R-module, and L asimple leftideal ofR.Then LE=Eand Eisfaithful. Proof. We have LE =L(RE)=(LR)E=RE =E. Suppose rxE =0 for some rxER.Then RrxRE=RrxE=O.But RrxR isatwo-sided ideal. Hence RrxR=0,and rx=O.This proves that Eisfaithful. Theorem 5.4. (RiefIel). Let Rbearing without two-sided ideals except 0 and R.Let Lbe anonzero leftideal, R' =EndR(L) and R" =EndR,(L). Then thenatural map A.:R R"isanisomorphism. Proof. The kernel ofAisatwo-sided ideal, soAisinjective. Since LR isatwo-sided ideal, wehave LR =Rand A(L)A(R)=A(R). For any x,yEL, andfER",wehavef(xy)=f(x)y, because right multiplication byyisan R-endomorphism ofL.Hence A(L) isaleftideal ofR", so R" =R"A(R)=R"A(L)A(R)=A(L)A(R)=A.(R), aswas tobeshown. InRieffel's theorem, wedonot need toassume that Lisasimple module. 656 SEMISIMPLICITY XVII,5 Ontheother hand, Lisanideal. Sothistheorem isnotequivalent with previous ones ofthe same nature. In7, weshall giveavery general condition under which thecanonical homomorphism R-+R" ofaring into thedouble endomorphism ring ofamodule isanisomorphism. This will cover allthepreviouscases. Aspointed out intheexample following Wedderburn's theorem, Rieffel's theorem applies togive another proof when Risafinite-dimensional algebra (with unit) over afield k. The next theorem givesaconverse, showing that matrix ringsover division algebras aresimple. Theorem 5.5. Let Dbe adivision ring, and Eafinite-dimensional vector space over D.LetR=EndD(E). Then Rissimple and Eisasimple R-module. Furthermore, D=EndR(E). Proof. Wefirst show that Eisasimple R-module. Let vEE,v=IO.Then vcan becompleted toabasis ofEover D,and hence, given wEE, there exists rJ..ERsuch that rJ..V =w.Hence Ecannot have any invariant subspaces other than 0oritself, and issimple over R.Itisclear that Eisfaithful over R.Let {Vl,...,vm}beabasis ofEover D.The map rJ.. (rJ..v l,..., rJ..l'm) ofRinto E(m) isanR-homomorphism ofRinto E(m), and isinjective. Given (Wl,...,wm)EE(m), there exists r:J.ERsuch thatrJ..Vi=Wiand hence RisR- isomorphic toE(m). This shows that R(as amodule over itself) isisomorphic toadirect sum ofsimple modules and istherefore semisimple. Furthermore, allthese simple modules areisomorphic toeach other, and hence Rissimple byTheorem 4.3. There remains toprove that D=EndR(E). We note that Eisasemisimple module over Dsince itisavector space, and every subspace admits acom- plementary subspace. We can therefore apply thedensity theorem (the roles ofRand Dare now permuted !).LetqJEEndR(E). Let vEE, v=f.O.Bythe density theorem, there exists anelement aEDsuch that qJ(v)=avo LetWEE. There exists anelement fER such thatf(v)=w.Then qJ(W)=qJ(f(v»)=f(qJ(v»=f(av)=af(v)=aWe Therefore qJ(w)=awforallWEE. This means thatqJED,and concludes our proof. Theorem 5.6. Let kbe afield and Eafinite-dimensional vector space of XVII,6THE JACOBSON RADICAL, BASE CHANGE, AND TENSOR PRODUCTS 657 dimension mover k.Let R =Endk(E). Then Risak-space, and dimkR =m2 . Furthermore, misthenumber ofsimple leftideals appearing inadirect sum decomposition ofRassuch asum. Proof. The k-space ofk-endomorphisms ofEisrepresented bythe space ofmxmmatrices ink,sothedimension ofRasak-space ism2 .Ontheother hand, theproof ofTheorem 5.5showed that RisR-isomorphicasanR-module tothedirect sum E(m). We know theuniqueness ofthedecompositionofa module into adirect sum ofsimple modules (Proposition 1.2), and this proves our assertion. Intheterminology introduced in 1,we seethat theinteger minTheorem 5.6isthelength ofR. We canidentify R=Endk(E) with thering ofmatrices Matm(k), once a basis ofEisselected. Inthat case, we can take thesimple leftideals tobethe ideals Li(i=1,..., m)where amatrix inLihas coefficients equal to0except inthei-thcolumn. Anelement ofLlthus looks like oall 0 o We seethat Risthedirect sum ofthe mcolumns. We also observe that Theorem 5.5implies thefollowing: Ifamatrix M EMatm(k) commutes with allelements ofMatm(k), then Misa scalar matrix. Indeed, such amatrix Mcan then beviewed asanR-endomorphism ofE, and weknow byTheorem 5.5that such anendomorphism liesink.Ofcourse, one can also verify thisdirectly byabrute force computation. 6. THE JACOBSON RADICAL, BASE CHANGE, AND TENSOR PRODUCTS Let Rbearing and letMbeamaximal leftideal. Then RIM isanR-module, and actually RIM issimple. Indeed, letJbe asubmodule ofRIM with ]=1=RIM. LetJbeitsinverse image inRunder thecanonical homomorphism. 658 SEMISIMPLICITY XVII,6 Then Jisaleftideal =1=Mbecause J=1=RIM, soJ=Rand J=O.Conversely, letEbe asimple R-module and letvEE,v=1=O.Then Rvisasubmodule =1=0 ofE,and hence Rv=E.LetMbethekernel ofthehomomorphismx xv. Then Misaleftideal, andMismaximal; otherwise there isaleftideal M'with R:)M':)MandM' =1=R, =1=M.Then RIM=EandRIM' isanon-zero homo- morphic image ofE,which cannot exist since Eissimple (Schur's lemma, Proposition 1.1). Thus weobtain abijection between maximal left ideals and simple R-modules (uptoisomorphism). We define theJacobson radical ofRtobetheleft ideal Nwhich isthe intersection ofallmaximal left ideals ofR .We may also denote N=Rad(R). Theorem 6.1. (a) For every simple R-module wehave NE=O. (b) Theradical Nisatwo-sided ideal, containing allnilpotent two-sided ideals. (c)Let Rbeafinite dimensional algebra overfield k.Itsradical is{OJ,ifand onlyifRissemisimple. (d)IfRisafinite dimensional algebra over afield k,then itsradical Nis nilpotent (i.e. Nr=0forsome positive integer r). These statements areeasy toprove, and hints will begiven appropriately. See Exercises 1through 5. Observe that under finite dimensionality conditions, theradical's being 0 givesus auseful criterion for aring tobesemisimple, which weshall use in the next result. Theorem 6.2. LetAbeasemisimple algebra, finite dimensional over afield k.LetKbe afinite separable extension ofk.Then K0kAisasemisimple over K. Proof. Inlight oftheradical criterion forsemisimplicity, itsuffices toprove that K0kAhas zero radical, anditsuffices todosofor aneven larger extension than K, sothat wemayassume KisGalois over k,saywith Galois group G. Then GoperatesonK0Aby u(x0a)=ax0afor xEK and aEA. LetNbetheradical ofK0A.Since Nisnilpotent, itfollows that uN isalso nilpotent foralluEG,whence uN=Nbecause Nisthemaximal nilpotent ideal (Exercise 5).Let{aI'. . .,am} be abasis ofAover k.Suppose Ncontains theelement =LXi0ai=1=0with XiEK. For every yEKtheelement (y0I)=LYx; 0a;also lies inN.Then trace«y 0I))=Lu=LTr(yx;) 0a;=L 10a;Tr(yx;) also lies inN,and lies in 10A=A,thus proving thetheorem. XVII,6THE JACOBSON RADICAL, BASE CHANGE, AND TENSOR PRODUCTS 659 Remark. For the case when Aisafinite extension ofk,compare with Exercises 1,2,3ofChapter XVI. Let Abe asemisimple algebra, finite dimensional over afield k.Then by Theorem 6.2 theextension ofscalars A0kkaissemisimple ifkisperfect.In general,analgebra Aover kissaid tobeabsolutely semisimple ifA0kkais semisimple. We now look atsemisimple algebras over analgebraically closed field. Theorem 6.3. Let A,Bbesimple algebras, finite dimensional over a field kwhich isalgebraically closed. Then A0kBisalso simple.We have A=Endk(V) and B=Endk(W) where V,Ware finite dimensional vector spaces over k,and there isanatural isomorphism A0kB=Endk(V 0kW)=Endk(V) 0kEndk(W). Proof. The formula isaspecial case ofTheorem 2.5ofChapter XVI, and theisomorphisms A=Endk(V), B=Endk(W) exist byWedderburn's theorem oritscorollaries. Let Abe analgebraover kand letFbeanextension field ofk.We denote byAFtheextension ofscalars AF=A0kF. Thus AFisanalgebraover F.As anexercise, prove thatifkisthe center ofA, then Fisthe center ofAF.(Here weidentify Fwith 10F.) LetA,Bbealgebrasover k.We leave tothereader theproof that forevery extension field Fofk,wehave anatural isomorphism (A0kB)F=AF0FBF. Weapply the above considerations tothe tensor product ofsemisimple algebras. Theorem 6.4. LetA,Bbeabsolutely semisimple algebras finite dimensional over afield k.Then A0kBisabsolutely semisimple. Proof. LetF=ka .Then AFissemisimple byhypothesis,soitisadirect product ofsimple algebras, which arematrix algebras, and inparticularwe can apply Theorem 6.3 toseethat AF0FBFhas noradical. Hence A0kBhas no radical (because ifNisitsradical, then N0kF=NFisanilpotent ideal of AF0FBF),whence A0kBissemisimple byTheorem 6.1(c). Remark. We have proved the above tensor product theorems rapidly in special cases, which arealready important invarious applications. For amore general treatment, Irecommend Bourbaki's Algebra, Chapter VIII, which gives anexhaustive treatment oftensor products ofsemisimple andsimple algebras. 660 SEMISIMPLICITY XVII,7 7. BALANCED MODULES Let Rbearing and Eamodule. WeletR'(E)=EndR(E) and R"(E)=EndR,(E). Let A:R R"bethenatural homomorphism such that Ax(V)=xvfor xER and vEE.IfAisanisomorphism, weshall saythat Eisbalanced. Weshall say that Eisagenerator (forR-modules) ifevery module isahomomorphic image ofa(possibly infinite) direct sum ofEwith itself. Forexample, Risagenerator. More interestingly, inRieffel's Theorem 5.4, theleft ideal Lisagen- erator, because LR=Rimplies that there isasurjective homomorphism Lx...xL Rsince we can write 1asafinite combination 1=xlal+· ··+xna nwith XiELand aiER. The map(xl'. . .,Xn) xlal+···+xna nisaR-homomorphism ofleftmodule ontoR. IfEisagenerator, then there isasurjective homomorphism en) R(we can take nfinite since Risfinitely generated, byone element 1). Theorem 7.1. (Morita). LetEbeanR-module. Then Eisagenerator if andonlyifEisbalanced andfinitely generated projective over R'(E). Proof. We shall prove half ofthetheorem, leaving theother half tothe reader, using similar ideas (see Exercise 12). Sowe assume that Eisagenerator, and weprove that itsatisfies theother properties byarguments due toFaith. Wefirst prove that forany module F,REBFisbalanced. Weidentify Rand Fasthesubmodules REB0and 0 FofREBF,respectively. For WEF, let«Pw:R F Fbethe map «Pw(x+v)=xw. Then anyfER"(R EBF) commutes with'7TI, '7T2, and each «Pw. From this we see at once that f(x+v)=f(I)(x+v)and hence that REBFisbalanced. Let Ebe agen- erator, andE(n) Rasurjective homomorphism. Since Risfree, we canwrite E(n)=REBFfor some module F, sothat En) isbalanced, Let gER'(E). Then g(n) commutes with every element 'P=('PU)inR'(E(n» (with components 'PUER'(E», and hence there issome XERsuch that g(n)=Ain).Hence g=Ax,thereby proving that Eisbalanced, since Aisobviously injective. Toprove that Eisfinitely generated over R'(E), wehave R'(E)(n) HomR(E(n), E) HomR(R, E)(f)HomR(F, E) asadditive groups. This relation also obviously holds asR'-modules ifwe define theoperation ofR'tobecomposition ofmappings (on theleft). Since HomR(R, E)isR'-isomorphic toEunder themap h h(1),itfollows that Eis anR'-homomorphic image ofR,(n), whence finitely generated over R'. Wealso seethat Eisadirect summand ofthe free R'-module R'(n) and istherefore projectiveover R'(E). This concludes theproof. XVII, Ex EXERCISES 661 EXERCISES The radical 1.(a)LetRbearing. Wedefine theradical ofRtobetheleftideal Nwhich istheinter- section ofallmaximal leftideals ofR.Show that NE=0forevery simple R-module E.Show that Nisatwo-sided ideal. (b)Show that theradical ofR/NisO. 2.Aring issaid tobeArtinian ifevery descending sequence ofleftideals JI::>J2::>... with Ji=1=J;+ Iisfinite. (a)Show that afinite dimensional algebra over afield is Artinian. (b)IfRisArtinian, show that everynon-zero left ideal contains asimple left ideal. (c)IfRisArtinian, show that every non-empty setofideals contains a minimal ideal. 3.Let RbeArtinian. Show that itsradical is0ifandonly ifRissemisimple. [Hint: Get aninjection ofRinto adirect sum EBR/M;where {M;} isafinite setofmaximal left ideals.] 4.Nakayama's lemma. Let Rbeanyring and M afinitely generated module. LetN betheradical ofR.IfNM=Mshow that M=O.[Hint: Observe that theproof ofNakayama's lemma stillholds.] 5.(a) LetJbeatwo-sided nilpotent ideal ofR.Show that Jiscontained intheradical. (b)Conversely,assume that RisArtinian. Show that itsradical isnilpotent, i.e., that there exists anintegerr>1such that Nr=O.[Hint: Consider thedescending sequence ofpowers Nr, andapply Nakayama toaminimal finitely generated left ideal LCN°Osuch that N°OL =1=O. 6.Let Rbeasemisimple commutative ring. Show that Risadirect product offields. 7.LetRbeafinite dImensional commutative algebra over afield k.IfRhas nonilpotent element =1=0,show that Rissemisimple. 8.(Kolchin) Let Ebeafinite-dimensional vector space over afield k.Let Gbeasub- group ofGL(E) such that every element AEGis oftype I+Nwhere Nisnilpotent. Assume Ei=O.Show that there exists anelement vEE,vi=0such that Av =vforall AEG.[Hint: First reduce thequestion tothe case when kisalgebraically closed by showing that theproblem amounts tosolving linear equations. Secondly, reduce itto the case when Eisasimple k[G]-module. Combining Burnside's theorem with the fact thattr(A)=tr(/) forallAEG,show that ifAoEG,Ao=I+N,then tr(NX)=0 forallXEEndk(E), and hence that N =0,Ao=I.] Semisimple operations 9.Let Ebeafinite dimensional vector space over afield k.Let Rbeasemisimple sub- algebra ofEndk(E). Let a,bER.Assume that Ker bE::>Ker aE, where bEismultIplicatIon bybonEandsimilarly foraE. Show that there exists an element SERsuch that sa=b.[Hint: Reduce toRsimple. Then R =Endo(E o) and E=Eg'). Let vl'...,VrEEbe aD-basls foraE. Define sbyS(avi)=bViand 662 SEMISIMPLICITY XVII, Ex extend sbyD-linearity. Then saE=bE, sosa==b.] 10.Let Ebe afinite-dimensional vector space over afield k.Let AEEndk(E). We say that Aissemisimple ifEisasemisimple A-space, orequivalently, letRbethek-algebra generated byA,then Eissemisimpleover R.Show that Aissemisimple ifandonly ifitsminimal polynomial has nofactors ofmultiplicity> lover k. 11.Let Ebeafinite-dimensional vector space over afield k,and letSbeacommutative setofendomorphisms ofE.Let R==k[S]. Assume that Rissemisimple. Show that every subset ofSissemisimple. 12.Prove that anR-module Eisagenerator ifandonly ifitisbalanced, andfinitely generated projectiveover R'(E). Show that Theorem 5.4 isaconsequence ofTheorem 7.1. 13.Let Abeaprincipal ring with quotient field K.Let Anben-space over A,and let T==AnEBAnEB...EBAn bethedirect sum ofAnwith itselfr times. Then Tisfreeofrank nrover A.Ifwe view elements ofAn ascolumn vectors, then Tisthe space ofnxrmatrices over A.Let M==Matn(A) bethering ofnxnmatrices over A,operating ontheleftofT.Bya lattice LinTwe mean anA-submodule ofrank nrover A.Prove that any such lattice which isM-stable isM-isomorphic toTitself. Thus there isjustoneM-isomorphism class oflattices. [Hint: LetgEM bethematrix with 1intheupper left corner and oeverywhere else, sogisaprojection ofAnon aI-dimensional subspace. Then multi- plication ontheleftg:T-+A,maps Tonthespace ofnxrmatrices with arbitrary first row and 0everywhere else. Furthermore, forany lattice LinTtheimage gLisa lattice inA"that isafree A-submodule ofrank r.Byelementary divisors there exists anrxrmatrix Qsuch that gL==A,Q (multiplicationontheright). Then show thatTQ==Landthatmultiplication byQontheright isanM-isomorphism ofTwith L.] 14.LetFbe afield. Let n==n(F)bethe vector space ofstrictly upper triangularnxn matrices over F.Show that nisactually analgebra, and allelements ofnarenilpo- tent (some positive integral power is0). 15.Conjugation representation. Let Abethemultiplicative group ofdiagonal matrices in Fwith non-zero diagonal components. For aEA,theconjugation action ofaon Matn(F) isdenoted byc(a),soc(a)M==aMa-1forM EMatn(F). (a)Show that n isstable under this action. (b)Show that nissemisimple under this action. More precisely, for I<i<j<n,letEijbethematrix with (ij)-component I,and allother components O.Then these matrices Eijform abasis for nover F,and each Eijisan eigenvector fortheconjugation action, namely for a==diag(al,.. .,an),wehave aEija-1=(ai/a) )Eij, sothecorresponding characterXuisgiven byXij(a)==ai/a). (c)Show thatMatn(F) issemisimple, and infact isequal tobEBnEBtn,where bisthe space ofdiagonal matrices. CHAPTER XVIII Representations ofFinite Groups The theory ofgroup representationsoccurs inmany contexts. First, itis developed foritsown sake: determine allirreducible representations ofagiven group. Seeforinstance Curtis-Reiner's Methods ofRepresentation Theory (Wiley- Interscience, 1981). Itisalso used inclassifying finite simple groups. Butalready inthis book wehave seen applications ofrepresentations toGalois theory and thedetermination oftheGalois group over therationals. Inaddition, there isan analogous theory fortopological groups. Inthis case, theclosest analogy iswith compact groups, and thereader will find aself-contained treatment ofthecompact case entirely similar to5ofthischapter inmybook SL2(R)(Springer Verlag), Chapter II,2.Essentially, finite sums arereplaced byintegrals, otherwise the formalism isthe same. The analysiscomes only intwoplaces. One ofthem is toshow that every irreducible representation ofacompact group isfinite dimen- sional; theother isSchur's lemma. The details ofthese extra considerations are carried outcompletely intheabove-mentioned reference. Iwas careful towrite up5with theanalogy inmind. Similarly, readers will findanalogous material oninduced representations in SL2(R), Chapter III,2(which isalso self-contained). Examples ofthegeneral theorycome invarious shapes. Theorem 8.4 may beviewed asanexample, showing how acertain representationcan beexpressed asadirect sum ofinduced representations from I-dimensional representations. Examples ofrepresentations ofS3and S4aregiven intheexercises. The entire last section works outcompletely thesimple characters forthegroup GL2(F) when Fisafinite field, and shows how these characters essentiallycome from induced characters. Forother examples alsoleading into Liegroups,seeW.Fulton and J.Harris, Representation Theory, Springer Verlag 1991. 663 664 REPRESENTATIONS OF FINITE GROUPS XVIII, 91 1. REPRESENTATIONS AND SEMISIMPLICITY Let Rbe acommutative ring and Gagroup. We form the group algebra R[G]. Asexplained inChapter II,3itconsists ofallformal linear combinations Lau(j UEG with coefficients auER,almost allofwhich areo.The product istaken inthe natural way, (Lau(j)(Lbtt)=La(JbraT. UEG tEG U,t Let Ebe anR-module. Every algebra-homomorphism R[G] EndR(E) induces agroup-homomorphism G AutR(E), and thus arepresentation ofthering R[G] inEgives rise toarepresentation of thegroup. Given such representations, wealso saythatR[G], orG,operate on E.We note that therepresentation makes Einto amodule over thering R[G]. Conversely, givenarepresentation ofthegroup, say p:G AutR(E), we can extend ptoarepresentation ofR[G] asfollows. Let a=Laua- and xEE. We define p(rx)x=Laup((j)x. Itisimmediately verified that phas been extended toaring-homomorphism of R[G] into EndR(E). We saythat pisfaithful onGifthemap p:G AutR(E) isinjective. The extension ofptoR[G] may not befaithful, however. Given arepresentation ofGonE,weoften write simplya-xinstead ofp(a-)x, whenever wedeal with afixed representation throughoutadiscussion. AnR-module E,together with arepresentation p,will becalled aG-module, orG-space, oralso a(G,R)-module ifwewish tospecify thering R.IfE,F areG-modules, werecall that aG-homomorphismf: E FisanR-linear map such thatf(ax)=a-f(x) forallxEEand a-EG. Given aG-homomorphism f:E F, we note that thekernel offisaG- submodule ofE,and that theR-factor module FIf(E) admits anoperation ofG inaunique way such that thecanonical map F FIf(E) isaG-homomorphism. Byatrivial representation p:G AutR(E), weshall mean therepresentation such that p(G)=1.Arepresentation istrivial ifandonly ifax=xforall xEE .We also say inthat case that Goperates trivially. XVIII, 1 REPRESENTATIONS AND SEMISIMPLICITY 665 We make Rinto aG-module bymaking GacttriviallyonR. We shall now discuss systematically therepresentations which arise from a given one, onHorn, thedual, and thetensor product. This pattern will berepeated later when wedeal with induced representations. First, HomR(E, F)isaG-module under theaction defined forfEHomR(E, F) by ([a-]f)(x)=a-f(a--1x). The conditions for anoperationaretrivially verified. Note the a--1inside the expression.Weshall usually omit parentheses, andwrite simply [a-]f(x) forthe left-hand side. We notethatf isaG-homomorphism ifandonly if[a-]f=ffor all a-EG. We areparticularly concerned when F=R(sowith trivial action), inwhich case HomR(E, R)=EVisthedual module. Intheterminology ofrepresentations, ifp:G AutR(E) isarepresentation ofGonE,then theaction wehave just described givesarepresentation denoted by pV:G AutR(Ev), and called thedual representation (also called contragredient (ugh!) inthe Iiterature ). Supposenow that themodules E,Farefree and finite dimensional over R. Let pberepresentation ofGonE.LetMbethematrix ofp(a-)with respectto abasis, and letMVbethematrix ofpV(a-)with respect tothedual basis. Then itisimmediately verified that (1) MV=tM-1 . Next weconsider the tensor product instead ofHorn. LetE,E'be(G,R)- modules .We can form their tensor product E0E',always taken over R.Then there isaunique action ofGonE0E'such that for a-EGwehave a-(x0x')=ax0ax'. Suppose that E,Farefinite free over R.Then theR-isomorphism (2) EV0F=HomR(E, F) ofChapter XVI, Corollary 5.5, isimmediately verified tobeaG-isomorphism. Whether Eisfree ornot, wedefine theG-invariant submodule ofEtobe invG(E)=R-submodule ofelements xEEsuch that ax=xforall a-EG.If E,Farefree then wehave anR-isomorphism (3) invG(EV0F)=HomG(E, F). 666 REPRESENTATIONS OFFINITE GROUPS XVIII, 1 Ifp:G AutR(E) andp': G AutR(E')arerepresentations ofGonE and E'respectively, then wedefine their sum pEBp'tobetherepresentation onthedirect sum EEBE',with a-EGacting componentwise. Observe that G-iso- morphism classes ofrepresentations have anadditive monoid structure under this direct sum, and also have anassociative multiplicative structure under the tensor product. With thenotation ofrepresentations,wedenote thisproduct by p0p'.This product isdistributive with respect totheaddition (direct sum). IfGisafinite group, and EisaG-module, then we can define thetrace TrG: E Ewhich isanR-homomorphism, namely TrG(x)=L(JX. tTEG We observe that TrG(x) lies ininvG(E), i.e. isfixed under theoperation of allelements ofG.This isbecause tTrG(x)=Ltax, tTEG andmultiplying bytontheleftpermutes theelements ofG. Inparticular, iff:E FisanR-homomorphism ofG-modules, then TrG(f): E FisaG-homomorphism. Proposition 1.1. Let Gbeafinite group and letE',E,F,F'beG-modules. Let E' E F!.F' beR-homomorphisms, and assume that cp, «/1areG-homomorphisms. Then TrG(t/J0f0qJ)=t/J0TrG(f)0qJ. Proof Wehave TrG(t/J0f0qJ)=L(J(t/J0f0qJ)=L(at/J)0(af)0(aqJ) tTEG tTEG =t/J0(Laf)0qJ=t/J0TrG(f)0qJ. tTEG Theorem 1.2. (Maschke). Let Gbeajinite group oj'order n,and letkbea field whose characteristic does notdivide n.Then thegroup ringk[G] is semisimple. Proof Let Ebe aG-module, and FaG-submodule. Since kisafield, there exists ak-subspace F'such that Eisthek-direct sum ofFand F' .We let thek-linear map n:E Fbetheprojection onF.Then n(x)=xforallxEF. XVIII, 2 CHARACTERS 667 Let 1 lp=-TrG(n). n We have then twoG-homomorphisms oFbE qJ such thatjistheinclusion, and lp0j=ideItfollows that EistheG-direct sum ofFand Ker lp,thereby proving that kEG] issemisimple. Except in7wedenote byGafinite group, and wedenote E,Ffinite dimensional k-spaces, where kisafield ofcharacteristic notdividing #(G). Weusually denote #(G) by n. 2. CHARACTERS Let p:kEG] Endk(E) be arepresentation. Bythe character Xpofthe representation, weshall mean thek-valued function Xp:kEG] k such thatXp(rx)=trp(rx) forall rxEkEG]. The trace here isthetrace ofanendo- morphism,asdefined inChapter XIII, 93.Ifweselect abasis forEover k,itis the trace ofthematrix representing p(rx), i.e.,the sum ofthediagonal elements. Wehave seenpreviously that thetrace does notdepend onthechoice ofthebasis. Wesometimes write XEinstead ofXp. We also call Etherepresentation space ofp. Bythetrivial character weshall mean thecharacter oftherepresentation of Gonthek-space equal tokitself, such that ax=xforallxEk.Itisthefunction taking thevalue 1onallelements ofG.Wedenote itbyXooralso byIGifwe need tospecify thedependence onG. We observe that characters arefunctions onG,and that the values ofa character onelements ofkEG] aredetermined byitsvalues onG(the extension from Gtok[G]being byk-linearity). We saythat tworepresentations p,lpofGonspaces E,Fareisomorphic if there isaG-isomorphism between Eand F.We then seethat ifp,lpareiso- morphic representations, then their characters areequal. (Put inanother way, ifE,Fare G-spaces and areG-isomorphic, thenXE=XF.) Ineverything that follows, we areinterested only inisomorphism classes ofrepresentations. 668 REPRESENTATIONS OFFINITE GROUPS XVIII, 2 IfE,Fare G-spaces, then their direct sum E(f)Fisalso aG-space, theopera- tion ofGbeing componentwise. Ifx(f)yEE(f)Fwith xEEand YEF,then (J(x (f)y)=ax(f)ay. Similarly, the tensor product E(8)kF=E(8)FisaG-space, theoperation ofGbeing given bya(x @y)=ax(8)ay. Proposition 2.1. IfE,Fare G-spaces, then XE+XF=XECfJFandXEXF=XE@F. IfXVdenotes thecharacter ofthedual representationonEV ,then XV(a-)=X(a--1) X(a-)ifk=C. Proof The first relation holds because thematrix ofanelement ainthe representation E(f)Fdecomposes into blocks corresponding totherepresenta- tion inEand therepresentation inF.Astothesecond, if{va isabasis ofEand {wj}isabasis ofFover k,then weknow that {Vi(8) Wj}isabasis [email protected] (aiv)bethematrix ofawith respect toour basis ofE,and (bhJitsmatrix with respect toour basis ofF.Then a(Vi @wj)=(JVi(8)aWj=Laivvv (8)LbjJJw JJ v JJ =LaivbjJJVv(8) W JJ. V,JJ Bydefinition, wefind XE@F(a)=LLaiibjj=XE(a)XF(a), ij thereby proving thestatement about tensor products. The statement forthechar- acter ofthedual representation follows from theformula forthematrix tM-1 given in 1.The value givenasthecomplex conjugate incase k=Cwill be proved later inCorollary 3.2. Sofar, wehave defined thenotion ofcharacter associated with arepresenta- tion. Itisnow natural toform linear combinations ofsuch characters with more general coefficients than positive integers. Thus byacharacter ofGweshall mean afunction onGwhich can bewritten asalinear combination ofcharacters ofrepresentations with arbitrary integer coefficients. The characters associated with representations will becalled effective characters. Everything wehave defined ofcourse dependsonthefield k,and weshall add over ktoourexpressions ifweneed tospecify thefield k. XVIII, 2 CHARACTERS 669 We observe that thecharacters form aring inview ofProposition 2.1. For most ofourwork wedonotneed themultiplicative structure, only theadditive one. Byasimple orirreducible character ofGone means thecharacter of a simple representation (Le., thecharacter associated with asimple k[G]-module). Taking into account The.orem 1.2, and theresults ofthepreceding chapter concerning the structure ofsimple andsemisimple modules over asemisimple ring (Chapter XVII, 4) weobtain: Theorem 2.2. There areonlyafinite number ofsimple characters ofG (over k).The characters ofrepresentations ofGarethelinear combinations ofthe simple characters with integer coefficients>o. We shall use thedirect product decomposition ofasemisimple ring. We have s kEG]=nRi i= 1 where each Riissimple, and wehave acorresponding decomposition oftheunit element ofk[G]: 1=el+...+es, where eiistheunit element ofRi,andeiej=0ifi=Ij.Also, RiRj=0ifi=f.j. We note that s=s(k)depends onk. IfLidenotes atypical simple module forRi(sayoneofthesimple leftideals), weletXibethecharacter oftherepresentationonLi. Weobserve that Xi(ex)=Oforall exERjifi=Ij.This isafundamental relation oforthogonality, which isobvious, butfrom which allourother relations will follow. Theorem 2.3. Assume that khascharacteristic O.Then every effective char- acter has aunique expressionasalinear combination s X=LniXi, i=1niEZ,ni>0, where Xl'...,Xsarethesimple characters ofGover k.Two representationsare isomorphic ifandonlyiftheir associated characters areequal. 670 REPRESENTATIONS OFFINITE GROUPS XVIII, 2 Proof. Let Ebetherepresentation space ofX.Then byTheorem 4.4of Chapter XVII, s EE9niLi. i= 1 The sum isfinite because we assume throughout that Eisfinite dimensional. Since eiacts asaunit element onLi,wefind Xi(ei)=dimkLi. We have already seen thatXi(ej)=0ifi=1=j.Hence x(ei)=nidimkLi. Since dimk Lidepends only onthe structure ofthegroup algebra,wehave recovered themultiplicitiesnb...,ns.Namely, niisthenumber oftimes that Lioccurs (up toanisomorphism) intherepresentation space ofX,and isthe value ofx(ei) divided bydimkLi(we areincharacteristic 0).This proves our theorem. As amatter ofdefinition, inTheorem 2.3 wecall nithemultiplicity ofXiinX. Inboth corollaries, wecontinue toassume that khascharacteristic O. Corollary 2.4. Asfunctions ofGinto k,thesimple characters Xl'..·,Xs arelinearly independent over k. Proof. Suppose thatLaiXi=0with aiEk.Weapply thisexpression toej and get o=(LaiXi)(ej)=ajdimkLj' Hence aj=0forallj. Incharacteristic 0wedefine thedimension ofaneffective character tobe thedimension oftheassociated representation space. Corollary 2.5. Thefunction dim isahomomorphism ofthemonoid ofeffective characters into Z. XVIII, 3 1-DIMENSIONAL REPRESENTATIONS 671 Example. Let Gbe acyclic group oforder equal toaprime number p. Weform thegroup algebra Q[G]. Let (Jbeagenerator ofG.Let I2 p-l+(J+a +...+a e2= 1-el. el= p Then Tel=elforanyTEGand consequently ei=et.Itthen follows that e=e2and ele2=O.The field Qet isisomorphic toQ.Let w =ae2. Then wp=e2. LetQ2=Qe2.Since w=Ie2,and satisfies theirreducible equation Xp-1+...+ 1=0 over Q2' itfollows that Q2(W) isisomorphic tothefield obtained byadjoining aprimitive p-th root ofunity totherationals. Consequently, Q[G] admits the direct product decomposition Q[G] QxQ«() where (isaprimitive p-th root ofunity. Asanother example, letGbeany finite group, and let 1 el= -La. ntTEG Then foranyTEGwehave Tel=el,andei=el.Ifwelete'l=1-elthen e,?=e'l'ande'lel=ele'l=O.Thus forany field k(whose characteristic does notdivide theorder ofGaccording toconventions inforce), we seethat kEG]=kelxk[G]e'l isadirect product decomposition. Inparticular, therepresentation ofGonthe group algebra k[G]itself contains aI-dimensional representationonthe component kel,whose character isthetrivial character. 3. 1-DIMENSIONAL REPRESENTATIONS Byabuse oflanguage,even incharacteristic p>0,wesaythat acharacter is I-dimensional ifitisahomomorphism G k*. Assume that EisaI-dimensional vector space over k.Let p:G Autk(E) bearepresentation. Let{v}beabasis ofEover k.Then foreach aEG,wehave av =X«(J)v 672 REPRESENTATIONS OFFINITE GROUPS XVIII, 3 for some element X«(J)Ek,andx(a) =I0since ainduces anautomorphism ofE. Then for! EG, !(JV =X«(J)!V=x(a)x('r)v=X«(JT)V. We seethat X:G k*isahomomorphism, and that our I-dimensional char- acter isthe same type ofthing that occurred inArtin's theorem inGalois theory. Conversely, letX:G k*beahomomorphism. Let EbeaI-dimensional k-space, with basis {v}, and define a(av)=ax(a)vforall aEk.Then we see at once that thisoperation ofGonEgivesarepresentation ofG,whose associated character isX. Since Gisfinite, wenote that X«(J)"=x(a")=X(I)=1. Hence the values ofI-dimensional characters are n-th roots ofunity. The I-dimensional characters form agroup under multiplication, and when Gisa finite abelian group,wehave determined itsgroup ofI-dimensional characters inChapter I,9. Theorem 3.1. Let Gbe afinite abeUan group, and assume that kisalge- braically closed. Then every simple representation ofG isI-dimensional. The simple characters ofGarethehomomorphisms ofGinto k*. Proof. The group ring k[G] issemisimple, commutative, and isadirect product ofsimple rings. Eah simple ring isaring ofmatrices over k(byCorollary 3.6Chapter XVII), and can becommutative ifandonly ifitisequal tok. For every I-dimensional character XofGwehave X«(J)-1=X«(J-l). Ifkisthefield ofcomplex numbers, then X«(J)=X«(J)-1=x(a- 1). Corollary 3.2. Let kbealgebraically closed. Let Gbeafinite group. For anycharacter Xand (JEG,thevalue X«(J) isequal toasumofroots ofunity with integer coefficients (i.e. coefficients inZorZlpZ depending onthechar- acteristic ofk). Proof. LetHbethesubgroup generated bya.Then Hisacyclic subgroup. Arepresentation ofGhaving character Xcan beviewed asarepresentation for Hbyrestriction, having the same character. Thus our assertion follows from Theorem 3.1. XVIII, 4 THE SPACE OF CLASS FUNCTIONS 673 4. THE SPACE OF CLASS FUNCTIONS Byaclass function ofG(over k,orwith values ink), weshall mean afunction f:G ksuch thatf(a-Ta--1)=f(T)forall a-, TEG.Itisclear that characters areclass functions, because forsquare matrices M,M' wehave tr(MM'M-1)=tr(M'). Thus aclass function may beviewed asafunction onconjugacy classes. We shall always extend thedomain ofdefinition ofaclass function tothe group ring, bylinearity. If rx=Laqa, qEG andfisaclass function, wedefine f(rx)=Laqf«(J). qEG Let aoEG.IfaEG,wewrite a aoif(Jisconjugate toao,that is,ifthere exists anelement tsuch that (J0=t(Jt- 1.Anelement ofthegroup ring oftype y=L(J q-qo will also becalled aconjugacy class. Proposition 4.1. Anelement ofk[G] commutes with every element ofGif andonlyifitisalinear combination ofconjugacy classes with coefficients ink. Proof Let rx=Laqa and assume rxt =trxforalltEG.Then qEG Laqt(Jt-l=Laqa. qEG qEG Henceaqo=aqwhenever (Jisconjugate to(Jo,and this means that wecanwrite rx=LayY y where the sum istaken over allconjugacy classes y. Remark. We note that theconjugacy classes infact form abasis ofthe center ofZ[G] over Z,and thus playa universal role inthetheory ofrep- resentations. We observe that theconjugacy classes arelinearly independentover k, and form abasis forthe center ofk[G]over k. 674 REPRESENTATIONS OFFINITE GROUPS XVIII, 4 Assume for the restofthis section that kisalgebraically closed. Then s kEG]=nRi i=1 isadirect product ofsimple rings, and each Riisamatrix algebraover k.Ina direct product, thecenter isobviously theproduct ofthe centers ofeach factor. Let usdenote bykitheimage ofkinRi,inother words, k.=ke.I n where eiistheunit element ofRi.Then the center ofk[G]isalso equal to s nkj i=1 which iss-dimensional over k. IfLiisatypical simple leftideal ofRi,then RiEndk(L i). Welet di=dimk Li. Then s dl=dimkRiand Ldl=n. i=1 We also have thedirect sum decomposition R. Ldi)I I asa(G,k)-space. The above notation will remain fixed from now on. We can summarize some ofour results asfollows. Proposition 4.2. Letkbealgebraically closed. Then thenumber ofconjugacy classes ofG isequal tothenumber ofsimple characters ofG, bothofthese being equal tothenumber sabove. The conjugacy classes Yl,...,Ysand theunit elements el'...,esform bases ofthe center ofk[G]. The number ofelements inYiwill bedenoted byhi.The number ofelements inaconjugacy class Ywill bedenoted byhy.Wecallitthe class number. The center ofthegroup algebra will bedenoted byZk(G). XVIII, THE SPACE OF CLASS FUNCTIONS 675 We can view kEG]asaG-module. Itscharacter will becalled theregular character, andwill bedenoted byXregorrGifweneed tospecify thedependence onG.Therepresentation onkEG] iscalled theregular representation. From our direct sum decomposition ofkEG] weget s Xreg=LdiXi. i= 1 Weshall determine thevalues oftheregular character. Proposition 4.3. LetXregbetheregular character. Then Xreg«(J)=0if(JEG, (J=I1 Xreg(l)=n. Proof Let 1=(Jl,...,(Inbetheelements ofG.They form abasis ofkEG] over k.The matrix of 1istheunit nxnmatrix. Thus our second assertion follows. If(J=I1,then multiplication by(Jpermutes(Jl'...,(Jn'and itisim- mediately clear that alldiagonal elements inthematrix representing(Jare O. This proves what wewanted. We observe that wehave two natural bases forthe center Zk(G) ofthe group ring. First, theconjugacy classes ofelements ofG.Second, theelements el,...,es(i.e. theunit elements oftherings Ri).We wish tofind therelation between these, inother words, wewish tofind thecoefficients ofeiwhen ex- pressed interms ofthegroup elements. The next proposition does this. The values ofthese coefficients will beinterpreted inthe next section asscalar products. This willclarify their mysterious appearance. Proposition 4.4. Assume again that kisalgebraically closed. Let ei=Latt, tEGatEk. Then 1_1di-1at=-Xreg(eit)= -Xi(t ). n n Proof We have foralltEG: Xreg(eir:-1)=Xre g(Lauar:-l )=LaUXreg«(Jr:-l). UEG uEG 676 REPRESENTATIONS OFFINITE GROUPS XVIII, 4 ByProposition 4.3, wefind xreg(eit- 1)=nat. Ontheother hand, s Xreg(eit-l)=LdjXj(eit-l)=diXi(ei!-l)=dixl t-l). j=1 Hence diXi(t- 1)=nat foralltEG.This provesourproposition. Corollary 4.5. Each eican beexpressed interms ofgroup elements with coefficients which lieinthefield generated over theprime field bym-th roots ofunity, ifm isanexponent for G. Corollary 4.6. The dimensions diare notdivisible bythecharacteristic ofk. Proof Otherwise, ei=0,which isimpossible. Corollary 4.7. The simple charactersXb. . .,Xsarelinearly independent over k. Proof The proof inCorollary 2.4applies, since we now know that the characteristic does notdivide di. Corollary 4.8. Assume inaddition that khascharacteristic O.Then d;jn foreach i. Proof. Multiplyingourexpression for eibynidi' and alsobye;,wefind n-1 dei= i...JXi((J)aei. i tTEG Let(beaprimitive m-th root ofunity, and letMbethemodule over Zgen- erated bythefinite number ofelements (Vaei(v=0,...,m-1and aEG). Then from thepreceding relation, we see atonce that multiplication bynidi maps Minto itself. Bydefinition, weconclude that nidi isintegralover Z, and hence liesinZ,asdesired. Theorem 4.9. Let kbealgebraically closed. LetZk(G) bethe center of kEG], and letXk(G) bethek-space ofclass functions onG.Then Zk(G) and Xk(G) arethedual spaces ofeach other, under thepairing (I,rx) f(rx). XVIII,9 5 ORTHOGONALITY RELATIONS 677 The simple characters and theunit elements et,...,esform orthogonal bases toeach other. We have X.(e.)=..d.I] I) I. Proof. The formula has been proved intheproof ofTheorem 2.3. The two spaces involved here both have.;: dimension s,and d;=f.0ink.Our prop- osition isthen clear. 5. ORTHOGONALITY RELATIONS Throughout thissection, weassume that kisalgebraically closed. IfRisasubring ofk,wedenote byXR(G) theR-module generated over R bythecharacters ofG.Itistherefore themodule offunctions which arelinear combinations ofsimple characters with coefficients inR.IfRistheprime ring (i.e.theintegers Zortheintegers mod pifkhascharacteristic p),then wedenote XR(G) byX(G). We shall now define abilinear maponX(G) xX(G). Iff,9EX(G), we define 1 <f,g)= -Lf(a)g((J-1).nO'EG Theorem 5.1. Thesymbol <1,g)forf,9EX(G) takes onvalues intheprime ring. Thesimple characters form anorthonormal basisfor X(G), inother words <XhXj)=ij. For each ring Rck,thesymbol has aunique extension toanR-bilinear form XR(G)xXR(G) R,given bythe same formula asabove. Proof ByProposition 4.4, wefind di-1) Xj(ei)= - i...JXi(a Xj«(J).nO'EG Ifi=Ijweget0ontheleft-hand side, sothatXiandXjareorthogonal. Ifi=j wegetdiontheleft-hand side, and weknow that di=f.0ink,byCorollary 4.6. Hence <XhXi)=1.Since every element ofX(G) isalinear combination of simple characters with integer coefficients, itfollows that the values ofour bilinear mapareintheprime ring. The extension statement isobvious, thereby proving our theorem. 678 REPRESENTATIONS OFFINITE GROUPS XVIII, 5 Assume that khascharacteristic O.Let mbeanexponent forG,and letR contain them-th roots ofunity. IfRhas anautomorphism oforder 2such that itseffect on aroot ofunity is,,-1,then weshall call such anautomorphism aconjugation, and denote itbya a. Theorem 5.2. Let khave characteristic 0,and letRbeasubring containing them-th roots ofunity, andhaving aconjugation. Then thebilinear form on X(G) has aunique extension toahermitian form XR(G) xXR(G) R, given bytheformula 1 - <1,g)=-Lf«(J)g(a).nUEG The simple characters constitute anorthonormal basis ofXR(G) with respect tothisform. Proof The formula given inthe statement ofthetheorem gives the same value asbefore forthesymbol <1,g)when1, glieinX(G). Thus theextension exists, and isobviously unique. We return tothe case when khasarbitrary characteristic. LetZ(G) denote the additive group generated bytheconjugacy classes Yl,...,Ysover theprime ring. Itisofdimension s.Weshall define abilinear map onZ(G)xZ(G).Ifrx=Lau(Jhascoefficients intheprime ring,wedenote by rJ..-theelement Laua- 1. Proposition 5.3. For rx,pEZ(G),wecandefine asymbol <rx,P)byeither one ofthefollowing expressions, which areequal: liS <rx,P)=-Xreg(rxP-)=-LXv(rx)Xv(P-).n nv= 1 The values ofthesymbol lieintheprime ring. Proof Each expression islinear initsfirst and second variable. Hence toprove their equality, itwill suffice toprove that thetwoexpressionsareequal when wereplacerxbyeiand Pbyanelement tofG.Butthen, ourequality is equivalent to S Xreg(eit-l)=LXv(ei)Xv(t-l). v== 1 Since Xv(ei)=0unless v=i,we seethat theright-hand side ofthis lastrelation isequaltodiXi(T-1).Our twoexpressions areequal inview ofProposition 4.4. XVIII, 5 ORTHOGONALITY RELATIONS 679 The fact that thevalues lieintheprime ring follows from Proposition 4.3: The values oftheregular character ongroup elements areequal to0orn,and hence incharacteristic 0,areintegers divisible byn. Aswith XR(G), we usethenotation ZR(G) todenote theR-module generated byYl,".' Ysover anarbitrary subring Rofk. Lemma 5.4. Foreach ring Rcontained ink,thepairing ofProposition 5.3 has aunique extension toamap ZR(G)xZ(G) R which isR-linear initsfirst variable. IfRcontains them-th roots ofunity, where misanexponent forG,and also contains Iln, then eiEZR(G)for alli. The class number hiisnotdivisible bythecharacteristic ofk,and wehave s1 ei=L(ej,Yv)- hYV. v=l v Proof We note that hiisnotdivisible bythecharacteristic because itis theindex ofasubgroup ofG(theisotropy group ofanelement inYiwhen G operates byconjugation), and hence hidivides n.The extension ofourpairing asstated isobvious, since )'1'...,)'Sform abasis ofZ(G)over theprime ring. Theexpression ofeiinterms ofthis basis isonlyareinterpretation ofProposition 4.4interms ofthepresent pairing. Let Ebeafree module over asubring Rofk,and assume that wehave a bilinear symmetric (orhermitian) form onE.Let{Vb...,vs}beanorthogonal basis forthis module. If v=a1v1+...+asvs with aiER,then wecall a1,...,astheFourier coefficients ofvwith respect to our basis. Interms oftheform, these coefficients aregiven by (V,Vi) a.= I(Vi'Vi) provided (Vi'Vi) #-O. We shall seeinthe next theorem that theexpression for eiinterms of Yb...,YsisaFourier expansion. Theorem 5.5. The conjugacy classes Yl, ..., Ysconstitute anorthogonal basis forZ(G). We have <Yi'Yi)=hi'For each ring Rcontained ink,the bilinear mapofProposition 5.3has aunique extension toaR-bilinear map ZR(G) xZR(G) R. 680 REPRESENTATIONS OFFINITE GROUPS XVIII, 5 Proof We usethelemma. Bylinearity, theformula inthelemma remains valid when wereplace Rbyk,and when wereplace eibyany element ofZk(G), in particular when wereplace eibyfi.But{ft,...,Ys}isabasis ofZk(G), over k. Hence wefind that <Y;,Yi)=hiand <Yi,Yj)=0ifi=f.j,aswas toshown. Corollary 5.6. IfGiscommutative, then 1n _1{oif aisnotequaltot -LXv«(J)Xv(t )= 1.f. I nv=1I (JISequa to 'C. Proof. When Giscommutative, each conjugacy class hasexactly one ele- ment, and thenumber ofsimple characters isequal totheorder ofthegroup. Weconsider the case ofcharacteristic 0forourZ(G) justaswedidforX(G). Let khave characteristic 0,and Rbeasubring ofkcontaining them-th roots of unity, andhavingaconjugation. Let (1.=Laa(Jwith aaER.Wedefine aEG a=LQaa-1 . aEG Theorem 5.7. Let khave characteristic 0,and letRbeasubring ofk,con- taining them-th roots ofunity, andhaving aconjugation. Then thepairing .of Proposition 5.3has aunique extension toahermitian form ZR(G)xZR(G)R given bytheformulas lIS_ <(1.,P)=-Xreg«(1.p)=-LXv«(1.)Xv(P). n nv= 1 The conjugacy classes Yl, ...,Ysform anorthogonal basis forZR(G). IfR contains Iln,then el,.. .,eslieinZR(G)and alsoformanorthogonal basisfor ZR(G). Wehave <ei,ei)=df/n. Proof The formula given inthe statement ofthetheorem gives the same value asthesymbol <(1.,P)ofProposition 5.3when (1.,PlieinZ(G). Thus the extension exists, and isobviously unique. Using thesecond formula inPropo- sition 5.3,defining thescalar product, andrecalling thatXv(ei)=0ifv=f.i,we seethat 1 - <e;,ei)=-Xi(ei)Xi(ei),n whence our assertion follows. XVIII, 5 ORTHOGONALITY RELATIONS 681 Weobserve that theFourier coefficients ofeirelative tothebasis rl,...,rs arethe same with respect tothebilinear form ofTheorem 5.5, orthehermitian form ofTheorem 5.7. This comes from thefact that rb...,YslieinZ(G), and form abasis ofZ(G)over theprime ring. Weshall now reprove andgeneralize theorthogonality relations byanother method. LetEbeafinite dimensional (G,k)-space,sowehave arepresentation G Autk(E). After selectingabasis ofE,wegetarepresentation ofGbydxdmatrices. If {Vl' ..., Vd}isthebasis, then wehave thedual basis {Ab ..., A.d}such that Ai(Vj)=ij.Ifanelement aofGisrepresented byamatrix (pij(a),then each coefficientPij«(J)isafunction ofa,called theij-coefficient function. We canalso write pij(a)=Aj(avi). But instead ofindexing elements ofabasis orthedual basis, wemay justas well work with any functional AonE,and any vector v.Then wegetafunction a A(av)=PA,v(a), which will also becalled acoefficient function. Infact, one canalways complete v=Vltoabasis such that A=Alisthefirst element inthedual basis, butusing thenotation PA,visinmany respectsmore elegant. Weshall constantly use: Schur's Lemma. LetE,Fbesimple (G,k)-spaces, and let qJ:EF beahomomorphism. Then either qJ=0orqJisanisomorphism. Proof Indeed, thekernel ofq>and theimage ofqJaresubspaces,sothe assertion isobvious. We usethe same formula asbefore todefine ascalar product onthespace of allk-valued functions onG,namely 1 <f,g)=-Lf(a)g(a-1).nO'EG Weshall derive various orthogonality relations among coefficient functions. Theorem 5.8. LetE,Fbesimple (G,k)-spaces. Let Abeak-linear functional onE,letxEEand YEF.IfE,Fare notisomorphic, then IA(ax)a-1 Y=o. O'EG 682 REPRESENTATIONS OFFINITE GROUPS XVIII, 5 IfJ1isafunctional onFthen thecoefficient functions PA,xandPjj,yare ortho- gonal, that is LA«(JX)J1«(J-1y)=o. tTEG Proof. The map xLA«(JX)(J-1yisaG-homomorphism ofEinto F,so Schur's lemma concludes theproof ofthefirst statement. The second comes by applying thefunctionalJ1. As acorollary,we seethatifX, «/1aredistinct irreducible characters ofG over k,then (X, «/1)=0, that isthecharacters areorthogonal. Indeed, thecharacter associated with a representation Pisthe sum ofthediagonal coefficient functions, d X=LPii' i=1 where disthedimension oftherepresentation. Two distinct characters cor- respond tonon-isomorphic representations,sowe canapply Proposition 5.8. Lemma 5.9. Let Ebeasimple (G,k)-space. Then anyG-endomorphism of Eisequal toascalar multiple oftheidentity. Proof. The algebra EndG,k(E)isadivision algebra bySchur's lemma, and isfinite dimensional over k.Since kisassumed algebraically closed, itmust beequal tokbecause any element generates acommutative subfield over k. This proves thelemma. Lemma 5.10. Let Ebearepresentation space for Gofdimension d.Let A beafunctional onE,and letxEE.Letq>A,xEEndk(E) betheendomorphism such that q>A,x(Y)=A(Y)X. Thentr(qJ;.,x)=A(X). Proof. Ifx=0the statement isobvious. Let x=IO.IfA(X) =f.0wepick abasis ofEconsisting ofxand abasis ofthekernel ofA.IfA(X)=0,wepick a basis ofEconsisting ofabasis forthekernel ofA,and one other element. In either case itisimmediate from thecorresponding matrix representing qJA,xthat the trace isgiven bytheformula asstated inthelemma. Theorem 5.11. Letp:G-+Autk(E) be asimple representation ofG,of dimension d.Then thecharacteristic ofk does notdivide d.Let x,yEE.Then foranyfunctionals A,J1onE, nLA(crX)J1(cr-lY)= dA.(Y)J1(x). tTEG XVIII, 5 ORTHOGONALITY RELATIONS 683 Proof Itsuffices toprove that nLl(o-x)q-1y= dl(y)x. tTEG For fixed ythemap xLl«(JX)(J-1y tTEG isimmediately verified tobeaG-endomorphism ofE,soisequal tocIfor some cEkbyLemma 5.9. Infact, itisequal to Lp((J- 1)0lpA,y0p((J). tTEG The trace ofthisexpression isequal ton.tr(lpA,Y) byLemma 5.10, and also todc. Taking A,ysuch thatl(y)=1shows that thecharacteristic doesntdivide d, and then wecan solve for casstated inthetheorem. Corollary 5.12. LetXbethecharacter oftherepresentation ofGonthe simple space E.Then <X,X>=1. Proof This follows immediately from thetheorem, and theexpression of Xas X=PI 1+.. .+Pdd. Wehave now recovered thefactthat thecharacters ofsimple representations areorthonormal. Wemay then recover theidempotents inthegroup ring, that is,ifXl'...,Xsarethesimple characters, wemay now define di -1ei= - I..JXi((J)(J. ntTEG Then theorthonormality ofthecharacters yields theformulas: s Corollary 5.13.x;(ej)=ijdiandXreg=LdiXi' i=1 Proof The first formula isadirect application oftheorthonormality ofthe characters. The second formula concerning theregular character isobtained bywriting Xreg=LmjXj j 684 REPRESENTATIONS OFFINITE GROUPS XVIII, 5 with unknown coefficients. Weknow thevaluesXreg(l)=nandXreg«(J)=0if (J=I1.Taking thescalar product ofXregwithXifori=1,...,simmediately yields thedesired values forthecoefficients mj' Since acharacter isaclass function, one seesdirectly that each eiisalinear combination ofconjugacy classes, and soisinthecenter ofthegroup ringk[G]. Now letEibearepresentation space ofXi'and letPibetherepresentation ofGorkEG] onEi.For exEkEG] weletPi(ex): Ei Eibethemap such that Pi(ex)X=exxforallxEEi. Proposition 5.14. We have p;(ei)=idaf1:d pi(ej)=0ifi=1=j. Proof The map x e;xisaG-homomorphism ofE;into itself since eiisin the center ofkEG]. Hence byLemma 5.9this homomorphism isascalar multiple oftheidentity. Taking thetrace andusing theorthogonality relations between simple characters immediately gives thedesired value ofthis scalar. We now find that s Lei=1 i=1 because thegroup ring kEG] isadirect sum ofsimple spaces, possibly with multiplicities, and operates faithfully onitself. Theorthonormality relations also allow ustoexpandafunction inaFourier expression, relative tothecharacters ifitisaclass function, and relative tothe coefficient functions ingeneral. We state this intwo theorems. Theorem 5.15. Letfbeaclass function onG.Then s f=L<f,Xi)Xi. i=1 Proof The number ofconjugacy class isequal tothenumber ofdistinct characters, and these arelinearly independent, sothey form abasis fortheclass functions. The coefficients aregiven bythestated formula, asone seesbytaking thescalar product offwith any characterXjandusing theorthonormality. Theorem 5.16. Letp(i)be amatrix representation ofGonEirelative toa choice ofbasis, andletpi! Jlbethecoefficient functions ofthismatrix, i=1,..., s and v,J1=1,..., dieThen thefunctions p!Jlformanorthogonal basisforthe space ofaUfunctions onG,and hence foranyfunctionfonGwehave _1(i) (i)f-d<I,PV,JJ)PV,Jl' i=1v,Jli XVIII, 5 ORTHOGONALITY RELATIONS 685 Proof That thecoefficient functions form anorthogonal basis follows from Theorems 5.8and 5.11. Theexpression offinterms ofthis basis isthen merely thestandard Fourier expansion relative toany scalar product. This concludes theproof. Supposenow for concreteness that k=Cisthecomplex numbers. Recall that aneffective character Xisanelement ofX(G), such thatif s X=2:miXi i= 1 isalinear combination ofthesimple characters with integral coefficients, then wehave mi>0foralli.Inlight oftheorthonormality ofthesimple characters, wegetforallelements XEX(G)therelations s IIxII2=(X,X)=2:ml and mi=(X,Xi).i= 1 Hence weget(a)ofthe next theorem. Theorem 5.17. (a)LetXbeaneffective character inX(G). Then Xissimple over CifandonlyifIIXII2=1,oralternatively, 2: /X(a-) /2=#(G). aEG (b) Let X,«/1beeffective characters inX(G), and letE,Fbetheir representation spacesover C.Then (X, «/1)G=dimHomG(E, F). Proof. The first part has been proved, andfor(b), let «/1=LqiXi.Then by orthonormality,weget (X, «/1)G=Lmiqi. ButifEiistherepresentation space ofXiover C,then bySchur's lemma dim HomG(E i,Ei)=1and dim HomG(E i,Ej)=0fori=1=j. Hence dim HomG(E, F)=Lmiqi, thus proving (b). Corollary 5.18 With theabove notation and k=Cfor simplicity,wehave: (a) Themultiplicity ofIGinEVFisdimk invG(EvF). (b) The(G,k)-space Eissimple ifandonlyifIGhasmultiplicity1inEVE. Proof Immediate from Theorem 5.17 and formula (3)of91. Remark. The criterion ofTheorem 5.17(a) isuseful intesting whether a representation issimple. Inpractice, representationsareobtained byinducing from I-dimensional characters, and such induced representations dohave aten- dency tobeirreducible. We shall see aconcrete case in 12. 686 REPRESENTATIONS OFFINITE GROUPS XVIII, 6 6. INDUCED CHARACTERS The notation isthe same asinthepreceding section. However, wedon't need alltheresults proved there; allweneed isthebilinear pairingonX(G), and its extension to XR(G)xXR(G) R. The symbol < ,>may beinterpreted either asthebilinear extension, orthe hermitian extension according toTheorem 5.2. Let Sbeasubgroup ofG.We have anR-linear map called therestriction res:XR(G) XR(S) which toeach class function onGassociates itsrestriction toS.Itisaring- homomorphism. We sometimes letfs denote therestriction off toS. Weshall define amap intheopposite direction, ind:XR(S) XR(G), which wecall theinduction map. IfgEXR(S),weextend gtogsonGby letting gs(a-)=0ifa- S.Then wedefine theinduced function gG(u)=ind(g)(u)= (5 1)2:g/...TUT-1). .'rEG Then ind(g) isaclass function onG.Itisclear thatind¥isR-linear. Since wedeal with two groups Sand G,weshall denote thescalar product by< ,>sand < ,>Gwhen itistaken with these respective groups. The next theorem shows among other things that therestriction and transfer areadjoint toeach other with respect toourform. Theorem 6.1. Let Sbeasubgroup ofG. Then thefollowing rules hold: (i)(Frobenius reciprocity) ForfEXR(G), and 9EXR(S)wehave (ind(g), f)G=(g,Resfj(f»s. (ii)Ind(g)f=ind(gfs). (iii)1fT CSCGaresubgroups ofG, then ind0ind=ind¥. (iv)Ifa-EGand gUisdefined bygU(TU)=g(T), where TU=a--1Ta-, then ind¥(g)=ind<T(gU). (v)If«/1isaneffective character ofSthenindfj( «/1)iseffective. XVIII, 6 INDUCED CHARACTERS 687 Proof. Let usfirst prove (ii).We must show thatgGf=(gfs)G.We have (gGf)(T)= (S 1)L9S<UTU-I)!(T)= (S 1)L9S(UTU-I)!(UTU-I). .UEG.UEG The lastexpression just obtained isequalto(gfs)G, thereby proving (ii). Let us sum over! inG.Theonly non-zero contributions inourdouble sum will come from those elements ofSwhich can beexpressed intheform a!(J-l with (J,!EG. The number ofpairs «(J,!)such that (J!(J-1isequal toafixed element ofGis equal ton(because foreveryA.EG,«(JA., A- I!A.) isanother such pair, and the total number ofpairs isn2).Hence ourexpression isequal to 1 (G:1)(S:1);../(A)!(A). Our first rule then follows from thedefinitions ofthescalar products inGand S respectively. Now letg=«/1be aneffective character ofS,and letf=Xbe asimple character ofGFrom (i)wefind that theFourier coefficients ofgG areintegers >0because resy(x)isaneffective character ofS.Therefore thescalar product <«/1,resr (X» s is>O.Hence t/JGisaneffective character ofG,thereby proving (v). Inorder toprove thetransitivity property, itisconvenient tousethefol- lowing notation. Let{c}denote the setofright cosets ofSinG .For each right coset c,we select afixed coset representative denoted byc.Thus ifc1,...,Crare these representatives, then r G=Uc=USc =USCi. c c i==l Lemma 6.2. Let 9beaclass functiononS.Then r indff(g)()=EgS(CiC;-I). ;=1 Proof. We cansplit the sum over allaEGinthedefinition oftheinduced function into adouble sum r L=LL tTEG tTES i= 1 688 REPRESENTATIONS OFFINITE GROUPS XVIII, 7 and observe that each termgs(a-cc-la--I) isequal togS(cc-l) ifa-ES,because gisaclass function. Hence the sum over a-ESisenough tocancel thefactor 1/(S :1)infront, togive theexpression inthelemma. IfTcScGaresubgroups ofG,and if G=USCi and S=UTJj aredecompositions intoright cosets, then {ajcJform asystem ofrepresentatives fortheright cosets ofTinG.From this thetransitivity property (iii) isobvious. Weshall leave (iv)asanexercise (trivial, using thelemma). 7. INDUCED REPRESENTATIONS Let Gbe agroup and Sasubgroup offinite index. LetFbe anS-module. Weconsider thecategory ewhose objectsareS-homomorphisms cp:F Eof Finto aG-module E.(We note that aG-module Ecan beregardedasanS- module byrestriction.) Ifcp':F E'isanother object ine,wedefine amorphism cp' cpinetobeaG-homomorphism 17:E' Emaking thefollowing diagram commutati ve: E' jF"E Auniversal object ineisdetermined uptoaunique G-isomorphism. Itwill bedenoted by ind¥:F ind¥(F). Weshall prove below that auniversal object always exists. IfqJ:F Eisa universal object,wecall Eaninduced module. Itisuniquely determined, uptoa unique G-isomorphism makingadiagram commutative. Forconvenience, we shall select one induced module such thatqJisaninclusion. Weshall then call thisparticular module ind¥(F) theG-module induced byF.Inparticular, given anS-homomorphism cp:F Einto aG-module E,there isaunique G-homo- morphism cp*:ind(F)Emaking thefollowing diagram commutative: . ndGind¥ (F) 17 jF (()*=indf(({)E XVIII, 7 INDUCED REPRESENTATIONS 689 The association cpind(cp)then induces anisomorphism HomG(indr(F), E)=Homs(F, res(E», for anS-module Fand aG-module E.We shall seeinamoment thatind isa functor from Mod(S) toMod(G), and the above formula may bedescribed as saying that induction istheadjoint functor ofrestriction. One also calls this relation Frobenius reciprocity formodules, because Theorem 6.1(i)isa corollary. Sometimes, ifthereference toFasanS-module isclear, weshall omit the subscript S,and write simply indG(F) fortheinduced module. Letf: F' FbeanS-homomorphism. If cp:F' ind(F') isaG-module induced byF', then there exists aunique G-homomorphism indr(F') indr(F) making thefollowing diagram commutative: F Ij, FGlps )indy(F') ,, ,,, , .,.jindf(f) )indG(F)cps Itissimply theG-homomorphism corresponding tothe universal property fortheS-homomorphism qJ0f,represented byadashed line inourdiagram. Thusind isafunctor, from thecategory ofS-modules tothecategory ofG- modules. From theuniversality anduniqueness oftheinduced module, wegetsome formal properties: indcommutes with direct sums: Ifwehave anS-direct sum F F',then ind(F EBF')=ind(F) EBind(F'), thedirect sum ontheright being aG-direct sum. Iff, g:F' -+Fare S-homomorphisms, then indr(f+g)=ind(f)+indr(g). 1fTcScGaresubgroups ofG, andFisaT-module, then indy0ind(F)=ind¥(F). 690 REPRESENTATIONS OFFINITE GROUPS XVIII, 7 Inallthree cases, theequality between theleftmember and theright member ofourequations follows atonce byusing theuniqueness oftheuniversal object. Weshall leave theverifications tothereader. Toprove theexistence oftheinduced module, weletM(F) betheadditive group offunctions f:G Fsatisfying (Jf()=f«(J) for (JESand EG.Wedefine anoperation ofGonM(F) byletting «(Jf)()=f((J) for (J, EG.Itisthen clear thatM(F) isaG-module. Proposition 7.1. Let qJ:F-+M(F) besuch that qJ(x)=qJxisthemap ()={o iftFJS qJxt.f S tX 1tE . ThenqJisanS-homomorphism, qJ:FM(F) isuniversal, and qJisinjective. The image ofqJconsists ofthose elements fEM(F) such thatf(t)=°if tFJs. ProofLet (JESand xEF.Let tEG.Then «(JqJx)( t)=qJx(t(J). IftES,then this lastexpression isequal toqJux(t).IftFJS,then t(JFJS,and hence both qJux(t) and qJx(t(J) areequal toO.ThusqJisanS-homomorphism, anditisimmediately clear thatqJisinjective. Furthermore, iffEM(F) issuch thatf(t)=0if!FJS,then from thedefinitions, weconclude thatf=qJxwhere x=f(I). There remains toprove thatqJisuniversal. Todothis, weshall analyzemore closely thestructure ofM(F). r Proposition 7.2. Let G=USCibeadecomposition ofGintoright cosets. i=1 LetF1betheadditive group offunctions inM(F) having value 0atelements EG, FJS.Then r M(F)=EBCi-1Fl' i=1 thedirect sumbeing taken asanabelian group. Proof ForeachfE M(F), leth bethefunction such that {o if FJSCiJi()= f():)l.f):S- ECi. XVIII, 7 INDUCED REPRESENTATIONS 691 For all (JESwehave fi«(Jc i)=(CifiX(J). Itisimmediately clear that Cifiliesin Fl,and r f=LCi-l(Cifi). i= 1 Thus M(F) isthe sum ofthesubgroups ci-1Fl'Itisclear that this sum is direct, asdesired. We note that{cII,. . .,C;-I}form asystem ofrepresentatives for theleft cosets ofSinG.Theoperation ofGonMb(F) isdefined bythepresceding direct sumdecomposition. We seethat Gpermutes thefactors transitively. The factor F1isS-isomorphic totheoriginal module F,asstated inProposition 7.1. Suppose that instead ofconsidering arbitrary modules, westart with acom- mutative ringRandconsider only R-modules Eonwhich wehave arepresentation ofG,Le. ahomomorphism G AutR(E), thus giving rise towhat wecall a (G,R)-module. Then itisclear that allourconstructions and definitions can be applied inthis context. Therefore ifwehave arepresentation ofSonanR-module F,then weobtain aninduced representation ofGonindy(F). Then wedeal with thecategory eofS-homomorphisms ofan(S,R)-module into a(G,R)-module. Tosimplify thenotation, wemay write "G-module" tomean "(G,R)-module" when such aring Renters asaring ofcoefficients. Theorem 7.3. Let{Ab...,Ar}beasystem ofleft coset representatives ofS in G.There exists aG-module Econtaining FasanS-submodule, such that r E=EBAiF i=1 isadirect sum (asR-modules). Letcp:F Ebetheinclusion mapping. Then cpisuniversal inour category e,i.e.Eisaninduced module. Proof Bytheusual set-theoretic procedure ofreplacing F1byFinM(F), obtain aG-module Econtaining FasaS-submodule, andhaving thedesired direct sum decomposition. Letq/:F E'be anS-homomorphism into a G-module E'.Wedefine h:E E' bytherule h(AIX l+...+ArXr)=Allp'(Xl) +...+Arlp'(Xr) forXiEF.This iswell defined since our sum forEisdirect. We must show that hisaG-homomorphism. Let (JEG.Then (JA.i=AO'(i)'!0',i where (J(i) issome index depending on (Jand i,andto', iisanelement ofS,also 692 REPRESENTATIONS OFFINITE GROUPS XVIII, 7 dependingon (J,i.Then h(aAi xi)=h(Au(i)tu,ixi)=Au(i)lp'('ru,ixi). Since q/isanS-homomorphism,we seethat thisexpression isequal to Au(i)tu,ilp'(Xi)=(Jh(Aixi). Bylinearity,weconclude that hisaG-homomorphism,asdesired. Inthe next propositionwereturn tothe case when Risour field k. Proposition 7.4. Let t/Jbethecharacter oftherepresentation ofSonthe k-space F.LetEbethespace ofaninduced representation. Then thecharacter XofEisequal totheinduced character .pG, i.e. isgiven bytheformula x()=Lt/JO(CC-l), c where the sum istaken over theright cosets cofS inG,Cisafixed coset repre- sentative for c,and t/J0istheextension oft/JtoGobtained bysetting t/Jo((J)=0 ifaftS. Proof Let{Wl,...,wm}beabasis forFover k.Weknow that E=EBC-1F. Let abeanelement ofG.The elements {c(J- lWj}c,jform abasis forEover k. Weobserve that caca-1isanelement ofSbecause SC(J =Sca =Scu . We have a(cu-1Wj)=C-l(ca cu- l)Wj. Let (caea- 1)Jlj bethecomponents ofthematrix representing theeffect ofc(Jca-1onFwith respect tothebasis {wb. ..,wm}.Then theaction ofaonEisgiven by (J(cu-1 Wj)=C-1L(caca-1)JljWJl Jl =L(c(Jca- l)Jlj(C- lWJl). Jl Bydefinition, x(a)=LL(ca c(J-l)jj. cu=cj XVIII, 7 INDUCED REPRESENTATIONS 693 But C(J=Cifandonly ifc(Jc-1ES.Furthermore, t/J(C(JC- 1)=L(C(JC-1)jj' j Hence x(a)=Lt/Jo(C(Jc-l), c aswas tobeshown. Remark. Having givenanexplicit description oftherepresentation space for aninduced character, wehave insome sense completed the more elementary partofthetheory ofinduced characters. Readers interested inseeinganapplication canimmediately read 12. Double eosets Let Gbe agroup and letSbe asubgroup. Toavoid superscriptswe use the following notation. Letl'EG .We write [y]S=ySy-l and S[y]=y-1Sy. Weshall suppose that Shasfinite index. We letHbe asubgroup. Asubset ofG oftheform HI'S iscalled adouble coset. Aswith cosets, itisimmediately verified that Gisadisjoint union ofdouble cosets .We let{I'} be afamily of double coset representatives,sowehave thedisjoint union G=UHI'S." For each l'wehave adecomposition intoordinary cosets H=UT..)H n[y]S), 'Ty where{T,,}isafinite family ofelements ofH,dependingon 1'. Lemma 7.5. The elements{T"y} formafamily ofleft coset representatives for SinG,.that is, wehave adisjoint union G=UT"YS.",'Ty Proof. First wehave byhypothesis G=UUT,,(Hn[y]S)yS," Ty and soevery element ofGcan bewritten intheform T,,1'SlY-II's 2=T"l'swith sl'S2,SES. Ontheother hand, theelementsT"l'represent distinct cosets ofS,because if T"yS=T,,'1"S,then l'=1",since theelements l'represent distinct double cosets, 694 REPRESENTATIONS OFFINITE GROUPS XVIII, 7 whenceTyand T"I'represent the same coset ofySy-I, and therefore areequal. This proves thelemma. LetFbe anS-module. Given yEG, wedenote by[y]F the[y]S-module such that forysy-IE[y]S, theoperationisgiven by ysy-I·[y]x=[y]sx. This notation iscompatible with thenotation thatifFisasubmodule ofaG- module E,then wemay form yFeither according totheformal definition above, oraccording totheoperation ofG.The two arenaturally isomorphic (essentially equal). We shall write [y]:F--+yF or[y]F fortheabove isomorphism from theS-module Ftothe[y]S-module yF.IfSI isasubgroup ofS,then byrestriction Fisalso anSI-module, and we use[y] also inthis context, especially forthesubgroup Hn[y]S which iscontained in [y]S. Theorem 7.6. Applied totheS-module F,wehave anisomorphism ofH- modules G.dG LD.dH [y]S[] resH0Ins=Q7InHn[y]s0resHn[y]S0y "I where thedirect sum istaken over double coset representatives y. Proof. The induced module ind¥(F)issimply thedirect sum ind(F)=E9TyyF y,Toy byLemma 7.5, which givesuscoset representatives ofSinG,and Theorem 7.3.Ontheother hand, foreach y,themodule E9TyyF Toy isarepresentation module fortheinduced representation from Hn[y]SonyF toH.Taking thedirect sum over y,wegettheright-hand side oftheexpression inthetheorem, and thus prove thetheorem. Remark. The formal relation ofTheorem 7.6 isone which occurred in Artin's formalism ofinduced characters andL-functions; cf.theexercises and [La70], Chapter XII,3. Forapplications tothecohomology ofgroups,see [La96]. The formalism alsoemerged inMackey's work [Ma51], [Ma53], which we shall now consider more systematically. The rest ofthis section isdue toMackey. For more extensive results and applications,see Curtis-Reiner [CuR 81], especially Chapter1.See also Exercises 15,16, and 17. Todeal more systematically with conjugations,wemake some general func- torial remarks. LetEbe aG-module. Possibly one may have acommutative ring Rsuch that Eisa(G,R)-module. We shall deal systematically with thefunctors XVIII, 7 INDUCED REPRESENTATIONS 695 HomG, EV ,and the tensor product. Let A:EAE byaR-isomorphism. Then interpreting elements ofGasendomorphisms ofE weobtain agroup AGA-IoperatingonAE. We shall also write [A]G instead of AGA-I.LetEI' E2be(G,R)-modules. LetAl:E; A;E; beR-isomorphisms. Then wehave anatural R-isomorphism (1)A2HomG(E I,E2)All=HomA2GXjl(AIEI,A2E2)' andespecially [A]HomG(E, E)=Hom[A]G(AE, AE). As aspecialcase ofthegeneral situation, letH,Sbesubgroups ofG,and let FI,F2be(H,R)- and (S,R)-modules respectively, and letu, TEG.Suppose that u-ITlies inthedouble coset D=HyS. Then wehave anR-isomorphism (2) Hom[0"]Hn[T]s([u]F1,[T]F 2)=HomHn['Y]s(f}, [y]F 2). This isimmediate byconjugation, writingT=uhys with hEH, sES,conjugating first with [ah]-l, and then observing that for sES,and anS-module F,we have [s]S=S,and[s-I]F isisomorphic toF.Inlight of(2), we see that the R-module ontheleft-hand side depends onlyonthedouble coset. Let Dbe a double coset .We shall use thenotation MD(F I,F2)=HomHn ['Y]S(FI'[y]F 2) where yrepresents thedouble coset D.With this notation wehave: Theorem 7.7.LetH, Sbesubgroups offinite index inG.LetFI'F2be (H,R)and (S,R)-modules respectively. Then wehave anisomorphism ofR- modules HomG(ind(FI)' ind¥(F 2»=EBMD(F I,F2), D where thedirect sum istaken over alldouble cosets HyS=D. Proof. We have theisomorphisms: HomG(ind(FI)' indf(F 2»=HomH(F I,resfi0indf(F 2» =EBHomH(F I,indJin['Y]s0res}J['Y]S0[y]F 2) 'Y =EBHomHn ['Y]s(F I,[y]F 2) 'Y byapplying thedefinition oftheinduced module inthefirst and third step, and applying Theorem 7.6 inthe second step. Each term inthe lastexpression is what wedenoted byMD(F I,F2)ifyisarepresentative forthedouble coset D. This proves thetheorem. 696 REPRESENTATIONS OFFINITE GROUPS XVIII, 7 Corollary 7.8. Let R=k=C.Let S,Hbesubgroups ofthefinite group G.LetD=HyS range over thedouble cosets, with representatives y.Let X beaneffective character ofHand «/1aneffective character ofS.Then <ind«x), ind(<</1»G=2:<X,[y]«/1)Hn['Y]S. 'Y Proof. Immediate from Theorem 5.17(b)and Theorem 7.7, taking dimen- sions ontheleft-hand side and ontheright-hand side. Corollary 7.9. (Irreducibility ofthe induced character). Let Sbe a subgroup ofthefinite group G.LetR=k=C.Let t/Jbeaneffective character ofS.Thenind (<</1) isirreducible ifandonlyif«/1isirreducible and <<</1,[y]«/1)sn['Y]S=0 forallyEG,y S. Proof. Immediate from Corollary 7.8 and Theorem 5.17(a). Itisofcourse trivial thatif«/1isreducible, then soistheinduced character. Another way tophrase Corollary7.9isasfollows. LetF,F'berepresentation spaces forS(over C). WecallF,F'disjoint ifnosimple S-spaceoccurs both inFand F'.Then Corollary 7.9 can bereformulated: Corollary 7.9'. Let Sbe asubgroup ofthefinite group G.Let Fbe an (S,k)-space (with k=C). Then ind(F) issimple ifandonlyifFissimple andforallyEGand y S,theSn[y]S-modules Fand[y]F aredisjoint. Next wehave thecommutation ofthedual and induced representations. Theorem 7.10. Let Sbeasubgroup ofG and letFbeafinite free R-module. Then there isaG-isomorphism ind(Fv)=(ind (F»v. Proof. LetG=UA;Sbe aleft coset decomposition. Then, asinTheorem 7 .3,we can express therepresentation space forind (F) as ind(F)=E9A;F. We may select Al=1(unit element ofG). There isaunique R-homomorphism f:FV(indy(F»v such that forcpEFVand xEFwehave {o ifi=1= 1 f(cp)(A;x)= ().f. 1cpX 1I= , which isinfact anR-isomorphism ofFVon(AIF)v.We claim that itisanS- XVIII, 7 INDUCED REPRESENTATIONS 697 homomorphism. This isaroutine verification, which wewrite down. We have {o ifi=1= 1 f([a-]cp)(A;x)= ((-1 ».f. 1 a-cpa- x11= . On theother hand, note that ifa-ESthen a--IAlESsoa--IAIXEAIF for xEF;butifa- S,then a--IA; Sfori=1= 1soa--IA;X AIF. Hence {o ifi=1= 1 [a-](f(cp»(Alx)=a-f(cp)(a--IA;X)= ((-I).f. 1 a-cpa- x11= . This proves thatfcommutes with theaction ofS. Bytheuniversal property oftheinduced module, itfollows that there isa unique (G,R)-homomorphism ind(f):ind(Fv) (ind(F»v , which must beanisomorphism becausefwasanisomorphismonitsimage, the AI-component oftheinduced module. This concludes theproof ofthetheorem. Theorems and definitions with Hom have analogues with thetensor product. We start with theanalogue ofthedefinition. Theorem 7.11. Let Sbe asubgroup offinite index inG.LetFbe anS- module, and EaG-module (over thecommutative ring R). Then there isan isomorphism indy(ress(E) 0F)=E0ind(F). Proof. The G-module ind(F) contains Fasasummand, because itisthe direct sumE9A;Fwith left coset representatives A;asinTheorem 7.3. Hence wehave anatural S-isomorphism f:ress(E) 0F E0AIF CE0ind(F). taking therepresentative Altobe 1(the unit element ofG).Bytheuniversal property ofinduction, there isaG-homomorphism ind(f):ind¥(ress(E) 0F) E0ind (F), which isimmediately verified tobeanisomorphism,asdesired. (Note that here itonly needed toverify thebijectivity inthis last step, which comes from the structure ofdirect sum asR-modules.) Before going further, wemake some remarks onfunctorialities. Supposewe have anisomorphism G=G', asubgroup HofGcorrespondingtoasubgroup H'ofG'under theisomorphism, and anisomorphism F=F'from anH-module FtoanH'-module F'commuting with theactions ofH,H'. Then wegetan isomorphism ind(F)=ind:(F'). 698 REPRESENTATIONS OFFINITE GROUPS XVIII, 7 Inparticular,wecould take a-EG,letG'=[a-]G=G,H'=[a-]H and F'=[a-]F. Next wedeal with theanalogue ofTheorem 7.7 .Wekeep the same notation asinthat theorem and thediscussion preceding it.With the two subgroups H and S,wemay then form the tensor product [a-]Fl 0[T]F 2 with a-, TEG.Supposea--1TED for some double coset D=HyS. Note that [a-]F 10[T]F 2isa[a-]H n[T]S-module. Byconjugationwehave anisomorphism (3) indfu]Hn[T]s([a-]F10[T]F 2)=indJ]n['Y]s (F 10[y]F 2). Theorem 7.12. There isaG-isomorphism ind(Fl) 0ind¥(F 2)=E9indn ['Y]s(F10[y]F 2), 'Y where the sum istaken over double coset representatives y. Proof. We have: ind(Fl) 0ind¥(F 2)=ind(Fl 0resHind¥(F 2» byTheorem 7.11 =E9ind(Fl 0indZn['Y]s resH n([ y]F 2) byTheorem 7.6 'Y =ind0ndZn[YIS (resf1n[YIS(Fj)0res}Js[YIS([ 'Y]F 2»))byTheorem 7.7 =E9indn ['Y]s(F10[y]F 2) bytransitivity ofinduction 'Y where weview Fln[y]F 2asanHn[y]S-module inthis last line. This proves thetheorem. General comment. This section hasgivenalotofrelations fortheinduced representations. Inlight ofthecohomology ofgroups, each formula may be viewed asgivinganisomorphism offunctors indimension 0,and therefore gives rise tocorresponding isomorphisms forthehigher cohomology groups Hq. The reader mayseethisdeveloped further than theexercises in[La96]. [CuR 81]Bibliography C.W. CURTIS and I.REINER, Methods ofRepresentation Theory, John Wiley and Sons, 1981 S.LANG, Topics incohomology ofgroups, Springer Lecture Notes 1996 S.LANG, Algebraic Number Theory, Addison-Wesley, 1970, reprinted by Springer Verlag, 1986 G.MACKEY, Oninduced representations ofgroups, Amer. J.Math. 73(1951), pp.576-592 G.MACKEY, Symmetric andanti-symmetric Kronecker squares ofinduced representations offinite groups, Amer. J.Math. 75(1953), pp.387-405[La96] [La70] [Ma 51] [Ma 53] XVIII, 8 POSITIVE DECOMPOSITION OFTHE REGULAR CHARACTER 699 The next three sections, which areessentially independent ofeach other, give examples ofinduced representations. Ineach case, weshow that certain representations areeither induced from certain well-known types, orarelinear combinations with integral coefficients ofcertain well-known types. The most striking feature isthat weobtain allcharacters aslinear combinations ofin- duced characters arising from I-dimensional characters. Thus thetheory of characters istoalarge extent reduced tothestudy ofI-dimensional, orabelian characters. 8. POSITIVE DECOMPOSITION OF THE REGULAR CHARACTER Let Gbeafinite group and letkbethecomplex numbers. WeletIGbethe trivial character, and rGdenote theregular character. Proposition 8.1. LetHbeasubgroup ofG,and let «/1beacharacter ofH. Let «/1Gbetheinduced character. Then themultiplicity ofIHin«/1isthe same asthemultiplicity ofIGin«/1G . Proof ByTheorem 6.1(i), wehave <«/1,IH)H=<«/1G ,IG)G. These scalar productsareprecisely themultiplicities inquestion. Proposition 8.2. The regular representation istherepresentation induced bythetrivial character onthetrivial subgroup ofG. Proof This follows atonce from thedefinition oftheinduced character «/1G(T)=2: «/1H( a-Ta--1), (TEG taking t/J=1onthetrivial subgroup. Corollary 8.3. Themultiplicity of1Gintheregular character rGisequal to1. Weshall now investigate thecharacter UG=rG-IG. Theorem 8.4. (Aramata). The characternUG isalinear combination with positive integer coefficients ofcharacters induced byI-dimensional characters ofcyclic subgroups ofG. Theproof consists oftwopropositions, which giveanexplicit description of theinduced characters. Iamindebted toSerre fortheexposition, derived from Brauer's. 700 REPRESENTATIONS OFFINITE GROUPS XVIII, 8 IfAisacyclic group oforder a,wedefine thefunction eAonAbythecondi- tions: {aif(Jisagenerator ofA e«(J)=A0otherwise. We letAA=q>(a)r A-eA(where q>istheEuler function), and AA=0ifa=1. The desired result iscontained inthefollowing twopropositions. Proposition 8.5. Let Gbeafinite group oforder n.Then nUG=LAX , the sumbeing taken over allcyclic subgroups ofG. Proof Given two class functions X,tfJonG,wehave the usual scalar prod uct : 1 - <tfJ,X)G=-LtfJ((J)X( (J). n(1EG Let tfJbeany class function onG.Then: <tfJ,nUG)=<tfJ,nrG)-<tfJ,nlG) =ntfJ(l)-LtfJ«(J). (1EG Ontheother hand, using thefact that theinduced character isthetranspose of therestriction, weobtain L<t/J,AX)=L<t/JIA, AA) A A =L<tfJIA,q>(a)r A-eA) A 1 =Lq>(a)tfJ(l)-L-LatfJ(a) A Aa(1gen A =ntfJ(l)-LtfJ«(J). (1EG Since thefunctions ontheright and leftoftheequality sign inthestatement ofour proposition have the same scalar product with anarbitrary function, they are equal. This proves ourproposition. Proposition 8.6. IfA=I{I}, thefunction AAisalinear combination ofir- reducible nontrivial characters ofAwith positive integral coefficients. XVIII, 8 POSITIVE DECOMPOSITION OFTHE REGULAR CHARACTER 701 Proof. IfAiscyclic ofprime order, then byProposition 8.5, weknow that AA=nuA,and our assertion follows from thestandard structure oftheregular representation. Inorder toprove theassertion ingeneral, itsuffices toprove that theFourier coefficients ofAAwith respect toacharacter ofdegree1areintegers>O.Let tfJbeacharacter ofdegree 1.Wetake thescalar product with respect toA,and obtain: <tfJ,AA>=q>(a)tfJ( 1)-LtfJ((J) 0'gen =q>(a)-LtfJ«(J) 0'gen =L(1-tfJ«(J». 0'gen The sumLtfJ(a)taken over generators ofAisanalgebraic integer, and isinfact arational number (for any number ofelementary reasons), hence arational integer. Furthermore, iftfJisnon-trivial, allreal parts of 1-tfJ((J) are> 0ifa=f.idand are0ifa=ideFrom thelast twoinequalities,weconclude that the sums must beequal toapositive integer. IftfJisthetrivial character, then the sum isclearly O.Our proposition isproved. Remark. Theorem 8.4 andProposition 8.6 arose inthe context ofzeta functions andL-functions, inAramata' sproof that the zeta function ofanumber field divides the zeta function of afinite extension [Ar31], [Ar33]. See also Brauer [Br47a], [Br47b]. These results were also used byBrauer inshowing anasymptotic behavior inalgebraic number theory, namely 10g(hR) logD1I2 for[k:Q]/log D 0, where histhenumber ofideal classes inanumber field k,Ristheregulator, and Distheabsolute value ofthediscriminant. For anexposition ofthisappli- cation, see[La70], Chapter XVI. Bibliography [Ar31] H.ARAMATA, Uber dieTeilbarkeit derDedekindschen Zetafunktionen, Proc. Imp. Acad. Tokyo 7(1931), pp.334-336 [Ar33] H.ARAMATA, Uber dieTeilbarkeit derDedekindschen Zetafunktionen, Proc. Imp. Acad. Tokyo 9(1933), pp.31-34 [Br47a] R.BRAUER, Onthe zeta functions ofalgebraic number fields, Amer. J.Math. 69(1947), pp.243-250 [Br47b] R.BRAUER, OnArtin's L-series with general group characters, Ann. Math. 48 (1947), pp.502-514 [La70] S.LANG, Algebraic Number Theory, Springer Verlag (reprinted from Addison- Wesley, 1970) 702 REPRESENTATIONS OFFINITE GROUPS XVIII, 9 9. SUPERSOLVABLE GROUPS Let Gbeafinite group. Weshall saythat Gissupersolvable ifthere exists a sequence ofsubgroups {I}CGlcG2C...cGm=G such that each Giisnormal inG,and Gi+I/G iiscyclic ofprime order. From thetheory ofp-groups,weknow that every p-group issuper-solvable, and soisthedirect product ofap-group with anabelian group. Proposition 9.1. Every subgroup and every factor group ofasuper-solvable group issupersolvable. Proof Obvious, using thestandard homomorphism theorems. Proposition 9.2. Let Gbe anon-abelian supersolvable group. Then there exists anormal abeUan subgroup which contains the center properly. Proof Let Cbethe center ofG,and letG=GIC. LetHbe anormal subgroup ofprime order inGand letHbeitsinverse image inGunder the canonical map G GIC. Ifaisagenerator ofH,then aninverse imageaofa, together with C,generate H. Hence Hisabelian, normal, and contains the center properly. Theorem 9.3. (Blichfeldt). LetGbeasupersolvable group, letkbealge- braically closed. Let Ebe asimple (G,k)-space. Ifdimk E>1,then there exists aproper subgroup HofGand asimple H-space Fsuch that Eisinduced byF. Proof Since asimple representation ofanabelian group isI-dimensional, ourhypothesis implies that Gisnotabelian. Weshall firstgive theproof ofourtheorem under theadditional hypothesis that Eisfaithful. (This means that ax=xforallxEEimplies a=1.)Itwill beeasy toremove this restriction attheend. Lemma 9.4. LetGbeafinite group, and assume kalgebraically closed. Let Ebeasimple, faithful G-space over k.Assume that there exists anormal abeUan subgroup HofGcontaining the center ofGproperly. Then there exists a proper subgroup HIofGcontaining H,and asimple HI-space Fsuch that E istheinduced module ofFfrom HI toG. Proof Weview EasanH-space. Itisadirect sum ofsimple H-spaces, and since Hisabelian, such simple H-space isI-dimensional. Let vEEgenerateaI-dimensional H-space. Lett/Jbeitscharacter. If WEE also generates aI-dimensional H-space, with the same character t/J,then XVIII, 9 SUPERSOLVABLE GROUPS 703 foralla,bEkand tEHwehave t(av +bw)=t/J(t)(av +bw). Ifwedenote byF",thesubspace ofEgenerated byallI-dimensional H-sub- spaces having thecharacter t/J,then wehave anH-direct sum decomposition E=E8F",. '" Wecontend that E=1=F",. Otherwise, let vEE,v=I0,and (JEG.Then (J-1V isaI-dimensional H-space byassumption, and has character t/J.Hence for tEH, t((J-1v)=t/J(t)(J-1V «(Jt(J-l)V=(Jt/J(t)(J-IV=t/J(t)v. This shows that (Jt(J-l and thave the same effect ontheelement vofE.Since Hisnot contained inthe center ofG,there exist tEHand (JEGsuch that (Jt(J-1=It,and wehave contradicted theassumption that Eisfaithful. Weshall prove that Gpermutes thespaces F",transitively. Let vEF",.For any tEHand (JEG,wehave t((Jv)=(J((J-1t(J)v=(Jt/J((J-1t(J)v=t/J(1(t)(Jv, where t/J(1isthefunction onHgiven byt/J(1(t)=t/J«(J-lt(J). This shows that a maps F'"into F"'a.However, bysymmetry,we seethat (J-l maps F"'ainto F"', and thetwo maps (J,(J-1give inverse mappings between F"'aandF",.Thus G permutes the spaces {F",}. LetE'=GFt/Jo=La-Ft/Jofor some fixed t/1o.Then E'isaG-subspace ofE, and since Ewas assumed tobesimple, itfollows that E'=E.This proves that the spaces {Ft/J}arepermuted transitively. LetF=F"'tfor some fixed t/Jl' Then FisanH-subspace ofE.LetHlbe thesubgroup ofallelements tEGsuch that tF =F.Then H1=IGsince E=IF",.We contend that Fisasimple HI-subspace, and that Eistheinduced space ofFfrom H1toG. To seethis, letG=UHie beadecomposition ofGinterms ofright cosets ofHI. Then theelements {c-l}form asystem ofleft coset representatives of H1.Since E=L(JF (1eG itfollows that E=Lc-1F. c Wecontend that this last sum isdirect, and that Fisasimple HI-space. 704 REPRESENTATIONS OFFINITE GROUPS XVIII, 10 Since Gpermutes thespaces {F",},we seebydefinition thatHIistheisotropy group ofFfortheoperation ofGonthis setofspaces, and hence that theelements oftheorbit areprecisely {c-1F},ascranges over allthe cosets. Thus thespaces {c-lF}aredistinct, and wehave adirect sum decomposition E=EBc-1F. c IfWisaproper HI-subspace ofF,thenEBc-1Wisaproper G-subspace ofE, contradicting thehypothesis that Eissimple. This proves our assertions. We can now apply Theorem 7.3toconclude that Eistheinduced module from F,thereby proving Theorem 9.3,incase Eisassumed tobefaithful. Suppose now that Eisnotfaithful. LetGobethenormal subgroup ofG which isthekernel oftherepresentation G-+Autk(E). Let G=GIGo. Then Egivesafaithful representation ofG.AsEisnotI-dimensional, then Gisnot abelian and there exists aproper normal subgroup HofGand asimple H-space Fsuch that E=indF). LetHbetheinverse image ofHinthenatural map G G.Then H ::)Go, and Fisasimple H-space. Intheoperation ofGasapermutation group ofthe k-subspaces {aF}UEG,weknow that Histheisotropy group ofonecomponent. Hence Histheisotropy group inGofthis same operation, and hence applying Theorem 7.3again,weconclude that Eisinduced byFinG,i.e. E=ind(F), thereby proving Theorem 9.3. Corollary 9.5. Let Gbeaproduct ofap-group and acyclic group, and letk bealgebraically closed. IfEisasimple (G,k)-space and isnotI-dimensional, then Eisinduced byaI-dimensional representation ofsome subgroup. Proof Weapply thetheorem step bystep using thetransitivity ofinduced representations until wegetaI-dimensional representation ofasubgroup. 10. BRAUER'S THEOREM Weletk=Cbethefield ofcomplex numbers. We letRbe asubring ofk. Weshall deal with XR(G), i.e. theringconsisting ofalllinear combinations with coefficients inRofthesimple characters ofGover k.(ItisaringbyProposition 2.1.) XVIII, 10 BRAUER'S THEOREM 705 LetH ={Hex} beafixed family ofsubgroups ofG,indexed byindices {}. WeletVR(G) betheadditive subgroup ofXR(G) generated byallthefunctions which areinduced byfunctions inXR(H ex)for some Hexinourfamily. Inother words, VR(G)=Linda(XR(Ha».a Wecould also saythat VR(G) isthesubgroup generatedover Rbyallthechar- acters induced from allthe Hex. Lemma 10.1. VR(G) isanideal inXR(G). Proof This isimmediate from Theorem 6.1. For many applications, thefamily ofsubgrou pswill consist of"elementary" subgroups: Let pbeaprime number. Byap-elementary groupweshall mean theproduct ofap-group and acyclic group (whose order may beassumed prime top,since we can absorb thep-part ofacyclic factor into thep-group). An element (JEGis said tobep-regular ifitsperiod isprime top,andp-singular ifitsperiod isapower ofp.Given xEG,we canwrite inaunique way x=aT where aisp-singular, !isp-regular, and a,!commute. Indeed, ifprm istheperiod ofx,with mprime top,then 1=vpr+J.1mwhence x=(xm)Jl(xpr)V and wegetour factorization. Itisclearly unique, since the factors have tolieinthecyclic subgroup generated byx.Wecall thetwo factors thep-singular andp-regular factors ofxrespectively. The above decomposition also shows: Proposition 10.2. Every subgroup and every factor group ofap-elementary group isp-elementary. IfSisasubgroup ofthep-elementary group PxC, where Pisap-group, and Ciscyclic, oforder prime top,then S=(SnP)x(SnC). Proof Clear. Our purpose istoshow, among other things, thatifourfamily {Hex} issuch that every p-elementary subgroup ofGiscontained insome Hex, then VR(G)=XR(G) for every ring R.Itwould ofcourse suffice todoitforR=Z,butfor our pur- poses, itisnecessary toprove theresult firstusingabigger ring. The main result iscontained inTheorems 10.11 and 10.13, due toBrauer. We shall give an exposition ofBrauer-Tate (Annals ofMath., July 1955). We letRbetheringZ[(] where (isaprimitive n-th root ofunity. There exists abasis ofRasaZ-module, namely 1,(,...,(N-1for some integer N. This isatrivial fact, and wecan take Ntobethedegree oftheirreducible poly- nomial of(over Q.This irred ucible polynomial hasleading coefficient 1,and 706 REPRESENTATIONS OFFINITE GROUPS XVIII, 10 hasinteger coefficients, sothefact that 1,(,...,(N-1 form abasis ofZ[(] follows from theEuclidean algorithm. Wedon't need to know anythingmore about thisdegree N. Weshall prove ourassertion first fortheabove ring R.The rest then follows byusing thefollowing lemma. Lemma 10.3. IfdEZand the constant function d.lGbelongs toVRthen d.lGbelongs toVz. Proof Wecontend that 1,(,...,(N-1arelinearly independentover Xz(G). Indeed, arelation oflinear dependence would yield sN- 1 LLCvjXv(j=0 v= 1j=0 with integers CvjnotallO.But thesimple characters arelinearly independent over k.The above relation isarelation between these simple characters with coefficients inR,and wegetacontradiction. Weconclude therefore that VR=Vz Vz(...(f)VZ(N-1 isadirect sum (ofabelian groups), and our lemma follows. Ifwe can succeed inproving that the constant function 1Glies inVR(G), then bythelemma, weconclude that itliesinVz(G),and since Vz(G)isanideal, that Xz(G)=Vz(G). Toprove ourtheorem, weneed asequence oflemmas. Two elements x,x'ofGaresaid tobep-conjugate iftheir p-regular factors areconjugate intheordinary sense. Itisclear thatp-conjugacyisanequivalence relation, and anequivalence class will becalled ap-conjugacy class, orsimplya p-class. Lemma 10.4. LetfEXR(G), and assume thatf«(J)EZfor all(JEG. Then fisconstant mod ponevery p-class. Proof Let x=at,where aisp-singular, and tisp-regular, and a,tcom- mute. Itwill suffice toprove that f(x)=f(t) (mod p). LetHbethecyclic subgroup generated byx.Then therestriction offtoH can bewritten fH=Lajt/Jj XVIII, 10 BRAUER'S THEOREM 707 withajER,and t/Jjbeing thesimple characters ofH,hence homomorphisms of Hinto k*. For some power prwehave xpr=tpr ,whencet/JJ{x)11'"=t/JJ{t)pr,and hence f(x)pr=f(t)pr (mod pR). We now usethefollowing lemma. Lemma 10.5. Let R=Z[(] beasbefore. IfaEZand aEpRthen aEpZ. Proof This isimmediate from thefact that Rhas abasis over Zsuch that 1isabasis element. Applying Lemma 10.5, weconclude thatf(x)=f(t) (mod p),because bpr=b(mod p)forevery integer b. Lemma 10.6. Let tbep-regular inG,and letTbethecyclic subgroup generated byt.Let Cbethesubgroup ofGconsisting ofallelements com- muting with t.LetPbeap-Sylow subgroup ofC. Then there exists anelement .pEXR(TxP)such that theinduced function f=«/phasthefollowing properties: (i)f«(J)EZforall (JEG. (ii)f(a)=0if(Jdoes notbelong tothep-class oft. (iii)f(t)=(C:P) =1=o(mod p). Proof We note that thesubgroup ofGgenerated byTand Pisadirect pro- duct TxP.Lett/J1,...,t/Jrbethesimple characters ofthecyclic group T,and assume that these areextended toTxPbycomposition with theprojection: TxPTk*. Wedenote theextensions again byt/Jl'...,t/Jr'Then welet r t/J=Lt/Jv(t)t/J v. v= 1 Theorthogonality relations forthesimple characters ofTshow that t/J(ty)=t/J(t)=(T:1)for YEP t/J«(J)=0if aETP, and (JfttP. We contend that .pGsatisfies ourrequirements. First, itisclear that .plies inXR(TP). 708 REPRESENTATIONS OFFINITE GROUPS XVIII, 10 We have for UEG: G _1 '" -I _1 ( «/1(u)- (TP:1)xfb«/11P(xax )- (P:1)J.Lu) where J.L(u)isthenumber ofelements xEGsuch that xax-Ilies inTP. The number J1«(J) isdivisible by(P:1)because ifanelement xofGmoves (Jinto tP byconjugation, sodoes every element ofPx. Hence thevalues of«/1GlieinZ. Furthermore, J1«(J) =I0only if(Jisp-conjugate tot,whence ourcondition (ii)follows. Finally,wecan have Xtx-1=tywith YEP only ify=1(because theperiod oftisprime top).Hence J1(t)=(C:1),and ourcondition (iii) follows. Lemma 10.7. Assume that thefamily ofsubgroups {Ha} covers G(i.e. every element ofG liesinsome Ha).Iffisaclassfunction onGtaking itsvalues in Z,and such that allthevalues aredivisible byn=(G:1),thenfbelongs to VR(G). Proof. Let ybeaconjugacy class, and letpbeprime ton.Every element ofGisp-regular, and allp-subgroups ofGaretrivial. Furthermore, p-conjugacy isthe same asconjugacy. Applying Lemma 10.6, wefind that there exists in VR(G) afunction taking thevalue 0onelements (JFJy,andtakinganintegral value dividingnonelements ofy. Multiplying thisfunction bysome integer,we find that there exists afunction inVR(G)taking thevalue nforallelements ofy, and thevalue 0otherwise. The lemma then follows immediately. Theorem 10.8. (Artin). Every character ofGisalinear combination with rational coefficients ofinduced characters from cyclic subgroups. Proof InLemma 10.7, let{Ha} bethefamily ofcyclic subgroups ofG.The constant function n.1Gbelongs toVR(G). ByLemma 10.3, thisfunction belongs toVz(G), and hence nXz(G)cVz(G). Hence 1 Xz(G)c-Vz(G),n thereby proving thetheorem. Lemma 10.9. Let pbeaprime number, and assume that every p-elementary subgroup ofG iscontained insome Ha. Then there exists afunctionfE VR(G) whose values areinZ,and =1(mod pr). Proof Weapply Lemma 10.6again. For each p-class y,wecanfind afunc- tionfyinVR(G), whose values are0onelements outside y,and=1=0mod pfor elements ofy.Letf=Lfy,the sum being taken over allp-classes. Then f(a) 1=0(modp) forall (JEG.Taking f(p-l)pr-I gives what wewant. XVIII, 10 BRAUER'S THEOREM 709 Lemma 10.10. Let pbeaprime number and assume that every p-elementary subgroup ofGiscontained insome Hex. Let n=noprwhere noisprime top. Then theconstant function no.1G belongs toVz(G). Proof ByLemma 10.3, itsuffices toprove that no.lG belongs toVR(G). LetfbeasinLemma 10.9. Then no.1G=no(lG-f)+nof Since no(1G-f)has values divisible bynop'==n,itliesinVR(G)byLemma 10.7. Ontheother hand, nofEVR(G)because fEVR(G).This proves ourlemma. Theorem 10.t1.(Brauer). Assume thatfor every prime number p,every p-elementary subgroup oj'Giscontained insome Hex. Then X(G)=Vz(G). Every character ofGisalinear combination, with integer coefficients, of characters induced from subgroups Hex. Proof Immediate from Lemma 10.10, since we can find functions no.1G in Vz(G) with norelatively prime toanygiven prime number. Corollary 10.12. Aclass function fonGbelongs toX(G) ifandonlyifits restriction toHexbelongs toX(Hex)for each. Proof Assume that therestriction offtoHexisacharacter onHexforeach. Bythetheorem, we canwrite IG=L Cainda( tPa) a where CexEZ,andt/lexEX(H c).Hence f=L Cainda(tPafHa)'a using Theorem 6.1. IffH OtEX(H ex),weconclude thatfbelongs toX(G). The converse isofcourse trivial. Theorem 10.13. (Brauer). Every character ofGisalinear combination with integer coefficients ofcharacters induced byI-dimensional characters of subgroups. Proof ByTheorem 10.11, and thetransitivity ofinduction, itsuffices to prove that every character ofap-elementary group has theproperty stated in thetheorem. But wehave proved this inthepreceding section, Corollary 9.5. 710 REPRESENTATIONS OFFINITE GROUPS XVIII, 11 11. FIELD OF DEFINITION OF A REPRESENTATION We goback tothegeneral case ofkhaving characteristic prime to#G. Let Ebe ak-space and assume wehave arepresentation ofGonE.Letk'be an extension field ofk.Then Goperatesonk'0kEbytherule (J(a (8)x)=a(8)(JX for aEk'and xEE.This isobtained from thebilinear map ontheproduct k'xEgiven by (a,x) a(8)(JX. Weview E' =k'(8)kEastheextension ofEbyk',and weobtain arepresentation ofGonE'. Proposition 11.1. Let thenotation beasabove. Then thecharacters ofthe representations ofGonEand onE'areequal. Proof. Let{v1,...,vm}beabasis ofEover k.Then {I(8) Vl'...,1(8)vm} isabasis ofE'over k'.Thus thematrices representinganelement aofGwith respect tothetwo bases areequal, andconsequently the traces areequal. Conversely, letk'be afield and kasubfield. Arepresentation ofGon a k'-space E'issaid tobedefinable over kifthere exists ak-space Eand arepre- sentation ofGonEsuch that E'isG-isomorphic tok'(8)kE. Proposition 11.2. Let E,Fbesimple representation spaces for thefinite group Gover k.Let k'beanextension ofk.Assume that E,Fare not G- isomorphic. Then nok'-simple component ofEk,appears inthedirect sum decomposition ofFk'into k'-simple subspaces. Proof. Consider thedirect product decomposition s(k) kEG]=nRJl(k) Jl=l over k,into adirect product ofsimple rings. Without loss ofgenerality,wemay assume that E,Fare simle leftideals ofk[G],andthey willbelong todistinct factors ofthisproduct byassumption. We now take the tensor product with k',getting nothing else butk'[G]. Then weobtain adirect product decomposi- tion over k'.SinceRv(k)RJl(k)=0ifv=IJ1,thiswillactually begiven byadirect XVIII, 11 FIELD OFDEFINITION OF AREPRESENTATION 711 product decomposition ofeach factorRJl(k): s(k) m(Jl) k'[G]=nnRJli(k'). Jl=l i=l Say E=Lvand F=LJlwith v=IJ1.ThenRJlE=O.HenceRJliEk'=0for each i=1,..., m(J1). This implies that nosimple component ofEk' can be G-isomorphic toanyone ofthesimple leftideals ofRJli'and proves what we wanted. Corollary 11.3. Thesimple characters Xl'...,Xs(k)ofGover karelinearly independent over any extension k'ofk. Proof. This follows atonce from theproposition, together with thelinear independence ofthek'-simple characters over k'. Propositions 11.1 and 11.2 areessentially general statements ofanabstract nature. The next theorem uses Brauer's theorem initsproof. Theorem 11.4. (Brauer). Let Gbe afinite group ofexponent m.Every representation ofG over thecomplex numbers (or analgebraically closed field ofcharacteristic 0)isdefinable over thefield Q((m)where (misaprimitive m-throotofunity. Proof. LetXbethecharacter ofarepresentation ofGover C,i.e.aneffec!ive character. ByTheorem 10.13, we can write X=Cjind('h),]CjEZ, the sum being taken over afinite number ofsubgroups Sj,and«/Ijbeinga1- dimensional character ofSj.Itisclear that each«/Ijisdefinable over Q((m). Thus theinduced character«/Ifisdefinable over Q((m). Eacht/Jfcan bewritten «/If=LdjlLX IL, ILILEZ where {XJl}arethesimple characters ofGover Q«(m). Hence x=(Cjdjll )Xw The expression ofXasalinear combination ofthesimple characters over kis unique, and hence thecoefficient Icjd jJl j is>O.This proves what wewanted. 712 REPRESENTATIONS OFFINITE GROUPS XVIII, 12 12. EXAMPLE: GL 2OVER AFINITE FIELD Let Fbe afield .We view GL2(F) asoperatingonthe 2-dimensional vector space V=F2 .We letFabethealgebraic closure asusual, and welet va=Fax Fa=Fa0V(tensor productover F).Bysemisimple,wealways mean absolutely semisimple, i.e.semisimpleover thealgebraic closure Fa. An element aEGL2(F)iscalled semisimple ifvaissemisimple over Fa[aJ.Asub- group iscalled semisimple ifallitselements aresemisimple. LetKbe aseparable quadratic extension ofF.Let{WI' W2} be abasis ofK. Then wehave theregular representation ofKwith respect tothis basis, namely multiplication representing K* asasubgroup ofGL2(F). The elements ofnorm 1correspond preciselytotheelements ofSL2(F)intheimage ofK*. Adifferent choice ofbasis ofKcorresponds toconjugation ofthisimage inGL2(F). Let CK denote one ofthese images. Then CKiscalled anon-split Cartan subgroup. The subalgebra F[CKJ CMat2(F) isisomorphic toKitself, and theunits ofthealgebra aretherefore theelements ofCK=K*. Lemma 12.1. The subgroup CKisamaximal commutative semisimple subgroup. Proof. IfaEGL2(F) commutes with allelements ofCKthen amust liein F[CK]'forotherwise {I,a}would belinearly independentover F[CK]'whence Mat2(F)would bec.ommutative, which isnot the case. Since aisinvertible, a isaunit inF[CKJ,soaECK,aswas tobeshown. Bythesplit Cartan subgroup we mean thegroup ofdiagonal matrices ()Witha,dEF*. We denote thesplit Cartan byA,orA(F) ifthereference toFisneeded. ByaCartan subgroupwe mean asubgroup conjugate tothesplit Cartan or toone ofthesubgroups CKasabove. Lemma 12.2. Every maximal commutative semisimple subgroup ofGL2(F) isaCartan subgroup, andconversely. Proof. Itisclear that thesplit Cartan subgroup ismaximal commutative semisimple. Suppose that Hisamaximal commutative semisimple subgroup of GL2(F). IfHisdiagonalizable over F,then Hiscontained inaconjugate ofthe split Cartan. Ontheother hand, suppose Hisnotdiagonalizable over F.Itis diagonalizable over theseparable closure ofF,and the two eigenspaces of XVIII, 12 EXAMPLE: GL 2OVER AFINITE FIELD 713 dimension 1give rise totwo characters «/1,«/1':H FS* ofHinthemultiplicative group oftheseparable closure. For each element aEHthevalues «/1(a)and «/1'(a)aretheeigenvalues ofa,andfor some element aEHthese eigenvaluesaredistinct, otherwise Hisdiagonalizable over F. Hence thepair ofelements «/1(a), «/1'(a)areconjugate over F.The image «/1(H) iscyclic, andif«/1(a)generates thisimage, then we seethat «/1(a)generates a quadratic extension KofF.The map a «/1(a) with aEH extends toanF-linear mapping, also denoted by «/1,ofthealgebra F[H] into K. Since F[H] issemisimple, itfollows that «/1:F[H] Kisanisomorphism. Hence «/1maps HintoK*, and infact maps Honto K*because Hwas taken to bemaximal. This proves thelemma. Intheabove proof, thetwo characters «/1,«/1'arecalled the(eigen)characters oftheCartan subgroup. Inthesplit case, ifahasdiagonal elements, a,dthen wegetthe two characters such that «/1(a)=aand «/1'(a)=d.Inthesplit case, the values ofthe characters areinF.Inthenon-split case, these values are conjugate quadratic over F,and lieinK. Proposition 12.3. LetHbeaCartan subgroup ofGL 2(F)(split ornot). Then Hisofindex 2initsnormalizer N(H). Proof. Wemay view GL2(F) asoperatingonthe2-dimensional vector space va=Fa EBFa, over thealgebraic closure Fa. Whether Hissplit ornot, the eigencharactersaredistinct (because oftheseparability assumption inthe non- split case), and anelement ofthenormalizer must either fixorinterchange the eigenspaces. Ifitfixes them, then itlies inHbythemaximality ofHinLemma 12.2. Ifitinterchanges them, then itdoes notlieinH,and generatesaunique coset ofNIH, sothat Hisofindex 2inN. Inthesplit case, arepresentative ofNIAwhich interchanges theeigenspaces isgiven by w=(). Inthenon-split case, let a-:K Kbethenon-trivial automorphism. Let {a,a-a} be anormal basis. With respect tothisbasis, thematrix ofa-isprecisely thematrix w=(). Therefore again inthis case we seethat there exists anon-trivial element inthe 714 REPRESENTATIONS OFFINITE GROUPS XVIII, 12 normalizer ofA.Note that itisimmediate toverify therelation M(a-)M(x)M(a--1)=M(ax), ifM(x) isthematrix associated with anelement xEK. Since theorder ofanelement inthemultiplicative group ofafield isprime tothecharacteristic, weconclude: IfFhascharacteristic p,then anelement offinite order inGL2(F)issemisimple ifandonlyifitsorder isprimetop. Conjugacy classes We shall determine theconjugacy classes explicitly.Wespecialize thesit- uation, and from now on welet: F=finite field with qelements; G=GL2(F); Z=center ofG; A=diagonal subgroup ofG; C=K*=anon-split Cartan subgroup ofG. Up toconjugacy there isonlyonenon-split Cartan because over afinite field there isonlyone quadratic extension (in agiven algebraic closure Fa) (cf. Corollary 2.7ofChapter XIV). Recall that #(G)=(q2-1)(q2-q)=q(q+I)(q-1)2. This should have been worked out asanexercise before. Indeed, FxFhasq2 elements, and#(G) isequaltothenumber ofbases ofFxF.There areq2-1 choices for afirst basis element, and then q2-qchoices for asecond (omitting (0,0)thefirst time, and allchosen elements the second time). This gives the value for#(G). There aretwo cases fortheconjugacy classes ofanelement a. Case 1. The characteristic polynomial isreducible, sotheeigenvalues lie inF.Inthis case, bytheJordan canonical form, such anelement isconjugate toone ofthematrices (),(),()with d*a. These arecalled central, unipotent, orrational notcentral respectively. Case 2. The characteristic polynomial isirreducible. Then aissuch that F[a]=E,where Eisthequadratic extension ofFofdegree 2.Then {I,a}is abasis ofF[a] over F,and thematrix associated with aunder therepresentation bymultiplicationonF[a]is (0-b ), 1-a XVIII, 12 EXAMPLE: GL 2OVER AFINITE FIELD 715 where a,barethecoefficients ofthecharacteristic polynomial X2+ax+b. We then have thefollowing table. Table 12.4 class #ofclasses #ofelements inthe class ()q-1 1 (:)q-1 q2-1 ()1 q2+q -(q-l)(q-2)2 with a=1=d aEC-F*1 q2_q -(q-l)q2 Ineach case onecomputes thenumber ofelements inagiven class astheindex ofthenormalizer oftheelement (orcentralizer oftheelement). Case 1istrivial. Case 2can bedone bydirect computation, since thecentralizer isthen seen to consist ofthematrices G),XEF, with x=1=O.The third and fourth cases can bedone byusing Proposition 12.3. Asforthenumber ofclasses ofeach type, thefirst and second cases correspond todistinct choices ofaEF* sothenumber ofclasses isq-1ineach case. In thethird case, theconjugacy class isdetermined bytheeigenvalues. There are q-1possible choices fora,and then q-2possible choices ford.But the non-ordered pair ofeigenvalues determines theconjugacy class, soone must divide (q-1)(q-2)by2togetthe number ofclasses. Finally, inthe case ofanelement inanon-split Cartan, wehave alreadyseen that if(Jgenerates Gal(K/F), then M(ax) isconjugate toM(x) inGL 2(F). But onthe other hand, suppose x,x'EK*andM(x), M(x') areconjugate inGL2(F)under agiven regular representation ofK* onKwith respect to agiven basis. Then this conjugation induces anF-algebra isomorphismonF[C K],whence anautomor- phism ofK,which istheidentity, orthenon-trivial automorphismu.Consequently thenumber ofconjugacy classes forelements ofthefourth type isequal to #(K)-#(F) q2-q 2 2 which gives thevalue inthetable. 716 REPRESENTATIONS OFFINITE GROUPS XVIII, 12 Borel subgroup and induced representations We let: U=group ofunipotent elements(); B=Borel subgroup=UA=AU. Then #(B)=q(q-1)2=(q-I)(q2-q).We shall construct representations ofGbyinducing characters from B,andeventuallyweshall construct allirre- ducible representations ofGbycombining theinduced representations inasuitable way. Weshall deal with four types ofcharacters. Except inthefirst type, which isI-dimensional and therefore obviously simple,weshall prove that theother types aresimple bycomputing induced characters. Inone case weneed tosubtract aone-dimensional character. Intheother cases, theinduced character will turn out tobesimple. The procedure will besystematic. We shall giveatable of values foreach type. Weverify ineach case that forthecharacter Xwhich we want toprove simplewehave LIx(I3)12=#(G), {3EG and then apply Theorem 5.17(a) togetthesimplicity. Once wehave done this forallfour types, from thetables ofvalues we seethattheyaredistinct. Finally, thetotal number ofdistinct characters which wehave exhibited will beequal to thenumber ofconjugacy classes, whence weconclude that wehave exhibited allsimple characters. We now carry out this program. Imyself learned thesimple characters of GL2(F)from aone-page handout byTate in acourse atHarvard, giving the subsequent tables and thevalues ofthecharacters onconjugacy classes. Ifilled out theproofs inthefollowing pages. First type J.L:F* C*denotes ahomomorphism. Then weobtain thecharacter J.L0det: G C* , which isI-dimensional. Itsvalues onrepresentatives oftheconjugacy classes aregiven inthefollowing table. Table 12.5(1) X()()()d*aaEC-F* J..L0det J.L(a)2 J..L(a)2 J.L(ad) J.L0det(a) XVIII, 12 EXAMPLE: GL 2OVER AFINITE FIELD 717 The stated values arebydefinition. The last value can also bewritten J-L(det a)=J-L(NK/F(a», viewingaasanelement ofK*,because thereader should know from field theory that thedeterminant gives the norm. Acharacter ofGwill besaid tobeoffirst type ifitisequal toJ.L0detfor some J-L.There are q-1characters offirst type, because #(F*)=q-1. Second type Observe that wehave BIU=A.Acharacter ofAcan therefore beviewed asacharacter onBviaBIU .We let: «/IlL=resA (J-L0det), and view«/IlLtherefore asacharacter onB.Thus IL(:)=p.(ad). We obtain theinduced character «/I=ind(<</IIL). Then «/Iisnotsimple. ItcontainsJ-L0det, asone sees byFrobenius reciprocity: <indg",I"J.l0det)G=<I"J.l0det)B=B)E1J.l0det(p)12=1. #peB Characters X=«/I- J-L0detwill becalled ofsecond type. The values ontherepresentatives ofconjugacy classes are asfollows. Table 12.5(11) X()()()d*aaEC-F* «/I- J-L0det qJ-L(a)2 0 J-L(ad)- J.L0det(a) Actually,one computes thevalues oft/J,and one then subtracts thevalue of ()0del. For this case and thenext two cases, we use theformula fortheinduced function: ind}i(cp)(a)=#(){3GCPH({3aW1) where({)Histhefunction equal to({)onHand 0outside H.Anelement ofthe center commutes with all{3EG, sofor({)=t/JILthevalue oftheinduced character 718 REPRESENTATIONS OFFINITE GROUPS XVIII, 12 onsuch anelement is #(G) 2 _ 2 #(B)JL(a)-(q+l)JL(a) , which gives thestated value. For anelement u=(),theonly elements f3EGsuch thatf3u{rIlies inBaretheelements ofB(bydirect verification). Itisthen immediate that ind("',J()=p.(af, which yields thestated value forthecharacter X.Using Table 12.4, one finds atonce thatLIx({3) 12=#(G), and hence; Acharacter Xofsecond type issimple. The table ofvaluesalso shows that there areq-1characters ofsecond type. The next two types deal especially with theCartan subgroups. Third type «/1:A C*denotes ahomomorphism. Asmentioned following Proposition 12.3, therepresentativew=WA=w-1for N(A)I Aissuch that w()w=()=aWifa=(). Thus conjugation bywisanautomorphism oforder 2onA.Let[w] «/1bethe conjugate character; that is,([w]«/1)(a)=«/1(waw)=«/1(aW)for aEA.Then [w](f-l0det)==f-l0del. The characters f-l0det onAareprecisely those which are invariant under [w].The others can bewritten intheform '"()="'1(a)"'2(d), with distinct characters «/11' «/12:F* C*. Inlight oftheisomorphism BIU=A,weview «/1has acharacter onB.Then weform theinduced character «/1G=ind(<</1)=ind([w]«/1). With «/1such that [w] «/1=1=«/1,thecharacters X=«/1Gwill besaid tobeofthe third type. Here istheir table ofvalues. XVIII, 12 EXAMPLE: GL 2OVER AFINITE FIELD 719 Table 12.5(111) x()()a=()d*aaEC-F* t/1G(q+ 1)t/J(a) t/J(a) t/J(a)+t/J(aW) 0 t/J=1=[w]t/1 The first entryoncentral elements isimmediate. For thesecond, wehave already seen thatif{3EGis such thatconjugating 13()WIEB, then (3EB,and sotheformula ifP(a)=#:B)I3GI/IB(l3aWI) immediately gives thevalue of«/p onunipotent elements. For anelement ofA with a=1=d,there istheadditional possibility ofthenormalizer ofAwith the elements w,and the value inthetable then drops outfrom theformula. For elements ofthenon-split Cartan group, there isnoelement ofGwhich conjugates them toelements ofB, sothevalue inthelast column isO. Weclaim that acharacter X=t/JGofthird type issimple. Theproof againuses thetestforsimplicity, i.e.thatL 1x({3) 12=#(G). Observe that two elements a,a'EAareinthe same conjugacy class inGifandonly if a'=aora'=[w]a. This isverified bybrute force. Therefore, writing the sumL 1t/JG({3) 12for(3inthevarious conjugacy classes, andusing Table 12.4, wefind: L 1t/JG({3) 12=(q+1)2(q-I) {3EG +(q-1)(q2-I)+(q2+q)L 1t/J(a)+t/J(a") 12. uE(A -F*)/w The third term can bewritten (q2+q)aEF*(I/I(a)+I/I(aW»(I/I(a-I)+I/I(a-W» = 21 (q2+q)L(l+ 1+I/I(al-w) +I/I(aw-l». UEA-F* We write the sum over aEA-F* as asum for aEAminus the sum for 720 REPRESENTATIONS OFFINITE GROUPS XVIII, 12 aEF*.IfaEF*then al-w=aW-1=I.Byassumptionon «/1,thecharacter a «/1(al-w)for aEA isnon-trivial, and therefore the sum over aEAisequal toO.Therefore, putting these remarks together,wefind that thethird term isequal to 1 2(q2+q)[2(q-1)2-2(q-1)-2(q-1)]=q(q2-I)(q-3). Hence finally L It/P(J3) (2=(q+1)(q2-I)+(q-1)(q2-I)+q(q2-I)(q-3) (3EG =q(q-1)(q2-I)=#(G), thus proving that «/1Gissimple. Finally weobserve that there are!(q-I)(q-2)characters ofthird type. This isthenumber ofcharacters «/1such that [w] «/1=1=«/1,divided by2because each pair «/1and[w] «/1yields the same induced character «/1G .The table ofvalues shows that uptothis coincidence, theinduced characters aredistinct. Fourth type o:K* C*denotes ahomomorphism, which isviewed asacharacter on C=CK. ByProposition 12.3, there isanelement WEN(C) but wC, w=w-1 .Then a waw=[w]a isanautomorphism ofC,but x wxw isalso afield automorphism of F[C]=Kover F.Since [K:F]=2,itfollows thatconjugation bywistheauto- morphism a aq .As aresult weobtain theconjugate character [w]8such that ([w]0)(a)=8([w]a)=O(aW), and wegettheinduced character OG=indg(O)=indg([w]O). LetJ..L:F* C*denote ahomomorphismasinthefirst type. Let: A:F+ C*be anon-trivial homomorphism. (JL,,\)=thecharacter onZU such that (JL,A)((:))=JL(a)A(x). (JL,,\)G=indu(J..L, ,\). XVIII, 12 EXAMPLE: GL 2OVER AFINITE FIELD 721 Aroutine computation ofthe same nature that wehave hadpreviously gives the following values fortheinduced characters OGand (J.L,A)G. x()()()d*aaEC-F* OG (q2-q)O(a) 0 0 O(a)+O(aW) (J.L,A)G (q2-1)J-L(a)-JL(a) 0 0 These areintermediate steps. Note that adirect computation using Frobenius reciprocity shows that OGoccurs inthecharacter (res 0,A)G, where therestriction res0istothegroup F*, sores0isone ofour charactersJ-L.Thus wedefine: 0'=(res0,A)G-OG=([w]O)', which isaneffective character. Acharacter 0'issaid tobeoffourth type if0 issuch that 0=/;[w] O.These arethecharacters we arelooking for. Using the intermediate table ofvalues, one then finds thetable ofvalues forthose characters offourth type. Table 12.5(IV) x()()()d*aaEC-F* 0' (q-l)O(a)-O(a) 0-O(a)-O(aW)o=1=[w]O Weclaim that thecharacters 0'offourth typearesimple. Toprove this, weevaluate L I0'(13) I2=(q-1)iq-1)+(q-1)(q2-1) (3EG +4(q2-q)aE{;-F*18(a)+8(a") I2. We use the same type ofexpansionasforcharacters ofthird type, and thefinal value does turn out tobe#(G), thus proving that 0'issimple. The table also shows that thereare4#(C-F*)=4(q2-q)distinct characters offourth type. We thus come totheend result ofourcomputations. 722 REPRESENTATIONS OF FINITE GROUPS XVIII, Ex Theorem 12.6. The irreducible characters ofG=GL2(F) are asfollows. typenumber ofdimensionthat type IJ.L0det q-1 1 II«/1- J..L0det q-1 q III «/1Gfrom pairs «/1=1=[w]t/11 1 -(q-l)(q-2) q+2 IV ()'from pairs()=1=[w]()1 q-1 -(q-l)q2 Proof. We have exhibited characters offour types. Ineach case itisimme- diate from ourconstruction that wegetthestated number ofdistinct characters ofthegiven type. The dimensions asstated areimmediately computed from the dimensions ofinduced characters astheindex ofthesubgroup from which we induce, and ontwo occasions wehave tosubtract something which was needed tomake thecharacter ofgiven type simple.The end result isthe onegiven in the above table. The total number oflisted characters isprecisely equal tothe number ofclasses inTable 12.4, and therefore wehave found allthesimple characters, thus proving thetheorem. EXERCISES 1.The group 53.Let 83bethesymmetric groupon3elements, (a) Show that there arethree conjugacy classes. (b)There aretwo characters ofdimension 1,onS3/A3. (c)Let dj(i=1,2,3)bethedimensions oftheirreducible characters. Since Ldt=6,the third irreducible character has dimension 2.Show that the third representationcan berealized byconsideringacubic equation X3+aX +b=0,whose Galois group is83over afield k.Let Vbethek- vector space generated bythe roots. Show that this space is2-dimensional andgives thedesired representation, which remains irreducible after tensoring with ka . (d)LetG=S3.Write down anidempotent foreach oneofthesimple components ofC[G]. What isthemultiplicity ofeach irreducible representation ofGin theregular representationonC[G]? XVIII, Ex EXERCISES 723 2.The groups S4andA4. Let S4bethesymmetric groupon4elements. (a) Show that there are 5conjugacy classes. (b) Show that A4has aunique subgroup oforder 4,which isnotcyclic, and which isnormal inS4. Show that thefactor group isisomorphictoS3'so therepresentations ofExercise 1give rise torepresentations ofS4. (c)Using therelation 2:dr=#(S4)=24,conclude that there areonly two other irreducible characters of84,each ofdimension 3. (d)LetX4+a2X2+a.X+aobeanirreducible polynomial over afield k,with Galois group 54.Show that the roots generatea3-dimensional vector space Vover k,and that therepresentation ofS4onthis space isirreducible, so weobtain one ofthetwomissing representations. (e)Let pbetherepresentation of(d). Define p'by p'(a)=p(a) ifaiseven; p'(a)=-p(a)ifaisodd. Show thatp'isalso irreducible, remains irreducible after tensoring with ka, and isnon-isomorphic top.This concludes thedescription ofallirreducible representations ofS4. (f) Show that the 3-dimensional irreducible representations ofS4providean irreducible representation ofA4. (g) Show that allirreducible representations ofA4aregiven bytherepresentations in(f)and three others which areone-dimensional. 3.The quaternion group. LetQ={+1,+x,+y,+z}bethequaternion group, with x2=y2=Z2=-1and xy=-yx, xz=-zx, yz=-zy. (a) Show that Qhas 5conjugacy classes. LetA={+I}.Then Q/Aisoftype (2,2), and hence has 4simple characters, which can beviewed assimple characters ofQ. (b) Show that there isonlyone more simple character ofQ,ofdimension 2. Show that thecorresponding representationcan begiven byamatrix rep- resentation such that p(x)=(),-l (0 1 ) (0i )p(y)= -10'p(z)= iO. (c)LetHbethequaternion field, i.e. thealgebra over Rhaving dimension 4, with basis {I,x,y,z}asinExercise 3,and thecorresponding relations as above. Show that C(8)RH=Mat2(C)(2x2complex matrices). Relate this to(b). 4.Let Sbe anormal subgroup ofG.Let «/1be asimple character ofSover C.Show thatind( «/1)issimple ifandonly if«/1=[a] «/1forallaES. 5.LetGbeafinite group and Sanormal subgroup. Letpbeanirreducible representation ofGover C.Prove that either therestriction ofptoShasallitsirreducible components S-isomorphic toeach other, orthere exists aproper subgroup HofGcontaining S and anirreducible representation(JofHsuch that p:::::::ind(J). 6.Dihedral group D2n.There isagroup oforder 2n(neven integer>2)generated bytwo elements a, 7'such that 724 REPRESENTATIONS OFFINITE GROUPS XVIII, Ex an=1,T2=1, and TaT=a-1 . Itiscalled thedihedral group. (a) Show that there arefour representations ofdimension 1,obtained bythefour possiblevalues+1for aand T. (b)Letenbethecyclic subgroup ofD2ngenerated by a.For each integer r=0,. . .,n-1letI/Irbethecharacter ofensuch that t/Jr((J)=(r((=prim. n-th root ofunity) Let Xrbetheinduced character. Show that Xr=Xn-r' (c) Show that for0<r<n/2 theinduced character Xrissimple, ofdimension 2,and that one gets thereby(-I)distinct characters ofdimension 2. (d) Prove that thesimple characters of(a)and (c)give allsimple characters of D2n. 7.Let Gbe afinite group, semidirect product ofA,Hwhere Aiscommutative and normal. LetA"=Hom(A, C*) bethedual group. LetGoperate byconjugationon characters, sothat for aEG, aEA, wehave [a]I/I(a)=I/I(a-I aa). Let 1/11'. . .,I/Irberepresentatives oftheorbits ofHinA", and letH;(i=1,.. .,r) betheisotropy group ofI/Ii.Let Gi=AH;. (a)For aEAand hEHi,define I/Ii(ah)=I/I;(a). Show that1/1;isthus extended toacharacter onG;. Let (Jbe asimple representation ofH;(on avector space over C). From H;=G;/A,view (Jasasimple representation ofG;.Let P;,8=indg,( 1/1;0(J). (b) Show thatP;,8issimple. (c) Show thatP;,8=Pi:8' implies i=i'and (J=(J'. (d) Show that every irreducible representation ofGisisomorphic tosome P;,8 8.Let Gbe afinite group operatingon afinite setS.LetC[S] bethe vector space generated bySover C.Let 1/1bethecharacter ofthecorresponding representation ofGonC[S]. (a)Let aEG.Show that I/I(a)=number offixed points ofainS. (b) Show that (1/1,1G)G isthenumber ofG-orbits inS. 9.Let Abe acommutative subgroup ofafinite group G.Show that every irreducible representation ofGover Chasdimension-<(G:A). 10. LetFbe afinite field and letG=SL2(F). Let Bbethesubgroup ofGconsisting of allmatrices a==(:)ESL2(F), sod==a-I, LetJ.L:F* C* be ahomomorphism and letI/I#J.:B C* bethehomomorphism such that1/I#J.(a)=J.L(a). Show that the induced characterind(I/I#J.)issimple if 1L2=1=1. XVIII, Ex EXERCISES 725 11. Determine allsimple characters ofSL2(F), givingatable forthenumber ofsuch characters, representatives fortheconjugacy classes, aswas done inthetext forGL2, over thecomplex numbers. 12. Observe thatA5=SL2(F4)=PSL2(F5).As aresult, verify that there are 5conjugacy classes, whose elements have orders 1,2,3,5,5respectively, and write down explicitly thecharacter table forA5aswas done inthe text forGL2. 13. Let Gbe ap-group and letG Aut(V) bearepresentation on afinite dimensional vector space over afield ofcharacteristic p.Assume that therepresentation isirre- ducible. Show that therepresentation istrivial, i.e. Gacts astheidentityonV. 14. Let Gbeafinite group and letCbeaconjugacy class. Prove that thefollowing two conditions areequivalent. They define what itmeans fortheclass toberational. RAT 1.For allcharacters XofG,x(a)EQfor aEC. RAT2.For allaEC,andjprimetotheorder ofa,wehave ajEC. 15. Let Gbe agroup and letHI, H2besubgroups offinite index. Let PI' P2berepre- sentations ofHI, H2onR-modules FI,F2respectively. LetMG(F I,F2)betheR- module offunctions!: G HomR(F I'F2)such that !(h lah2)=P2(h 2)!(a)PI(h l) forallaEG,h;EH;(i=1,2).Establish anR-module isomorphism HomR(Fy, Ffj) MG(F., F2). ByFywehave abbreviated ind (F;).I 16.(a)Let GI,G2betwo finite groups with representationsonC-spaces EI,E2.Let EI0E2betheusual tensor productover C,but now prove that there isanaction ofGIxG2onthis tensor product such that (ai' a2)(x (8)y)=alx (8)a2Y foralEGI,a2EG2. This action iscalled the tensor product ofthe other two. IfPI' P2are the representations ofGI,G2onEI'E2respectively, then their tensor product is denoted byPI(8)P2.Prove: IfPI' P2areirreducible thenP2(8)P2isalso irreducible. [Hint: Use Theorem 5.17.] (b)Let XI' X2bethecharacters ofPI' P2respectively. Show that XI(8)X2isthe character ofthe tensor product. Bydefinition, XI(8)X2(al, a2)=XI(al) X2(a2). 17.With the same notation asinExercise 16,show that every irreducible representation ofGIxG2over Cisisomorphic toatensor product representationasinExercise 16.[Hint: Prove that ifacharacter isorthogonal toalltheproducts XI(8)X2of Exercise 16(b) then thecharacter is0.] Tensor product representations 18. Let Pbethenon-commutative polynomial algebra over afield k,innvariables. Let Xl'...,x,bedistinct elements ofPI(i.e. linear expressions inthevariables t1,...,tn) 726 REPRESENTATIONS OFFINITE GROUPS XVIII, Ex and letah. ..,arEk.If alX +...+arx;=0 forallintegersv=1,...,rshow that ai=0for i=1,..., r.[Hint: Take the homomorphism onthecommutative polynomial algebra and argue there.] 19.Let Gbeafinite setofendomorphisms ofafinite-dimensional vector space Eover the field k.For each aEG,letCfTbeanelement ofk.Show that if LC(1Tr(a)=0 (1eG forallintegers r>1,thenC(1=0forallaEG.[Hint: Use thepreceding exercise, and Proposition 7.2ofChapter XVI.] 20.(Steinberg). LetGbeafinite monoid, andk[G]themonoid algebraover afield k.Let G-+Endk(E) be afaithful representation (i.e.injective), sothat weidentify Gwith a multiplicative subset ofEndk(E). Show that Trinduces arepresentation ofGonTr(E), whence arepresentation ofk[G] onTr(E) bylinearity. IfexEk[G] andifTr(ex)=0for allintegers r>1,show that ex=O.[Hint: Apply thepreceding exercise.] 21.(Burnside). Deduce from Exercise 20thefollowing theorem ofBurnside: Let Gbe afinite group, kafield ofcharacteristic prime totheorder ofG,and Eafinite dimensional (G,k)-space such that therepresentation ofGisfaithful. Then every irreducible representation ofGappears with multiplicity>1insome tensor power Tr(E). 22. LetX(G) bethecharacter ring ofafinite group G,generated over Zbythesimple characters over C.Show that anelementf EX(G) isaneffective irreducible character ifandonly if(f,f)G=1andf(1)>O. 23. Inthis exercise, we assume the next chapteronalternating products. Let pbe an irreducible representation ofGon avector space Eover C.Then byfunctorialitywe have thecorresponding representations sr(p) and/,{(p)onther-thsymmetric power and r-thalternating power ofEover C.IfXisthecharacter ofp,weletsr(X) and /'{(X)bethecharacters ofsr(p) and/,{(p) respectively,onsr(E) and/,{(E). Let t be avariable and let 00 00 at(x)=Lsr(X) tr ,At(X)=L/'{(X)tr . r=O rO (a)Comparing with Exercise 24ofChapter XIV, prove that for xEGwehave at(x)(x)=det(l-p(X)t)-1 and A,(X)(x)=det(I +p(x)t). (b)For afunctionfonGdefine 1JIn(f) by1JIn(f)(x)=f(xn).Show that -d logIT,(X)=i1pn(X)t" and-log L,(X)=i1pn(X)(n.dtn=I dtn=I (c) Show that n nSn(x)=L 1JIr(X)sn- r(X) and r=100 n!\n(x)=L(-I)r-11JIr(X)!\n-r(X). r= 1 XVIII, Ex EXERCISES 727 24. Let Xbe asimple character ofG.Prove that 1JI'n(X) isalso simple. (The characters are over C.) 25. We now assume that you know 3ofChapter xx. (a) Prove that theGrothendieck ring defined there forModc(G) isnaturally isomorphic tothecharacter ringX(G). (b)Relate theabove formulas with Theorem 3.12ofChapter XX. (c) Read Fulton-Lang's Riemann-Roeh Algebra, Chapter I,especially 6, and show thatX(G) isaA-ring, with 1JI'n astheAdams operations. Note. Forfurther connections with homology and thecohomology ofgroups,see Chapter XX,3, and thereferences given attheendofChapter XX,3. 26.Thefollowing formalism istheanalogue ofArtin's formalism ofL-series innumber theory. Cf. Artin's "Zur Theorie der L-Reihen mit allgemeinen Gruppenchar- akteren", Collected papers, and also S.Lang, "L-series ofacovering", Proc. Nat. Aead Se.USA (1956). For theArtin formalism inacontext ofanalysis,seeJ.Jor- genson and S.Lang, "Artin formalism and heat kernels", J.reine angew. Math. 447 (1994) pp.165-200. Weconsider acategory with objects {V}. Asusual, wesaythat afinite group G operates onUifwearegivenahomomorphism p:G-+Aut( V). Wethen saythat Uisa G-object, and also that pisarepresentation ofGinU.We saythat Goperates trivially ifp(G)=ide For simplicity, weomit the pfrom thenotation. ByaG-morphism I:U-+Vbetween G-objects, one means amorphism suchthatf0(J=(J0ffor all (JEG. We shall assume that foreach G-object Vthere exists anobject U/Gonwhich G operates trivially, and aG-morphism nu, G:U-+V/Ghaving thefollowing universal property:Iff: U-+V'isaG-morphism, then there exists aunique morphism f/G: U/G-+U'/G making thefollowing diagram commutative: U j V/Gf)V' j )V'/GfiG Inparticular, ifHisanormal subgroup ofG,show thatG/H operates inanatural way onU/H. Let kbeanalgebraically closed field ofcharacteristic o.We assume givenafunctor Efrom ourcategory tothecategory offinite dimensional k-spaces. IfVisanobject in ourcategory, andI:V-+V'isamorphism, then wegetahomomorphism E(/)=f*:E(U)-+E(V'). (The reader may keep inmind thespecialcase when wedeal with thecategory of reasonable topological spaces, and Eisthehomology functor inagiven dimension.) IfGoperates onV,then weget anoperation ofGonE(V) byfunctoriality. Let Ube aG-object, and F:U-+VaG-morphism. IfPF(t)=n(t-(Xi)isthe characteristic polynomial ofthelinear map F*:E(V)-+E(V), wedefine ZF(t)=n(1-ait), 728 REPRESENTATIONS OFFINITE GROUPS XVIII, Ex and call this the zeta function ofF.IfFistheidentity, then ZF(t)=(1-t)B(U) where wedefine B(U) tobedimkE(U). LetX.beasimple character ofG.Letdxbethedimension ofthesimple representation ofGbelonging toX,and n=ord(G). Wedefine alinear maponE(U) byletting dx, -1 ex= -L'1.(0" )0"*. nCJeG Show thate;=ex'and that foranypositive integer J1wehave(ex0F*)Jl=ex0F:. IfPx(t)=n(t-Pj{X»isthecharacteristic polynomial ofex0F*,define LF(t, X,UIG)=n(1-Pj{X)t). Show that thelogarithmic derivative ofthis function isequal to 100 - -Ltr(e x0F:)tJl-I . NJl= 1 Define LF(t, X,VIG)forany character Xbylinearity. Ifwewrite V=UIGbyabuse of notation, then wealso write LF(t, X,UIV). Then forany X,X'wehave bydefinition, LF(t, X+X',VIV)=LF(t, X,VIV)LF(t, X',UIV). Wemake oneadditional assumption onthesituation: Assume that thecharacteristic polynomial of 1 -L0"*0F*nueG isequal tothecharacteristic polynomial ofFIG onE(UIG). Prove thefollowing statement: (a)IfG={I}then LF(t, 1,UIV)=ZF(t). (b) Let V=UIG. Then LF(t, 1,UIV)=ZF(t). (c)LetHbeasubgroup ofGand lett/Jbeacharacter ofH.LetW =VIH,and let t/JGbetheinduced character from HtoG.Then LF(t, t/J,U/W)=LF(t, t/JG,U/V). (d) LetHbenormal inG.Then GIH operates onUIH=W. Let1/1beacharacter ofGIH, and letXbethecharacter ofGobtained bycomposing 1/1with the canonical map G --+GIH. Let lp=FIH bethemorphism indu:ed on UIH=W. Then LqJ(t, t/J,WIV)=LF(t, X,UIV). (e)IfV= UIG and B(V)=dimkE(V), show that (1-t)B(V) divides (1-t)B(U). Use theregular character todetermine afactorization of(1-t)B(U). XVIII, Ex EXERCISES 729 27. Dothis exercise after you have read some ofChapter VII. Thepoint isthat forfields ofcharacteristic notdividing theorder ofthegroup, therepresentations can beobtained by"reducing modulo aprime". Let Gbe afinite group and letpbe aprimenot dividing theorder ofG.Let Fbe afinite extension oftherationals with ring of algebraic integers OF.Suppose that Fissufficiently largesothat allF-irreducible representations ofGremain irreducible when tensored with Qa=Fa. Let pbe a prime ofOFlying above p,and letopbethecorresponding local ring.· (a) Show that anirreducible (G,F)-space Vcan beobtained from a(G,op)- module Efree over0p,byextending thebase from 0ptoF,i.e.bytensoring sothat V=E0F(tensor productover 0p). (b) Show that thereduction mod pofEisanirreducible representation ofGin characteristic p.Inother words, letk=0/p=op/mpwherempisthemaximal ideal ofOpeLetE(p)=E0k(tensor productover op).Show that Goperates onE(p) inanatural way, and that thisrepresentation isirreducible. Infact, ifXisthecharacter ofGonV,show that Xisalso thecharacter onE,and that Xmod mpisthecharacter onE(p). (c) Show that allirreducible characters ofGincharacteristic pareobtained as in(b). CHAPTER XIX The Alternating Product Thealternating product hasapplications throughout mathematics. Indiffer- ential geometry,one takes themaximal alternating product ofthetangent space togetacanonical line bundle over amanifold. Intermediate alternating products give rise todifferential forms (sections ofthese productsover themanifold). In thischapter,wegive thealgebraic background forthese constructions. For areasonably self-contained treatment oftheaction ofvarious groups of automorphisms ofbilinear forms ontensor andalternating algebras, together with numerous classical examples, Irefer to: R.HOWE, Remarks onclassical invariant theory, Trans. AMS 313(1989), pp.539-569 1 DEFINITION AND BASIC PROPERTIES Consider thecategory ofmodules over acommutative ring R. We recall that anr-multilinear mapf:E(r) Fissaid tobealternating iff(xl'...,xr)=0whenever Xi=Xjfor some i=Ij. Let arbethesubmodule ofthetensor product Tr(E) generated byallelements oftype Xl(8)...(8)Xr where Xi=Xjfor some i=Ij.Wedefine /\r(E)=Tr(E)/a r. Then wehave anr-multilinear map E(r) /\r(E) (called canonical) obtained 731 732 THE ALTERNATING PRODUCT XIX, 1 from thecomposition E(r) -+Tr(E) Tr(E)/or=/\r(E). Itisclear that our map isalternating. Furthermore, itisuniversal with respect tor-multilinear alternating maps onE.Inother words, iff:E(r) Fissuch a map, there exists aunique linear mapf*:/\r(E)Fsuch that thefollowing diagram iscommutative: /\r(E) E(r)/ jf. F Our mapf*exists because we canfirst get aninduced map Tr(E) Fmaking thefollowing diagram commutative: Tr(E) E(r)/ jF and this induced map vanishes onOr,hence inducing ourf*. The image ofanelement (xl'...,Xr)EE(r) inthe canonical map into /\r(E)will bedenoted byXl 1\... 1\Xr'Itisalso theimage ofXl(8)...(8)Xrin thefactor homomorphism Tr(E) /\r(E). Inthis way,f\r becomes afunctor, from modules tomodules. Indeed, let u:E Fbe ahomomorphism. Given elements Xl"..,XrEE,we can map (Xl'. . .,Xr) U(XI) "·. ·1\U(X r)Ef\r(F). This map ismultilinear alternating, and therefore induces ahomomorphism f\r(u): f\r(E) f\r(F). The association Uf\r(u) isobviously functorial. Example. Open any book ondifferential geometry (complex orreal) and you will see anapplication ofthis construction when Eisthetangent space of apointon amanifold, orthedual ofthetangent space. When taking thedual, theconstruction gives rise todifferential forms. Welet/\(E)bethedirect sum 00/\(E)=ffi/\r(E). r=O XIX, 1 DEFINITION AND BASIC PROPERTIES 733 Weshall makeI\(E) into agraded R-algebra andcall itthealternating algebra ofE,oralso theexterior algebra, ortheGrassmann algebra. We shall first discuss thegeneral situation, with arbitrary graded rings. Let Gbe anadditive monoid again, and letA =EBArbe aG-graded reG R-algebra. Suppose given for each Arasubmodule ar,and let a=EB are reG Assume that aisanideal ofA.Then aiscalled ahomogeneous ideal, and wecan define agraded structure onAla. Indeed, thebilinear map ArxAs Ar+s sends arxAsinto ar+sandsimilarly, sends A,xasinto ar+s.Thus using repre- sentatives inAr,Asrespectively,we can define abilinear map Ar/arxAsla sAr+sla r+s, and thus abilinear map A/axAla A/a, which obviously makes Ala into a graded R-algebra. Weapply this toTr(E) and themodules ardefined previously. If Xi=Xj(i=Ij) inaproduct Xl1\... 1\X"then forany Yb. ..,YsEEwe seethat Xl 1\... 1\Xr1\Y11\... 1\Ys liesinar+s,andsimilarly fortheproductontheleft. Hence thedirect sumEBar isanideal ofT(E), and we can define anR-algebrastructure onT(E)/a. The productonhomogeneous elements isgiven bytheformula ((x 11\... 1\Xr),(y 11\... 1\Ys)) XI1\... 1\Xr1\YI1\... 1\YS. We usethesymbol1\also todenote theproduct in/\(E).This productiscalled thealternating productorexterior product. IfxEEand yEE,then x1\y=-y1\x,asfollows from thefact that (x+y)1\(x+y)=o. Weobserve that/\isafunctor from thecategory ofmodules tothecategory ofgraded R-algebras. Toeach linear mapf:E-+Fweobtain amap /\(f):/\(E) /\(F) which issuch that forXl'...,XrEEwehave /\(f)(Xl1\... 1\Xr)=f(xl)1\... 1\f(xr). Furthermore, /\(f)isahomomorphism ofgraded R-algebras. 734 THE ALTERNATING PRODUCT XIX,1 Proposition 1.1. Let Ebefree ofdimension nover R.Ifr»nthen /\r(E)=O.Let{VI'. . .,vn}be abasis ofEover R.If1<r<fl,then /\r(E)isfree over R,and theelements V;11\..·1\v;r' i1<"'<i r formabasis of/\r(E)over k.Wehave dimR/\r(E)=(). Proof. Weshall first proveour assertion when r=n.Every element ofE can bewritten intheformLaiVi'and hence using theformula x1\y= -Y1\X weconclude that Vl1\...1\Vngenerates /\n(E).Ontheother hand, weknow from thetheory ofdeterminants that given aER,there exists aunique multi- linear alternating form faonEsuch that fa(v1,...,Vn)=a. Consequently, there exists aunique linear map /\n(E)R taking the value aon v11\... 1\Vn.From this itfollows at once that Vl1\... 1\Vnisabasis of/\n(E)over R. We now prove our statement for 1<r<n.Suppose that wehave arelation o= a(' )v'1\... 1\v'i...J III Ir with i1<· · ·<irand a(OER.Select anyr-tuple (j)=(jl'. . .,jr)such that jl<...<jrandletjr+ 1'... ,jnbethose values ofiwhich donotappear among (jb... ,jr). Take thealternating product withVjr+11\..'1\Vjn.Then weshall have alternating products inthe sum with repeated components inalltheterms except the(j)-term, and thus weobtain o=a(' )v'1\... 1\v'1\... 1\v' .J 11 Jr In Reshuffling vh1\...1\vjninto Vl1\... 1\Vnsimply changes theright-hand sidebyasign. From what weproved atthebeginning ofthisproof, itfollows thataU)=O.Hence wehave proved our assertion for 1<r<n. When r=0,wedeal with theempty product, and 1isabasis for/\O(E)=R over R.We leave the case r>nasatrivial exercise tothereader. The assertion concerning thedimension istrivial, considering that there isa bijection between the setofbasis elements, and thesubsets ofthe setofintegers (1,...,n). XIX,1 DEFINITION AND BASIC PROPERTIES 735 Remark. Itispossible togive thefirst part oftheproof, for/\n(E), without assuming known theexistence ofdeterminants. One must then show that an admits aI-dimensional complementary submodule inTn(E). This can bedone bysimple means, which weleave asanexercise which thereader can look up inthe more general situation of4.When Risafield, this exercise iseven more trivial, since one canverify atonce that VI0.··0Vndoes notlieinan.This alternative approach tothetheorem then proves theexistence ofdeterminants. Proposition 1.2. Let o E' E E" 0 beanexact sequence offree R-modules offinite ranks r,n,and srespectively. Then there isanatural isomorphism cP:I\rE'0I\sE"1\nEe This isomorphism istheunique isomorphism having thefollowing property. For elements VI'. ..,vrEE'and WI'. ..,WsEE", letUI'...,Usbeliftings of WI,. . .,WsinE.Then CP«VI1\. · ·1\vr)0(wI1\· · ·1\ws»=VI1\. ·.1\Vr1\UI1\...1\us. Proof. The proof proceeds inthe usual two steps. First one shows the existence ofahomomorphism cPhaving thedesired effect. The value ontheright ofthe above formula isindependent ofthe choice ofu.,..., Uslifting WI'. ..,Wsbyusing thealternating property,soweobtain ahomomorphism cpo Selecting inparticular {V.,...,vr}and{WI'.."ws}tobebases ofE'and E" respectively,one then sees that cpisboth injective andsurjective. We leave the details tothereader. Given afree module Eofrank n,wedefine itsdeterminant tobe detE=I\maxE=I\nE. Then Proposition 1.2may bereformulated bytheisomorphism formula det(E') 0det(E")=det(E). IfR=kisafield, then wemay saythat det isanEuler-Poincare map onthe category offinite dimensional vector spaces over k. Example. Let Vbe afinite dimensional vector spaceover R.Byavolume onVwe mean anorm IIIIondet V.Since Visfinite dimensional, such anorm isequivalent toassigningapositive number ctoagiven basis ofdet(V). Such abasis can beexpressed intheform el1\·· ·1\en'where {e.,. . .,en}isabasis ofV.Then for aERwehave IIaeI1\...1\enII= IaIc. 736 THE ALTERNATING PRODUCT XIX, 1 Inanalysis, givenavolume asabove, one then defines aHaar measureJ..LonV bydefining the measure of asetStobe JL(S)=file)1\...1\enIIdXl...dxn, s where Xl'. . .,Xnarethecoordinates onVwith respect totheabove basis. As anexercise, show that theexpressionontheright istheindependent ofthechoice ofbasis. Proposition 1.2isaspecialcase ofthefollowingmore general situation. We consider againanexact sequence offree R-modules offinite rank asabove. With respecttothesubmodule E'ofE,wedefine /\7E=submodule of/\nE generated byallelements , , Xl/\.../\Xi/\Yi+l/\.../\Yn with x'l'...,x;EE'viewed assubmodule ofE. Then wehave afiltration /\iE ::::>/\i +1E. Proposition 1.3. There isanatural isomorphism /\iE' (8)/\n-iE" /\iEI/\i+ lEe Proof LetX'{,..., X_ibeelements ofE", and liftthem toelements Yl,...,Yn-i ofE.Weconsider themap (' ,,, ")' ,Xl,...,Xi,Xl"..,Xn-i Xl/\.../\Xi/\Yl/\.../\Yn-i with theright-hand side taken mod/\i+ lEeThen itisimmediate that this map factors through /\iE' (8)/\n-iE" /\iEI/\i+ lE, andpicking bases shows that one getsanisomorphismasdesired. Inasimilar vein, wehave: Proposition 1.4. LetE=E'EBE"beadirect sumoffinite free modules. Then for every positive integer n,wehave amodule isomorphism /\nE E9/\PE' (8)/\qE". p+q==n XIX, 1 DEFINITION AND BASIC PROPERTIES 737 Interms ofthealternating algebras, wehave anisomorphism I\E=1\£' 0suI\E". where 0su isthesuper product ofgraded algebras. Proof. Each natural injection ofE'andE"into Einduces anatural mapon thealternating algebras, and sogives thehomomorphism /\E' (8)/\E" /\E, which isgraded, Le.forp=0,...,nwehave /\PE' (8)/\n-PE"-+/\nE. Toverify that thisyields thedesired isomorphism,one can argue bypicking bases, which weleave tothereader. The anti-commutation rule ofthealternating product immediately shows that theisomorphism isanalgebra isomorphism for thesuper product I\E' 0su1\£". We end this section with comments onduality. InExercise 3,youwill prove: Proposition 1.5. LetEbefree ofrank nover R.For each positive integer r,wehave anatural isomorphism I\r(EV)=I\r(E)V. Theisomorphism isexplicitly described inthat exercise. Amore precise property than "natural" would bethat theisomorphism isfunctorial with respect tothe category whose objectsarefinite free modules over R,and whose morphisms areisomorphisms. Examples. Let Lbe afree module over Rofrank 1.We have the dual module LV=HomR(L, R),which isalso free ofthe same rank. For apositive integer m, wedefine LfS-m=(LV)fSm=LV0", 0LV(tensor product taken mtimes). Thus wehave defined thetensor product ofalinewith itself fornegative integers. We define LfZ)O =R.You caneasily verify that therule LfSp0LfSq=LfS(p+q) holds forallintegers p,qEZ,with anatural isomorphism. Inparticular, if q=-pthen weget Ritself ontheright-hand side. Now letEbe anexact sequence offree modules: E :0 Eo EI· · .EmO. 738 THE ALTERNATING PRODUCT XIX,2 We define thedeterminant ofthis exact sequence tobe det(E)=Q9det(E;)(-li. As anexercise, prove thatdet(E) has anatural isomorphism with R,functorial with respect toisomorphisms ofexact sequences. Examples. Determinants ofvector spacesorfree modules occur inseveral branches ofmathematics, e.g.complexes ofpartial differential operators, homol- ogytheories, thetheory ofdeterminant line bundles inalgebraic geometry, etc. Forinstance, givenanon-singular projective variety Vover C, one defines the determinant ofcohomology ofVtobe detH(V)=Q9detH;(V)(_l)i, where H;(V) arethecohomology groups. Then detH(V) isaone-dimensional vector spaceover C,but there isnonatural identification ofthis vector space with C,because apriori there isnonatural choice of abasis. For anotable application ofthedeterminant ofcohomology, following work ofFaltings,see Deligne, Ledeterminant delacohomologie, inRibet, K.(ed.), Current Trends inArithmetical Algebraic Geometry, Proc. Arcata 1985. (Contemporary Math. vol 67, AMS (1985), pp.93-178.) 2. FITTING IDEALS Certain ideals generated bydeterminants arecoming more and more into use, inseveral branches ofalgebra andalgebraic geometry. Therefore Iinclude this section which summarizes some oftheir properties. For amore extensive account, seeNorthcott's book Finite Free Resolutions which Ihave used, aswell astheappendix ofthepaper byMazur-Wiles: "Class Fields ofabelian extensions ofQ," which they wrote inaself-contained way. (Invent. Math. 76(1984), pp. 179-330.) Let Rbeacommutative ring. Let Abeapxqmatrix and Baqxsmatrix with coefficients inR.Let r>0beaninteger. Wedefine thedeterminant ideal Ir(A) tobetheideal generated byalldeterminants ofrxrsubmatrices ofA. This ideal may also bedescribed asfollows. LetS:bethe setofsequences J=(jl'...,jr)with 1<j1<j2<...<jr<p. Let A=(aij).Let 1<r<min(p, q).LetK =(kb. ..,kr)beanother element ofS:. Wedefine ahk 1ah k2 ahk r A(r) -ahk 1aj2k2aj2kr JK- ajrk 1ajrk2a.kJr r XIX,2 FITTING IDEALS 739 where thevertical bars denote thedeterminant. With J,Kranging over S: wemay view ArkastheJK-component ofamatrix A(r)which wecall ther-th exterior power ofA. One may also describe thematrix asfollows. Let{e.,. . .,ep}be abasis of RPand{u1,. . .,Uq}abasis ofRq. Then theelements e. 1\".1\ e.Jl Jr (j 1<j2<...<jr) form abasis for/,(RPandsimilarly for abasis of!\Rq.We may view Aasa linear map ofRPinto Rq,and thematrix Alr)isthen thematrix representing the exterior power /'(Aviewed asalinear map of/,(RPintoI\rRq.Onthewhole, thisinterpretatIon will not beespecially useful forcertain computations, butit does giveaslightly more conceptual context fortheexterior power. Just atthe beginning, thisinterpretation allows for animmediate proof ofProposition 2.1. For r=0wedefine A(0)tobethe 1x1matrix whose single entry isthe unit element ofR.Wealso note that A(1) =A. Proposition 2.1. LetAbeapxqmatrix and Baqxsmatrix. Then (AB)(r)=A(r)B(r) forr>o. Ifone uses thealternating productsasmentioned above, theproof simply says that thematrix ofthecomposite oflinear maps with respect tofixed bases istheproduct ofthematrices. Ifone does not usethealternating products, then one can prove theproposition byadirect computation which will belefttothe read er . We have formed amatrix whose entries areindexed byafinite setS:.For any finite set Sanddoubly indexed family (CJK) with J,KESwemay also define thedeterminant as det(cJK)=L£«(J)(nCJ,(1(J») (1 JeS where (Jrangesover allpermutations ofthe set. For r>0wedefine thedeterminant ideal Ir(A) tobetheideal generated by allthecomponents ofA(r), orequivalently byallrxrsubdeterminants ofA. We have bydefinition A(O) =Rand A(l) =ideal generated bythecomponents ofA. Furthermore Ir(A)=0for r>min(p, q) and theinclusions R=Io(A)::)Il(A)::)12(A)::)... 740 THE ALTERNATING PRODUCT XIX,2 ByProposition 10.1, wealso have (1) Ir(AB)cIr(A) nIr(B). Therefore, ifA=UBU' where U,U'aresquare matrices ofdeterminant 1,then (2) Ir(A)=Ir(B). Next, letEbeanR-module. LetXl'...,xqbegenerators ofE.Then we may form thematrix ofrelations (a1,...,aq)ERqsuch that q Laixi=O. i= 1 Suppose first wetake only finitely many relations, thus giving rise toapxq matrix A.Weform thedeterminant ideal Ir(A). We letthedeterminant ideals ofthefamily ofgenerators be: Ir(Xl'...,Xq)=Ir(X)=ideal generated byIlA) forallA. Thus wemay infact take theinfinite matrix ofrelations, and saythat Ir(x) is generated bythedeterminants ofallrxrsubmatrices. The inclusion relations of(1)show that R=Io(x)::)I1(x)::)I2(X)::)... Ir(x)=0if r>q. Furthermore, itiseasy toseethat ifweform asubmatrix Mofthematrix ofall relations bytaking onlyafamily ofrelations which generate theideal ofall relations inRq,then wehave Ir(M)=Ir(x). Weleave theverification tothereader. We cantake Mtobeafinite matrix when Eisfinitely presented, which happens ifRisNoetherian. Interms ofthisrepresentation ofamodule asaquotient ofRq, wegetthe following characterization. Proposition 2.2. Let Rq E 0bearepresentation ofEasaquotient of Rq,and letXl,. . .,xqbetheimages oftheunit vectors inRq. Then Ir(x) isthe ideal generated byallvalues A.(wb...,Wr) where wl,. ..,WrEKer(Rq E)and A.EL(Rq, R). Proof. This isimmediate from thedefinition ofthedeterminant ideal. XIX,2 FITTING IDEALS 741 The above propositioncan beuseful toreplaceamatrix computation bya more conceptual argument with fewer indices. The reader canprofitably trans- late some ofthefollowing matrix arguments inthese more invariant terms. We now change thenumbering, and lettheFitting ideals be: Fk(X)=Iq-k(x)for 0<k<q Fk(X)=R when k>q. Lemma 2.3. The Fitting ideal Fk(x) does notdependonthe choice of generators (x). Proof. LetYl,.. .,Ysbeelements ofE.Weshall prove that Ir(x)=Ir+s(x, y). The relations of(x,y)constitute amatrix oftheform all alq0 apq0 blq10 bsq0W =apl bll bslo o o 1 Byelementary column operations, we canchange this toamatrix () and such operations donotchange thedeterminant ideals by(2). Then we conclude that forallr>0wehave Ir(A)=Ir+s(W)cIr+s(x, y). This proves thatIr(x)cIr+s(x, y). Conversely, letCbe amatrix ofrelations between thegenerators (x,y). We also have amatrix ofrelations C blq1z= bsq0 Byelementary row operations, we canbring this matrix into the same shape 742 THE ALTERNATING PRODUCT XIX,2 asBabove, with some matrix ofrelations A'for(x),namely Z' =() Then Ir(A')=Ir+s(Z')=Ir+iZ)::)Ir+s(C), whence Ir+iC)cIr(x). Taking allpossible matrices ofrelations Cshows that Ir+s(x, y)cIr(x), which combined with theprevious inequality yields Ir+s(x, y)=Ir(x). Now given two families ofgenerators (x)and(y),wesimply put them side byside (x,y)and usethe new numbering fortheFktoconclude theproof of thelemma. Now letEbeafinitely generated R-module with presentation o K Rq E 0, where thesequence isexact andKisdefined asthekernel. Then Kisgenerated byq-vectors, and can beviewed asaninfinite matrix. The images oftheunit vectors inRq aregenerators (Xl'. . .,Xq).We define theFitting ideal ofthe module tobe Fk(E)=Fk(X). Lemma 2.3 shows that theideal isindependent ofthechoice ofpresentation. The inclusion relations of adeterminant ideal Ir(A) ofamatrix now translate into reverse inclusion relations fortheFitting ideals, namely: Proposition 2.4. (i)Wehave Fo(E)cF'l(E)cF2(E)c... (ii)IfEcan begenerated byqelements, then Fq(E)=R. (iii)IfEisfinitely presented then Fk(E) isfinitely generated forallk. This last statement merely repeats theproperty that thedeterminant ideals ofa matrix can begenerated bythedeterminants associated with afinite submatrix ifthe row space ofthematrix isfinitely generated. XIX,2 FITTING IDEALS 743 Example. Let E=Rqbethefree module ofdimension q.Then: F(E)={oif0<k<q kRifk>q. This isimmediate from thedefinitions and thefact that theonly relation ofa basis forEisthetrivial one. TheFitting ideal Fo(E) iscalled thezero-th orinitial Fitting ideal. Insome applications itistheonlyone which comes up,inwhich case itIScalled "the" Fitting ideal F(E) ofE.Itistheideal generated byallqxqdeterminants in thematrix ofrelations ofqgenerators ofthemodule. For any module EweletannR(E) betheannihilator ofEinR,that isthe setofelements aERsuch that aE=O. Proposition 2.5. Suppose that Ecan begenerated byqelements. Then (annR(E»qcF(E)cannR(E). Inparticular, ifEcan begenerated byoneelement, then F(E)=annR(E). Proof. LetXl'...,xqbegenerators ofE.Let al'...,aqbeelements ofR annihilating E.Then thediagonal matrix whose diagonal componentsare aI,...,aqisamatrix ofrelations, sothedefinition oftheFitting ideal shows that the determinant ofthis matrix, which istheproducta1...aqlies in Iq(E)cFo(E). This proves theinclusion annR(E)qcF(E). Conversely, letAbeaqxqmatrix ofrelations between Xl'...,Xq.Then det(A )Xi=0forallisodet(A)EannR(E). Since F(E) isgenerated bysuch determinants, wegetthe reverse inclusion which proves theproposition. Corollary 2.6. LetE=Riaforsome ideal a.Then F(E)=a. Proof. The module Riacan begenerated byone element sothecorollary isanimmediate consequence oftheproposition. Proposition 2.7. Let o E' E E" 0 beanexact sequence offinite R-modules. Forintegers m,n>0wehave Fm(E')F n(E")cFm+n(E). 744 THE ALTERNATING PRODUCT XIX,2 Inparticular (orF=F0' F(E')F(E")cF(E). Proof. We may assume E'isasubmodule ofE.Wepick generators Xl'...,xpofE'and elements Yl'...,YqinEsuch that their images y'{,...,y; inE"generate E". Then (x,y)isafamily ofgenerators forE.Suppose first that m<pand n<q.Let Abe amatrix ofrelations among y'{,...,y;with q columns. If(a1,...,aq)issuch arelation, then alYl +...+aqY qEE' sothere exist elements bl,...,bpERsuch that a. y.+b.x.=Oi...J I I i...J J J. Thus we can find amatrix Bwith pcolumns and the same number ofrows as Asuch that (B,A)isamatrix ofrelations of(x,y).LetCbeamatrix ofrelations of(xb...,xp).Then () isamatrix ofrelations of(x,y).IfD"isa(q-n)x(q-n)subdeterminant of Aand D'isa(p-m)x(p-m)subdeterminant ofCthen D"D'isa (p+q-m-n)x(p+q-m-n) subdeterminant ofthematrix () and D"D' EFm+n(E). Since Fm(E') isgenerated bydeterminants like D'and Fn(E") isgenerated bydeterminants likeD",this proves theproposition inthe present case. Ifm>pand n>qthen Fm+n(E)=Fm(E')=Fn(E")=Rsotheproposition istrivial inthis case. Saym<pand n>q.Then Fn(E")=R=Fq(E")and hence Fm(E')Fn(E")=Fq(E")Fm(E')cFp+n(E)cFm+n(E) where theinclusion follows from thefirst case. Asimilar argument proves theremaining case with m>pand n<q.This concludes theproof. Proposition 2.8. LetE',E"befinite R-modules. For anyintegern>0we have Fn(E' E")=LFr(E')F s(E"). r+s=n XIX,2 FITTING IDEALS 745 Proof. LetXl'. . .,Xpgenerate Eiand Y1,. . .,Yqgenerate E". Then (x,y) generate E'EBE".ByProposition 2.6 weknow theinclusion LFr(E')F s(E")cFn(E' E"), sowehave toprove the converse. Ifn>p+qthen we can take r>pand s>qinwhich case Fr(E')=Fs(E")=Fn(E)=R and we aredone. So we assume n<p+q.Arelation between (x,y)inthe direct sum splits into arelation for(x)and arelation for(y). The matrix of relations for(x,y)istherefore oftheform (A' 0 )C= 0A" where A'isthematrix ofrelations for(x)and A"thematrix ofrelations for (y). Thus Fn(E' E")=LIp+q-n(C) c where the sum istaken over allmatrices Casabove. LetDbea (p+q-n)x(p+q-n) subdeterminant. Then Dhas theform B'0D= oB" where B'isak'x(p-r)matrix, and B"isakIfx(q-s)matrix with some positive integers k',kIf,r,ssatisfying k'+kIf =p+q-nand r+s=n. Then D=0unless k'=p-rand kIf =q-s.Inthat case D=det(B')det(B")EFr(E')Fs(E"), which proves the reverse inclusion and concludes theproof oftheproposition. Corollary 2.9. Let s E=EBRla i i=1 where Qiisanideal. Then F(E)=al. ..as. Proof. This isreallyacorollary ofProposition 2.8 andCorollary 2.6. 746 THE ALTERNATING PRODUCT XIX,3 3. UNIVERSAL DERIVATIONS AND THE DE RHAM COMPLEX Inthissection, allrings R,A,etc. areassumed commutative. Let AbeanR-algebra and ManA-module. Byaderivation D:A M (over R)wemean anR-linear map satisfying theusual rules D(ab)=aDb +bDa. Note thatD(l)=2D(1) soD(l)=0,whence D(R)=o.Such derivations form anA-module DerR(A, M)in anatural way, where aDisdefined by(aD)(b)=aDb. Byauniversal derivation for Aover R,we mean anA-module Q,and a derivation d:AQ such that, givenaderivation D:A Mthere exists aunique A-homomorphism f:Q Mmaking thefollowing diagram commutative: Ad) Q\} M Itisimmediate from thedefinition that auniversal derivation (d,Q)isuniquely determined uptoaunique isomorphism. Bydefinition, wehave afunctorial isomorphism IDerR(A, M);:::::HomA(Q, M). I Weshall now prove theexistence ofauniversal derivation. Thefollowing general remark will beuseful. Let fl,f2:A B betwo homomorphisms ofR-algebras, and letJbeanideal inBsuch that J2 =O.Assume thatfl=f2mod J;this means thatfl(x)=f2(x) mod Jfor allxinA.Then D=f2-fl isaderivation. This fact isimmediately verified asfollows: f2(ab)=f2(a)f2(b)=[fl(a) +D(a)] [fl(b) +D(b)] =fl(ab) +fl(b)D(a) +fl(a)D(b). XIX,3 UNIVERSAL DERIVATIONS AND THE DERHAM COMPLEX 747 But theA-module structure ofJisgiven viaflorf2(which amount tothe same thing inlight ofourassumptionsonfl,f2),sothefact isproved. Letthetensor product betaken over R. Let mA:A(8)A Abethemultiplication homomorphism, such that mA(a (8)b)=ab. LetJ=Ker mA.Wedefine themodule ofdifferentials QA/R=JIJ2 , asanideal in(A(8)A)IJ2 .The A-module structure willalways begiven viathe embeddingonthefirst factor: A A(8)Abya a(8)1. Note that wehave adirect sum decomposition ofA-modules A(8)A=(A(8)1) J, and therefore (A(8)A)IJ2=(A(8)1) JIJ2 . Let d:A JIJ2betheR-linear mapa 1(8)a-a(8) 1mod J2. Takingfl:aa0 1andf2:a10a,we seethat d=f2-fl'Hence dis aderivation when viewed asamap into J/J2 . We note that Jisgenerated byelements oftheform LXidYi. Indeed, ifLXi(8)YiEJ,then bydefinition LXiYi=0,and hence LXi(8)Yi=Lxi(1(8)Yi-Yi(8)1), according totheA-module structure wehave putonA(8)A(operation ofAon theleftfactor.) Theorem 3.1. Thepair (JIJ2 ,d)isuniversal for derivations ofA.This means: Given aderivation D:A Mthere exists aunique A-linear map f:JIJ2Mmaking thefollowing diagram commutative. Ad) JIJ2\1 M 748 THE ALTERNATING PRODUCT XIX,3 Proof. There isaunique R-bilinear map f:A(8)A M given by x(8)yxDy, which isA-linear' byourdefinition oftheA-module structure onA(8)A.Then bydefinition, thediagram iscommutative onelements ofA,when wetakef restricted toJ,because f(1 (8)y-y(8)1)=Dy. Since JIJ2isgenerated byelements oftheform xdy,theuniqueness ofthemap inthediagram ofthe theorem isclear. This proves the desired universal property. Wemay write theresult expressed inthetheorem asaformula DerR(A, M) HomA(JIJ2 ,M). The reader willfind exercises onderivations which giveanalternative way of constructing the universal derivation, especially useful when dealing with finitely generated algebras, which arefactors ofpolynomial rings. Iinsert here without proofssome furtl:er fundamental constructions, im- portant indifferential andalgebraic geometry. Theproofs areeasy, andprovide . . nIce exercIses. Let R AbeanR-algebra ofcommutative rings. For i>0define iI\i 1QA/R=QA/R, whereQ/R=A. Theorem 3.2. There exists aunique sequence ofR-homomorphisms d.ni ni+1 i.I.A/R I.A/R such thatforWEQiand '1EQj wehave d(w/\'1)=dw/\'1+(-tyw /\d'1. Furthermore d0d=O. Theproof will beleft asanexercise. Recall that acomplex ofmodules isasequence ofhomomorphisms . 1di-I.di . 1 ... E'- E' E'+ such that di0di-1=O.One usually omits thesuperscript onthemaps d.With thisterminology,we see that thefl/Rform acomplex, called theDeRham complex. XIX,4 THE CLIFFORD ALGEBRA 749 Theorem 3.3. Letkbeafield ofcharacteristic 0,and letA=k[Xl'. ..,Xn] bethepolynomial ring innvariables. Then theDeRoom complex o-+k AQ/k.. .Q/k0 isexact. Again theproof will beleft as an exerCise. Hint: Use induction and integrate formally. Other results concerning connections will befound intheexercises below. 4. THE CLIFFORD ALGEBRA Let kbe afield. By analgebra throughout thissection, we mean ak-algebra given byaring homomorphismk Asuch that theimage ofkisinthe center ofA. Let Ebe afinite dimensional vector space over thefield k,and let9be a symmetric form onE.Wewould like tofind auniversal algebra over k,inwhich we can embed E,and such that thesquare inthealgebra corresponds tothevalue ofthequadratic form inE.More precisely, byaClifford algebra forg,we shall mean ak-algebra C(g), also denoted byCg(E),and alinear map p:E C(g) having thefollowing property: If«/1:E Lisalinear map ofE into ak-algebra Lsuch that «/J(x)2=g(x, x)·1 (1=unit element ofL) forallxEE,then there exists aunique algebra-homomorphism C(t/J)=t/J*:C(g)-+L such that thefollowing diagram iscommutative: EP)C(g)\/ L Byabstract nonsense, aClifford algebra for9isuniquely determined, uptoa unique isomorphism. Furthermore, itisclear that if(C(g), p)exists, then C(g) isgenerated bytheimage ofp,i.e.byp(E),asanalgebra over k. Weshall write p=Pgifitisnecessary tospecify thereference to9explicitly. 750 THE ALTERNATING PRODUCT XIX,4 We have trivially p(X)2=g(X, x)·1 forallxEE,and p(x)p(y)+p(y)p(x)=2g(x, y).1 asone sees byreplacingxbyx+yinthepreceding relation. Theorem 4.1. Let gbe asymmetric bilinear formon afinite dimensional vector space Eover k.Then theClifford algebra (C(g), p)exists. The map p ininjective, andC(g) has dimension 2nover k,ifn=dim E. Proof. LetT(E) bethetensor algebraasinChapter XVI, 7.Inthatalgebra, weletI9bethetwo-sided ideal generated byallelements x0x-g(x, x)·1for xEE. WedefineCg(E)=T(E)II g.Observe that Eisnaturally embedded inT(E) since T(E)=kE9EE9(E0E)E9. .. . Then thenatural embedding ofEinTEfollowed bythecanonical homomorphisms ofT(E) ontoCg(E)defines ourk-linear map p:ECg(E).Itisimmediate from theuniversal property ofthe tensor product thatCg(E)asjust defined satisfies theuniversal property of aClifford algebra, which therefore exists. The only problem istoprove that ithas thestated dimension over k. We first prove that the dimension is<2n .Wegiveaproof only when the characteristic ofkis =1=2and leave characteristic 2tothe reader. Let {V.,. . .,vn}be anorthogonal basis ofEasgiven byTheorem 3.1ofChapter XV. Lete;=o/(v;), where 0/:E Lisgivenasinthebeginning ofthe sec- tion. Let ci=g(v;, Vi). Then wehave therelations e=c.I "e.e.=-e.e. foralli=1=J. I] ]I. This immediately implies that thesubalgebra ofLgenerated by«/1(E) over kis generatedasavector spaceover kbyallelements e}1. · ·enwithVi=0or 1fori=1,..., n. Hence thedimension ofthissubalgebra is<2n.Inparticular, dimCg(E)<2n asdesired. There remains toshow that there exists atleast one «/1:E Lsuch that L isgenerated by«/1(E) asanalgebra over k,and has dimension 2n;forinthat case, thehomomorphism 0/*:Cg(E)Lbeing surjective, itfollows that dim Cg(E)::>2nand thetheorem will beproved. We construct Linthefollowing way.We first need some general notions. LetMbe amodule over acommutative ring. Leti,jEZ/2Z. Suppose M isadirect sum M=Mo E9M1where 0, 1areviewed astheelements ofZ/2Z. We then say that MisZ/2Z-graded. IfMisanalgebra over thering,wesay XIX,4 THE CLIFFORD ALGEBRA 751 itisaZ/2Z-graded algebra ifM;MjCM;+jforalli,jEZ/2Z. Wesimply saygraded, omitting theZ/2Z prefix when the reference toZ/2Z isfixed throughoutadiscussion, which will bethe case inthe rest ofthis section. LetA,Bbegraded modules asabove, with A=AoEBA1and B=BoEBBI. Then the tensor product A0Bhas adirect sum decomposition A0B=EBA;0Bj. ;,j Wedefine agradingonA0Bbyletting (A0B)oconsist ofthe sum over indices i,jsuch that i+j=0(inZ/2Z), and(A0B)I consist ofthe sum over the indices i,jsuch that i+j=1. Suppose thatA,Baregraded algebras over thegiven commutative ring. There isaunique bilinear map ofA0Binto itself such that (a0b)(a' 0b')=(-I)U aa'0bb' ifa'EA;and bEBj.Just asinChapter XVI, 6, one verifies associativity and thefact that thisproduct gives rise toagraded algebra, whose product iscalled thesuper tensor product, orsuper product. As amatter ofnotation, when we take thesuper tensor product ofAandB,weshall denote theresulting algebra by A0u B todistinguish itfrom theordinary algebra A0BofChapter XVI, 6. Next suppose that Ehasdimension lover k.Then thefactor polynomial ring k[X]I(x2 -CI)isimmediately verified tobetheClifford algebra inthis case. We lettlbetheimage ofXinthefactor ring,soCg(E)=k[t.J withtt=CI. The vector space Eisimbedded asktIinthedirect sum kEBktI. Ingeneralwe now take thesuper tensor product inductively: Cg(E)=k[t.J 0suk[t2] 0su·· ·0suk[t n],with k[t;]=k[X]/(x2-Ci). Itsdimension is2n.Then Eisembedded inCg(E) bythemap alvl+... +anv n altl EB...EBantn. The desired commutation rules among t;,tjareimmediately verified from the definition ofthesuper product, thus concluding theproof ofthedimension of theClifford algebra. Note that theproof givesanexplicit representation oftherelations ofthe algebra, which also makes iteasy tocompute inthealgebra. Note further that thealternating algebra of afree module isaspecial case, taking C;=0forall i.Taking thec;tobealgebraically independent shows that thealternating algebra isaspecialization ofthegeneric Clifford algebra, orthat Clifford algebrasare what one calls perturbations ofthealternating algebra. Just asforthealternating algebra,wehave immediately from theconstruction: Theorem 4.2. Let g,g'bysymmetric formsonE,E'respectively. Then we 752 THE ALTERNATING PRODUCT XIX,4 have analgebra isomorphism C(g E9g')=C(g)0suC(g'). Examples. Clifford algebras have hadincreasingly wide applications in physics, differential geometry, topology, group representations (finite groups andLiegroups), and number theory. First, intopology Irefer toAdams [Ad62] and[ABS 64]giving applications oftheClifford algebra tovarious problems intopology, notablyadescription oftheway Clifford algebras over the reals arerelated totheexistence ofvector fields onspheres. Themultiplication inthe Clifford algebra gives rise toamultiplicationonthesphere, whence tovector fields. [ABS 64] also givesanumber ofcomputations related totheClifford algebra and itsapplications totopology andphysics. Forinstance, letE=Rn and letgbethenegative ofthestandard dotproduct. Ormore invariantly, take forEann-dimensional vector spaceover R,and letgbe anegative definite symmetric form onE.Let Cn=C(g). The operation VI0. . .0VrVr0. . .0VI=(VI0. . ·0vr)*forViEE induces anendomorphism ofTr(E) for r>O.Since V0V-g(v, v).1(for VEE)isinvariant under this operation, there isaninduced endomorphism *:Cn Cn'which isactuallyaninvolution, that isx**=xand(xy)*=y*x* for xECn. We letSpin(n) bethesubgroup ofunits inCngenerated bytheunit sphere inE(i.e.the setofelements such that g(v,v)= -1),andlying inthe even part ofCn.Equivalently, Spin(n) isthe group ofelements xsuch that xx*=1.The name dates back toDirac who used this group inhisstudy ofelec- tron spin. Topologists and others view that groupasbeing theuniversal cover- inggroup ofthespecial orthogonal group SO(n)=SUn(R). An account ofsome ofthe results of[Ad 62] and [ABS 64]will also be found in[Hu75], Chapter 11.Second Irefer totwo works encompassing two decades, concerning theheat kernel, Dirac operator, index theorem, andnumber theory, ranging from Atiyah, Bott and Patodi [ABP 73] toFaltings [Fa91], see especially 4,entitled "The local index theorem forDirac operators". The vector spacetowhich thegeneral theory isapplied ismostly thecotangent spaceata pointon amanifold. Irecommend thebook [BGV 92], Chapter 3. Finally, Irefer toBrocker and Tom Dieck forapplications oftheClifford algebra torepresentation theory, starting with their Chapter I,6,[BtD 85]. Bibliography [Ad 62] [ABP 73]F.ADAMS, Vector Fields onSpheres, Ann. Math. 75(1962) pp.603-632 M.ATIY AH, R.BOTT, V.PATODI, Ontheheatequation and theindex theorem, Invent. Math. 19(1973) pp.270-330; erratum 38(1975) pp.277-280 XIX, Ex EXERCISES 753 [ABS 64] M.ATIYAH, R.BOTT, A.SHAPIRO, Clifford Modules, Topology Vol. 3, Supp.1(1964) pp.3-38 [BGV 92] N.BERLINE, E.GETZLER, and M.VERGNE, Heat Kernels and Dirac Oper- ators, Springer Verlag, 1992 [8tD 85] T.BROCKER and T. TOM DIECK, Representations ofCompact LieGroups, Springer Verlag 1985 [Fa91] G.FALTINGS, Lectures onthearithmetic Riemann-Roch theorem, Annals of Math. Studies 1991 [Hu 75] D.HUSEMOLLER, Fibre Bundles, Springer Verlag, Second Edition, 1975 EXERCISES 1.LetEbeafinite dimensional vector space over afield k.LetXl'...,xpbeelements ofE such that XlA...AXpi=0,andsimIlarly YIA.../\Yp=1=O.IfCEkand XlA...AXp=CYIA...AYp show that xI'. . .,xpand YI'. . .,Ypgenerate the same subspace. Thus non-zero decomposablevectors inI\PE up to non-zero scalar multiples correspond to p-dimensional subspaces ofE. 2.Let Ebe afree module ofdimension nove( thecommutative ring R.Letf:E-+E bealinear map. LetlX,(f)=trI\r(f),where I\r(f)IStheendomorphism ofI\'(E) into itself induced byf.We have lXo(f)=1, lXI(f)=tr(f), lXn(f)=detf, andlX,(f)=0ifr>n.Show that det(1+f)=LlXr(f). rO [Hint: Asusual, prove thestatement whenfISrepresented byamatrix with variable coefficients over theintegers.] Interpret thelXr(f)Interms ofthecoefficients ofthe characteristic polynomial off. 3.Let Ebe afinite dimensional free module over thecommutative ring R.Let EVbe itsdual module. For each integerr>1show thatI\rE andI\rEvaredual modules toeach other, under thebilinear map such that (VI 1\...1\vnv;1\...1\v;) det«Vi'vi») where (Vi'vi)isthevalue ofvionVi'asusual, for ViEEandvjEEV. 4.NotatIon beingasinthepreceding exercise, letFbeanother R-module which isfree, finite dimensional. Letf:E-+Fbealinear map. Relative tothebilinear map ofthe preceding exercise, show that thetranspose of1\1isI\r('!),i.e.isequal tother-th alternating product ofthetranspose off. 5.Let Rbe acommutative ring. IfEisanR-module, denote byL(E) themodule of 754 THE ALTERNATING PRODUCT XIX, Ex r-multilinear alternating maps ofEinto Ritself (i.e. ther-multilinear alternating forms onE).LetL(E)=R,and let 0() Q(E)=EBL(E). r=O Show that Q(E) isagraded R-algebra, themultiplication being defined asfollows. If OJEL(E) and t/JEL(E), and Vh...,Vr+sareelements ofE,then (OJAt/J)(v h..., vr+s)=I£(0')OJ(Vo- 1,...,vo-r)t/!(Vo-(r+ 1)'...,vo-s)' the sum being taken over allpermutations0'of(1,...,r+s)such that 0'1<...<ar and O'(r+1)<...<O's. Derivations Inthefollowing exercises onderivations, allringsareassumed commutative. Among other things, theexercises give another proof oftheexistence ofuniversal derivations. Let R-+Abe aR-algebra (ofcommutative rings, according toourconvention). Wedenote themodule ofuniversal derivations ofAover Rby(dA/R,Q/R)'but wedonot assume that itnecessarily exists. Sometimes wewrite dinstead ofdA/Rforsimplicity ifthereference toAIR isclear. 6.Let A=R[X cx]be apolynomial ring invariables Xcx'where a.ranges over some indexing set,possibly infinite. LetQbethefree A-module onthesymbols dXcx'and let d:A-+Q bethemapping defined by ofdf(X)=L-;-dXcx. cxuXcx Show that thepair (d,Q)isauniversal derivation (dA/R,Q/R). 7.Let A-+Bbeahomomorphism ofR-algebras. Assume that theuniversal derivations forAjR, BjR, andBjA exist. Show that one has anatural exact sequence: B(8)AQ/R-+Qi/R-+Qi/A-+O. [Hint: Consider thesequence 0-+ DerA(B, M)-+DerR(B, M)-+DerR(A, M) which you prove isexact. Use thefact that asequence ofB-modules N' -+N-+N" -+0 isexact ifandonly ifitsHorn into M ISexact forevery B-module M.Apply this tothe sequence ofderivations.] 8.Let R-+AbeanR-algebra, and letIbeanideal ofA.LetB=AjI. Suppose that the universal derivation ofAover Rexists. Show that theuniversal derivation ofBover R. XIX, Ex EXERCISES 755 also exists, and that there isanatural exact sequence 1112B(8)AQ/R Qj/Ro. [Hint: LetMbeaB-module. Show that thesequence o DerR(B, M) DerR(A, M) HomB(III2 ,M) isexact.] 9.Let R Bbe anR-algebra. Show that theuniversal derivation ofBover Rexists asfollows. Represent Bas aquotient ofapolynomial ring, possibly ininfinitely many variables. Apply Exercises 6and 7. 10. Let R AbeanR-algebra. LetSobeamultiplicative subset ofR,and Samultiplicative subset ofAsuch that Somaps into S.Show that theuniversal derivation ofS-1 Aover So1Ris(d,S-lQ/R)'where d(als)=(sdA/R(a)-adA/R(s»/s2 . 11.Let BbeanR-algebra and MaB-module. OnBffiMdefine aproduct (b,x)(b', y)=(bb', by+b'x). Show that BffiMisaB-algebra, ifweidentifyanelement bEBwith (b,0).For any R-algebra A,show that thealgebra homomorphisms HomA1g/R(A,BEBM)consist of pairs (cp,D), where qJ:A Bisanalgebra homomorphism, and D:A Misa derivation fortheA-module structure onMinduced bycp. 12. Let AbeanR-algebra. Let t;:A Rbeanalgebra homomorphism, which wecall an augmentation. LetMbeanR-module. Define anA-module structure onMvia t;,by a.x=f,(a)x for aEA and xEM. Write Me.todenote Mwith this new module structure. Let: Dere(A, M)=A-module ofderivations forthet;-module structure onM I=Ker t;. Then Derl;(A, M) isanAll-module. Note that there isanR-module direct sum de- composition A=RffiI.Show that there isanatural A-module isomorphism QA/RIIQA/R 1112 and anR-module isomorphism Der£(A, M) HomR(III2 ,M). Inparticular, let'1:A 1112betheprojection ofAon1112relative tothedirect sum decomposition A=REBI.Then'1istheuniversal t;-derivation. Derivations and connections 13. Let R Abeahomomorphism ofcommutative rings, soweview AasanR-algebra. 756 THE ALTERNATING PRODUCT XIX, Ex Let EbeanA-module. Aconnection onEisahomomorphism ofabelian groups V:E-+QIR(8)AE such that for aEAand xEEwehave V(ax)=aV(x) +da(8)x, where the tensor product istaken over Aunless otherwise specified. The kernel ofV, denoted byEv ,iscalled thesubmodule ofhorizontal elements, orthehorizontal submodule of(E, V). (a)For anyinteger i>1,define "/\" 1Q/R='QAIR. Show that Vcan beextended toahomomorphism ofR-modules Vi:Q/R (8)E-+Qi (8)E by Vlw (8)x)=dw(8)x+(-l)iwAV(x). (b) Define thecurvature oftheconnection tobethemap K =VI0V:E-+Q/R (8)AE. Show that KisanA-homomorphism. Show that Vi+10Vi(w (8)x)=WAK(x) for wEQ/Rand xEE. (c)LetDer(AjR) denote theA-module ofderivations ofAinto itself, over R. LetVbeaconnection onE.Show that Vinduces aunique A-linear map V:Der(AjR)-+EndR(E) such that V(D)(ax)=D(a)x +aV(D)(x). (d) Prove theformula [V(D 1),V(D 2)]-V([D 1,D2])=(D 1AD2)(K)... Inthisformula, thebracket isdefined by[I,g]=log-go1fortwo endo- morphisms I,gofE.Furthermore, theright-hand side isthecomposed mapping K2 DIAD 2E-+QAIR (8)E )A(8)E E. XIX, Ex EXERCISES 757 14.(a)For anyderivation Dofaring AInto itself, prove Leibniz's rule: D"(xy)=Jo()Di(X)Dn-i(y). (b)Suppose Ahascharacteristic p.Show that DPisaderivation. 15. LetAIR beanalgebra, and letEbeanA-module with aconnection V.Assume that R hascharacteristic p.Define tjJ:Der(AjR) EndR(E) by tjJ(D)=(V(D»P-V(DP). Prove that tjJ(D) isA-linear. [Hint: Use Leibniz's formula and thedefinition ofa connection.] Thus theimage oftjJisactually inEndA(E). Some Clifford exercises 16. LetCg(E)betheClifford algebraasdefined in4.Define F;(Cg)=(k+E);,viewing Easembedded inCg.Define thesimilar object F;(f\E) inthealternating algebra. Then F;+ 1:JF;inboth cases, and wedefine thei-thgraded module gr;=F;/F;_I. Show that there isanatural (functorial) isomorphism gr;(Cg(E)) gr;(f\E). 17.Suppose that k=R, soEisareal vector space, which we now assume ofeven dimension 2m. We also assume that 9isnon-degenerate. Weomit theindex 9since thesymmetric form isnow fixed, and wewrite C+, C-forthe spaces ofdegree 0 and 1respectively intheZ/2Z-grading. For elements x,yinC+ orC-,define their supercommutator tobe {x,y}=xy-(-1)(degx)(degy)yx. Show that F2m-tisgenerated bysupercommutators. 18. Still assuming 9non-degenerate, letJbe anautomorphism of(E,g)(i.e. g(Jx, Jy)=g(x, y)forallx,yEE)such that J2=-ide LetEc=C(8)RE bethe extension ofscalars from RtoC.Then Echas adirect sum decomposition Ec=Ec EBEc into theeigenspaces ofJ,with eigenvalues1and-1respectively. (Proof?) There isarepresentation ofEconf\Ec, i.e. ahomomorphism Ec Endc(E c)whereby anelement ofEcoperates byexterior multiplication, and anelement ofEcoperates byinner multiplication, defined asfollows. Forx'EEcthere isaunique C-linear map having theeffect r x'(x 11\...1\xr)= -22:(-1);-1(x',x;)XI1\...1\x;1\...1\Xr. ;= 1 758 THE ALTERNATING PRODUCT XIX, Ex Prove that under thisoperation, you getanisomorphism Cg(E)c Endc(AE c). [Hint: Count dimensions.] 19. Consider theClifford algebraover R.The standard notation isCnifE=Rnwith thenegative definite form, andCifE=Rnwith thepositive definite form. Thus dim Cn=dimC=2n . (a) Show that Ct::::::C C;::::::RxR 20. Establish isomorphisms: C(8)R C=CxC; C(8)R H=Mz(C); H(8)R H=M4(R)Cz::::::H(the division ring ofquatemions) C::::::Mz(R) (2x2matrices over R) where Md(F)=dxdmatrices over F.For thethird one, with HQS)H,define an isomorphism I:H0RH HomR(H, H)=M4(R) byI(x (8)y)(z)=xzy, where ify=Yo+Yt;+yzj+Y3k then y=Yo-Yt;-yzj-Y3k . 21. (a) Establish isomorphisms Cn+Z=C(8)Cz and C+z=Cn(8)C. [Hint: Let{e(,. . .,en+z} betheorthonormalized basis with e[=-1. Then for .thefirstisomorphism map e; e;QS)e(eZ for;=1,..., nand mapen+], en+z on 1QS)e]and 1QS)ezrespectively.] (b) Prove that Cn+8=CnQS)M(6(R) (which iscalled theperiodicity property). (c)Conclude that Cnisasemi -simple algebra over Rforall n. From (c) one can tabulate thesimple modules over Cn. See[ABS 64], reproduced inHusemoller [Hu75], Chapter 11,6. Part Four HOMOLOGICAL ALGEBRA Intheforties andfifties (mostly intheworks ofCartan, Eilenberg, MacLane, andSteenrod, see[CaE 57]), itwas realized that there was asystematic way of developing certain relations oflinear algebra, depending onlyonfairly general constructions which were mostly arrow-theoretic, and were affectionately called abstract nonsense bySteenrod. (For amore recent text, see[Ro79].) The results formed abody ofalgebra,some ofitinvolving homological algebra, which had arisen intopology, algebra, partial differential equations, andalgebraic geometry. Intopology,some ofthese constructions had been used inpart togethomology andcohomology groups oftopological spacesasinEilenberg-Steenrod [ES52]. Inalgebra, factor sets andl-cocycles had arisen inthetheory ofgroup extensions, and, forinstance, Hilbert's Theorem 90. More recently, homological algebra hasentered inthecohomology ofgroups and therepresentation theory ofgroups. See forexample Curtis-Reiner [CuR 81], and any book onthecohomology of groups, e.g.[La96], [Se64], and [Sh72]. Note that [La96]was written topro- vide background forclass field theory in[ArT 68]. From anentirely different direction, Leray developedatheory ofsheaves and spectral sequences motivated bypartial differential equations. The basic theory ofsheaves was treated inGodement's book onthesubject [Go 58]. Fundamental insights were also given byGrothendieck inhomological algebra [Gro 57], tobeapplied byGrothendieck inthetheory ofsheaves over schemes inthefifties and sixties. InChapter XX, Ihave included whatever isnecessary ofhomological algebra forHartshorne's use in[Ha77]. Both Chapters XX and XXI giveanappropriate background forthehomological algebra used inGriffiths- Harris [GrH 78], Chapter 5(especially 3and4), andGunning [Gu90]. Chapter XX carries outthegeneral theory ofderived functors. The exercises andChapter XXI may beviewed asproviding examples andcomputations inspecific concrete instances ofmore specialized interest. 759 760 HOMOLOGICAL ALGEBRA PART FOUR The commutative algebra ofChapter Xand thetwochaptersonhomological algebra inthis fourth part also provideanappropriate background forcertain topics inalgebraic geometry such asSerre's study ofintersection theory [Se65] , Grothendieck duality, and Grothendieck's Riemann-Roch theorem inalgebraic geometry. See forinstance [SGA 6]. Finally Iwant todraw attention tothe useofhomological algebra incertain areas ofpartial differential equations,asinthepapers ofAtiyah-Bott-Patodi and Atiyah-Singeroncomplexes ofelliptic operators. Readers can trace some ofthe literature from thebibliography given in[ABP 73]. The choice ofmaterial inthis partwas toalarge extent motivated byallthe above applications. For thischapter, considering thenumber ofreferences and cross-references given, thebibliography fortheentire chapter isplacedattheendofthechapter. CHAPTER XX General Homology Theory To alarge extent thepresent chapter isarrow-theoretic. There isasubstantial body oflinear algebra which can beformalized very systematically, and con- stitutes what Steenrod called abstract nonsense, butwhich providesawell-oiled machinery applicabletomany domains. References will begiven along theway. Most ofwhat weshall doappliestoabelian categories, which were mentioned inChapter III, endof 3.However, infirstreading, Irecommend that readers disregard any allusions togeneral abelian categories and assume that we are dealing with anabelian category ofmodules over aring, orother specific abelian categories such ascomplexes ofmodules over aring. 1. COMPLEXES Let Abearing. Byanopen complex ofA-modules, one means asequence ofmodules andhomomorphisms {(Ei ,di)}, £i-lEi! Ei+l where iranges over allintegers and dimaps Eiinto Ei+1,and such that di0di-1=0 foralli. One frequently considers afinite sequence ofhomomorphisms, say El...Er 761 762 GENERAL HOMOLOGY THEORY xx, 1 such that thecomposite oftwo successive ones is0,and one can make this sequence into acomplex byinserting 0ateach end: -+0-+0 E1 ...Er-+0 0 Such acomplex iscalled afinite orbounded complex. Remark. Complexescan beindexed with adescending sequence ofintegers, namely, di+1 di-+Ei+1----. EiEi-1 When that notation isused systematically, then one uses upper indices for complexes which areindexed with anascending sequence ofintegers: -+Ei-1EiEi+1 Inthisbook, Ishall deal mostly with ascending indices. Asstated intheintroduction ofthischapter, instead ofmodules over aring, wecould have taken objects inanarbitrary abelian category. The homomorphisms diareoften called differentials, because some ofthe firstcomplexes which arose inpracticewere inanalysis, with differential operators and differential forms. Cf. theexamples below. Wedenote acomplexasabove by(E,d).Ifthecomplex isexact, itisoften useful toinsert thekernels and cokernels ofthedifferentials inadiagram as follows, letting Mi=Ker di=1mdi- 1. Ei-2) Ei-l) )Ei)Ei+1 //\/ Mi-lMiMi+l/\//\ o 0 0 0 Thus bydefinition, weobtain afamily ofshort exact sequences o-+MiEiMi+1O. Ifthecomplex isnot exact, then ofcourse wehave toinsert both theimage of di-1and thekernel ofdieThe factor (Ker di)/(Im di-1) will bestudied inthe next section. Itiscalled thehomology ofthecomplex, and measures thedeviation from exactness. xx, 1 COMPLEXES 763 LetMbeamodule. By aresolution ofMwe mean anexact sequence En En-1-+...-+Eo M O. Thus aresolution isanexact complex whose furthest term ontheright before oisM.The resolution isindexed asshown. Weusually write EMforthepart of complex formed only oftheE;'s, thus: EMis:En-+En-l...Eo, stopping atEo. We then write Eforthecomplex obtained bysticking 0on theright: Eis: En En-l-+ ...EoO. Iftheobjects Eioftheresolution aretaken insome family, then theresolution is qualified inthe same wayasthefamily. For instance, ifEiisfree foralli>0 then wesaythat theresolution isafree resolution. IfEiisprojective forall i>0then wesaythat theresolution isprojective. And soforth. The same terminology isapplied totheright, with aresolution o-+M -+EO -+E1 ... En-1-+En, also written o M -+EM. Wethen write Eforthecomplex o-+EO ElE2 ... . See5forinjective resolutions. Aresolution issaid tobefinite ifEi(orEi)=0forallbut afinite number of indices i. Example. Every module admits afree resolution (on theleft). This isa simple application ofthenotion offree module. Indeed, letMbe amodule, and let{Xj}be afamily ofgenerators, withjinsome indexing setJ.For eachjlet Rejbe afree module over Rwith abasis consisting ofone elementej'Let F=EBRej jeJ betheir direct sum. There isaunique epimorphism FMO sending ejonxj,Now weletM1bethekernel, andagain represent M1asthe quotient of afree module. Inductively,we can construct the desired free resolution. 764 GENERAL HOMOLOGY THEORY xx, 1 Example. The Standard Complex. Let Sbe aset. For i=0,1,2,. . . letEibethefree module over Zgenerated by(i+ 1)-tuples (xo,. . .,x;)with Xo,. . .,X;ES.Thus such (i+ 1)-tuples form abasis ofE;over Z.There isa unique homomorphism d;+I: E;+I E; such that ;+ 1 d;+I(Xo,..., X;+I)=L(-I)j(xo,..., Xj,..., x;+I), j=O where thesymbol Xjmeans that this term istobeomitted. For i=0,wedefine do:Eo Ztobetheunique homomorphism such that do(xo)=1.The map do issometimes called theaugmentation, and isalso denoted by£.Then weobtain aresolution ofZbythecomplex E;+ 1 E;.. ·Eo--4Z o. The formalism oftheabove maps d;ispervasive inmathematics. See Exercise 2forthe useofthestandard complex inthecohomology theory ofgroups. For still another example ofthis same formalism, compare with theKoszul complex inChapter XXI, 4. Given amodule M, one may form Hom(E;, M)foreach i,inwhich case one gets coboundary maps 5i:Hom(E i,M) Hom(E;+I' M), 5(f)=fodi+l , obtained bycomposition ofmappings. This procedure will beused toobtain deri ved functors in6.InExercises 2through 6,youwill seehow thisprocedure isused todevelop thecohomology theory ofgroups. Instead ofusing homomorphisms,one mayuse atopological version with simplices, andcontinuous maps, inwhich case thestandard complex gives rise to thesingular homology theory oftopological spaces. See[GreH 81], Chapter 9. Examples. Finite free resolutions. InChapter XXI, you will find other examples ofcomplexes, especially finite free, constructed invarious ways with different tools. This subsequent entire chapter may beviewed asproviding examples forthe current chapter. Examples with differential forms. InChapter XIX,3, wegave the exam- pleofthedeRham complex inanalgebraic setting. Inthetheory ofdifferential manifolds, the deRham complex hasdifferential maps d;:{}i {};+1 , sending differential forms ofdegree itothose ofdegree i+1,and allows for thecomputation ofthehomology ofthemanifold. Asimilar situation occurs incomplex differential geometry, when themaps d;aregiven bytheDolbeault a-operators ai:{}P'; {}p,;+1 xx, 1 COMPLEXES 765 operatingonforms oftype (p,i).Interested readers can look upforinstance Gunning's book [Gu 90]mentioned intheintroduction toPart IV,Volume I,E. The associated homology ofthiscomplex iscalled theDolbeault ora-cohom- ology ofthecomplex manifold. Let usreturn tothegeneral algebraic aspects ofcomplexes and resolutions. Itisaninteresting problem todiscuss which modules admit finite resoutions, and variations onthis theme. Some conditions arediscussed later inthischapter and inChapter XXI. Ifaresolution o En-+En-l...-+Eo M 0 issuch that Em=0for m>n,then wesaythat theresolution haslength<n (sometimes wesayithaslength nbyabuse oflanguage). Aclosed complex ofA-modules isasequence ofmodules andhomomorph- isms {(Ei ,di)}where iranges over the setofintegers mod nfor some n>2 and otherwise satisfying the same propertiesasabove. Thus aclosed complex looks like this: ElE2-+...-+En Wecall ntheJength oftheclosed complex. Without fear ofconfusion, one can omit theindex iondiand write just d. We also write (E,d)forthecomplex {(Ei ,di)},oreven more briefly,wewrite simply E. Let(E,d)and(E',d')becomplexes (both openorboth closed). Let rbean integer. Amorphism orhomomorphism (ofcomplexes) f:(E',d') (E,d) ofdegree risasequence h:E'iEi+r ofhomomorphisms such that forallithefollowing diagram iscommutative: E,(i-1) d-jIi-1 )Ei-1+r jd Ei+r E,iI, Just aswewrite dinstead ofdi ,weshall also writejinstead ofj.Ifthe com- plexes areclosed, wedefine amorphism from one into theother only ifthey have the same length. Itisclear that complexes form acategory. Infactthey form anabelian category. Indeed, saywedeal with complexes indexed byZforsimplicity, and morphisms ofdegreeO.Saywehave amorphism ofcomplexes f:C-+C"or 766 GENERAL HOMOLOGY THEORY xx, 1 putting theindices: )Cn j)Cn-l j) )C" )C"n n-l We letC=Ker(C n-+C). Then thefamily (C) forms acomplex, which we define tobethekernel off.Weletthereader check thedetails that this and a similar definition for cokernel and finite direct sums make complexes of modules into anabelian category. Atthispoint, readers should refer toChapter III,9, where kernels and cokernels arediscussed inthis context. The snake lemma ofthatchapter will now become central tothe next section. Itwill beuseful tohave another notion todeal with objects indexed bya monoid. Let Gbe amonoid, which we assume commutative and additive to fittheapplications wehave inmind here. Let{MJieG beafamily ofmodules indexed byG.The direct sum M =EBMi ieG will becalled theG-graded module associated with thefamily {MJ ieG.Let {MJieG and{M}ieG befamilies indexed byG,and letM,M'betheir asso- ciated G-graded modules. LetrEG. ByaG-graded morphism/: M' Mof degreerweshall mean ahomomorphism such thatj'maps Minto Mi+rfor each iEG(identifying Miwith thecorresponding submodule ofthe direct sum onthei-thcomponent). Thusfisnothing else than afamily ofhomo- morphismsh :M-+Mi+r. If(E,d)isacomplexwemay view EasaG-graded module (taking thedirect sum ofthecomponents ofthecomplex), and wemay view dasaG-graded morphism ofdegree 1,letting GbeZorZlnZ. The most common case we en- counter iswhen G=Z.Then wewrite thecomplexas E=EBEi,and d:E-+E maps Einto itself. The differential disdefined asdioneach direct summand Ei,and hasdegree 1. Conversely, ifGisZorZlnZ, one may view aG-graded module asacom- plex, bydefining dtobethe zero map. Forsimplicity,weshall often omit theprefix" G-graded"infront oftheword "morphism ",when dealing with G-graded morphisms. XX,2 HOMOLOGY SEQUENCE 767 2. HOMOLOGY SEQUENCE Let(E,d)beacomplex. Welet Zi(E)=Ker di and callZi(E) themodule ofi-cycles. We let Bi(E)=1mdi-1 and callBi(E) the module ofi-boundaries. Wefrequently write Ziand Bi instead ofZi(E) andBi(E), respectively. We let Hi(E)=ZilBi=Kerdi/Im di-l , and callHi(E) thei-thhomology group ofthecomplex. The graded module associated with thefamily {Hi} will bedenoted byH(E), and will becalled the homology ofE.One sometimes writes H*(E) instead ofH(E). Iff:E' Eisamorphism ofcomplexes, sayofdegree 0,then wegetan induced canonical homomorphism Hi(f):Hi(E') Hi(E) oneach homology group. Indeed, from thecommutative diagram defininga morphism ofcomplexes,one sees atonce thatfmaps Zi(E') intoZi(E) andBi(E') intoBi(E), whence theinduced homomorphism Hi(f). Compare with thebegin- ning remarks ofChapter III,9. One often writes this induced homomorphism asfi*rather than Hi(f), andifH(E) denotes thegraded module ofhomologyas above, then wewrite H(f)=f*:H(E') H(E). We callH(f) the map induced byfonhomology. IfHi(f) isanisomorphism foralli,then wesay thatfisahomology isomorphism. Note thatiff: E' Eand g:E E" aremorphisms ofcomplexes, then it isimmediately verified that H(g)0H(f)=H(g0f) and H(id)=ide Thus Hisafunctor from thecategory ofcomplexes tothecategory ofgraded modules. We shall consider short exact sequences ofcomplexes with morphisms of degree 0: o E' E!!.E" 0, 768 GENERAL HOMOLOGY THEORY XX,2 which written outinfulllook like this: I I I 0)E,(i-1)Ei-l)E,,(i-l))0) IfI I 0)E,i)Ei g)E"i)0 IfI I 0)E,(i+1))Ei+1 g)E,,(i+1))0 I I I 0)E,(i+2))Ei+2)E,,(i+2))0 I I I One candefineamorphism lJ:H(E") H(E') ofdegree 1,inother words, afamily ofhomomorphisms lJi:H"i H,(i+1) bythesnake lemma. Theorem 2.1. Let o E' E E" 0 beanexact sequence ofcomplexes with 1,gofdegree O.Then thesequence H(E')f. )H(E) ) H(E") isexact. This theorem ismerelyaspecial application ofthesnake lemma. Ifonewrites outinfullthehomology sequence inthetheorem, then itlooks like this: H,i Hi H"i H,(i+1)Hi+1H,,(i+1) XX,3 EULER CHARACTERISTIC AND THE GROTHENDIECK GROUP 769 Itisclear that our map isfunctorial (in anobvious sense), and hence that ourwhole structure (H,)isafunctor from thecategory ofshort exact sequences ofcomplexes into thecategory ofcomplexes. 3. EULER CHARACTERISTIC AND THE GROTHENDIECK GROUP This section may beviewed asacontinuation ofChapter III,8, onEuler- Poincare maps. Consider complexes ofA-modules, forsimplicity. LetEbe acomplex such that almost allhomology groups Hiareequal toO. Assume that Eisanopen complex. AsinChapter III,8, letcpbe anEuler- Poincare mappingonthecategory ofmodules (Le. A-modules). We define the Euler-Poincare characteristicXcp(E) (or more briefly theEuler characteristic) with respect tocp,tobe Xcp(E)=L(-IYq>(Hi) provided q>(Hi)isdefined forallHi,inwhich case wesaythatXcpisdefined forthe complex E. IfEisaclosed complex,weselect adefinite order (E1 ,...,En)fortheintegers mod nand define theEuler characteristic bytheformula n Xcp(E)=L(_l)iq;(Hi) i==1 provided again allq>(Hi)aredefined. For anexample, thereader may refer toExercise 28ofChapter I. One may view Hasacomplex, defining dtobethe zero map. Inthat case, we seethatXcp(H)isthealternating sum given above. More generally: Theorem 3.1. Let Fbe acomplex, which isofeven length ifitisclosed. Assume that q>(Fi)isdefined foralli,q>(Fi)=0foralmost alli,andHi(F)=0 foralmost alli.ThenXcp(F)isdefined, and Xcp(F)=L(-1Yq;(Fi). i Proof LetZiand Bibethe groups ofi-cycles and i-boundaries inFi respectively. We have anexact sequence o-+Zi -+FiBi+1o. HenceXlp(F)isdefined, and q>(Fi)=q>(Zi) +q>(Bi+1). 770 GENERAL HOMOLOGY THEORY XX,3 Taking thealternating sum, ourconclusion follows atonce. Acomplex whose homology istrivial iscalled acyclic. Corollary 3.2. Let Fbeanacyclic complex, such that qJ(Fi)isdefined for alli,andequal to0foralmost alli.IfFisclosed, we assume that Fhas even length. Then XqJ(F)=O. Inmany applications,anopen complex Fissuch that Fi=0foralmost alli,and one can then treat thiscomplexasaclosed complex bydefiningan additional map going from azero onthefarright toazero onthefarleft. Thus inthis case, thestudy ofsuch anopen complex isreduced tothestudy ofa closed complex. Theorem 3.3. Let o E' E E" -+0 beanexact sequence ofcomplexes, with morphisms ofdegreeO.Ifthe com- plexes areclosed, assume that their length iseven. LetqJbeanEuler-Poincare mapping onthecategory ofmodules. IfXqJisdefined for twooftheabove three complexes, then itisdefined forthethird, and wehave XqJ(E)=XqJ(E')+XqJ(E"). Proof We have anexact homology sequence H,,(i-1)H,i Hi H"i H,(i+1) This homology sequence isnothing but acomplex whose homology istrivial. Furthermore, each homology group belonging say toEisbetween homology groups ofE'and E". Hence ifXqJisdefined forE'and E"itisdefined forE. Similarly fortheother twopossibilities. Ifourcomplexes areclosed ofeven length n,then thishomology sequence has even length 3n. We can therefore apply thecorollary ofTheorem 3.1togetwhat wewant. For certain applications, itisconvenient toconstruct auniversal Euler mapping. Letabethe setofisomorphism classes ofcertain modules. IfEisa module, let[E] denote itsisomorphism class. Werequire that asatisfy the Euler-Poincare condition, Le.ifwehave anexact sequence o E' E E" -+0, then [E] isinaifandonly if[E'] and[E"] areina.Furthermore, the zero module isina. XX,3 EULER CHARACTERISTIC AND THE GROTHENDIECK GROUP 771 Theorem 3.4. Assume that asatisfies theEuler-Poincare condition. Then there isamap }':a K(a) ofainto anabelian group K(a) having theuniversal property with respect to Euler-Poincare maps defined ona. Toconstruct this, letFab(a)bethefree abelian group generated bythe setof such [E]. LetBbethesubgroup generated byallelements oftype [E]-[E']-[E"], where o E' E E" -+0 isanexact sequence whose members areina.WeletK(ct) bethefactor group Fab(a)IB, andlet}': a K(a) bethenatural map. Itisclear that}' has the universal property. Weobserve thesimilarity ofconstruction with theGrothendieck group ofa monoid. Infact, thepresent group isknown astheEuler-Grothendieck group ofa,with Euler usually leftout. The reader should observe that theabove argumentsarevalid inabelian categories, althoughwestill used theword module. Just aswith theelementary isomorphism theorems forgroups,wehave theanalogue oftheJordan-Holder theorem formodules. Ofcourse inthe case ofmodules, wedon't have toworry about thenormality ofsubmodules. We now goalittle deeper intoK-theory. Letabe anabelian category. In firstreading,one may wish tolimit attention toanabelian category ofmodules over aring. Letebe afamily ofobjects ina.Weshall saythateisaK-family ifitsatisfies thefollowing conditions. K1.eisclosed under taking finite direct sums, and 0isine. K2.Given anobject Einathere exists anepimorphism L-+EO with Line. K3. Let Ebeanobject admittingafinite resolution oflengthn o-+Ln...Lo E 0 with LiEeforalli.If ONFn-l ...Fo-+EO isaresolution with Ninaand F0'...,Fn- 1ine,then Nisalso ine. 772 GENERAL HOMOLOGY THEORY XX,3 We note that itfollows from these axioms that ifFisineand F'isiso- morphic toF,then F'isalso ine,asone seesbylooking attheresolution o-+F' F-+0 0 andapplying K3.Furthermore, givenanexact sequence o F' F-+F" 0 with Fand F"ine,then F'isine,again byapplying K3. Example. One may take forathecategory ofmodules over acommutative ring, and forethefamily ofprojective modules. Later weshall also consider Noetherian rings, inwhich case one may take finite modules, and finite pro- jective modules instead. Condition K2will bediscussed in8. From now on we assume that eisaK-family. For each object EinQ,we let[E] denote itsisomorphism class. Anobject Eofawill besaid tohave finite e-dimension ifitadmits afinite resolution with elements ofe.We let a(e) bethefamily ofobjects inawhich areoffinite e-dimension. We may then form the K(a(e»=Z[a(e)]IR(a(e» where R(a(e» isthe group generated byallelements [E]-[E']-[E"] arising from anexact sequence o-+E' -+E E" -+0 ina(e).Similarlywedefine K(e)=z[(e)]IR(e), where R(e) isthegroup ofrelations generatedasabove, buttaking E',E,E" ineitself. There arenatural maps Ya(e): a(e) K(a(e» and re:e K(e), which toeach object associate itsclass inthecorresponding Grothendieck group. There isalso anatural homomorphism (:K(e) K(a(e» since anexact sequence ofobjects ofecan also beviewed asanexact sequence ofobjects ofa(e). XX,3 EULER CHARACTERISTIC AND THE GROTHENDIECK GROUP 773 Theorem 3.5. LetMEa(e)and supposewehave two resolutions LM M 0andL-+M 0, byfinite complexes LMandL ine.Then L(-l)iye(L i)=L(-I)iye(L;). Proof Take first thespecialcase when there isanepimorphism L-+LM, with kernel Eillustrated onthefollowing commutative and exact diagram. o)E)L j M j o)LM j )Mid j o)0 The kernel isacomplex o-+En En- 1...-+Eo-+0 which isexact because wehave thehomology sequence Hp(E)Hp(L')Hp(L)Hp-l(E) For p>1wehaveHp(L)=Hp(L')=0bydefinition, soHp(E)=0forp>1. And forp=0weconsider the exact sequence Hl(L) Ho(E) Ho(L') Ho(L) Now wehave Hl(L)=0,and Ho(L') Ho(L) corresponds totheidentity morphisms onMsoisanisomorphism. Itfollows that Ho(E)=0also. Bydefinition ofK-family, theobjects Epareine.Then taking theEuler characteristic inK(e)wefind X(L')-x(L)=x(E)=0 which proves our assertion inthespecialcase. The general case follows byshowing that given two resolutions ofMine we canalways find athird one which tops both ofthem. The pattern ofour construction will begiven byalemma. 774 GENERAL HOMOLOGY THEORY XX,3 Lemma 3.6. Given two epimorphisms u:M Nand v:M' Nina, there exist epimorphisms F MandF M'with Finemaking thefollowing diagram commutative. /F M M'N/ Proof Let E=M xNM',that isEisthekernel ofthemorphism M xM' -+N given by(x,y) ux-vy.(Elements are notreally used here, and wecould write formallyu-vinstead.) There issome Fineand anepimorphism F E-+O.Thecomposition ofthisepimorphism with thenatural projections ofEoneach factor givesuswhat wewant. We construct acomplex L'Mgivingaresolution ofMwith acommutative and exact diagram: o 1 LM 1 L'M IL I o)M lid ,M lid)0)0 )M)0 The construction isdone inductively, soweputindices: L. )Li-1 I 1 1 L' )L'I 1- 1 I I L )L-1 1 XX,3 EULER CHARACTERISTIC AND THE GROTHENDIECK GROUP 775 Suppose that wehave constructed uptoL'_ 1with thedesired epimorphisms on Li-IandL_ 1.We want toconstruct L'. Let B;=Ker(L i-1 Li-2)and similarly forBandB'. Weobtain thecommutative diagram: )B.1 r B' 1 j)L;-1 r )L'1- 1 j)L;-2 r )L'1-2 jL.I L 1)B 1)L-1)L;-2 IfB;' B;orBi' B;arenotepimorphisms, then wereplace L'_ 1by L;'_ 1(f)Li(f)Li. We lettheboundary map toLi'- 2be0onthe new summands, andsimilarly define themaps toLi-1andLi- 1tobe0onLand Li-1respectively. Without loss ofgeneralitywemay now assume that B' -+B. and B' BI I I I areepimorphisms. We then use the construction ofthepreceding lemma. We let E. =L.'BB' and E =B' ,L 1 IQl.ill I\J:7BiI. Then both EiandEihave natural epimorphismsonBi'. Then welet N. =E.D" EI I\Q7Vj1 and wefind anobject Li'inewith anepimorphism Li'-+Ni.This gives usthe inductive construction ofL"uptothevery end. Tostop the process,we use K3and take thekernel ofthelastconstructed Li'toconclude theproof. Theorem 3.7. The natural map £:K(e) K(a(e» isanisomorphism. Proof. The map issurjective because givenaresolution OFn-+...FoM-+O with FiEeforalli,theelement L(-l)iye(F;) 776 GENERAL HOMOLOGY THEORY XX,3 maps onra(elM)under £.Conversely, Theorem 3.5shows that theassociation ML(-I)iye(Fi) isawell-defined mapping. Since foranyLEe wehave ashort exact sequence o-+L-+L 0,itfollows that thismapping following(.istheidentity onK(e), so (.isamonomorphism. Hence (.isanisomorphism,aswas tobeshown. Itmay behelpful tothereader actually toseethe next lemma which makes theadditivity oftheinverse more explicit. Lemma 3.8. Given anexact sequence in<1(e) o-+M' -+M -+M" 0 there exists acommutative and exact diagram o)LM' j )M' j o)LM" j M". j o.0)LM j )M j o)0 o withfinite resolutions LM',LM,LM"ine. Proof We first show that we can find L',L,L"inetofitanexact and commutative diagram o)L')L)L" j j j )M')M)M" j j j 0 0 0)0 o )0 We first select anepimorphism L" -+M"with L"ine.ByLemma 3.6there exists LlEeandepimorphisms Ll M,Ll L"making thediagram com- mutative. Then letL2-+M'beanepimorphism with L2Ee,andfinally define L=Ll(f)L2.Then wegetmorphisms L-+Mand L L"inthe obvious way. LetL'bethekernel ofL L". Then L2cL'sowegetanepimorphism L' -+M'. XX,3 EULER CHARACTERISTIC AND THE GROTHENDIECK GROUP 777 This now allows ustoconstruct resolutions inductively until wehitthe n-th step, where nissome integer such that M,M"admit resolutions oflength nine.The lasthorizontal exact sequence that weobtain is o-+L Ln-+L-+0 andLcan bechosen tobethekernel ofL_1L_2.ByK3weknow that Lliesine,and thesequence O L"L" n n- 1 isexact. This implies that inthe next inductive step,we can takeL+ 1=o. Then oL +1 Ln +1-+0 0 isexact, and atthe next stepwejust take thekernels ofthevertical arrows to complete thedesired finite resolutions ine.This concludes theproof ofthe lemma. Remark. The argument intheproof ofLemma 3.8infact shows: If o M' M M" 0 isanexact sequence in(1,andifM,M"havefinite e-dimension, then sodoes M'. Inthecategory ofmodules, one has amore precise statement: Theorem 3.9. Let (1bethecategory ofmodules over aring. Let (Pbethe family ofprojective modules. Given anexact sequence ofmodules o E' E E" -+0 ifany twoofE',E,E"admit finite resolutions in(pthen thethird does also. Proofs inamore subtle case will begiven inChapter XXI, Theorem 2.7. Next weshall use the tensor product toinvestigatearing structure onthe Grothendieck group. We suppose forsimplicity that wedeal with anabelian category ofmodules over acommutative ring, denoted by(1,together with aK- familyeasabove, but we now assume that (1isclosed under thetensor product. Theonly propertiesweshall actuallyuseforthe next results arethefollowing ones, denoted byTG(for "tensor" and "Grothendieck" respectively): TG 1. There isabifunctorial isomorphism giving commutativity M@NN@M forallM,Nin(1;andsimilarly fordistributivityover direct sums, andassociativity. 778 GENERAL HOMOLOGY THEORY XX,3 TG 2. For allLinethefunctor M L(8)Misexact. TG 3.IfL,L'areinethen L(8)L'isine. Then wemay giveK(e) thestructure ofanalgebra bydefining cle(L) cle(L')=cle(L (8)L'). Condition TG 1implies that thisalgebra iscommutative, and wecall itthe Grothendieck algebra. Inpractice, there isaunit element, butifwewant one in thepresent axiomatization, wehave tomake itanexplicit assumption: TG 4. There isanobject Rinesuch that R(8)M MforallMina. Then cle(R) istheunit element. Similarly, condition TG 2shows that we can define amodule structure on K(a) over K(e) bythe same formula cle(L) cIa(M)=cIa(L(8)M), andsimilarly K(a(e» isamodule over K(e), where werecall thata(e) isthe family ofobjects inawhich admit finite resolutions byobjects ine. Since weknow from Theorem 3.7 thatK(e)=K(a(e», wealso have a ring structure onK(a(e» viathisisomorphism. We then can make theproduct more explicitasfollows. Proposition 3.10. LetMEa(e) and letNEa.Let o Ln...Lo M 0 beafinite resolution ofMbyobjects ine.Then cle(M) cla(N)=L(-IY cla(Li (8)N). =L(-l)i cla(Hi(K» where Kisthecomplex o Ln(8)N...Lo(8)N-+M(8)N 0 andHi(K) isthei-thhomology ofthiscomplex. Proof. The formulas areimmediate consequences ofthedefinitions, andof Theorem 3.1 . Example. Letabetheabelian category ofmodules over acommutative ring. Letebethefamily ofprojective modules. From 6onderived functors thereader will know that thehomology ofthecomplex KinProposition 3.10 isjustTor(M, N). Therefore theformula inthatpropositioncan also bewritten c1e(M) cla(N)=L(-l)i cla(Tori(M, N». XX,3 EULER CHARACTERISTIC AND THE GROTHENDIECK GROUP 779 Example. Letkbe afield. Let Gbeagroup. Bya(G,k)-module, weshall mean apair (E,p),consisting ofak-space Eand ahomomorphism p:G Autk(E). Such ahomomorphism isalso called arepresentation ofGinE.Byabuse of language,wealso saythat thek-space EisaG-module. The group Goperates onE,and wewrite (JXinstead ofp«(J)x. The field kwill bekept fixed inwhat follows. LetModk(G) denote thecategory whose objectsare(G,k)-modules. Amor- phism inModk(G) iswhat wecall aG-homomorphism, that isak-linear map f:E Fsuch thatf(ax)=a-f(x) forall a-EG.The group ofmorphisms in Modk(G) isdenoted byHomG. IfEisaG-module, and (JEG,then wehave bydefinition ak-automorphism (J:E E.Since Trisafunctor, wehave aninduced automorphism Tr«(J): Tr(E) Tr(E) foreach r,and thus Tr(E) isalso aG-module. Taking thedirect sum, we see that T(E) isaG-module, and hence that Tisafunctor from thecategory of G-modules tothecategory ofgraded G-modules. Similarly forI\r,sr,and1\,S. Itisclear that thekernel ofaG-homomorphism isaG-submodule, and that thefactor module ofaG-module byaG-submodule isagainaG-module sothe category ofG-modules isanabelian category. We can now apply thegeneral considerations ontheGrothendieck group which wewrite K(G)=K(Modk(G» forsimplicity inthepresent case. We have thecanonical map cl:Modk(G) K(G). which toeach G-module associates itsclass inK(G). IfE,Fare G-modules, then their tensor product over k,E(8)F,isalso a G-module. Here again, theoperation ofGonE(8)Fisgiven functorially. If aEG,there exists aunique k-linear map E(8)F E(8)Fsuch that for xEE, YEFwehave x(8)Y «(Jx) (8)«(JY). The tensor product induces alaw of compositiononModk(G) because the tensor products ofG-isomorphic modules areG-isomorphic. Furthermore alltheconditions TG 1through TG 4aresatisfied. Since kisa field, wefind also thattensoringanexact sequence ofG-modules over kwith any G-module over kpreserves the exactness, soTG 2issatisfied forall(G,k)- modules. Thus theGrothendieck group K(G) isinfact theGrothendieck ring, ortheGrothendieck algebra over k. 780 GENERAL HOMOLOGY THEORY XX,3 ByProposition 2.1 and Theorem 2.3ofChapter XVIII, wealso see: The Grothendieck ringofafinite group Gconsisting ofisomorphism classes of finite dimensional (G,k)-spacesover afield kofcharacteristic 0isnaturally isomorphic tothecharacter ringXz(G). We can axiomatize this alittle more. Weconsider anabelian category of modules over acommutative ring R,which wedenote byaforsimplicity. For two modules M,NinaweletMor(M, N)asusual bethemorphisms ina,but Mor(M, N)isanabeliansubgroupofHomR(M, N). Forexample, wecould take atobethecategory of(G,k)-modules asintheexamplewehave just discussed, inwhich case Mor(M, N)=HomG(M, N). Weletebethefamily offinite free modules ina.We assume that esatisfies TG 1,TG 2,TG 3,TG 4,and also that eisclosed under taking alternating pro- ducts, tensor products andsymmetric products. We letK=K(e). As wehave seen, Kisitself acommutative ring. Weabbreviate cle=cl. Weshall define non-linear maps Ai:K K using thealternating product. IfEisfinite free, welet Ai(E)=cl(f\iE). Proposition1.1ofChapter XIX can now beformulated fortheK-ringasfollows. Proposition 3.11. Let o E' E E" 0 beanexact sequence offinite free modules ina.Thenforevery integern>0 wehave " A"(E)=LAi(E')A"- i(E"). i=O As aresult oftheproposition,we can define amap At:K 1+tK[[t]] ofKinto themultiplicative group offormal power series with coefficients inK, and with constant term 1,byletting 00 At(X)=LAi(X)ti . i=O XX,3 EULER CHARACTERISTIC AND THE GROTHENDIECK GROUP 781 Proposition 1.4ofChapter XIX can beformulated bysaying that: The map At:K 1+tK[[t]] isahomomorphism. We note thatifLisfreeofrank 1,then AO(L)=ground ring; Al(L)=cl(L); Ai(L)=0for i>1. This can besummarized bywriting At(L)= 1+cl(L)t. Next wecando asimilar construction with thesymmetric product instead of thealternating product. IfEisafinite free module ineweletasusual: S(E)=symmetric algebra ofE; Si(E)=homogeneous component ofdegree iinS(E). Wedefine ai(E)=cl(Si(E» and thecorresponding power series at(E)=Lai(E)ti . Theorem 3.12. Let Ebe afinite free module ina,ofrank r.Then forall integersn> 1wehave r L(-l)iAi(E)a"- i(E)=0, i==0 where bydefinition aj(E)=0forj<o.F'urthermore at(E)A-t(E)=1, sothepower series a,(E) and A_,(E) are inverse toeach other. Proof. The first formula dependsontheanalogue forthesymmetric product and thealternating product oftheformula given inProposition 1.1ofChapter 782 GENERAL HOMOLOGY THEORY XX,4 XIX. Itcould beproved directly now, butthereader will find aproofasaspecial case ofthetheory ofKoszul complexes inChapter XXI, Corollary 4.14. The power series relation isessentiallyareformulation ofthefirst formula. From theabove formalism, itispossible todefine other maps besides AJand u; . Example. Assume that thegroup Gistrivial, andjust write Kfor the Grothendieck ring instead ofK(1).For xEKdefine t/J-t(x)=-,:,logAt(x)=-,A;(x)/At(x). Show that «/I-tisanadditive andmultiplicative homomorphism. Show that «/1t(E)=1+cl(E) t+cl(E)2 t2+·.. . This kind ofconstruction with thelogarithmic derivative leads totheAdams operations «/I;intopology andalgebraic geometry. See Exercise 22ofChapter XVIII. Remark. Ifithappens inTheorem 3.12 that Eadmits adecomposition into I-dimensional free modules intheK-group, then theproof trivializes byusing thefact that At(L)=1+cl(L)t ifLisI-dimensional. But intheexample of (G,k)-spaces when kisafield, this isingeneral notpossible, anditisalso not possible inother examples arising naturally intopology andalgebraic geometry. However, by"changing thebase," one can sometimes achieve this simpler situation, butTheorem 3.12 isthen used inestablishing thebasic properties. Cf. Grothendieck [SGA 6],mentioned intheintroduction toPartIV, andother works mentioned inthebibliography attheend, namely [Ma 69], [At61], [At67], [Ba68], [Bo62]. The lectures byAtiyah and Bott emphasize thetopological aspectsasdistinguished from thealgebraic-geometric aspects. Grothendieck [Gr 68]actually shows how the formalism ofChern classes from algebraic geometry andtopology also enters thetheory ofrepresentations oflinear groups. See also theexposition in[FuL 85], especially theformalism ofChapter I,6. Forspecial emphasis onapplications torepresentation theory,seeBrocker-tom ieck [BtO 85], especially Chapter II,7,concerning compact Lie groups. 4. INJECTIVE MODULES InChapter III,4, wedefined projective modules, which have anatural relation tofree modules. Byreversing thearrows, we can define amodule Qto beinjective ifitsatisfies anyone ofthefollowing conditions which areequivalent: I1.Given any module Mand asubmodule M', and ahomomorphism f:M' Q,there exists anextension ofthis homomorphism toM, XX,4 INJECTIVE MODULES 783 that isthere exists h:M Qmaking thefollowing diagram commuta- tive : o )M')M 11/ 12. The functor M HomA(M, Q)isexact. I3.Every exact sequence 0 Q M -+M" -+0splits. We prove theequivalence. General considerations onhomomorphisms asin Proposition 2.1, show that exactness ofthehomed sequence may failonly at onepoint, namely given o M' M M" 0, thequestion iswhether HomA(M, Q) HomA(M', Q) 0 isexact. But this isprecisely thehypothesisasformulated inI1,soI1implies I2isessentiallyamatter oflinguistic reformulation, and infactI1isequivalent toI2. Assume I2orI1,which weknow areequivalent. TogetI3isimmediate, by applying lIto thediagram: o)Q M id1/ Toprove the converse, weneed thenotion ofpush-out (cf. Exercise 52of Chapter I).Given anexact diagram 0)M')M j Q weform thepush-out: M')M j 1 Q)N =Q(f)M' M. 784 GENERAL HOMOLOGY THEORY XX,4 Since M' Misamonomorphism, itisimmediately verified from theconstruc- tion ofthepush-out that Q Nisalso amonomorphism. ByI3,thesequence OQN splits, and wecan now compose thesplitting map N Qwith thepush-out map M Ntogetthedesired h:M Q,thus proving I1. We saweasily that every module isahomomorphic image ofafree module. There isnoequally direct construction forthedual fact: Theorem 4.1. Every module isasubmodule ofaninjective module. The proof will begiven bydualizing thesituation, with some lemmas. We first look atthesituation inthecategory ofabelian groups. IfMisanabelian group, letitsdual group beM"=Hom(M, Q/Z). IfFisafree abelian group, itisreasonable toexpect, and infact itiseasily proved that itsdual F"isan injective module, since injectivity isthedual notion ofprojectivity. Furthermore, Mhas anatural map into thedouble dualM"", which isshown tobe amono- morphism. Now represent M" asaquotient ofafree abelian group, F M" o. Dualizing this sequence yieldsamonomorphism o M"" F", and since Misembedded naturallyas asubgroup ofM"", wegetthedesired embedding ofMasasubgroup ofF". This proof also works ingeneral, but there aredetails tobefilled in.First wehave toprove that thedual ofafree module isinjective, and second wehave tobecareful when passing from thecategory ofabelian groups tothecategory ofmodules over anarbitrary ring. We now carry out thedetails. We saythat anabehan group Tisdivisible ifforevery integer m,thehomo- morphism mT:x mx issurjective. Lemma 4.2. IfTisdivisible, then Tisinjective inthecategory ofabelian groups. Proof. LetM' cMbeasubgroup ofanabelian group, and letf:M' -+T be ahomomorphism. Let xEM. We want first toextend ftothemodule (M', x)generated byM'and x.Ifxisfree over M', then weselect any value tET,and itisimmediately verified thatfextends to(M', x)bygiving thevalue f(x)=f.Suppose that xistorsion with respect toM', that isthere isa positive integermsuch that mxEM'. Let dbetheperiod ofxmod M', so XX,4 INJECTIVE MODULES 785 dxEM', and disthe least positive integer such that dxEM'. Byhypothesis, there exists anelement UETsuch that du=f(dx). For anyinteger n,and ZEM' define f(z +nx)=.f(z) +nu. Bythedefinition ofd,and thefact that Zisprincipal, one sees that this value forfisindependent oftherepresentation ofanelement of(M', x)intheform z+nx, and then itfollows atonce that this extended definition offisa homomorphism. Thus wehave extended fto(M', x). The restoftheproof ismerelyanapplication ofZorn's lemma. Weconsider pairs (N,g)consisting ofsubmodules ofMcontaining M', and anextension 9 offtoN. We saythat (N,g)<(Nl'gl)ifNcN1and therestriction ofgl toNisg.Then such pairs areinductively ordered. Let(N,g)beamaximal element. IfN=IMthen there issome xEM, xFJNand we canapply thefirst part oftheproof toextend thehomomorphism to(N,x),which contradicts themaximality, and concludes theproof ofthelemma. Example. The abelian groups Q/Z andR/Z aredivisible, and hence are injective inthecategory ofabelian groups. We can prove Theorem 4.1 inthecategory ofabelian groups following the pattern described above. IfFisafree abelian group, then thedualF/\isadirect product ofgroups isomorphic toQ/Z, and istherefore injective inthecategory ofabelian groups byLemma 4.2. This concludes theproof. Next wemust make thenecessary remarks toextend thesystem tomodules. Let Abearing and letTbeanabelian group. Wemake Homz(A, T)into an A-module asfollows. Letf:A Tbeanabelian group homomorphism. For aEAwedefine theoperation (af)(b)=f(ba). The rules foranoperation arethen immediately verified. Then foranyA-module Xwehave anatural isomorphism ofabelian groups: Homz(X, T)-=+HomA(X, Hornz(A, T». Indeed, lettjJ:X TbeaZ-homomorphism. We associate with tjJthehomo- morphism f:X Homz(A, T) such that f(x)(a)=tjJ(ax). 786 GENERAL HOMOLOGY THEORY XX,4 The definition oftheA-module structure onHomz(A, T)shows thatfisan A-homomorphism, soweget anarrow from Homz(X, T)to HomA(X, Homz(A, T». Conversely, letf:X Homz(A, T)beanA-homomorphism. We define the corresponding t/Jby t/J(x)=f(x)(l). Itisthen immediately verified that these maps areinverse toeach other. Weshall apply this when Tisany divisible group, although wethink ofT asbeing Q/Z, and wethink ofthehomomorphisms into Tasrepresenting the dual group according tothepattern described previously. Lemma 4.3. IfTisadivisible abelian group, then Homz(A, T)isinjective in thecategory ofA-modules. Proof. Itsuffices toprove that if0 X Yisexact inthecategory of A-modules, then thedual sequence obtained bytaking A-homomorphisms into Homz(A, T)isexact, that isthetopmap inthefollowing diagram issurjective. HomA(Y, Homz(A, T» 1)HomA(X, Homz(A, T» 1? )0 Homz(Y, T) Homz(X, T))0 But wehave theisomorphisms described before thelemma, given bythevertical arrows ofthediagram, which iscommutative. The bottom map issurjective because Tisaninjective module inthecategory ofabelian groups. Therefore thetopmap issurjective, thus proving thelemma. Now weprove Theorem 4.1forA-modules. LetMbeanA-module. We can embed Minadivisible abelian group T, o MLT. Then weget anA-homomorphism M Homz(A, T) byxfx'where fx(a)=f(ax). One verifies atonce that xfxgives anem- bedding ofMinHomz(A, T),which isaninjective module byLemma 4.3. This concludes theproof ofTheorem 4.1. XX,5 HOMOTOPIES OFMORPHISMS OFCOMPLEXES 787 5. HOMOTOPIES OF MORPHISMS OF COMPLEXES The purpose ofthis section istodescribe acondition under which homo- morphisms ofcomplexes induce the same map onthehomology and toshow that this condition issatisfied inanimportant case, from which wederive applications inthe next section. The arguments areapplicable toany abelian category. The reader may pre- fertothink ofmodules, but we use alanguage which applies toboth, and isno more complicated than ifweinsisted ondealing only with modules. Let E={(E", d")} and E' ={(E'", d'")} betwocomplexes. Let f,9:E E' betwomorphisms ofcomplexes (ofdegree 0).We saythatfis homotopic to9 ifthere exists asequence ofhomomorphisms h":E" E'("-1) such that J,-d'("-1)h+h d" "-gn-"" +1. Lemma 5.1. Iff, 9arehomotopic, thenf,9induce the same homomorphism onthehomology H(E), that is H(f")=H(g"): H"(E)-+H"(E'). Proof. The lemma isimmediate, because J"-g"vanishes onthecycles, which arethekernel ofd",and thehomotopy condition shows that theimage of J"-g"iscontained intheboundaries, that is,intheimage ofd'("- 1). Remark. The terminology ofhomotopy isused because the notion and formalism first arose inthe context oftopology. Cf.[ES52] and[GreH 81]. Weapply Lemma 5.1 toinjective objects. Note that asusual thedefinition ofaninjective module applies without change todefine aninjective object in any abelian category. Instead of asubmodule inI1,we use asubobject,or equivalentlyamonomorphism.Theproofs oftheequivalence ofthethree con- ditions defininganinjective module depended onlyonarrow-theoretic juggling, andapply inthegeneralcase ofabelian categories. We saythat anabelian category hasenough injectives ifgiven anyobject M there eXIsts amonomorphism OM-+I 788 GENERAL HOMOLOGY THEORY XX,5 into aninjective object. Weproved in4that thecategory ofmodules over a ring hasenough injectives. We now assume that theabeUan categorywework with hasenough injectives. By aninjective resolution ofanobject Mone means anexact sequence o-+M /0 /1 -+/2 -+ such that each]n(n>0)isinjective. Given M,such aresolution exists. Indeed, themonomorphism o M ]0 exists byhypothesis. LetMO beitsimage. Again byassumption, there exists a monomorphism 0-+]olMo ]1, and thecorresponding homomorphism 1°-+]1 has kernel MO. So wehave constructed thefirst step oftheresolution, and the next steps proceed inthe same fashion. Aninjective resolution isofcourse notunique, butithas some uniqueness which wenow formulate. Lemma 5.2. Consider twocomplexes: )EO)£1)£2) ... o)M ! o)M')/0)]1)/2) ... Suppose that thetop row isexact, and that each In(n>0)isinjective. Let qJ:M -+M'be agiven homomorphism. Then there exists amorphism fof complexes suchthatf_ 1=qJ;and any two such arehomotopic. Proof Bydefinition ofaninjective, thehomomorphism M ]0viaM' extends toahomomorphism fo:EO -+/° , which makes thefirst square commute: . 1:0 M')]0 XX,5 HOMOTOPIES OFMORPHISMS OFCOMPLEXES 789 Next wemust construct fl. Wewrite thesecond square intheform o EOIM fo! 1°)11)El with the exact top row asshown. Again because 11isinjective,wecanapply the same argument and findfltomake thesecond square commute. And soon, thus constructing themorphism ofcomplexes f Suppose f,9are two such morphisms. We define ho:EO -+M' tobeO. Then thecondition for ahomotopy issatisfied inthefirst instance, when f-l =g-1 ==qJ. Next letd-1 :M -+EObetheembedding ofMinEO. Since 1°isinjective, we can extend dO:EO/1m d-l-+El toahomomorphism hI:El1°.Then thehomotopy condition isverified for fo-go. Since ho=0weactually have inthis case fo-go=hIdO, but thissimplification ismisleading fortheinductive step which follows. We assume constructed themaphn+1,and wewish toshow theexistence ofhn+2 satisfying fn+ 1-gn+ 1=d,nhn+I+hn+2dn+l . Since 1mdn=Ker dn+I,wehave amonomorphism En+l/Im dnEn+2.By thedefinition ofaninjective object, which inthis case isIn+1,itsuffices toprove that In+I-gn+1-d,nhn+1vanishes ontheimage ofdn , and tousetheexact diagram: o)En+111m dn fn+I-gn +I! In+1.En+2 toget the existence ofhn+2:En+2-+In+1extending fn+1-gn+1.But we have : (fn+ 1-gn+ 1-d,nhn+l)dn =(f,.+ 1-gn+l)dn-d,nhn+Idn 790 GENERAL HOMOLOGY THEORY XX,6 -(I, )dnd,n(f, d,(n-l) h)-n+ 1-gn+ 1- n-gn- n =(fn+ 1-gn+l)dn-d,n(fn-gn) =0byinduction because d'd' =0 because 1,9are homomorphisms of complexes. This concludes theproof ofLemma 5.2. Remark. Dually, letPM' M' 0be acomplex with piprojective for i>0,andletEM M Obearesolution. Letq>: M' Mbeahomomorphism. Then ({Jextends toahomomorphism ofcomplex P E.Theproof isobtained byreversing arrows inLemma 5.2. The books onhomological algebra that I know ofinfact carry out theprojective case, and leave theinjectivecase tothe reader. However, one ofmymotivations istodohere what isneeded, for instance in[Ha77], Chapter III, onderived functors, as apreliminarytothe cohomology ofsheaves. For anexample ofprojective resolutions using free modules, seeExercises 2-7, concerning thecohomology ofgroups. 6. DERIVED FUNCTORS Wecontinue towork inanabelian category. Acovariant additive functor F:a(B issaid tobeleft exact ifittransfdrms anexact sequence o M' M M" into anexact sequence 0F(M') F(M)-+F(M"). We remind the reader that Fiscalled additive ifthemap Hom(A', A)-+Hom(F A',FA) isadditive. We assume throughout that Fisleft exact unless otherwise specified, and additive. We continue toassume that our abelian category hasenough in- jectives. Given anobject M,let o M ]0 ]1 -+]2 beaninjective resolution, which weabbreviate by o M ]M, where ]Misthecomplex ]0 11]2. Welet] bethecomplex o ]0 ]1 ]2 XX,6 DERIVED FUNCTORS 791 Wedefine theright-derived functor R"F by R"F(M)=H"(F(I», inother words, then-th homology ofthecomplex oF(IO) F(Il)F(I2) Directly from thedefinitions and themonomorphism M 1o,we seethat there isanisomorphism ROF(M)=F(M). This isomorphism seems atfirst todepend ontheinjective resolution, and so dothefunctors R"F(M) forother n.However, from Lemmas 5.1and 5.2 we seethatgiven twoinjective resolutions ofM,there isahomomorphism between them, and that any twohomomorphisms arehomotopic. Ifwe apply thefunctor Ftothese homomorphisms and tothehomotopy, then we seethat thehomology ofthecomplex F(I) isinfact determined uptoaunique isomorphism. One therefore omits theresolution from thenotation and from thelanguage. Example 1. Let Rbe aring and leta=Mod(R) bethecategory ofR- modules. Fix amodule A.The functor M Hom(A, M)isleftexact, Le.given anexact sequence 0 M' M M", thesequence o Hom(A, M') Hom(A, M) Horn (A,M") isexact. Itsright derived functors aredenoted byExtn(A, M)forMvariable. Similarly, for afixed module B,thefunctor X Horn (X,B)isright exact, and itgives rise toitsleft derived functors. For theexplicit mirror image of theterminology,seetheendofthis section. Inany case, wemay consider Aas variable. In8weshall gomore deeply into this aspect oftheformalism, by dealing with bifunctors. Itwill turn outthat Extn(A,B)has adual interpretation as aleft derived functor ofthefirst variable andright derived functor ofthe second variable. SeeCorollary 8.5. Intheexercises, you will prove that Ext1(A,M) isinbijection with iso- morphism classes ofextensions, ofMbyA,thatis,isomorphism classes ofexact sequences o A E M o. The name Ext comes from thisinterpretation indimension 1. For thecomputation ofExt; incertain important cases, seeChapter XXI, Theorems 4.6 and4.11, which serve asexamples forthegeneral theory. Example 2. Let Rbecommutative. The functor M A0Misright exact, inother words, thesequence A0M' A0M A0M" 0 isexact. Itsleft derived functors aredenoted byTorn(A, M)forMvariable. 792 GENERAL HOMOLOGY THEORY XX,6 Example 3. Let Gbe agroup and letR=Z[G] bethegroup ring. Let Ci bethecategory ofG-modules, Le. Ci=Mod(R), also denoted byMod(G). For aG-module A,letAGbethe submodule (abelian group) consisting ofthose elements vsuch that xv=vforallxEG.Then A AGisaleft exact functor from Mod(R) into thecategory ofabelian groups. Itsleftderived functors give rise tothecohomology ofgroups. Some results from thisspecial cohomology will becarried out intheexercises, asfurther examples ofthegeneral theory. Example 4. Let Xbe atopological space (we assume the reader knows what thisis).Byasheaf ofabelian groupsonX,we mean thedata: (a)For every opensetUofXthere isgivenanabelian group (U). (b)For every inclusion VCUofopensets there isgivenahomomorphism res:(U) (V), called therestriction from UtoV,subject tothefollowing conditions: SH 1.(empty set)=O. SH 2.res istheidentity (U) (U). SH3.IfWeve Uareopen sets, then resw0res=res. SH 4.Let Ube anopenset and{\';} be anopen covering ofU.Let sE(U). Iftherestriction ofstoeach \';is0,then s.O. SH5.Let Ube anopen setand let{\';} beanopen covering ofU.Suppose given s;E(\';) foreach i,such thatgiven i,jtherestrictions ofs; andSjto\';nVjareequal. Then there exists auniqueSE(U)whose restriction to\';isS;foralli. Elements of(U) arecalled sections of over U.Elements of(X) arecalled global sections. Just asforabelian groups, itispossible todefine thenotion of homomorphisms ofsheaves, kernels, cokernels, and exact sequences. The asso- ciation (X)(global sections functor) isafunctor from thecategory of sheaves ofabelian groups toabelian groups, and this functor isleft exact. Its right derived functors arethebasis ofcohomology theory intopology andalgebraic geometry (among other fields ofmathematics). The reader will find aself- contained brief definition ofthebasic properties in[Ha77], Chapter II, 1,as well asaproof that these form anabelian category. For amore extensi vetreatment Irecommend Gunning's [Gu 91], mentioned intheintroduction toPart IV, notably Volume III,dealing with thecohomology ofsheaves. We now return tothegeneral theory ofderived functors. Thegeneral theory tells usthat these derived functors donotdependontheresolution byprojectives orinjectives according tothevariance. As weshall also seein8, one can even useother special types ofobjects such asacyclicorexact (tobedefined), which giveseven more flexibility inthewaysone has tocompute homology. Through certain explicit resolutions, weobtain means ofcomputing thederived functors XX,6 DERIVED FUNCTORS 793 explicitly. Forexample, inExercise 16,you will see that thecohomology of finite cyclic groupscan becomputed immediately byexhibitingaspecific free resolution ofZadaptedtosuch groups. Chapter XXI will contain several other examples which show how toconstruct explicit finite free resolutions, which allow thedetermination ofderived functors invarious contexts. The next theorem summarizes thebasic properties ofderived functors. Theorem 6.1. Letabeanabelian category with enough injectives, and let F:a illbeacovariant additive left exact functor toanother abelian cate- gory (B.Then: (i)For each n>0,R"F asdefined above isanadditive functor from a to(B.Furthermore, itisindependent, uptoaunique isomorphism of functors, ofthechoices ofresolutions made. (ii) There isanatural isomorphism F ROF. (iii) Foreach short exact sequence o M' M M" 0 andforeach n>0there isanatural homomorphism lJ" :R"F(M") R"+1F(M) such that weobtain along exact sequence: R"F(M') R"F(M) R"F(M") R"+1F(M')-+. (iv) Given amorphism ofshort exact sequences )M I)0 o)M' I)M" I o)N')N)N")0 thelJ'sgive acommutative diagram: R"F(M") I R"F(N")b" )R"+1F(M') I )R"+1F(N')b" (v)For each injective object IofAandforeach n>0wehaveR"F(I)=O. Properties (i),(ii),(iii), and(iv)essentially saythat R"Fisadelta-functor ina sense which will beexpanded inthe next section. The lastproperty (v)will be discussed after wedeal with thedelta-functor part ofthetheorem. 794 GENERAL HOMOLOGY THEORY XX,6 We now describe how toconstruct the<5-homomorphisms. Given ashort exact sequence,wecanfind aninjective resolution ofM',M,M"separately, but they don't necessarily fitinanexact sequence ofcomplexes. Sowemust achieve this toapply theconsiderations of91.Consider thediagram: o0 0 0 j 1 j )M')M)M" j j j )]'0)X)]"0)0 o)o. Wegivemonomorphisms M' ]'0andM" ]"0intoinjectives, and wewant to find Xinjective with amonomorphism M Xsuch that thediagram isexact. Wetake Xtobethedirect sum x =]'0 EE>]"0. Since ]'0isinjective, themonomorphism M' ]'0can beextended toahomo- morphism M ]'0. We take thehomomorphism ofMinto ]'0 EE>]"0which comes from this extension onthefirst factor 1'0,and isthecomposite map M M" /"0 onthesecond factor. Then M Xisamonomorphism. Furthermore ]'0 X isthemonomorphism onthefirstfactor, and X ]"0istheprojection onthe second factor. So wehave constructed thediagram wewanted, giving the beginning ofthecompatible resolutions. Now wetake thequotient homomorphism, defining thethird row, toget an exact diagram: o0 0 0 j j j )M')M)M" j j j )]'0)]0)]"0 j j j )N')N)N" j j j 0 0 0)0 o )0 o )0 XX,6 DERIVED FUNCTORS 795 where welet1° =X,andN',N,N" arethecokernels ofthevertical maps by definition. The exactness oftheN-sequence isleft asanexercise tothereader. Wethen repeat theconstruction with theN-sequence, andbyinduction construct injective resolutions o0 0 0 j j j )M' )M)M" j j j )l,)1M I")M")0 o )0 oftheM-sequence such that thediagram oftheresolutions isexact. We now apply thefunctor Ftothisdiagram. Weobtain ashort sequence of complexes: oF(I') F(I) F(I") 0, which isexact because I=I'(f)I"isadirect sum and Fisleftexact, soFcom- mutes with direct sums. We are now inaposition toapply theconstruction of 1togetthecoboundary operator inthehomology sequence: R"F(M')-+R"F(M) R"F(M") R"+1F(M'). This islegitimate because theright derived functor isindependent ofthechosen resolutions. Sofar, wehave proved (i),(ii),and(iii). Toprove (iv), that isthenaturality of thedelta homomorphisms, itisnecessary togothroughathree-dimensional commutative diagram. Atthispoint, Ifeelitisbest toleave this tothereader, since itisjust more ofthe same routine. Finally, thelast property (v)isobvious, forifIisinjective, then we can usetheresolution OIIO tocompute thederived functors, from which itisclear that R"F=0for n>O. This concludes theproof ofTheorem 6.1. Inapplications, itisuseful todetermine thederived functors bymeans of other resolutions besides injectiveones (which are useful for theoretical purposes, but notforcomputational ones). Letagain Fbealeftexact additive functor. Anobject Xiscalled F-acyclic ifRnF(X)=0forall n>O. 796 GENERAL HOMOLOGY THEORY XX,6 Theorem 6.2. Let o M -+XO Xl X2 ... bearesolution ofMbyF-acyclics. Let o M ]0 -+]1 ]2... beaninjective resolution. Then there exists amorphism ofcomplexes XM ]M extending theidentity onM,and thismorphism induces anisomorphism H"F(X) H"F(I)=R"F(M) forall n>O. Proof The existence ofthemorphism ofcomplexes extending theidentity onMismerely Lemma 5.2. The usual proof ofthetheorem viaspectralse- quencescan beformulated independently inthefollowing manner, shown to mebyDavid Benson. We need alemma. Lemma 6.3. Letyi(i>0)beF-acyclic, and suppose thesequence o-+yO yl y2... isexact. Then oF(YO) F(yl) F(y2)... isexact. Proof Since Fisleftexact, wehave anexact sequence oF(YO) F(yl) F(y2). We want toshow exactness atthe nextjoint. Wedraw thecokernels: o)yO)yl)y2)y3\/\/ Zl Z2/\/\000 SoZl=Coker(Yo yl); Z2=Coker(yl y2); etc. Applying Fwehave anexact sequence oF(YO) F(yl) F(Zl) RIF(YO)=O. XX,6 DERIVED FUNCTORS 797 SoF(Zl)=Coker(F(YO)-+F(y1».We now consider the exact sequence O-+ZlY2-+Y3 givingthe exact sequence o-+F(Zl) F(y2) F(y3) bythe left-exactness ofF,and proving what wewanted. But we can now continue byinduction because Z1isF-acyclic, bythe exact sequence oRnF(yl)-+R"F(Zl)-+R"+IF(Yo)=O. This concludes theproof ofLemma 6.3. We return totheproof ofTheorem 6.2. Theinjective resolution o M 1M can bechosen such that thehomomorphisms X" I"aremonomorphisms for n>0,because thederived functor isindependent ofthechoice ofinjective resolution. Thus wemay assume without loss ofgenerality that wehave an exact diagram: oo j )XO j )1° j )yO j o)M idj o)M oo j )Xl j )J1 j )yl j oo j )X2 j )12 j )y2 j o defining Y"astheappropriate cokernel ofthevertical map. Since X"and I"areacyclic, soisY"from the exact sequence RkF(I") RkF( Y") Rk+1F(X"). Applying Fweobtain ashort exact sequence ofcomplexes oF(X) F(I) F(Y) O. 798 GENERAL HOMOLOGY THEORY XX,6 whence thecorresponding homology sequence H"-lF(Y) H"F(X) H"F(I)-+H"F(Y). Both extremes are0byLemma 6.3, sowegetanisomorphism inthemiddle, which bydefinition istheisomorphism H"F(X) R"F(M), thus proving thetheorem. Left derived functors We conclude this section byasummary oftheproperties ofleftderived functors. Weconsider complexes going theother way, X"...-+X2Xl XoMO which weabbreviate by XM M O. Wecall such acomplexaresolution ofMifthesequence isexact. Wecall ita projective resolution ifX"isprojective forall n>O. Given projective resolutions XM,YM,and ahomomorphism lp:M M' there always exists ahomomorphism XM YM,extending lp,and any two such arehomotopic. Infact, one need only assume that XMisaprojective resolution, and that YM,isaresolution, notnecessarily projective, fortheproof togothrough. LetTbeacovariant additive functor. Fix aprojective resolution ofanob- jectM, PM M O. Wedefine theleftderived functor L"Tby LnT(M)=H"(T(P», where T(P) isthecomplex T(P")-+ ...T(P 2)T(P l)T(Po) O. The existence ofhomotopies shows thatL"T(M) isuniquely determined up toaunique isomorphism ifonechanges theprojective resolution. Wedefine Ttoberight exact ifanexact sequence M' M -+M" 0 XX,7 DELTA-FUNCTORS 799 yieldsanexact sequence T(M') T(M) T(M") O. IfTisright exact, then wehave immediately from thedefinitions LoT(M) M. Theorems 6.1and 6.2then goover tothis case with similar proofs. One has toreplace "injectives" by"projectives" throughout, and inTheorem 6.1, thelastcondition states that for n>0, L"T(P)=0 ifPisprojective. Otherwise, itisjustaquestion ofreversing certain arrows intheproofs. For anexample ofsuch left derived functors, see Exercises 2-7 concerning the cohomology ofgroups. 7. DELTA-FUNCTORS Inthissection, weaxiomatize theproperties stated inTheorem 6.1following Grothendieck. Leta, CBbeabelian categories. A(covariant) -functor from atoCBisa family ofadditive functors F={F"}"o,and toeach short exact sequence o M' M M" 0 anassociated family ofmorphisms ":F"(M") F"+l(M') with n>0,satisfying thefollowing conditions: DEL I.For each short exact sequenceasabove, there isalong exact sequence oFO(M') FO(M) FO(M") Fl(M')-+. .. -+F"(M') F"(M) F"(M") F"+l(M') DEL 2.For each morphism ofone short exact sequenceasabove into another 0 N' N N" 0,the'sgiveacommutative diagram: F"(M") ! F"(N")lJ )F"+l(M') ! lJ)F"+l(N'). 800 GENERAL HOMOLOGY THEORY XX,7 Before going anyfurther, itisuseful togive another definition. Many proofs inhomology theoryaregiven byinduction from one index tothe next. Itturns outthat theonly relevant data forgoing upbyone index isgiven intwo succes- sivedimensions, and that theother indices areirrelevant. Therefore wegeneral- izethenotion of<5-functor asfollows. Ao-functor defined indegrees 0,1isapair offunctors (F\Fl) and to each short exact sequence o A' -+A-+A" -+0 anassociated morphism <5:FO(A") Fl(A") satisfying the two conditions asbefore, butputtingn=0,n+1=1,and for- getting about allother integersn.Wecould also useany two consecutive posi- tive integers toindex the<5-functor, orany sequence ofconsecutive integers >O.Inpractice, only the case ofallintegers>0occurs, butforproofs, itis useful tohave theflexibility provided byusing only two indices, say0,1. The b-functor Fissaid tobeuniversal, ifgiven any other <5-functor Gofa into CB,andgiven anymorphism offunctors fo:FO GO, there exists aunique sequence ofmorphisms in:F" Gn forall n>0,which commute with the <5"foreach short exact sequence. Bythedefinition ofuniversality,a<5-functor Gsuch that GO =FOisuniquely determined uptoaunique isomorphism offunctors. Weshall giveacondition for afunctor tobeuniversal. Anadditive functor Fofainto CBiscalled erasable iftoeach object Athere exists amonomorphismu:A Mfor some Msuch thatF(u)=O.Inpractice, iteven happens thatF(M)=0,but wedon't need itintheaxiomatization. Linguistic note. Grothendieck originally called thenotion "effaceable" in French. Thedictionary translation is"erasable," asIhave used above. Ap- parently people who didnotknow French have used theFrench word inEnglish, butthere isnoneed forthis, since theEnglish word isequally meaningful and convenient. We saythefunctor iserasable byinjectives ifinaddition Mcan betaken to beinjective. xx, 7 DELTA-FUNCTORS 801 Example. Ofcourse, aright derived functor iserasable byinjectives, and aleftderived functor byprojectives. However, there aremanycases when one wants erasability byother types ofobjects. InExercises 9and 14,dealing with thecohomology ofgroups, youwill seehow one erases thecohomology functor with induced modules, orregular modules when Gisfinite. Inthecategory of coherent sheaves inalgebraic geometry, one erases thecohomology with locally free sheaves offinite rank. Theorem 7.1. LetF={F"} beacovariant 1unctor from Ciinto (B.ifF" is erasable foreach n>0,then Fisuniversal. Proof Given anobject A,we erase itwith amonomorphism u,and geta short exact sequence: o-+A M X-+O. Let Gbeanother -functor with given fo:FO -+GO. We have anexact com- mutative diagram FO(M))FO(X){), )F1(A))0 fO ) fO), I I :fI? I ,I, GO(M))GO(X){)G)Gl(A) Wegetthe0onthetopright because oftheerasability assumption that Fl(cp)=O. We want toconstruct fl(A):F1(A)-+Gl(A) which makes thediagram commutative, isfunctorial inA,and also commutes with the. Commutativity intheleftsquare shows that KerFiscontained in thekernel ofG0fo. Hence there exists aunique homomorphism fl(A): F1(A) Gl(A) which makes theright square commutative. We aregoing toshow thatfl(A) satisfies thedesired conditions. The restoftheproof then proceeds byinduction following the same pattern. Wefirst prove thefunctoriality inA. Let u:A Bbeamorphism. Weform thepush-out Pinthediagram AcP )M u) ) B)p 802 GENERAL HOMOLOGY THEORY XX,7 Sinceq>isamonomorphism, itfollows that B Pisamonomorphism also. Then weletP Nbeamonomorphism which erases Pl'This yieldsacom- mutative diagram o)M vj)0)A uj)X wj o)B)N)f)0 where B Nisthecomposite B P N,andfisdefined tobethecokernel ofB N. Functoriality inAmeans that thefollowing diagram iscommutative. Pl(A) f1(A)j Gl(A)Fl(U))Fl(B) jft(B) )Gl(B)Fl(u) This square istheright-hand side ofthefollowingcube: DFF'(A) DrF'(B)FO(X) W).1;)(X) F()(Y) f>GG1(A) GO(X)----I,(B) G'(B) GO(Y) Allthe faces ofthecube arecommutative except possibly theright-hand face. Itisthen ageneral factthatifthetopmaps here denoted byFareepimorphisms, xx, 7 DELTA-FUNCTORS 803 then theright-hand side iscommutative also. This can beseen asfollows. We start withfl(B)Fl(U)F' We then usecommutativity onthetopofthecube, then thefront face, then theleftface, then thebottom, andfinally theback face. This yields fl(B)Fl(U)F=Gl(u)fl (A)F. Since Fisanepimorphism,we can cancel Ftogetwhat wewant. Second, wehave toshow thatft commutes with. Let o A' A A" 0 beashort exact sequence. The same push-out argumentasbefore shows that there exists anerasing monomorphism 0 A' -+Mand morphisms v,w making thefollowing diagram commutative: o)A' idl )A'All ) Iw)A Iv)0 o)M)X)0 Here Xisdefined astheappropriate cokernel ofthebottom row. We now consider thefollowing diagram: GO(X)FO(A ") .ro fJF GO(A")/6F"'"FI(A') ft(A') fJG GI(A') ..FO(",) Fo(X) to Our purpose istoprove that theright-hand face iscommutative. Thetriangles ontop and bottom arecommutative bythedefinition ofa-functor. The 804 GENERAL HOMOLOGY THEORY XX,7 left-hand square iscommutative bythehypothesis that /0isamorphism offunctors. The front square iscommutative bythe definition offl(A'). Therefore wefind: fl(A')F=fl(A')FFo(w) =FfoFO(w) =FGO(w)fo =Ffo(top triangle) (front square) (left square) (bottom triangle). This concludes theproof ofTheorem 7.1, since instead ofthepair ofindices (0,1)wecould have used (n, n+1). Remark. Themorphismfl constructed inTheorem 7.1depends functori- allyonfo inthefollowingsense. Supposewehave three delta functors F,G,H defined indegrees 0,1.Suppose given morphisms fo:FO -+GO and go: GO HO. Suppose that theerasing monomorphismserase both Fand G.Then we can constructfl and glbyapplying thetheorem. Ontheother hand, thecomposite gofo=ho:FO HO isalso amorphism offunctors, and thetheorem yields theexistence ofamorph- Ism hl:FlHl such that (ho,hl)isa-morphism. Byuniqueness,wetherefore have h1=g1fl' This iswhat wemean bythefunctorial dependence asmentioned above. Corollary 7.2. Assume that Cihasenough injectives. Thenfor anyleftexact junctor F:Ci CB,thederived functors R"Fwith n>0formauniversal 1unctor with F ROF,which iserasable byinjectives. Conversely, if G={G"}"o isauniversal 1unctor, then GOisleft exact, and the G" are isomorphic toR"GO foreach n>o. Proof. IfFisaleft exact functor, then the{R"F}"oform a-functor byTheorem 6.1. Furthermore, foranyobject A,letu:A Ibeamonomor- phism ofAinto aninjective. Then R"F(I)=0for n>0byTheorem 6.1(iv), soR"F(u)=O.Hence R"F iserasable forall n>0,and we canapply Theorem 7.1. Remark. Asusual, Theorem 7.1applies tofunctors with different variance. Suppose {F"} isafamily ofcontra variant additive functors, with nranging over XX,7 DELTA-FUNCTORS 805 asequence ofconsecutive integers, sayforsimplicityn>O.We saythat Fisa contravariant t>-functor ifgivenanexact sequence o-+M' M M" -+0 then there isanassociated family ofmorphisms bn:Fn(M') Fn+l(M') satisfying DEL 1and DEL 2with M'interchanged with Mil and N'inter- changed with Nil. We saythat Fiscoerasable iftoeach object Athere exists an epimorphism u:M -+Asuch that F(u)=O.We say that Fisuniversal if given any other b-functor Gofainto CBandgivenamorphism offunctors fo:FO-+GO there exists aunique sequence ofmorphisms fn:Fn Gn forall n>0which commute with bforeach short exact sequence. Theorem 7.1'. Let F ={Fn}(nranging over aconsecutive sequence of integers>0)be acontravariant b{unctor from ainto CB,and assume that Fniscoerasable forn>1.Then Fisuniversal. Examples ofb-functors with thevariances asinTheorems 7.1and 7.1'will begiven inthe next section inconnection with bifunctors. Dimension shifting Let F={Fn}be acontravariant delta functor with n>O.Let 8be a family ofobjects which erases Fnforall n>1,that isFn(E)=0for n> 1and EE8.Then such afamily allows ustodowhat iscalled dimension shiftingas follows. Given anexact sequence OQE-+MO with EE8,wegetfor n>1anexact sequence o=Fn(E) Fn(Q)-+Fn+l(M)-+Fn+l(E)=0, and therefore anisomorphism Fn(Q) Fn+l(M), which exhibits ashift ofdimensions byone. More generally: Proposition 7.3. Let o Q En-l...Eo-+M 0 806 GENERAL HOMOLOGY THEORY XX,8 beanexact sequence, such that EiE8.Then wehave anisomorphism FP(Q) FP+n(M) for p>1. Proof LetQ=Qn. Also without loss ofgenerality, take p=1.We may insert kernels and cokernels ateach step asfollows: En-1)En-2/\/\/ Qn Qn- I Qn-2 Q1//\// o 0 0 o...0) ... )Eo/\ M o Then shifting dimension with respect toeach short exact sequence, wefind isomorphisms F1(Qn) F2(Qn_ 1)...Fn+l(M). This concludes theproof. One says that Mhas F-dimension <difFn(M)=0for n>d+1.By dimension shifting, we seethat ifMhas F-dimension <d,then Qhas F- dimension <d-ninProposition 7.3. Inparticular, ifMhasF-dimension n, then QhasF-dimension O. The reader should rewrite allthisformalism bychanging notation, using for Fthestandard functors arising from Hom inthefirst variable, onthecategory ofmodules over aring, which hasenough projectives toerase theleftderived functors of A Hom(A, B), forBfixed. Weshall study thissituation, suitably axiomatized, inthe next sec- tion. 8. BIFUNCTORS Inanabelian categoryone often deals with Hom, which can beviewed asa functor intwovariables; and also the tensor product, which isafunctor intwo variables, buttheir variance isdifferent. Inany case, these examples lead tothe notion ofbifunctor. This isanassociation (A,B) T(A, B) XX,8 BIFUNCTORS 807 where A,Bareobjects ofabelian categories aand CBrespectively, with values insome abelian category. This means that Tisfunctorial ineach variable, with theappropriate variance (there arefour possibilities, with covariance and con- travariance inallpossible combinations); and if,say, Tiscovariant inall variables, wealso require that forhomomorphisms A' Aand B' Bthere isacommutative diagram T(A', B') j T(A, B'))T(A', B) j )T(A, B). Ifthevariances areshuffled, then the arrows inthediagramaretobereversed in theappropriate manner. Finally, werequire that asafunctor ineach variable, Tisadditive. Note that Hom isabifunctor, contravariant inthefirst variable and covari- antinthesecond. The tensor product iscovariant ineach variable. The Hom functor isabifunctor Tsatisfying thefollowing properties: HOM 1.Tiscontravariant andleft exact inthefirst variable. HOM 2. Tiscovariant andleft exact inthesecond variable. HOM 3. Foranyinjective object Jthefunctor A T(A, J) isexact. They are theonly properties which will enter into consideration inthis section. There isapossible fourth one which might come inother times: "OM 4. Foranyprojective object Qthefunctor B T(Q, B) isexact. But weshall deal non-symmetrically, and view Tas afunctor ofthe second variable, keeping thefirst onefixed, inorder togetderived functors ofthesecond variable. Ontheother hand, weshall also obtain a<5-functor ofthefirst variable byusing thebifunctor, even though this <5-functor isnot aderived functor. IfCBhasenough injectives, then wemay form theright derived functors with respect tothesecond variable B R"T(A, B), also denoted byR"TA(B), 808 GENERAL HOMOLOGY THEORY XX,8 fixing A,andviewing Basvariable. IfT=Hom, then thisright derived functor iscalled Ext, sowehave bydefinition Extn(A, X)=RnHom(A, X). Weshall now giveacriterion tocompute theright derived functors interms oftheother (first) variable. We saythat anobject AisT-exact ifthefunctor B T(A, B)isexact. ByaT-exact resolution ofanobject A,we mean aresolu- tion Ml-+Mo-+A-+O where MnisT-exact forall n>O. Examples. Letaand (Bbethecategories ofmodules over acommutative ring. LetT =Hom. Then aT-exact object isbydefinition aprojective module. Now letthetranspose ofTbegiven by tT(A, B)=T(B, A). Then atT-exact object isbydefinition aninjective module. IfTisthe tensor product, such thatT(A, B)=A0B,then aT-exact object iscalled flat. Remark. Inthecategory ofmodules over aring, there areenough pro- jectives andinjectives. But there areother situations when this isnot the case. Readers who want to seeallthis abstract nonsense inaction may consult [GriH 78], [Ha77], nottospeak of[SGA 6]andGrothendieck's collected works. Itmay genuinely happen inpractice that CBhasenough injectives butadoes not have enough projectives,sothesituation isnotallsymmetric. Thus thefunctor A RnT(A, B)forfixed Bisnot aderived functor inthevariable A.Inthe above references, wemay take forathecategory ofcoherent sheaves on a variety, and for (Bthecategory ofallsheaves. We letT=Hom. Thelocally free sheaves offinite rank areT-exact, and there areenough ofthem ina.There areenough injectives inCB.And soitgoes. Thebalancing actbetween T-exacts on one side, andinjectivesontheother isinherent tothesituation. Lemma 8.1. Let Tbeabifunctor satisfying HOM 1,HOM 2.Let AEa, and letMA A 0,that is -+Ml-+Mo-+AO beaT-exact resolution ofA.LetFn(B)=Hn(T(M, B»)for BE (B. Then F isab-functor and FO(B)=T(A, B).Ifinaddition Tsatisfies HOM 3, then Fn(J)=0forJinjective and n>1. XX,8 BIFUNCTORS 809 Proof Given anexact sequence o B' B B" 0 weget anexact sequence ofcomplexes o-+T(M, B') T(M, B)-+T(M, B") 0, whence acohomology sequence which makes Finto at5-functor. For n=0 weget1"O(B)=T(A, B)because X T(X, B)iscontravariant and left exact forXEa.IfBisinjective, then F"(B)=0for n>1bynOM 3,because X T(X, B)isexact. This proves thelemma. Proposition 8.2. Let Tbeabifunctor satisfying nOM 1,HOM 2,nOM 3. Assume that CBhasenough injectives. Let AEa.Let MA-+A-+O beaT-exact resolution ofA.Then the twot5{unctors B R"T(A, B) and BH"(T(M, B» areisomorphic asuniversal t5{unctors vanishing oninjectives, forn>1,and such that ROT(A, B)=HO(T(M), B)=T(A, B). Proof This comes merely from theuniversality ofat5-functor erasable byinjectives. We now look atthefunctoriality inA. Lemma 8.3. Let Tsatisfy HOM 1,nOM 2,and HOM 3.Assume that CBhasenough injectives. Let o A' A A" 0 beashort exact sequence. Then forfixed B,wehave along exact sequence o T(A", B) T(A, B) T(A', B) -+RIT(A", B) RIT(A, B) RIT(A', B) such that theassociation A R"T(A, B) isal>junctor. 810 GENERAL HOMOLOGY THEORY XX,8 Proof Let0 B IBbeaninjective resolution ofB.From the exactness ofthefunctor A T(A, J),forJinjectiveweget ashort exact sequence of complexes o T(A", IB)T(A, IB)T(A', IB) o. Taking theassociated long exact sequence ofhomology groups ofthese com- plexes yields the sequence oftheproposition. (The functorality isleft to thereaders.) IfT=Hom, then the exact sequence looks like oHom(A", B) Hom(A, B) Hom(A', B) Ext1(A", B) Extl(A, B)-+Extl(A', B) and soforth. Weshall saythat ahasenough T-exacts ifgiven anobject Ainathere isa T-exact Mand anepimorphism M A O. Proposition 8.4. Let Tsatisfy nOM 1,nOM 2,HOM 3.Assume that CB hasenough injectives. Fix BECB.Then theassociation A.-...R"T(A, B) isacontravariant 1unctoronawhich vanishes onT-exacts, forn>1.If ahasenough T-exacts, then thisfunctor isuniversal, coerasable byT-exacts, withvalue ROT(A, B)=T(A, B). Proof ByLemma 8.3 weknow that theassociation isa-functor, and it vanishes onT-exacts byLemma 8.1. The last statement isthen merely an application oftheuniversality oferasable -functors. Corollary 8.5. Leta=CBbethecategory ofmodules over aring. Forfixed B,letext"(A, B)betheleftderived functor ofA Hom(A, B),obtained by means ofprojective resolutions ofA.Then ext"(A, B)=Ext"(A, B). Proof. Immediate from Proposition 8.4. Thefollowing proposition characterizes T-exacts cohomologically. XX,8 BIFUNCTORS 811 Proposition 8.6. Let Tbeabifunctor satisfying nOM 1,HOM 2,HOM 3. Assume that CBhasenough injectives. Then thefollowing conditions are equivalent: TE 1. AisT-exact. TE2.For every Band every integern>1,wehave R"T(A, B)=O. TE3. For every BwehaveRIT(A, B)=o. Proof Let o B ]0 ]1 beaninjective resolution ofB.Bydefinition, R"T(A, B)isthen-thhomology of thesequence o T(A, ]0)-+T(A, ]1) T(A, ]2) IfAisT-exact, then this sequence isexact for n>1,sothehomology is0and TE 1implies TE 2.Trivially, TE2implies TE3.Finallyassume TE3.Given anexact sequence o B' -+B B" -+0, wehave thehomology sequence o T(A, B') T(A, B) T(A, B")-+R1T(A, B'). IfR1T(A, B')=0,then bydefinition AisT-exact, thusproving theproposition. Weshall saythat anobject AhasT-dimension <dif R"T(A, B)=0 for n>dand allB. Then theproposition states inparticular that AisT-exact ifandonlyifAhas T-dimension O. Proposition 8.7. Let Tsatisfy nOM 1,HOM 2,nOM 3.Assume that CB hasenough injectives. Suppose that anobject Aadmits aresolution o Ed-+Ed-l...Eo A 0 where Eo,...,EdareT-exact. Then AhasT-dimension <d.Assume this isthe case. Let o Q Ld-1...Lo A-+0 bearesolution where Lo,...,Ld_1areT-exact. Then QisT-exact also. Proof. Bydimension shiftingweconclude that Qhas T-dimension 0, whence QisT-exact byProposition 8.6. 812 GENERAL HOMOLOGY THEORY XX,8 Proposition 8.7,like others, isused inthe context ofmodules over aring. Inthat case, we can take T=Hom, and RnT(A, B)=Extn(A, B). For Atohave T-dimension <dmeans that Extn(A, B)=0 for n>dand allB. Instead ofT-exact, one can then read projective intheproposition. Let usformulate theanalogous result for abifunctor that willapply tothe tensor product. Consider thefollowing properties. TEN 1. Tiscovariant andright exact inthefirst variable. TEN 2. Tiscovariant andright exact inthesecond variable. TEN 3. Foranyprojective object Pthefunctor A T(A, P) isexact. AsforHom, there isapossible fourth property which willplaynorole inthis section: TEN 4.For anyprojective object Qthefunctor B T(Q, B) isexact. Proposition 8.2'. Let Tbe abifunctor satisfying TEN 1,TEN 2,TEN 3. Assume that CBhasenough projectives. Let AEa.Let MA-+A-+O beaT-exact resolution of'A. Then the twob{unctors B LnT(A, B) and B Hn(T(M, B» areisomorphic asuniversal b-functors vanishing onprojectives, and such that LoT(A, B)=Ho(T(M), B)=T(A, B). Lemma 8.3'. Assume that Tsatisfies TEN 1,TEN 2,TEN 3.Assume that CBhasenough projectives. Let o A' -+A A" 0 XX,8 BIFUNCTORS 813 beashort exact sequence. Then forfixed B,wehave along exact sequence: -+LlT(A', B) LlT(A, B)-+LlT(A", B)-+ -+T(A', B) T(A, B)-+T(A", B) 0 which makes theassociation A LnT(A, B)a1unctor. Proposition 8.4'. Let Tsatisfy TEN 1,TEN 2,TEN 3.Assume that CBhas enough projectives. Fix BE CB.Then theassociation A LnT(A, B) isacontravariant b-functor onawhich vanishes onT-exacts for n>1.Ifa hasenough T-exacts, then thisfunctor isuniversal, coerasable byT-exacts, with thevalue LoT(A, B)=T(A, B). Corollary 8.8. Ifthere isabifunctorial isomorphism T(A, B) T(B, A), andifBisT-exact, thenfor allA,LnT(A, B)=0for n>1.Inshort, T-exact implies acyclic. Proof LetMA=PAbe aprojective resolution inProposition 8.2'. By hypotheses, X T(X, B)isexact soHn(T(P, B)=0for n>1; sothe corollary isaconsequence oftheproposition. The above corollary isformulated soastoapply tothe tensor product. Proposition 8.6'. Let Tbe abifunctor satisfying TEN 1,TEN 2,TEN 3. Assume that CBhasenough projectives. Then thefollowing conditions are equivalent: TE 1. AisT-exact. TE2.For every Band every integer n>1wehave LnT(A, B)=O. TE3.For every B,wehave LlT(A, B)=o. Proof Werepeat theproof of8.6 sothereader can seethe arrows pointing indifferent ways. Let Ql Qo B 0 beaprojective resolution ofB.Bydefinition, LnT(A, B)isthen-thhomology ofthesequence T(A, Ql) T(A, Qo) o. 814 GENERAL HOMOLOGY THEORY XX,9 IfAisT-exact, then this sequence isexact for n>1,sothehomology is0,and TE 1implies TE 2.Trivially, TE 2implies TE 3.Finally,assume TE 3.Given anexact sequence OB'-+B-+B"O wehave thehomology sequence -+L1T(A, B") T(A, B') T(A, B) T(A, B")-+O. IfLlT(A, B,,)is0,then bydefinition, AisT-exact, thusproving theproposition. 9. SPECTRAL SEQUENCES This section isincluded forconvenience ofreference, and has two purposes: first, todraw attention toanalgebraic gadget which has wide applications in topology, differential geometry, and algebraic geometry,seeGriffiths-Harris, [GrH 78]; second, toshow that thebasic description ofthisgadget inthecontext inwhich itoccurs most frequentlycan bedone injustafew pages. Intheapplications mentioned above, one deals with afiltered complex (which weshall define later), and acomplex may beviewed asagraded object, with adifferential dofdegree 1.Tosimplify thenotation atfirst, weshall deal with filtered objects and omit thegrading index from thenotation. This index isirrelevant fortheconstruction ofthespectral sequence, forwhich wefollow Godement. SoletFbeanobject with adifferential (i.e.endomorphism) dsuch that d2=O.We assume that Fisfiltered, that isthat wehave asequence F=FO::::>F1 ::::>F2 ::::>...::::>F" ::::>F"+1={0}, and that dFP cFP.This data iscalled afiltered differential object. (We assume that thefiltration ends with 0after afinite number ofsteps forconvenience.) One defines theassociated graded object GrF=EBGrPFwhere GrPF=FP/FP+l. PO Infact, GrFisacomplex, with adifferential ofdegree 0induced byditself, and wehave thehomology H(GrPF). Thefiltration {FP} also induces afiltration onthehomology H(F, d)=H(F); namelywelet H(F)P=image ofH(FP) inH(F). XX,9 SPECTRAL SEQUENCES 815 Since dmaps FPinto itself, H(FP) isthehomology ofFPwith respect tothe restriction ofdtoFP,and ithas anatural image inH(F) which yields thisfiltra- tion. Inparticular, wethen obtain agraded object associated with thefiltered homology, namely GrH(F)=EBGrPH(F). Aspectral sequence isasequence {Er, dr}(r>0)ofgraded objects Er=EBE po together with homomorphisms (also called differentials) ofdegree r, d.Ep-+Ep+r r.r r satisfying d;=0,and such that thehomology ofErisEr+l'that is H(Er)=Er+1. Inpractice, oneusually hasEr=Er+ 1=...for r>ro.This limit object is called E00'and one says that thespectral sequence abuts toE00.Actually, tobe perfectly strict, instead ofequalitiesone should really begiven isomorphisms, butforsimplicity,we useequalities. Proposition 9.1. Let Fbeafiltered differential object. Then there exists a spectral sequence {Er} with: Eg=FPIFP+ 1; Ef=H(GrPF); E=GrPH(F). Proof. Define z={xEFPsuch that dx EFP+r} E=Z/[dz-=-r-1)+zlJ. The definition ofE:makes sense, sinceZisimmediately verified tocontain dz-=-:r-1)+Zl.Furthermore, dmaps Z:intoz:+r, and hence includes a homomorphism d.EpEp+r r.r-+ r. Weshall now compute thehomology and show that itiswhat wewant. First, forthecycles: Anelement xEZ:representsacycle ofdegree pinEr ifandonly ifdx EdZ:ll+Z;t+l,inother words dx =dy+z, withYEZ:l and zEZ:l+l. 816 GENERAL HOMOLOGY THEORY XX,9 Write x=y+u,sodu=z.Then uEFPand du EFP+r+ 1,that isuEZ+ l'It follows that p-cycles ofEr=(Z+ 1+Zl)/(dZl+1+Zl). Ontheother hand, thep-boundaries inErarerepresented byelements of dz-r, which contains dZr+1.Hence p-boundaries ofEr=(dz-r +Zl)/(dZr+1+Z:l). Therefore HP(Er)=(Z:+ 1+Z:l)/(dZ:-r+Z:l) =Z;+I/(Z;+1n(dZf-r +Z;/)). Since Zpdzp-rdZpZp+1-Zp+l r+1::) r anr+1nr-1- r , itfollows that HP(Er)=Z:+1/(dZ:-r+Z+I)=E+ l' thus proving theproperty ofaspectral sequence. Remarks. Itissometimes useful inapplications tonote therelation dZ:1r-1)+Z:l=Z:n(dFP-r+1+FP+ 1). The verification isimmediate, butGriffiths-Harris use theexpressiononthe right indefining thespectral sequence, whereas Godement uses theexpression ontheleft aswehave done above. Thus thespectral sequence may also be defined by IE=Zmod(dFrr+1+p+ 1). I This istobeinterpreted inthe sense that Zmod Smeans (Z+S)/S orZ/(Z nS). The term EgisFPIFP+1immediately from thedefinitions, and bythe general property already proved, wegetEf=H(FPIFP+ 1).AstoE, for rlarge wehaveZ=ZP =cycles inFP,and E=ZPI(ZP+1+(dFOnFP» XX,9 SPECTRAL SEQUENCES 817 which isindependent ofr,and isprecisely GrPH(F), namely thep-graded component ofH(F), thus proving thetheorem. The differential d1can bespecifiedasfollows. Proposition 9.2. The homomorphism d.EpEp+ 11.11 isthecoboundary operator arising from theexact sequence oFP+IIFp+2 FPIFp+2 FPIFP+l 0 viewing each term asacomplex with differential induced byd. Proof Indeed, thecoboundary :E=H(FPIFP+l)H(Fp+lIFp+2)=El{+l isdefined on arepresentative cyclezbydz,which isthe same way that wede- fined d1. Inmost applications, thefiltered differential object isitself graded, because itarises from thefollowing situation. LetKbeacomplex, K =(KP, d)with p>0and dofdegree 1.Byafiltration FK, also called afiltered complex,we mean adecreasing sequence ofsubcomplexes K =FOK ::)F1K ::)F2K ::)...::)F"K ::)F"+1K ={O}. Observe that ashort exact sequence ofcomplexes o K' K K" 0 gives rise toafiltration K ::)K' ::){O},viewing K'asasubcomplex. Toeach filtered complex FKweassociated thecomplex GrFK =GrK =E8GrPK, PO where GrPK =FPKIFp+1K, and thedifferential istheobvious one. The filtration FPKonKalso induces a filtration FPH(K) onthecohomology, by FPHq(K)=FPzqIFPBq. 818 GENERAL HOMOLOGY THEORY XX,9 The associated graded homology is whereGrH(K)=EBGrPHq(K), p,q GrPHq(K)=FPHq(K)IFP+1Hq(K). Aspectral sequence isasequence {En dr}(r>0)ofbigraded objects Er=EBE:,q p,q0 together with homomorphisms (called differentials) dr:E:'qE:+r,q-r+ 1satisfying d2=0r , and such that thehomology ofErisEr +l'that is H(Er)=Er+ 1. Aspectral sequence isusually represented bythefollowing picture: (p ,q) e EP.q, e(p+r,q-r+ I) Inpractice,oneusually hasEr=Er+1= ...for r>ro. This limit object iscalled E00'and one says that thespectral sequence abuts toE00. Proposition 9.3. LetFKbeafiltered complex. Then there exists aspectral sequence {Er} with: Eg,q=FPKp+qIFP+IKp+q; El{,q=Hp+q(GrPK); Eq=GrP(Hp+q(K». The lastrelation isusually written Er=>H(K), and wesaythat thespectral sequence abuts toH(K). XX,9 SPECTRAL SEQUENCES 819 The statement ofProposition 9.3ismerelyaspecialcase ofProposition 9.1, taking into account theextra graduation. One ofthemain examples isthespectral sequence associated with adouble complex K =EBKp.q p,q0 which isabigraded object, together with differentials d':KP,q KP+l,qand d":KP,q KP,q+1 satisfying d,2 =d,,2=0and d'd" +d"d' =o. We denote thedouble complex by(K,d',d"). The associated single complex (Tot(K), D)(Tot fortotal complex), abbreviated K*, isdefined by K" =EBKP,q and D=d'+d". p+q==" There aretwo filtrations on(K*, D)given by 'FPK" =EBKP',q p'+q=" p'p "FqK" =EBKP,q". P+q'=" q'q There aretwospectral sequences {'Er}and{"Er}, both abutting toH(Tot(K». Forapplications,see[GrH 78], Chapter 3,5;andalso, forinstance, [FuL 85], Chapter V.There aremany situations when dealing with adouble complex directly isauseful substitute forusing spectral sequences, which arederived from double complexes anyhow. Weshall now derive theexistence ofaspectral sequence inone ofthe most important cases, theGrothendieck spectral sequence associated with the com- posite oftwofunctors. We assume that ourabeUan category hasenough injectives. Let C=EBCPbe acomplex, and suppose CP =0ifp<0forsimplicity. Wedefine injective resolution ofCtobearesolution o-+C ]0 ]1 ]2 -+... written briefly o C]e such that each]jisacomplex, ]j=EB]j,P,with differentials dj,P:]j,P]j,p+ 1 820 GENERAL HOMOLOGY THEORY XX,9 and such that /j.Pisaninjective object. Then inparticular, foreach pweget aninjective resolution ofCP,namely: o CP1°'P11,P-+... Welet: zj,P=Ker dj,P=cycles indegree p Bj,P=1mdj,P-1=boundaries indegree p Hj,p=Zj'PIBj,p=homology indegree p. Wethen getcomplexes o ZP(C) Zo,p -+Z1,p -+ o BP(C) BO,p B1,p o HP(C) HO,p -+Hl,p We say that theresolution 0 C Ieisfully injective ifthese three com- plexes areinjective resolutions ofZP(C),BP(C)and HP(C)respectively. Lemma 9.4. Let o M' M Mil 0 beashort exact sequence. Let o M' 1M,and 0 Mil 1M" beinjective resolutions ofM'andMil. Then there exists aninjective resolution o M 1M ofMandmorphisms which make thefollowing diagram exact and commutative: o. l' )M' 1 o. l )M 1 o)0)IM" 1 ·Mil 1 o)0 o Proof Theproof isthe same asatthebeginning oftheproof ofTheorem 6.1. XX,9 SPECTRAL SEQUENCES 821 Lemma 9.5. Given acomplex Cthere exists afully injective resolution ofc. Proof We insert thekernels and cokernels inC,giving rise totheshort exact sequences with boundaries BPandcycles ZP: o BP -+ZP -+HP -+0 o-+ZP-Icp-I-+BP -+O. Weproceed inductively. We start with aninjective resolution of o ZP-1-+CP-1BP -+0 using Lemma 9.4. Next let O-+HPIHP beaninjective resolution ofHP. ByLemma 9.4there exists aninjective resolu- tion O-+ZPIzp which fitsinthemiddle oftheinjective resolutions wealready have forBPand HP. This establishes theinductive step, and concludes theproof. Given aleft exact functor Gonanabelian category with enough injectives, wesay that anobject XisG-acyclic ifRPG(X)=0for p>1.Ofcourse, ROG(X)=G(X). Theorem 9.6. (Grothendieck spectral sequence). Let T:a(B and G:(Be becovariant left exact functors such thatifIisinjective ina,then T(I) is G-acyclic. Thenforeach Ainathere isaspectral sequence {Er(A)}, such that E,q(A)=RPG(RqT(A» and Ef,q abuts (with respect top)toRp+q(GT)(A), where qisthegrading index. Proof LetAbeanobject ofa,and let0 A-+CAbeaninjective resolu- tion. Weapply Ttoget acomplex TC: 0 TCo -+TCI-+TC2 ByLemma 9.5there exists afully injective resolution o TC ITC which hasthe2-dimensional representation: 822 GENERAL HOMOLOGY THEORY XX,9 o1 )12.1 1 )12.0 1 )TC2 1 o1 )10'1 1 )10'° 1 )TCO 1 o1 )ILl 1 )11,0 1 )TCI 1 oo o Then GIisadouble complex. LetTot(GI) betheassociated single complex. We now consider each ofthe two possible spectral sequences insuccession, which wedenote by1E:'qand2E:'q. The first one istheeasiest. For fixed p,wehave aninjective resolution o TCP -+Ifc where wewrite Ifcinstead ofITCP' This isthep-th column inthediagram. By definition ofderived functors, GIPisacomplex whose homology isRqG, in other words, taking homology with respect tod"wehave "HP,q(GI)=Hq(GIP)=(RqG)(TCP). Byhypothesis, CPinjective implies that (RqG)(TCP)=0for q>O.Since G isleftexact, wehave ROG(TCP)=TCP. Hence weget {G(CP)ifq=0 "HP,q(GI)= o ifq>0.' Hence the non-zero terms are onthep-axis, which looks like oGT(CO) GT(Cl)GT(C2) Taking,HP weget lE.q(A)={R OP(G1)(A) ifq=0 ifq>O. This yields H"(Tot(GI» R"(GT)(A). XX,9 SPECTRAL SEQUENCES 823 The second onewill usethefullstrength ofLemma 9.5,which had not been used inthefirst part oftheproof,soitisnow important that theresolution ITCisfully injective. Wetherefore have injective resolutions OZP(TC) lZO,p lZl,p lZ2,p OBP(TC)-+ IBo,p-+ IB1,p IB2,p OHP(TC) IHo,p IHl,p IH2,p and the exact sequences o1zq,pIq,P1Bq+1,P0 o1Bq,P 1zq,P1Hq.P0 split because oftheinjectivity oftheterms. Wedenote byI(p)thep-th row ofthe double complex I={Iq,P}. Then wefind: ,Hq,P(GI)=Hq(GI(P»=Glzq,PIG1Bq,P =G'Hq,P(I)bythefirstsplit sequence bythesecond split sequence because applying thefunctor Gtoasplit exact sequence yieldsasplit exact sequence. Then 2E.q="HP('Hq,P(GI)=HP(G1Hq,P(I». Bythefullinjectivity oftheresolutions, thecomplex' Hq,P(I) with p>0isan injective resolution of Hq(TC)=(RqT)(A). Furthermore, wehave HP(G'Hq,P)=RPG(RqT(A), since aderived functor isthehomology ofaninjective resolution. This proves that (RPG)RqT(A» abuts toR"(GT)(A), and concludes theproof ofthetheorem. Just toseethespectral sequence atwork, wegive oneapplication relating ittotheEuler characteristic discussed in93. Let (1have enough injectives, and let T:(1CB beacovariant left exact functor. Letabeafamily ofobjects in(1giving rise toaK-group. More precisely, inashort exact sequence in(1,iftwooftheobjects lieina' then sodoes thethird. We also assume that theobjects ofahave finite RT-dimension, which means bydefinition that ifAEathenRiT(A)=0 824 GENERAL HOMOLOGY THEORY XX,9 forallisufficiently large. Wecould take 3'ainfact tobethefamily ofallobjects inawhich have finite RT-dimension. Wedefine theEuler characteristic associated with TonK(3' a)tobe 00 XT(A)=L(-l)icl(RiT(A». i=0 The cldenotes the class intheK-group K(3'<8) associated with some family 3'<8ofobjects inCB,and such thatRiT(A)E3'<8forallAE3'a. This isthemini- mum required fortheformula tomake sense. Lemma 9.7. The map XTextends toahomomorphism K(3'a)-+K(3'<8). Proof Let o A' A A" 0 beanexact sequence in3'.Then wehave thecohomology sequence RiT(A') RiT(A) RiT(A") Ri+1T(A') inwhich allbut afinite number ofterms are O.Taking thealternating sum inthe K-group shows that XTisanEuler-Poincare map, and concludes theproof. Note that wehave merely repeated something from 93,inajazzed upcontext. Inthe next theorem, wehave another functor G:CB e, and wealso have afamily 3'giving rise toaK-group K(3'e). We suppose that wecanperform theabove procedure ateach step, and also need some condition sothat wecanapply thespectral sequence. So,precisely,we assume: CHAR 1.For alli,RiT maps 3'ainto 3'<8, RiG maps 3'<8into 3'e, and Ri(GT) maps 3'ainto 3'. CHAR 2. Each subobject ofanelement of3'alies in3'aand has finite RT- and R(GT)-dimension; each subobject ofanelement of 3'<8liesin3'<8and hasfinite RG-dimension. Theorem 9.8. Assume that T:a-+CBand G:CB esatisfy theconditions CHAR 1and CHAR 2.Also assume that Tmaps injectives toG-acyclics. Then XG0XT=XGT' XX,9 SPECTRAL SEQUENCES 825 Proof ByTheorem 9.6, theGrothendieck spectral sequence ofthe com- posite functor implies theexistence ofafiltration ...cFPR"(GT)(A)CFP+1R"(GT)(A)c... ofR"(GT)(A), such that FP+IIFP E"-p. Then 00 XGT(A)=L(-I)" cl(R"(GT)(A» "=0 00 00 =L(-I)n Lcl(E"-P) "=0 p=o 00 =L(-I)" cl(E). "=0 Ontheother hand, 00 XT(A)=L(-I)q cl(RqT(A» q=O and so 00 XG0XT(A)=L(-I)qXG(RqT(A» q=O 00 00 =L(-I)q L(-I)P cl(RPG(RqT(A» q=O p==O 00 " =L(-I)" Lcl(RPG(R"-PT(A») "=0 p=O 00 =L(-I)"cl(E). "==0 Since Er+1isthehomology ofEr,weget 00 00 00 L(-1)" cl(E)=L(-I)" cl(E 3)= ...=L(-1)" cl(E). "=0 "=0 "=0 This concludes theproof ofthetheorem. 826 GENERAL HOMOLOGY THEORY XX, Ex EXERCISES 1.Prove that theexample ofthestandard complex given in Iisactuallyacomplex, and isexact, soitgivesaresolution ofZ.[Hint: Toshow that thesequence ofthe standard complex isexact, choose anelement zESand define h:E; E;+Ibyletting h(xo,. . .,x;)=(z,xo,. ..,x;). Prove that dh+hd=id,and that dd=O.Exactness follows atonce.] Cohomology ofgroups 2.Let Gbe agroup. Use Gasthe setSinthestandard complex. Define anaction of Gonthestandard complex Ebyletting x(xo,. . .,x;)=(xxo,. . .,xx;). Prove that each E;isafree module over the group ring Z[G]. Thus ifwe let R=Z[G]bethegroup ring, and consider thecategory Mod( G)ofG-modules, then thestandard complex givesafree resolution ofZinthis category. 3.The standard complex Ewas written inhomogeneous form, sotheboundary maps have acertain symmetry. There isanother complex which exhibits useful features asfollows. LetF;bethe free Z[G]-module having for basis i-tuples (rather than (i+I)-tuples) (XI'. ..,x;). Fori=0wetake Fo=Z[G] itself. Define theboundary operator bytheformula ;-1 d(Xl,.. .,Xi)=Xl(X2,. ..,X;)+L(-I)j(Xl,. ..,XjXj+ I,.. .,Xi) j=I +(-1);+I(Xl,...,X;). Show that E=F(ascomplexes ofG-modules) viatheassociation (XI'... ,x;)(I,xl,xlx2'... ,XIX2.'.x;), and that theoperator dgiven forFcorresponds totheoperator dgiven forEunder thisisomorphism. 4.IfAisaG-module, letAGbethesubmodule consisting ofallelements vEAsuch that xv=vforall XEG.Thus AGhastrivial G-action. (This notation isconvenient, but isnotthe same asfortheinduced module ofChapter XVIII.) (a)Show that ifHq(G, A) denotes the q-th homology of the complex HomG(E, A),then IfO(G, A)=AG. Thus theleftderived functors ofA AG arethehomology groups ofthecomplex HomG(E, A), orforthat matter, ofthecomplex Hom(F, A), where FisasinExercise 3. (b) Show that the group ofI-cycles ZI(G, A)consists ofthose functions f:G Asatisfying f(x) +xf(y)=f(xy) forallx,yEG. Show that thesubgroup ofcoboundaries BI(G, A)consists ofthose functions ffor which there exists anelement aEAsuch thatf(x)=xa-a.The factor group isthenHl(G, A). SeeChapter VI, 10forthedetermination ofaspecial case. XX, Ex EXERCISES 827 (c) Show that the group of2-cocycles Z2(G, A)consists ofthose functions f:G Asatisfying xf(y, z)-f(xy, z)+f(x, yz)-f(x, y)=o. Such 2-cocyclesarealso called factor sets, andtheycan beused todescribe isomorphism classes ofgroup extensions, asfollows. 5.Group extensions. Let Wbe agroup and Aanormal subgroup, written multipli- catively. Let G=WjAbethefactor group. Let F;G Wbe achoice ofcoset representatives. Define f(x, y)=F(x)F(y)F(xy)-I. (a) Prove thatf isA-valued, andthatf: GxG Aisa2-cocycle. (b)Given agroup Gand anabelian group A, weview anextension Wasan exact sequence lAWG1. Show thatiftwo such extensions areisomorphic then the2-cocycles associated tothese extensions asin(a)define the same class inHI(G,A). (c) Prove that themap which weobtained above from isomorphism classes of group extensions toH2(G, A)isabijection. 6.Morphisms ofthecohomology functor. Let A:G' Gbeagroup homomorphism. Then Agives rise toanexact functor <1>,\:Mod(G) Mod(G'), because every G-module can beviewed asaG'-module bydefining theoperation of a'EG'tobea'a=A(a')a. Thus weobtain acohomology functor HG'0<1>,\. LetG'be asubgroup ofG.Indimension 0,wehave amorphism offunctors A* :Hg Hg,0<1>,\given bytheinclusion AG AG'=<I>,\(A)G'. (a) Show that there isaunique morphism of5-functors A*:HG HG'0<1>,\ which has theabove effect onHg.We have thefollowing important special cases. Restriction. LetHbeasubgroup ofG.LetAbeaG-module. Afunction from Ginto Arestricts toafunction from Hinto A.Inthis way,wegeta natural homomorphism called therestriction res:Hq(G, A) Hq(H, A). Inflation. Suppose that Hisnormal inG.LetAHbethesubgroup ofA consisting ofthose elements fixed byH.Then itisimmediately verified that AHisstable under G,and soisaGjH-module. TheinclusionAH Ainduces ahomomorphism H1;(U)=uq:Hq(G, AH) Hq(A). Define theinflation inf/H:Hq(GjH, AH) Hq(G, A) 828 GENERAL HOMOLOGY THEORY XX, Ex asthecomposite ofthefunctorial morphism Hq(G/H,AH) Hq(G,AH) followed bytheinduced homomorphismuq=H'b(u)asabove. Indimension 0,theinflation gives theidentity (AH)G/H =AG. (b) Show that theinflation can beexpressedonthe standard cochain complex bythenatural map which toafunction ofG/HinAHassociates afunction ofGinto AHCA. (c) Prove that thefollowing sequence isexact. oHI(G/H, AH) HI(G, A) HI(H, A). (d)Describe how one getsanoperation ofGonthecohomology functor HG"by conjugation" andfunctoriality. (e)In(c), show that theimage ofrestriction ontheright actually lies in HI(H, A)G (the fixed subgroup under G). Remark. There isananalogous result forhigher cohomology groups, whose proof needs aspectral sequence ofHochschild-Serre. See [La96], Chapter VI,2,Theorem 2.Itisactually this version forH2which isapplied toH2(G,K*), when KisaGalois extension, and isused inclass field theory [ArT 67]. 7.Let Gbe agroup, Banabelian group andMG(B)=M(G, B)the setofmappings from Ginto B.For xEGandfE M(G, B)define ([x]f)(y)=f(yx). (a) Show that B MG(B) isacovariant, additive, exact functor from Mod(Z) (category ofabelian groups) into Mod(G). (b)Let G'be asubgroup ofGand G=UxjG'acoset decomposition. For fEM(G, B)letfjbethefunction inM(G', B)such thatfj(y)=f(xjY). Show that themapfOfj J isaG'-isomorphism from M(G, B)tof1M(G', B).j 8.For each G-module AEMod(G), define EA: A M(G, A)bythe condition EA(a)=thefunction fasuch thatfz(a)=aafor aEG.Show that a faisa G-module embedding, and that the exact sequence EAo A M(G, A) XA=cokerEA 0 splitsover Z.(Infact, themapf f(e) splits theleft side arrow.) 9.Let BEMod(Z). Let Hqbetheleftderived functor ofA AG. (a) Show thatHq(G, MG(B))=0forallq>O.[Hint: use acontracting homotopy s:Cr(G, MG(B)) Cr-I(G, MG(B)) by (Sf)X2' ,xr(x)=ix,X2, ,xr(l). . Show thatf=sdf+dsf.] Thus MGerases thecohomology functor. (b)Also show that forallsubgroups G'ofGone hasHq(G', MG(B))=0for q>O. 10. Let Gbe agroup and Sasubgroup. Show that thebifunctors (A,B) HomG(A, Mb(B)) and(A,B) Homs(A, B) onMod( G)xMod(S) with value inMod(Z) areisomorphic. The isomorphism is givenbythemaps cp (a ga)' for cpEHoms(A, B), where ga(a)=cp(ua), gaEMb(B). XX, Ex EXERCISES 829 The inverse mapping isgiven by ff(l)withf EHomG(A, Mb(B». Recall thatMb(B) was defined inChapter XVIII, 7fortheinduced representation. Basically you should already know theabove isomorphism. II. Let Gbe agroup and Sasubgroup. Show that themap Hq(G, Mb(B» Hq(S, B)forBEMod(S), obtained bycomposing therestriction res with theS-homomorphism ff(I),is anisomorphism for q>O.[Hint: Use theuniqueness theorem forcohomology functors. ] 12. Let Gbe agroup. Let e:Z[G] Zbethehomomorphism such thate(Ln(x)x)= Ln(x). LetIGbeitskernel. Prove thatIGisanideal ofZ[G]and that there isan isomorphism offunctors (on thecategory ofgroups) G/GC=IG/lb, by xGC (x-I)+lb. 13. LetA EMod(G) and aEHl(G, A).Let{a(x)}XEG beastandard I-cocycle representing a.Show that there exists aG-homomorphismf: IG Asuch thatf(x-I)=a(x), sofE(Hom(1 G,A»G. Show that thesequence o A=Hom(Z, A) Hom(Z[G], A) Hom(/ G,A) 0 isexact, and that if5isthecoboundary for thecohomology sequence, then 5(f)=-a. Finite groups We now turn tothe case offinite groups G.For such groups and aG-module Awe have thetrace TG:A A defined by TG(a)=L aa. UEG We define amodule AtobeG-regular ifthere exists aZ-endomorphismu:A Asuch that idA=TG(u). Recall that theoperation ofGonEnd(A) isgiven by [a]f(a)=af(a-1a) for aEG. 14. (a) Show that aprojective object inMod(G) isG-regular. (b) Let Rbe acommutative ring and letAbeinModR(G) (the category of(G,R)- modules). Show that AisR[G]-projective ifandonly ifAisR-projective and R[G]-regular, meaning that idA=TG(u) for some R-homomorphismu:A A. 15. Consider the exact sequences: E(I) 0 IG Z[G] Z 0 E' (2) 0 Z Z[G] JG 0 where thefirst one defines IG,and thesecond isdefined bytheembedding e' :Z Z[G] such that e'(n)=n(La), i.e. onthe"diagonal". The cokernel ofe'isJGbydefinition. (a) Prove that both sequences (I)and(2)split inMod(G). 830 GENERAL HOMOLOGY THEORY XX, Ex (b)Define MG(A)=Z[G] 0A(tensor productover Z)forAEMod(G). Show thatMG(A) isG-regular, and that one getsexact sequences (IA)and (2A)by tensoring (I)and(2)with A.As aresult one getsanembedding t;=t;'(8)id :A=Z(8)A Z[G](8)A. 16.Cyclic groups. Let Gbe afinite cyclic group oforder n.Let abeagenerator ofG. LetKi=Z[G] fori>O.Let t;:KO Zbetheaugmentationasbefore. For iodd >I,letdi:Ki Ki-I bemultiplication byI-a. For ieven>2,letdibe multiplication byI+a+.. .+an-I. Prove that Kisaresolution ofZ.Conclude that: For iodd: Hi(G, A)=AG/TGA where TG:a (I+a+... +an-I)a; For ieven>2:Hi(G, A)=AT/(I-a)A, where ATisthekernel ofTGinA. 17. Let Gbe afinite group. Show that there exists a5-functor Hfrom Mod(G) to Mod (Z) such that: (I) HOis(isomorphic to)thefunctor A AG/TGA. (2)Hq(A)=0ifAisinjective and q>0,andHq(A)=0ifAisprojective and q isarbitrary. (3)Hiserased byG-regular modules. Inparticular, Hiserased byMG. The 5-functor ofExercise 17iscalled thespecial cohomology functor. Itdiffers from theother oneonly indimension O. 18. LetH=HGbethespecial cohomology functor for afinite group G.Show that: HO(/G)=0;HO(Z)=HI(/)=Z/nZ where n=#(G); HO(Q/z)=HI(Z)=H2(/)=0 HI(Q/z)=H2(Z)=H3(/)=G"=Hom(G, Q/Z) bydefinition. Injectives 19.(a)Show that Ifanabelian group TisinjectiveInthecategory ofabelian groups, then ItisdivIsible. (b) Let Abeaprincipal entIre ring. Define thenotIon ofdivisibility byelements ofAfor modules inamanner analogous tothat forabelian groups. Show that anA- module ISinjective Ifandonly ifItisA-divIsible. [The proof forZshould work inexactly the same way.] 20. Let Sbe amultiplicative subset ofthecommutative Noetherian ring A.If/isan injective A-module, show thatS-I/ isaninjective S-IA-module. 21. (a)Show that adirect sum ofprojective modules ISprojective. (b)Show that adirect product ofInjective modules ISInjective. 22. Show that afactor module, direct summand, direct product, and direct sum ofdivIsible modules aredivIsible. 23. LetQbe amodule over acommutative ring A.Assume that forevery leftideal Jof A,every homomorphism cp:J Qcan beextended toahomomorphism ofAinto Q.Show that Qisinjective. [Hint: Given M' CMandf:M' Q,letXoEM andXott.M'. LetJbetheleft ideal ofelements aEAsuch that axoEM'. Let cp(a)=f(axo) and extendcptoA, ascan bedone byhypothesis. Then show that XX, Ex EXERCISES 831 one can extend ftoMbytheformula f(x'+bxo)=f(x') +cp(b), forx'EMand bEA.Then useZorn's lemma. This isthe same pattern ofproofas theproof ofLemma 4.2.] 24.Let 0-+11-+12-+13-+0 beanexact sequence ofmodules. Assume that 11,12areinjective. (a)Show that thesequence splits. (b)Show that 13isinjective. (c)IfIisinjectIve and 1=MEBN,show that Misinjective. 25.(Do this exercise after you have read about Noetherian rings.) Let AbeaNoetherian commutative ring, and letQbeaninjective A-module. Let abeanIdeal ofA,and let Q(Q) bethe subset ofelements xEQsuch that anx =0for some n,dependingon x. Show that Q(Q) isinjective. [Hint: Use Exercise 23.] 26. Let Abe acommutative ring. Let EbeanA-module, and letE"=Homz(E, Q/Z) bethedual module. Prove thefollowing statements. (a)Asequence O-+N-+M-+E-+O isexact ifandonly ifthedual sequence o E" M" N" 0 isexact. (b) Let Fbeflat and 1injective inthecategory ofA-modules. Show that HomA(F, I)isinjective. (c) Eisflatifandonly ifE"isinjective. 27. Extensions ofmodules. LetM,Nbemodules over aring. Byanextension ofM byNwe mean anexact sequence (*) ONEM O. We shall now define amap from such extensions toExtI(M, N). LetPbeprojective, with asurjective homomorphism onto M, sowegetanexact sequence (**) 0 K P-4M 0 where Kisdefined tobethekernel. Since Pisprojective, there exists ahomomorphism u:P E,anddependingon uaunique homomorphismv:K Nmaking the diagram commutative: o-----. K-----. P M----+ 0 vjujidj o-----. N---+ E---+ M---+ 0 832 GENERAL HOMOLOGY THEORY XX, Ex Ontheother hand, wehave the exact sequence (***) 0 Hom(M, N) Hom(P, N) Hom(K, N) Extl(M, N) 0, with the last term ontheright being equalto0because Extl(P,N)=O.Tothe extension (*) weassociate theimage ofvinExtl(M, N). Prove that this association isabijection between isomorphism classes ofextensions (i.e. isomorphism classes ofexact sequencesasin(*)), andExt1(M, N). [Hint: Construct aninverse asfollows. Given anelement eofExtl(M, N),usinganexact sequence (**), there issome element vEHom(K, N)which mapson ein(***). Let Ebethepush-out ofvand w.Inother words, letJbethe submodule ofNEBP consisting ofallelements (v(x), -w(x)) with xEK,and letE=(NEBP)/J. Show that themap y (y,0)mod Jgivesaninjection ofNinto E.Show that themap NEBP Mvanishes onJ,and sogivesasurjective homomorphism E M O. Thus weobtain anexact sequence (*); that is,anextension ofMbyN.Thus toeach element ofExt1(M,N) wehave associated anisomorphism class ofextensions ofM byN.Show that the mapswehave defined areinverse toeach other between iso- morphism classes ofextensions and elements ofExt1(M,N).] 28. Let Rbeaprincipal entire ring. Let aER.For every R-module N,prove: (a)Extl(R/aR,N)=N/aN. (b)For bERwehave Ext1(R/aR, R/bR)=R/(a, b),where (a,b)istheg.c.d ofaandb,assuming ab =1=O. Tensor product ofcomplexes. 29. LetK =EBKpand L=EBLqbetwo complexes indexed bytheintegers, and with boundary maps lower indices byI.Define K(8)Ltobethedirect sum ofthemodules (K(8)L)n, where (K(8)L)n=EBKp(8)Lq. p+q=n Show that there exist unique homomorphisms d=dn:(K(8)L)n (K(8)L)n- 1 such that d(x (8)y)=d(x) (8)y+(-1)Px 0d(y). Show that K(8)Lwith these homomorphismsISacomplex, that isdad =O. 30. LetK,Lbedouble complexes. Wewrite K;andL;fortheordinary column complexes ofKand Lrespectively. Letcp:K Lbe ahomomorphism ofdouble complexes. Assume that each homomorphism tn..K. L.T'l. I I isahomology isomorphism. (a) Prove that Tot( cp):Tot(K) Tot(L) isahomology isomorphism. (Ifyou want toseethis worked out, cf.[FuL 85], Chapter V,Lemma 5.4.) (b) Prove Theorem 9.8using (a)instead ofspectral sequences. XX, Ex [ArT 68] [At61] [At67] [ABP 73] [Ba68] [8069] [BtD 85] [CaE 57] [CuR 81] [ES52] [FuL 85] [Go58] [GreH 81] [GriH 78] [Gro 57] [Gro 68] [Gu 91] [Ha77] [HiS 70] [La96] [Man 69] [Mat 70] [No 68] [No 76] [Ro79] [Se64]EXERCISES 833 Bibliography E.ARTIN andJ. TATE, Class Field Theory, Benjamin, 1968; Addison-Wesley, 1991 M.ATIYAH, Characters and cohomology offinite groups, Pub. IHES 9 (1961), pp.5-26 M.ATIYAH, K-theory, Benjamin, 1967; reprinted Addison-Wesley, 1991 M.ATIY AH, R.BOTT, and R.PATODI, Ontheheat equation and theindex theorem, Invent. Math. 19(1973), pp.279-330 H.BASS, Algebraic K-theory, Benjamin, 1968 R.BOTT, Lectures onK(X), Benjamin, 1969 T.BROCKER and T. TOM DIECK, Representations ofCompact LieGroups, Springer Verlag, 1985 H.CARTAN and S.ElLENBERG, Homological Algebra, Princeton University Press, 1957 C.CURTIS and I.REINER, Methods ofRepresentation Theory, John Wiley & Sons, 1981 S.ElLENBERG and N. STEENROD, Foundations ofAlgebraic Topology, Princeton University Press, 1952 W. FULTON and S.LANG, Riemann-Roch algebra, Springer Verlag, 1985 R.GODEMENT, Theorie desfaisceaux, Hermann Paris, 1958 M. GREENBERG and J.HARPER, Algebraic Topology: AFirst Course, Ben- jamin-Addison- Wesley, 1981 P.GRIFFITHS and J.HARRIS, Principles ofalgebraic geometry, Wiley Inter- science 1978 A.GROTHENDIECK, Sur quelques points d'algebre homologique, Tohoku Math. J.9(1957) pp. 119-221 A.GROTHENDIECK, Classes deChern etrepresentations lineaires des groupes discrets, Dixexposessurlacohomologie etale desschemas, North-Holland, Amsterdam, 1968 R.GUNNING, Introduction toholomorphic functions ofseveral variables, Vol. IIIWadsworth &Brooks/Cole, 1990 R.HARTSHORNE, Algebraic Geometry, Springer Verlag, 1977 P.J.HILTON andU.STAMMBACH, ACourse inHomological Algebra, Graduate Texts inMathematics, Springer Verlag, 1970. S.LANG, Topics incohomology ofgroups, Springer Lecture Notes, 1996 J.MANIN, Lectures ontheK-functor inAlgebraic Geometry, Russian Math Surveys 24(5) (1969) pp. 1-89 H. MATSUMURA, Commutative Algebra, Second Edition, Benjamin- Cummings, 1981 D.NORTHCOTT, Lessons onRings, Modules andMultiplicities, Cambridge University Press, 1968 D.NORTHCOTT, Finite Free Resolutions, Cambridge University Press, 1976 J.ROTMAN, Introduction toHomological Algebra, AGademic Press, 1979 J.-P. SERRE, Cohomologie Galoisienne, Springer Lecture Notes 5,1964 834 GENERAL HOMOLOGY THEORY [Se65] [SGA 6] [Sh72]XX, Ex J.-P.ERRE, Algebre locale, multiplicites, Springer Lecture Notes 11(1965) Third Edition 1975 P.BERTHELOT, A.GROTHENDIECK, L.ILLUSIE etal.Theorie desintersections ettheoreme deRiemann-Roch, Springer Lecture Notes 146, 1970 S.SHATZ, Profinite groups, arithmetic and geometry, Ann. ofMath Studies, Princeton University Press 1972 CHAPTER XX I Finite Free Resolutions This chapter putstogether specific computations ofcomplexes andhomology. Partly these provide examples forthegeneral theory ofChapter XX, andpartly they provide concrete results which have occupied algebraists for acentury. They have oneaspect incommon: thecomputation ofhomology isdone bymeans ofafinite free resolution, i.e.afinite complex whose modules arefinite free. The first section shows ageneral technique (themapping cylinder) whereby thehomology arising from some complexcan becomputed byusing another complex which isfinite free. One application ofsuch complexes hasalready been given inChapter X,putting together Proposition 4.5followed byExercises 10-15 ofthatchapter. Then wegotomajor theorems, going from Hilbert's Syzygy theorem, from acentury ago, toSerre's theorem about finite free resolutions ofmodules over polynomial rings, and theQuillen-Suslin theorem. We also include adiscussion ofcertain finite free resolutions obtained from theKoszul complex. These apply, among other things, totheGrothendieck Riemann-Roch theorem ofalgebraic geometry. Bibliographical references refer tothelistgiven attheend ofChapter XX. 1. SPECIAL COMPLEXES Asinthepreceding chapter, wework with thecategory ofmodules over a ring, but the reader will notice that thearguments hold quite generally inan abelian category. Insome applicationsone determines homology from acomplex which is notsuitable forother types ofconstruction, likechanging thebase ring. Inthis section, wegiveageneral procedure which constructs another complex with 835 836 FINITE FREE RESOLUTIONS XXI, 1 better properties than thefirst one, while giving the same homology. For an application toNoetherian modules, seeExercises 12-15 ofChapter X. Letf:K-+Cbe amorphism ofcomplexes. We saythatj'isahomology isomorphism ifthenatural map H(f): H(K)-+H(C) isanisomorphism. The definition isvalid inanabelian category, butthereader may think ofmodules over aring, orabelian groupseven. Afamily 3'ofobjects will becalled sufficient ifgivenanobject Ethere exists anelement Fin3'and anepimorphism F E 0, and if3'isclosed under taking finite direct sums. Forinstance, wemayusefor 3'thefamily offree modules. However, inimportant applications,weshall deal with finitely generated modules, inwhich case 3'might betaken asthefamily of finite free modules. These areinfact theapplications Ihave inmind, which resulted inhaving axiomatized thesituation. Proposition 1.1. Let Cbe acomplex such that HP(C) =I0only for o<p<n.Let 3'be asufficient family ofprojectives. There exists a complex o-+KO K1...Kn0 such that: KP =I0onlyfor 0<p<n; KPisin3'forallp>1; and there exists ahomomorphism ofcomplexes f:K C which isahomology isomorphism. Proof Wedefinefm bydescending induction onm: bm+1 )Km+lK )Km+2 J/m+1 J/m+2)Km JIm )Cm)Cm+1 b+1)Cm+2 We suppose that wehave defined amorphism ofcomplexes with p>m+1 such thatHP(f)isanisomorphism forp>m+2,and fm+ 1:zm+l(K) Hm+l(C) XXI, 1 SPECIAL COMPLEXES 837 isanepimorphism, where Zdenotes thecycles, that isKer. Wewish tocon- struct Kmandfm,thus propagating tothe left. First let m>O.Let Bm+1be thekernel of Ker;+l Hm+l(C). LetK'bein3'with anepimorphism ':K' Bm+1 . LetK" -+Hm(C)beanepimorphism with K"in3',and let f":K" zm(c) beanylifting, which exists since K"isprojective. Let Km=K' EE>K" and define m: Km -+Km+1tobe'onK'and 0onK". Then fm+ 10'(K')cC(Cm)' and hence there existsf':K' -+cmsuch that c0f'=fm +10'. We now define fm:Kmcmtobef'onK'andf"onK". Then wehave defined amorphism ofcomplexes truncated down tomasdesired. Finally, ifm=-1, wehave constructed down toKO,o,andjwith KO HO(C) 0 exact. The last square looks likethis, defining K-1=o. o)CObO=b' )'K' CK1 Ifl )Clo)K' EE>K" 1'\/1" Wereplace KObyKOI(Ker onKerfo). Then HO(f) becomes anisomorphism, thus proving theproposition. We want tosaysomething more about KO. For this purpose,wedefine a new concept. Let3'beafamily ofobjects inthegiven abelian category (think ofmodules infirstreading). Weshall saythat 3'iscomplete ifitissufficient, and forany exact sequence o F' F F" 0 with F"and Fin3'then F'isalso in3'. 838 FINITE FREE RESOLUTIONS XXI, 1 Example. InChapter XVI, Theorem 3.4 weproved that thefamily offinite flatmodules inthecategory offinite modules over aNoetherian ring iscomplete. Similarly, thefamily offlatmodules inthecategory ofmodules over aring is complete. We cannot getaway with justprojectivesorfree modules, because inthe statement oftheproposition,KOisnotnecessarily free but we want to include itinthefamilyashaving especially nice properties. Inpractice, the family consists oftheflatmodules, orfinite flatmodules. Cf.Chaper X,Theorem 4.4, andChapter XVI, Theorem 3.8. Proposition 1.2. Letj':K Cbe amorphism ofcomplexes, such that KP, HP(C) are#0onlyfor p=1,...,n. Let3'beacompletefamily,andassume that KP, CP are in3'forallp,except possibly for KO.Iffisahomology isomorphism, then KOisa/so in3'. Before giving theproof,wedefine anewcomplex called themapping cylinder ofanarbitrary morphism ofcomplexes fbyletting MP =KP(f)CP-1 anddefining M:MP MP+1by M(X, y)=(x,fx-y). Itistrivially verified that Misthen acomplex, i.e.0 =O.IfC'isthe com- plex obtained from Cbyshifting degrees byone(and making asign change inc),soC'P =CP- 1,then wegetanexact sequence ofcomplexes o C' M K 0 and hence themapping cylinder exact cohomology sequence HP(K) HP+l(C') " HP(C))HP+l(M))HP+l(K))HP+2(C') /I HP+l(C) and one sees from thedefinitions that thecohomology maps HP(K)-+HP+l(C') HP(C) arethe ones induced byf:K C. We now return totheassumptions ofProposition 1.2, sothat these mapsare isomorphisms. We conclude thatH(M)=O.This implies that the sequence o-+KO -+M1-+M2 ...M"+10 isexact. Now each MP isin3'byassumption. Inserting the kernels and cokernels ateach step and using induction together with thedefinition ofa complete family,weconclude that KOisin3',aswas tobeshown. XXI,2 FINITE FREE RESOLUTIONS 839 Inthe next proposition,wehave axiomatized thesituation sothat itis applicable tothetensor product, discussed later, and tothe case when thefamily 3'consists offlatmodules, asdefined inChapter XVI. Noknowledge ofthis chapter isneeded here, however, since theaxiomatization usesjust thegeneral language offunctors and exactness. Let3'beacomplete family again, and letTbeacovariant additive functor onthegiven category. We saythat 3'isexact forTifgivenanexact sequence o-+F' -+F-+F" 0 in3',then oT(F') T(F) T(F") 0 isexact. Proposition 1.3. Let 3'be acomplete family which isexact for T.Let f:K Cbeamorphism ofcomplexes, such that KPand CPare in3'forall p,and KP,HP(C) are zeroforallbut afinite number ofp.Assume thatfisa homology isomorphism. Then T(f): T(K) T(C) isahomology isomorphism. Proof. Construct themapping cylinder Mforf.Asintheproof ofPropo- sition 1.2, wegetH(M)=0soMisexact. We then start inductively from the right with zeros. WeletZPbethecycles inMPand usetheshort exact sequences o ZP MP zp+10 together with thedefinition ofacomplete family toconclude that ZPisin3'for allp.Hence theshort sequences obtained byapplying Tare exact. ButT(M) isthemapping cylinder ofthemorphism T(f):T(K) T(C), which istherefore anisomorphism,asone sees from thehomology sequence of themapping cylinder. This concludes theproof. 2. FINITE FREE RESOLUTIONS The first part ofthis section develops thenotion ofresolutions for acase somewhat more subtle than projective resolutions, andgivesagood example for theconsiderations ofChapter xx. Northcott in[No 76]pointed outthat minor adjustments ofstandard proofs also applied tothenon-Noetherian rings, only occasionally slightly less tractable than theNoetherian ones. 840 FINITE FREE RESOLUTIONS XXI,2 Let Abearing. Amodule Eiscalled stably freeifthere exists afinite free module Fsuch that E(f)Fisfinite free, and thus isomorphic toA(n) for some positive integern.Inparticular, Eisprojective andfinitely generated. Wesaythat amodule Mhas afinite free resolution ifthere exists aresolution o-+En...Eo M 0 such that each Eiisfinite free. Theorem 2.1. LetMbeaprojective module. Then Misstably freeifand onlyifMadmits afinite free resolution. Proof. IfMisstably free then itistrivial that Mhas afinite freeresolution. Conversely assume the existence oftheresolution with the above notation. We prove that Misstably freebyinduction on n.The assertion isobvious if n=O.Assume n>1.Insert thekernels and cokernels ateach step, inthe manner ofdimension shifting. Say M1=Ker(E o P), giving rise tothe exact sequence o M1Eo M o. Since Misprojective, this sequence splits, and Eo M(f)MI. But M1has a finite free resolution oflength smaller than theresolution ofM, sothere exists afinite free module Fsuch that M1(f)Fisfree. Since Eo(f)Fisalso free, this concludes theproof ofthetheorem. Aresolution o En...Eo M 0 iscalled stably freeifallthemodules Ei(i=0,...,n)arestably free. Proposition 2.2. LetMbeanA-module. Then Mhas afinite free resolution oflength n>1ifandonlyifMhas astably free resolution oflength n. Proof. One direction istrivial, sowesuppose given astably free resolution with theabove notation. Let 0<i<nbesome integer, and letF;,Fi+1be finite free such that Ei(f)Fiand Ei+1 Fi+1are free. Let F=FiFi+1. Then wecanform anexact sequence o En.. .Ei+1(f)F Ei(f)F...EO M -+0 intheobvious manner. Inthis way,wehave changed two consecutive modules intheresolution tomake them free. Proceeding byinduction, we can then make Eo, E1free, then El,E2free, and soontoconclude theproof ofthe proposition. XXI,2FINITE FREE RESOLUTIONS 841 The next lemma isdesigned tofacilitate dimension shifting. We saythat two modules Ml'M2arestably isomorphic ifthere exist finite free modules F1,F2such that M1Fl M2 F2. Lemma 2.3. LetM1bestably isomorphic toM2.Let ONl El-+Ml O 0N2E2-+M20 beexact sequences, where M1isstably isomorphic toM2,and El,E2are stably free. Then N1isstably isomorphic toN2. Proof. Bydefinition, there isanisomorphism M1 F1 M2(f)F2. We have exact sequences ONl-+El Fl-+Ml (f)Fl-+O o N2 E2(f)F2 M2(f)F2-+0 BySchanuel's lemma (seebelow) weconclude that Nl(f)E2(f)F2 N2(f)El Fl. Since El'E2,Fl'F2arestably free, we can add finite free modules toeach side sothat thesummands ofN1and N2become free, andbyadding I-dimensional free modules ifnecessary,we can preserve theisomorphism, which proves that N1isstably isomorphic toN2. Westill have totake care ofSchanuel's lemma: Lemma 2.4. Let OKP-+MO o K' P' M -+0 beexact sequences where P,P'areprojective. Then there isanisomorphism K(f)P' K'(f)P. Proof. Since Pisprojective, there exists ahomomorphism P P'making theright square inthefollowing diagram commute. o.P !w)0 )K u!)M Id )M)0j.P' o)K' 842 FINITE FREE RESOLUTIONS XXI,2 Then one can find ahomomorphism K K'which makes the left square commute. Then weget anexact sequence o K P(f)K' P' 0 byx (ix,ux)for xEKand (y,z) wy-jz. We leave theverification of exactness tothereader. Since P'isprojective, thesequence splits thus proving Schanuel's lemma. This also concludes theproof ofLemma 2.3. The minimal length ofastably free resolution ofamodule iscalled its stably free dimension. Toconstruct astably free resolution ofafinite module, weproceed inductively. Thepreceding lemmas allow ustocarry outtheinduc- tion, and also tostop theconstruction ifamodule isoffinite stably free dimen- SIon. Theorem 2.5. LetMbeamodule which admits astably free resolution of length n o En...Eo-+M -+O. Let Fm-+...FoMO beanexact sequence with Fistably freefor i=0,...,m. (i)Ifm<n-1then there exists astably free Fm+1such that the exact sequence can becontinued exactly to Fm+1...F0 M o. (ii)Ifm=n-1,letFn=Ker(F n-1 Fn-2).Then Fnisstably free and thus o-+Fn-+Fn- 1...F0 M 0 isastably free resolution. Remark. IfAisNoetherian then ofcourse (i)istrivial, and we can even pick Fm+1tobefinite free. Proof. Insert thekernels and cokernels ineach sequence, say Km=Ker(E m-+Em- 1)if m=f.0 Ko=Ker(E oM), and define K:nsimilarly. ByLemma 2.3, Km isstably isomorphic toK:n, say Km(f)FK(f)F' with F,F'finite free. XXI,2 FINITE FREE RESOLUTIONS 843 Ifm<n-1,then Km isahomomorphic image ofEm+1;soboth Km(f)F andK(f)F'arehomomorphic images ofEm+1(f)F.Therefore Kisahomo- morphic image ofEm +1(f)Fwhich isstably free. WeletFm+1=Em +1(f)Fto conclude theproof inthis case. Ifm=n-1,then we can take Kn=En. Hence Km(f)Fisstably free, and soisK(f)F'bytheisomorphism inthefirstpart oftheproof. Itfollows trivially thatKisstably free, andbydefinition, K=Fm+1inthis case. This concludes theproof ofthetheorem. Corollary 2.6. If0 M1 E M 0isexact, Mhasstably free dimen- sion<n,and Eisstably free, then M1hasstably free dimension<n-1. Theorem 2.7. Let o M' M M" 0 beanexact sequence. Ifany twoofthese modules have afinite free resolution, then sodoes thethird. Proof. Assume M' andMhave finite free resolutions. Since Misfinite, it follows that M" isalso finite. Byessentially the same construction asChapter XX, Lemma 3.8, we can construct anexact and commutative diagram where E',E,E" arestably free: oo j . 11 )E j )M j oo j ) 1'; ·E" j )Mil j o)0 oo j )M' r )E' j )M' j oo)0 )0 We then argue byinduction onthestably free dimension ofM. We see that M1hasstably free dimension <n-1(actuallyn-1,but wedon't care), andM; hasfinite stably free dimension. Byinduction we arereduced tothe case when Mhasstably free dimension 0,which means that Misstably free. Since byassumption there isafinite free resolution ofM',itfollows that M" also has afinite freeresolution, thus concluding theproof ofthefirst assertion. 844 FINITE FREE RESOLUTIONS XXI,2 Next assume that M', Mil have finite free resolutions. Then Misfinite. Ifboth M'and Mil have stably free dimension 0,then M', Mil areprojective and M M'(f)Mil isalso stably free and we are done. We now argue by induction onthemaximum oftheir stably free dimension n,and we assume n>1.We can construct anexact and commutative diagramasintheprevious case with E',E,E"finite free(we leave thedetails tothereader). But themaxi- mum ofthestably free dimensions ofM'1 andM'; isatmost n-1,and soby induction itfollows that MIhasfinite stably free dimension. This concludes the proof ofthesecond case. Observe that thethird statement hasbeen proved inChapter XX, Lemma 3.8 when AisNoetherian, taking for(i,theabelian category offinite modules, and for CCthefamily ofstably free modules. Mitchell Stokes pointedout tomethat the statement isvalid ingeneral without Noetherian assumption, and can be provedasfollows. We assume thatM,M"have finite free resolutions. Wefirst show thatM'isfinitely generated. Indeed, suppose first thatMisfinite free. We have two exact sequences o M' M M" 0 o K" F" M" 0 where F"isfinite free, andK"isfinitely generated because oftheassumption that M" has afinite free resolution. That M' isfinitely generated follows from Schanuel's lemma. IfMisnotfree, one can reduce thefinite generation ofM' tothe case when Misfree byapull-back, which weleave tothereader. Now suppose that thestably free dimension ofM" ispositive. We use the same exact commutative diagramasintheprevious cases, with E',E,E"finite free. The stably free dimension ofM'; isone less than that ofM", and we are done byinduction. This concludes theproof ofTheorem 2.7. This also concludes ourgeneral discussion offinite free resolutions. For more information cf.Northcott's book onthesubject. We now come tothesecond part ofthissection, which provides anapplica- tion topolynomial rings. Theorem 2.8. Let Rbeacommutative Noetherian ring. Let xbeavariable. Ifevery finite R-module has afinite free resolution, then every finite R[x]-module has afinite free resolution. Inother words, inthecategory offinite R-modules, ifevery object isof finite stably freedimension, then the same property applies tothecategory of finite R[x]-modules. Before proving thetheorem, westate theapplicationwe have inmind. Theorem 2.9. (Serre). Ifkisafield and XI,. . .,xrindependent vari- ables, then every finite projective module over k[XI'. . .,xr]isstably free, or equivalently admits afinite free resolution. XXI,2 FINITE FREE RESOLUTIONS 845 Proof. Byinduction andTheorem 2.8 weconclude that every finite module over k[Xb. . .,xr]isoffinite stably free dimension. (We areusing Theorem 2.1.) This concludes theproof. The rest ofthis section isdevoted totheproof ofTheorem 2.8. LetMbe afinite R[x]-module. ByChapter X,Corollary 2.8, Mhas afinite filtration M=Mo::)M l::)...::)Mn=O such that each factor MilM i+1isisomorphic toR[x]IPi for some prime Pi. Inlight ofTheorem 2.7,itsuffices toprove thetheorem incase M=R[x]/p where Pisprime, which we now assume. Inlight ofthe exact sequence o-+P-+R[x]-+R[x]IP o. and Theorem 2.7, we note that Mhas afinite free resolution ifandonly ifP does. Let p=PnR.Then pisprime inR.Suppose there issome M =R[x]IP which does notadmit afinite freeresolution. Among allsuch Mweselect onefor which theintersection pismaximal inthefamily ofprime ideals obtained as above. This ispossible inlight ofone ofthe basic properties characterizing Noetherian rings. LetRo=Rip soRoisentire. LetPo=PlpR[xJ. Then wemay view M asanRo[x]-module, equal toRoIPo. Letfl'...' inbeafinite setofgenerators forPo, and letfbe apolynomial ofminimal degree inPo. LetKo bethe quotient field ofRo. Bytheeuclidean algorithm,we canwrite h=qif+rifor i=1,...,n with qi,riEKo[x] and deg ri<degf. Letdobeacommon denominator for thecoefficients ofallqi,ri.Then do=f.0and doh=qf +r where q=doqiandr=dorilieinRo[x]. Since degfisminimal inPoit follows thatr=0foralli,so doPocRo[x]f=(f). LetNo=Po/(f),soNoisamodule over Ro[x], and we can also view No asamodule over R[x]. When soviewed, wedenote NobyN.LetdERbeany element reducing todomod p.Then dFJpsince do=f.O.The module Nohas afinite filtration such that each factor module ofthefiltration isisomorphic to some Ro[x]/Qo where Qoisanassociated prime ofNo. LetQbetheinverse image ofQoinR[x]. These prime ideals Qareprecisely theassociated primes ofNinR[x]. Since dokills Noitfollows that dkills Nand therefore dliesin every associated prime ofN.Bythemaximality property intheselection ofP, 846 FINITE FREE RESOLUTIONS XXI,3 itfollows that everyone ofthefactor modules inthefiltration ofNhas afinite free resolution, andbyTheorem 2.7itfollows that Nitself has afinite free resolution. Now weview Ro[x]asanR[x]-module, viathecanonical homomorphism R[x]-+Ro[x]=R[x]/pR[x]. Byassumption, phas afinite free resolution asR-module, say o-+En...-+Eo-+p o. Then wemay simply form themodules Ei[X] intheobvious sense toobtain a finite free resolution ofp[x]=pR[x]. From the exact sequence opR[x] R[x] Ro[x]-+0 weconclude thatRo[x] has afinite free resolution asR[x]-module. Since Roisentire, itfollows that theprincipal ideal(f)inRo[x]isR[x]- isomorphic toRo[x], and therefore has afinite free resolution asR[x]-module. Theorem 2.7applied tothe exact sequence ofR[x]-modules o-+(f) Po N -+0 shows that Pohas afinite free resolution; and further applied tothe exact sequence OpR[x]PPoO shows that Phas afinite free resolution, thereby concluding theproof of Theorem 2.8. 3. UNIMODULAR POLYNOMIAL VECTORS Let Abeacommutative ring. Let(fl,. ..,f,,)beelements ofAgenerating theunit ideal. Wecall such elements unimodular. Weshall saythat they have theunimodular extension property ifthere exists amatrix inGLn(A) with first column '(fl,.. .,f,,).IfAisaprincipal entire ring, then itisatrivial exercise to prove that this isalways the case. Serre originally asked thequestion whether itistrue for apolynomial ringk[x 1,. ..,Xr]over afield k.The problemwas solved byQuillen and Suslin. Wegive here asimplification ofSuslin's proof by Vaserstein, also using aprevious result ofHorrocks. The method isbyinduc- tion onthenumber ofvariables, insome fashion. We shall writef=t(11'...,j)forthecolumn vector. We first remark thatfhas theunimodular extension property ifandonly ifthe vector obtained byapermutation ofitscomponents hasthisproperty. Similarly, we can make XXI,3 UNIMODULAR POLYNOMIAL VECTORS 847 theusual rowoperations, adding amultiple gj;tojj(j=Ii),andfhas theuni- modular extension property ifand only ifanyone ofitstransforms byrow operations hastheunimodular extension property. Wefirst prove thetheorem inacontext which allows theinduction. Theorem 3.1. (Horrocks). Let(0,m)bealocal ring and letA=o[x] bethepolynomial ring inone variable over o.Letfbeaunimodular vector inA(n)such that some component hasleading coefficient1.Thenfhas the unimodular extension property. Proof. (Suslin). Ifn= 1or2then thetheorem isobvious even without assuming that 0islocal. So we assume n>3and do aninduction ofthe smallest degree dofacomponent offwith leading coefficient 1.First wenote that bytheEuclidean algorithm and row operations, wemay assume thatfl hasleading coefficient 1,degree d,and that degj; <dforj=I1.Sincefis unimodular, arelation Lgij;= 1shows that notallcoefficients off2,. ..,j" can lieinthemaximal ideal m.Without lossofgenerality,wemayassume that some coefficient off2does notlieinmand soisaunit since 0islocal. Write fl(x)=xd+ad_l xd-l+... +ao with aiEO, f2(x)=bsxs +...+bowith biE0,S<d-1, sothat some biisaunit. Let abetheideal generated byallleading coefficients ofpolynomials 91fl+g2f2ofdegree<d-1.Then acontains allthe co- efficients bi,i=0,...,s.One sees thisbydescending induction, starting with bswhich isobvious, and then usingalinear combination xd-Sf2(x)-bsfl(x). Therefore aistheunit ideal, and there exists apolynomial 91fl+92f2of degree<d-1andleading coefficient 1.Byrowoperations, wemay now get apolynomial ofdegree<d-1andleading coefficient 1assome component inthei-thplace for some i=I1,2.Thus ultimately, byinduction, wemay assume that d=0inwhich case thetheorem isobvious. This concludes the proof. Over any commutative ring A,fortwo column vectors f,9wewritef----g over Atomean that there exists MEGLn(A) such that f=Mg, and wesay thatfisequivalent to9over A.Horrocks' theorem states that a unimodular vectorfwith one component having leading coefficient 1iso[x]- equivalent tothefirst unit vector e1 .We areinterested ingettingasimilar descent over non-local rings. We can writef=f(x), and there isanatural "constant" vector f(O) formed with theconstant coefficients. As acorollary of Horrocks' theorem, weget: 848 FINITE FREE RESOLUTIONS XXI,3 Corollary 3.2. Let 0be alocal ring. Letfbe aunimodular vector in o[x](n) such that some component hasleading coefficient 1.Thenff(O) overo[x]. Proof. Note thatf(O)Eo(n)has onecomponent which isaunit. Itsuffices toprove that over any commutative ring Rany element cER(n) such that some component isaunit isequivalentover Rtoe1,and this isobvious. Lemma 3.3. Let Rbeanentire ring, and letSbe amultiplicative subset. Let x,ybeindependent variables. Iff(x)"-'f(O)over S-1R[x], then there exists cESsuch thatf(x+cy)"-'f(x)over R[x, y]. Proof. Let M EGLn(S-1R[x]) besuch thatf(x)=M(x)f(O). Then M(x)- If(x)=f(O) isconstant, and thus invariant under translation x x+y. Let G(x, y)=M(x)M(x +y)-l. Then G(x,y)f(x +y)=f(x). We have G(x, 0)=Iwhence G(x, y)=I+yH(x, y) with H(x, y)ES-1R[x, y]. There exists cESsuch that cH hascoefficients in R.Then G(x, cy)hascoefficients inR.Since detM(x) isconstant inS-1R,it follows that detM(x +cy) isequal tothis same constant and therefore that detG(x, cy)=1.This proves thelemma. Theorem 3.4. Let Rbeanentire ring, and letfbeaunimodular vector in R[x](n), such that one component hasleading coefficient 1.Then f(x)"-'f(O) over R[x]. Proof. LetJbethe setofelements cERsuchthatf(x +cy)isequivalent tof(x) over R[x, y].Then Jisanideal, forifCEJand aERthen replacing y byayinthedefinition ofequivalence shows thatf(x +cay) isequivalent to f(x) over R[x, ay], soover R[x, y].Equally easily, one sees that ifc,c'EJ then c+c'EJ.Now letpbeaprime ideal ofR.ByCorollary 3.2 weknow thatf(x) isequivalent tof(O) over Rp[x], andbyLemma 3.3 itfollows that there exists cERand cftpsuch thatf(x +cy) isequivalent tof(x) over R[x, y]. Hence Jisnotcontained inp,and soJisunit ideal inR,sothere exists aninvertible matrix M(x, y)over R[x, y]such that f(x +y)=M(x, y)f(x). Since thehomomorphic image ofaninvertible matrix isinvertible, wesubstitute ofor xinthis lastrelation toconclude theproof ofthetheorem. Theorem 3.5. (Quillen-Suslin). Letkbeafield and letfbeaunimodular vector ink[x 1,. . .,xr](n).Thenfhas theunimodular extension property. XXI,3 UNIMODULAR POLYNOMIAL VECTORS 849 Proof. Byinduction on r.Ifr= 1then k[x1]isaprincipal ring and the theorem islefttothereader. Assume thetheorem forr-1variables with r>2, and put R =k[xl,...,X r-1]. Weviewfasavector ofpolynomials inthelastvariable Xrand want toapply Theorem 3.4. We can dosoifsome component ofjhas leading coefficient 1in thevariable Xr.We reduce thetheorem tothis case asfollows. Theproof ofthe Noether Normalization Theorem (Chapter VIII, Theorem 2.1) shows thatifwe let \J =XJr r Y.=x.-x'"I I I r then thepolynomial vector f(x1,. ..,Xr)=g(Y1,...,Yr) has onecomponent with Yr-Ieading coefficient equal to1.Hence there exists a matrix N(y)=M(x) invertible over R[xr]=R[Yr] such that g(Yb...' Yr)=N(Yl'...' Yr)g(Yb...' Yr-l, 0), and g(Yb...,Yr- l'0)isunimodular ink[Yb...,Yr-l](n). We can therefore conclude theproof byinduction. We now give other formulations ofthe theorem. First werecall that a module Eover acommutative ring Aiscalled stably free ifthere exists afinite free module Fsuch that E(f)Fisfinite free. We shall saythat acommutative ring Ahas theunimodular column exten- sion property ifevery unimodular vectorfEA(n)has theunimodular extension property, forallpositive integersn. Theorem 3.6. LetAbeacommutative ring which has theunimodular column extension property. Then every stably free module over Aisfree. Proof. Let Ebestably free. We use induction ontherank ofthe free modules Fsuch that E(f)Fisfree. Byinduction, itsuffices toprove that if E(f)Aisfree then Eisfree. Let E(f)A=A(n)and let p:A(n) A betheprojection. Let u1be abasis ofAover itself. Viewing Aasadirect summand inE(f)A=A(n)wewrite U1=t(a1b.. .,an 1)with ailEA. 850 FINITE FREE RESOLUTIONS XXI,4 Then ulisunimodular, and byassumptionu1isthefirst column ofamatrix M =(aij)whose determinant isaunit inA.Let uj=Mejforj=1,..., n, where ejisthej-th unit column vector ofA(n).Note that u1isthefirst column ofM.Byelementary column operations,wemay change Msothat ujEEfor j=2,...,n.Indeed, ifpej=cu1forj>2weneed only replaceejbyej-ce1 . Without loss ofgeneralitywemay therefore assume that u2 ,...,unlieinE. Since Misinvertible over A,itfollows that Minduces anautomorphism of A(n) asA-module with itself by XMX. Itfollows immediately from theconstruction and thefact that A(n)=E(f)A that Mmaps thefree module with basis {e2 ,. ..,en}onto E.This concludes theproof. Ifwe now feed Serre's Theorem 2.9into thepresent machinery consisting oftheQuillen-Suslin theorem andTheorem 3.6, weobtain thealternative version oftheQuillen-Suslin theorem: Theorem 3.7. Let kbeafield. Then every finite projective module over the polynomial ring k[xI'. . .,xr]isfree. 4. THE KOSZUL COMPLEX Inthis section, wedescribe afinite complex built out ofthealternating product ofafree module. This givesanapplication ofthealternating product, and also gives afundamental construction used inalgebraic geometry, both abstract and complex, asthereader canverify bylooking atGriffiths-Harris [GrH 78],Chapter V,3;Grothendieck's [SGA 6];Hartshorne [Ha77],Chapter III,7;andFulton-Lang [FuL 85],Chapter IV,2. We know from Chapter XX that afree resolution ofamodule allows usto compute certain homologyorcohomology groups of afunctor. Weapply this now toHom and also tothetensor product. Thus wealso getexamples ofexplicit computations ofhomology, illustrating Chapter XX, by means oftheKoszul complex. We shall also obtain aclassical application byderiving theso-called Hilbert Syzygy theorem. LetAbe aring (always assumed commutative) andMamodule. Asequence ofelements XI'. . .,XrinAiscalled M-regular ifMI(x.,. . .,xr)M=1=0,ifXI XXI,4 THE KOSZUL COMPLEX 851 isnotdivisor ofzero inM, and fori>2,Xiisnotdivisor of0in MI(xb...' Xi-l)M. Itiscalled regular when M =A. Proposition 4.1. LetI=(Xl"..,xr)begenerated byaregular sequence inA.Then 1112isfree ofdimension roverAll. Proof. LetXibetheclass ofXimod 12 .Itsuffices toprove thatXl'...,Xr arelinearly independent. Wedothisbyinduction on r.For r=1,ifax =0, then ax =bx2for some bEA,sox(a-bx)=o.Since xisnot zero divisor inA, wehave a=bx soa=O. Now suppose theproposition true fortheregular sequence Xl'...,Xr-l. Suppose r LQiXi=0 In 1112 . i=1 Wemayassume thatLaixi=0inA;otherwise Laixi=LYiXi with YiEIand we canreplace aibyai-Yiwithout changing ai. Since Xrisnot zero divisor inAI(x l'. ..,Xr-1)there exist biEAsuch that r- 1 r- 1 r- 1 arx r+Laix i=0=>ar=Lbix i=>L(ai+bixr)x i=O. i=1 i=l i=l Byinduction, r- 1 aj+bjxrE LAx i i=1u=1,..., r-1) soajEIforallj,soaj=0forallj, thus proving theproposition. LetK,Lbecomplexes, which wewrite asdirect sums K =EBKpand L=EBLq with p,qEZ. Usually, Kp=Lq=0forp,q<o.Then the tensor product K(8)Listhecomplex such that (K(8)L)n=EBKp(8)Lq;p+q=n and for UEKp'VELqthedifferential isdefined by d(u (8)v)=du(8)v+(-I)Pu (8)dv. (Carry outthedetailed verification, which isroutine, that thisgivesacomplex.) 852 FINITE FREE RESOLUTIONS XXI,4 LetAbeacommutative ring and xEA.Wedefine thecomplex K(x) tohave Ko(x)=A,Kl(x)=Ael, where elisasymbol, Ael isthefreemodule ofrank 1 with basis {el}, and theboundary map isdefined bydel=x,sothecomplex can berepresented bythesequence o )Ael II )K1(x)d )A " )Ko(x))0)0 o More generally, forelements Xl'...,XrEAwedefine the Koszul complex K(x)=K(Xl'...,xr)asfollows. Weput: Ko(x)=A; K}(x)=free module Ewith basis {el'.. .,er}; Kp(x)=free module I'fE with basis{e;t1\...1\e;p}' i}<· ··<ip; Kr(x)=free module /'{E ofrank 1with basis e}1\...1\er- Wedefine theboundary maps bydei=Xiand ingeneral d:Kp(x)Kp-1(x) by p d(e.1\'.. 1\e.)=L(-1)j-1x.e.1\... 1\€':.1\...1\e. .11 Ip . IJ 11 IJ Ip}=l Adirect verification shows that d2=0,sowehave acomplex oKr(x) ...Kp(x)... Kl(X) A 0 The next lemma shows the extent towhich thecomplex isindependent ofthe ideal I=(xb...,xr)generated by(x). Let I=(xl'...,Xr)::)I'=(y1,...,Yr) betwo ideals ofA .We have anatural ring homomorphism can:All' All. Let{e'b...,e}beabasis forK1(y),and let Yi=LcijXjwithcijEA. We define 11:Kl(y) Kl(x)by fle= c..e.1 i..J I}} XXI,4 THE KOSZUL COMPLEX 853 and fp=fl1\... 1\fl, prod ucttaken ptimes. Let D =det(cij)bethedeterminant. Then forp=rwegetthat f,.:K,(y) Kr(x) ismultiplication byD. Lemma 4.2. Notation asabove, thehomomorphismsfp defineamorphism of Koszul complexes: o )Kr(Y)---+... )Kp(Y)---+. . .---+Kl(y)---+ A All'---+ 0 jf,=Dy. y' Id jean o---+Kr(x)---+...---+Kp(x)---+...---+ Kl(X)---+ A---+All---+ 0 anddefineanisomorphism ifDisaunit inA,forinstance if(y)isapermutation of(x). Proof. Bydefinition f(e1\... 1\e)=(c..e.)1\... 1\(c..e.)II IP '11 )rpJ ). j=l j=l Then fd(e11\... 1\ep) =f(t(-l)k-1 Yikei,/\.../\ /\.'./\eip) =L(-l)k-l yik(Icide j)/\.../\I/\.../\(ICiPjej) k j= 1 k j= 1 =L(-l)k-l(.Icidej)1\...1\(.I.cikjXjej )1\... 1\(ICiPjej) )=1 )=1 j=l'-v-" omitted =dlf(e1\.'. 1\e) II rp using Yik=LCikjXj.This concludes theproof thatthefpdefine ahomomorphism ofcomplexes. Inparticular, if(x)and(y)generate the same ideal, and thedeterminant D isaunit (i.e. thelinear transformation going from (x)to(y) isinvertible over thering), then the two Koszul complexesareisomorphic. 854 FINITE FREE RESOLUTIONS XXI,4 The next lemma givesus auseful way ofmaking inductions later. Proposition 4.3. There isanatural isomorphism K(xl'...,Xr) K(X1)(8)...(8)K(Xr). Proof. Theproof will beleft asanexercise. LetI=(x1,.. .,xr)betheideal generated byXl'...,Xr.Then directly from thedefinitions we seethat theO-th homology oftheKoszul complex issimply AlIA. More generally, letMbeanA-module. Define theKoszul complex ofMby K(x; M)=K(x l,..., Xr;M)=K(x l,..., xr)(8)AM Then thiscomplex looks like o Kr(x) 0M· · .K2(x)0AM M(r) M o. We sometimes abbreviateHp(x; M)forHpK(x; M). The first and lasthomology groupsarethen obtained directly from thedefinition ofboundary.We get Ho(K(x; M»=MIIM; Hr(K(x); M)={vEMsuch that xiV=0foralli=1,...,r}. Inlight ofProposition 4.3, westudy generally what happens toatensor product ofanycomplex with K(x), when xconsists of asingle element. Let YEA andletCbeanarbitrary complex ofA-modules. Wehave anexact sequence ofcomplexes (1) o C C0K(y) (C0K(y»/C 0 made explicitasfollows. oj )(C n+1(8)A) (C n(8)K1(y» j )(C n(8)A) (C n-1(8)K1(y) j )(C n-1(8)A) (C n-2(8)K1(y» jj )Cn(8)K1(y) jd.@Id )Cn-1(8)K1(y» jd._I@,d )Cn-2(8)K1(y) j)0 o1 )Cn+1 j )Cn j )Cn-l jo )0 )0 XXI,4 THE KOSZUL COMPLEX 855 We note that C0K1(y) isjust Cwith adimension shift byoneunit, inother words (2) (C0K1(Y»n+l=Cn0K1(y). Inparticular, (3) Hn+1(C 0K(y)/C)=Hn(C). Associated with anexact sequence ofcomplexes, wehave thehomology sequence, which inthis case yields thelong exact sequence )Hn+1(C))Hn+l(C (8)K1(y» )Hn+1(C(8)K(y)jC) n Hn(C)o )Hn(C) which wewrite stacked upaccording totheindex: Hp+1(C) Hp+1(C) Hp+1(C0K(y» Hp(C) Hp(C) Hp(C0K(y» ending inlowest dimension with(4) (5) H1(C) H1(C 0K(y» Ho(C) Ho(C). Furthermore, adirect application ofthedefinition oftheboundary map and the tensor product ofcomplexes yields: Theboundary maponHp(C)(p>0)isinduced bymultiplication by(-1)Py: (6) a=(-I)Pm(y):Hp(C) Hp(C). Indeed, write (C0K(y»p=(Cp0A)E9(Cp-10K1(y»=CpE9Cp-1. Let(v,w) ECpEBCp-1with vECpand wECp-1.Then directly from the definitions, (7) d(v, w)=(dv +(-I)p-lyw, dw). To see(6), onemerely follows upthedefinitions oftheboundary, takingan element wECp=Cp0K1(y), lifting back to(0,w),applying d,andlifting back toCpoIfwe start with acycle, Le. dw=0,then themap iswell defined onthehomology class, with values inthehomology. Lemma 4.4. LetyEA and letCbeacomplexasabove. Then m(y) annihilates Hp(C0K(y» forallp>o. Proof. If(v,w)isacycle, Le.d(v, w)=0,then from (7) weget atonce that(yv, yw)=d(O, (-I)Pv), which proves thelemma. 856 FINITE FREE RESOLUTIONS XXI,4 Intheapplicationswehave inmind, welety=xrand C=K(x"...,xr-I;M)=K(XI,...,X r-I)0M. Then weobtain: Theorem 4.5.(a) There isanexact sequence with mapsasabove: HpK(x.,..., xr-l; M) HpK(XI'...' xr-l; M) HpK(XI'...'xr;M) m(xr)...HI(XI,...,xr;M) Ho(x.,..., Xr-l; M)----+HO(xI'...' xr-l; M). (b)Every element ofI=(X.,. . .,xr)annihilatesHp(x; M)forp>o. (c)IfI=A,thenHp(x; M)=0forallp>o. Proof. This isimmediate from Proposition 4.3 and Lemma 4.4. We define theaugmented Koszul complex tobe o Kr(x; M)·..KI(x; M)=M(r) M MI1M o. Theorem 4.6. LetMbeanA-module. (a)Letxl'. ..,Xrbe aregular sequence for M. ThenHpK(x; M)=0for p>o.(Of course, HoK(x; M)=MIIM.) Inother words, theaugmented Koszul complex isexact. (b)Conversely, suppose Aislocal, and Xl,. . .,Xrlieinthemaximal ideal of A.Suppose Misfinite over A,and also assume thatHIK(x; M)=O.Then (xl,. . .,xr)isM-regular. Proof. Weprove (a)byinduction onr.Ifr=1thenHI(x; M)=0directly from thedefinition. Supposer>1.We use the exact sequence ofTheorem 4.5(a). Ifp> 1thenHp(x; M)isbetween twohomology groups which are0,so Hp(x; M)=O.IfP=1,we use thevery end ofthe exact sequence ofTheorem 4.5(a), noting thatm(x r)isinjective,sobyinduction wefindHI(x;M)=0also, thus proving (a). Asto(b),byLemma 4.4 and thehypothesis,weget anexact sequence m(xr)HI(x.,. . .,xr-l;M)----+HI(x.,. . .,xr-l; M) HI(x; M)=0, som(x r)issurjective. ByNakayama's lemma, itfollows that HI(XI,..., Xr-l; M)=O. Byinduction (X.,. . .,xr-l) isanM-regular sequence. Looking again atthetail endoftheexact sequenceasin(a)shows that XrisMI(xI,. . .,xr-I)M-regular, whence proving (b)and thetheorem. We note that (b), which uses only thetriviality ofHI(and not allHp)is due toNorthcott [No 68], 8.5, Theorem 8.By(a), itfollows thatHp=0for p>o. XXI,4 THE KOSZUL COMPLEX 857 Animportant specialcase ofTheorem 4.6(a) iswhen M=A,inwhich case we restate thetheorem intheform: LetXl'.. .,Xrbe aregular sequence inA.Then K(x 1,. . .,xr)isafree resolution ofAII: oKr(x)...Kl(X) AAll o. Inparticular, All hasTor-dimension <r. For theHom functor, wehave: Theorem 4.7. LetXI'. ..,Xrbearegular sequence inA.Then there isan isomorphism lfJx,M: Hr(Hom(K(x), M» MllM tobedescribed below. Proof. The module Kr(x) isI-dimensional, with basis el1\...1\ere Dependingonthis basis, wehave anisomorphism Hom(Kr(x), M) M, wherebyahomomorphism isdetermined byitsvalue atthebasis element inM. Then directly from thedefinition oftheboundary map drintheKoszul complex, which is r d.e11\·· ·1\e"" (-1Y.-1x.e11\· ··1\e.1\...1\er' r J ) r )=1 we seethat Hr(Hom(Kr(x), M) Hom(Kr(x), M)ldr-1Hom(Kr- 1(x),M) MllM. This proves thetheorem. The reader who has read Chapter XX knows that thei-thhomology group ofHom(K(x), M)iscalled Exti(AI I,M), determined uptoaunique isomorphism bythecomplex, since two resolutions ofAIIdiffer byamorphism ofcomplexes, and two such morphisms differ byahomotopy which induces ahomology iso- morphism. Thus Theorem 4.7givesanisomorphism lfJx,M: Extr(AIl, M) MIIM. Infact, weshall obtain morphisms oftheKoszul complex from changing the sequence. We goback tothehypothesis ofLemma 4.2. 858 FINITE FREE RESOLUTIONS XXI,4 Lemma 4.8. IfI=(x)=(y)where (x),(y) are tworegular sequences, then wehave acommutative diagram MIIM7 )Extr(A/I,M)D=del(e'j) MIIM where allthemapsareisomorphisms ofAll-modules. The factthat wearedealing withAII-modules isimmediate since multiplication byanelement ofAcommutes with allhomomorphisms insight, andIan- nihilates All. ByProposition 4.1, weknow that1112isafree module ofrank roverAII. Hence /\r(III2) isafree module ofrank 1,with basis Xl/\.../\xr(where the bar denotes residue class mod 12).Taking thedual ofthisexterior product,we seethat under achange ofbasis, ittransforms according tothe inverse ofthedeterminant mod 12 .This allows ustogetacanonical isomorphismasinthe next theorem. Theorem 4.9. Letxl,.. .,Xrbearegular sequence inA,and letI=(x). LetMbeanA-module. Let t/Jx,M: MIIM (MIIM) (8)/\r(III2)dual betheembedding determined bythebasis (Xl/\.../\xr)dual of/\r(II12)dual. Then thecomposite isomorphism Extr(AII, M) MIIM (MIIM) (8)/\r(III2)dual isafunctorial isomorphism, independent ofthechoice ofregular generators forI. We also have theanalogue ofTheorem 4.5 inintermediate dimensions. Theorem 4.10. LetXl'. ..,XrbeanM-regular sequence inA.LetI=(x). Then Exti(AII, M)=0for i<r. Proof. For theproof, we assume that the reader isacquainted with the exact homology sequence. Assume byinduction that Exti(AII, M)=0for XXI,4 THE KOSZUL COMPLEX 859 i<r-1.Then wehave the exact sequence o=Exti-l(AI1, MlxlM) Exti(AI1, M) Exti(AI1, M) fori<r.But XlE1somultiplication byXlinduces 0onthehomology groups, which gives Exti(AI1, M)=0asdesired. LetLN N 0beafree resolution ofamodule N.Bydefinition, Tort(N, M)=i-thhomology ofthecomplex L(8)M. This isindependent ofthechoice ofLNuptoaunique isomorphism. We now want todoforTor what wehave just done forExt. Theorem 4.11. LetI=(xl,. ..,xr)beanideal ofAgenerated byaregular sequence oflength r. (i)There isanatural isomorphism Tort(AI1, All) !\/I(1112), for i>O. (ii)Let Lbe afree All-module, extended naturally toanA-module. Then Tort(L, All) L(8)!\/I(1112), for i>o. These isomorphisms will follow from the next considerations. First we useagain that theresid ueclasses Xl'...,xrmod 12form abasis of 1112over All. Therefore wehave aunique isomorphism ofcomplexes qJx:K(x) (8)All !\(II12)=E8!\i(1112) with zero differentials ontheright-hand side, such that - - e.1\... 1\e. x.1\... 1\X. .11 Ip 11 Ip Lemma 4.12. LetI=(x) ::)I'=(y)betwo ideals generated byregular sequences oflength r.Letf:K(y) K(x) bethemorphism ofKoszul complexes defined inLemma 4.2. Then thefollowing diagram iscommutative: K(y) (8)All'qJy)!\A/I,(/'ll'2) f @can! !canonical horn K(x) (8)AII qJx)!\A/I(I 112) 860 FINITE FREE RESOLUTIONS XXI,4 Proof. We have qJx0(f(8)can)(e11\... 1\ep(8)1) r r =c..X. 1\... 1\'c. .x.i...J 111 J IpJ J j=2 j=l =Y-.1\... 1\Y-. =can(fn(e1\... 1\e».II Ip 't'y II Ip This proves thelemma. Inparticular, ifI'=Ithen wehave thecommutative diagram K(y) j @dj)1\(1/12) K(x) which shows that the identification ofTori(AII, All) with l\i(1112)via the choices ofbases iscompatible under oneisomorphism oftheKoszul complexes, which providearesolution ofAll. Since any other homomorphism ofKoszul complexes ishomotopic tothis one, itfollows that this identification does not dependonthechoices made and proves thefirst part ofTheorem 4.11. The second part follows atonce, because wehave Tort(A/I, L)=Hi(K(x) (8)L)=Hi«K(x) (8)AAll) (8)A/IL =1\/1(1112) (8)L. This concludes theproof ofTheorem 4.11. Example. Letkbeafield and letA=k[x 1,...,xr]bethepolynomial ring inrvariables. LetI=(xl'...,xr)betheideal generated bythevariables. Then AII=k,and therefore Theorem 4.11 yields fori>0: Tort(k, k)I\l(1112) Tor(L, k) L(8)1\(1112) Note that inthepresent case, wecanthink ofII12asthevector space over kwith basis Xl'...,xr.Then Acan beviewed asthesymmetric algebra SE, where E isthis vector space. We cangiveaspecific example oftheKoszul complex inthis context asinthe next theorem, given for afree module. XXI,4 THE KOSZUL COMPLEX 861 Theorem 4.13. Let Ebeafinitefree module ofrank rover thering R.For each p=1,..., rthere isaunique homomorphism dp:!\pE(8)SE -+!\p-1E(8)SE such that di«x 1/\.../\Xp)(8)y) p =L(-l)i-l(X l/\.../\ /\.../\ Xp)(8)(Xi(8)y) i=1 where XiEEand YESE. This gives theresolution o-+!\rE(8)SE!\r-1E(8)SE...-+!\0E(8)SE R 0 Proof. The above definitions aremerely examples oftheKoszul complex forthesymmetric algebra SEwith respect totheregular sequence consisting of some basis ofE. Since dpmaps !\PE (8)SqE into!\p-1E(8)sq+1E,we candecompose this complex into adirect sum corresponding toagiven graded component, and hence: Corollary 4.14. For each integern>1,wehave anexact sequence o!\rE(8)sn-rE.. .!\1E(8)sn-1E snE 0 whereSjE =Oforj<O. Finally, wegiveanapplication toaclassical theorem ofHilbert. Thepoly- nomial ring A=k[x 1,...,xr]isnaturally graded, bythedegrees ofthehomo- geneous components. Weshall consider graded modules, where thegrading isin dimensions >0,and we assume that homomorphisms aregraded ofdegree o. Sosuppose Misagraded module (and thus Mi=0fori<0)andMisfinite over A.Then we can find agraded surjective homomorphism Lo M 0 where Loisfinite free. Indeed, letWl,...,Wnbehomogeneous generators ofM. Let el,...,enbebasis elements for afree module Loover A.Wegive Lothe grading such thatifaEAishomogeneous ofdegree dthen aeiishomogeneous of degree deg aei=dega+deg Wi. Then thehomomorphism ofLoonto Msending ei Wiisgradedasdesired. 862 FINITE FREE RESOLUTIONS XXI,4 The kernel M1isagraded submodule ofLo.Repeating theprocess,wecanfind a surjective homomorphism L1M1O. We continue inthis way toobtain agraded resolution ofM. We want this resolution tostop, and thepossibility ofitsstopping isgiven bythenext theorem. Theorem 4.15. (Hilbert Syzygy Theorem). Let kbeafield and A=k[x 1,...,Xr] thepolynomial ring inrvariables. LetMbeagraded module over A,and let o K Lr- 1...Lo M 0 beanexact sequence ofgraded homomorphisms ofgraded modules, such that Lo,...,Lr- 1arefree. Then Kisfree. If'M isinaddition finite over Aand Lo,...,Lr-1arefinite free, then Kisfinite free. Proof. From theKoszul complexweknow thatTori(M, k)=0fori>r and allM.Bydimension shifting, itfollows that Tori(K, k)=0for i>O. The theorem isthen aconsequence ofthe next result. Theorem 4.16. LetFbeagraded finite module over A=k[x l,..., xr].If Tor1(F,k)=0then Fisfree. Proof. The method isessentially todo aNakayama type argument inthe case ofthenon-local ring A.First note that F@k=FIIF where I=(Xl'...,xr).Thus F@kisnaturally anAll=k-module. Let Vl,...,Vnbehomogeneous elements ofFwhose residue classes mod IFform a basis ofFIIF over k.LetLbeafree module with basis et, ..., en.Let LF bethegraded homomorphism sending ei Vifori=1,...,n.Itsuffices to prove that this isanisomorphism. Let Cbethecokernel, sowehave the exact sequence L F C o. Tensoring with kyields the exact sequence L@k F@k C@k O. XXI,4THE KOSZUL COMPLEX 863 Since byconstruction the map L(8)k F(8)kissurjective, itfollows that C(8)k=O.But Cisgraded,sothe next lemma shows that C=O. Lemma 4.17. Let Nbe agraded module over A=k[x 1,...,xr].Let I=(xl'...,Xr).IfNilN =0then N =O. Proof. This isimmediate byusing thegrading, looking atelements ofN ofsmallest degree ifthey exist, andusing thefact that elements ofIhave degree >o. We now getanexact sequence ofgraded modules O-+E-+LFO and wemust show that E=O.But the exact homology sequence and our as- sumption yields o=Torl(F, k) E(8)k-+L(8)k F(8)k O. Byconstruction L(8)k F(8)kisanisomorphism, and hence E(8)k=O. Lemma 4.17 now shows that E=O.This concludes theproof ofthesyzygy theorem. Remark. Theonly place intheproof where weused that kisafield isinthe proof ofTheorem 4.16 when wepicked homogeneous elements Vb...,VninM whose residue classes mod 1M form abasis ofMIIMover AlIA. Hilbert's theorem can begeneralized bymaking theappropriate hypothesis which allows ustocarry outthis step,asfollows. Theorem 4.18. Let Rbeacommutatvelocal ring andletA=R[Xb.. .,xr] bethepolynomial ring inrvariables. LetMbeagraded finite module over A, projective over R.Let o K Lr-1...Lo-+M 0 beanexact sequence ofgraded homomorphisms ofgraded modules such that Lo,...,Lr- 1arefinite free. Then Kisfinite free. Proof Replace kbyReverywhere intheproof ofthe Hilbert syzygy theorem. We usethefact that afinite projective module over alocal ring isfree. Not aword needs tobechanged intheabove proof with thefollowing exception. We note that theprojectivity propagates tothekernels and cokernels inthe given resolution. Thus Finthe statement ofTheorem 4.16 may beassumed projective, and each graded component isprojective. Then F'IIF'isprojective over AIIA =R,and soiseach graded component. Since afinite projective mod uleover alocal ring isfree, and onegets thefreeness byliftingabasis from the resid ueclass field, wemay pick VI'...,Vnhomogeneous exactlyaswedidinthe proof ofTheorem 4.16.This concludes theproof. 864 FINITE FREE RESOLUTIONS XXI, Ex EXERCISES For exercises Ithrough 4ontheKoszul complex,see[No68], Chapter 8. I.Let0 M' M Mil 0beanexact sequence ofA-modules. Show thattensoring with theKoszul complex K(x) onegetsanexact sequence ofcomplexes, andtherefore anexact homology sequence o HrK(x; M') HrK(x; M) HrK(x; M")... ...HpK(x; M') HpK(x; M) HpK(x; M")... ...HoK(x; M') HoK(x; M) HoK(x; M") 0 2.(a) Show that there isaunique homomorphism ofcomplexes !:K(x; M) K(xl'. . .,xr-I;M) such that for vEM: {e.1\".1\ e.(8)xv f,(e;1\." 1\ei(8)v)=I) Ipr p ) pe. 1\...1\e.0vI) Ipifip=r ifip=r. (b) Show that! isinjective ifXrisnot adivisor ofzero inM. (c)For acomplex C,denote byC(-I) thecomplex shifted byoneplace totheleft, soC(-I)n=Cn-Iforall n.LetM=M/xrM. Show that there isaunique homomorphism ofcomplexes g:K(XI"..'xr-I,I;M) K(Xl'...' xr-I;M)(-I) such that for vEM: {e.1\...1\e. (8)v g(e. 1\...1\C.(8)V)=I) Ip-) P I) Ip0ifip=r ifip<r. (d)IfXrisnot adivisor of0inM,show that thefollowing sequence isexact: f g- o K(x; M) K(x I'. . .,Xr-I, I;M) K(x I'. . .,xr-I;M)(-I) O. Using Theorem 4.5(c), conclude that forallp>0,there isanisomorphism HpK(x; M) HpK(x.,..., xr-I;M). 3.Assume AandMNoetherian. LetIbeanideal ofA.Letai'. . .,akbeanM-regular sequence inI.Show that this sequencecan beextended to amaximal M-regular sequence aI'. . .,aqinI,inother words anM-regular sequence such that there is noM-regular sequence ai'. . .,aq+IinI. 4.Again assume Aand MNoetherian. LetI=(xI'. . .,xr)and let aI'. . .,aqbe a maximal M-regular sequence inI.Assume 1M =1=M.Prove that Hr-q(x; M)=1=0butHp(x; M)=0forp>r-q. [See [No 68], 8.5Theorem 6.The result issimilar totheresult inExercise 5,and generalizes Theorem 4.5(a). See also [Mat 80], pp. 100-103. The result shows that XXI, Ex EXERCISES 865 allmaximal M-regular sequences inMhave the same length, which iscalled the I-depth ofMand isdenoted bydepth[(M). For theproof, let 5bethemaximal integer such that HsK(x; M)=1=O.Byassumption, Ho(x; M)=M/IM=1=0, so 5exists. We have toprove that q+ 5=r.First note that ifq=0then 5=r.Indeed, if q=0then every element ofIiszero divisor inM,whence Iiscontained inthe union oftheassociated primes ofM,whence insome associated prime ofM.Hence Hr(x; M)=1=O. Next assume q>0andproceed byinduction. Consider the exact sequence OMMM/aIM 0 where thefirst map ism(a l).Since IannihilatesHp(x; M)byTheorem 4.5(c), we getanexact sequence oHp(x; M) Hp(x; M/aIM) Hp_l(x; M) O. Hence Hs+I(x;M/aiM)=1=0,butHp(x;M/aiM)=0forp>S+2.From thehypothesis thatai'. . .,aqisamaximal M-regular sequence, itfollows atonce that a2,. . .,aq ismaximal M/aIM-regular inI,sobyinduction, q-1=r-(5+1)and hence q+5=r,aswas tobeshown.] 5.Thefollowing exercise combines some notions ofChapter XX onhomology, and some notions covered inthischapter and inChapter X, 5.LetMbeanA-module. LetAbeNoetherian, Mfinite module over A,andIanideal ofAsuch that 1M i=M. Let rbeaninteger>1.Prove that thefollowing conditions areequivalent: (i)Exti(N, M)=0foralli<randallfinite modules Nsuch thatsupp(N)c(I). (ii)Exti(A/I, M)=0foralli<r. (iii) There exists afinite module Nwith supP(N)=(I) such that Exti(N, M)=0foralli<r. (iv) There exists anM-regular sequence a1,...,a,inI. [Hint: (i) (ii) (iii) isclear. For(iii) (iv), first note that o=ExtO(N, M)=Hom(N, M). Assume supp(N)=(I). Find anM-regular element inI.Ifthere isnosuch element, then Iiscontained inthe setofdivisors of0ofMinA,which istheunion ofthe as- sociated primes. Hence IcPfor some associated prime P.This yields aninjection A/PcM, so oi=HomAp(Ap/PAp,M). Byhypothesis, Npi=0soNp/PN p=1=0,and Np/PN pisavector space over Ap/P Ap, sothere exists anon-zero Ap/P Aphomomorphism Np/PN p-+Mp, soHomAp(N p,Mp) =1=0,whence Hom(N, M) =1=0,acontradiction. This proves the existence ofoneregular element a1. 866 FINITE FREE RESOLUTIONS XXI, Ex Now letM1=Mja 1M. The exact sequence Ql0-+ M --.M -+MjalM-+0 yields the exact cohomology sequence -+Exti(N, M)-+Exti(N, Mja 1M)-+Exti+I(N,M)--. soExti(N, Mja1 M)=0fori<r-1.Byinduction there exists anMI-regular se- quence a2,...,a,and we aredone. Last, (iv) (i). Assume the existence oftheregular sequence. Byinduction, Exti(N, aiM)=0fori<r-1.We have anexact sequence fori<r: o-+Exti(N, M) Exti(N, M) But supp(N)=(ann(N)) C(l), soICrad(ann(N)), soatisnilpotentonN. Hence atisnilpotentonExti(N, M), soExti(N, M)=O.Done.] See Matsumura's [Mat 70], p.100, Theorem 28. The result isuseful inalgebraic geometry, with for instance M=Aitself. One thinks ofAastheaffine coordinate ring ofsome variety, and onethinks oftheequationsai=0asdefining hypersurface sections ofthisvariety, and thesimultaneous equations at= ...=ar=0asdefiningacomplete intersection. The theorem givesacohomological criterion interms ofExtfortheexistence ofsuch acomplete intersection. APPENDIX 1 The Transcendence of eand n Theproof which weshall give here follows theclassical method ofGelfond andSchneider, properly formulated. Itisbased onatheorem concerning values offunctions satisfying differential equations, andithad been recognized forsome time that such values aresubject tosevere restrictions, invarious contexts. Here, wedeal with the most general algebraic differential equation. We shall assume that thereader isacquainted with elementary facts con- cerning functions ofacomplex variable. Letfbe anentire function (Le. a function which isholomorphic onthecomplex plane). For our purposes,we sayfisoforder <pifthere exists anumber C> 1such that foralllarge Rwe have If(z)1<CRP wheneverIzI<R.Ameromorphic function issaid tobeoforder <pifitisa quotient ofentire functions oforder <p. Theorem. LetKbeafinite extension oftherational numbers. Letfl'...,fN bemeromorphic functions oforder <p.Assume that thefieldK(fl'...,fN) hastranscendence degree>2over K,and that thederivative D=dldz maps theringK[fl'...,fN]intoitself. Let Wl,...,Wmbedistinct complex numbers notlying among thepoles oftheJi,such that Ji(w v)EK foralli=1,...,Nand v=1,..., m.Then m<lOp[K:Q]. Corollary 1.(Hermite-Lindemann). IflJ.isalgebraic (over Q)and =f.0, then e(1istranscendental. Hence 1tistranscendental. 867 868 THE TRANSCENDENCE OF eAND 'IT APPENDIX 1 Proof Suppose that rxand eaarealgebraic. LetK =Q(rx,ea).The two functions zand eZarealgebraically independentover K(trivial), and thering K[z,eZ]isobviously mapped into itself bythederivative. Our functions take on algebraic values inKatrx,2rx,...,mrxforany m,contradiction. Since e2ni=1, itfollows that 2ni istranscendental. Corollary 2.(Gelfond-Schneider). Ifrxisalgebraic =I0, 1andiff3is algebraic irrational, then rxfJ=efJlog aistranscendental. Proof WeproceedasinCorollary 1,considering thefunctions efJtand et which arealgebraically independent because pisassumed irrational. Welook atthenumbers log rx,210g rx,...,mlogrxtoget acontradiction asinCorollary 1. Before giving themain arguments proving thetheorem, westate some lemmas. The first two, due toSiegel, have todowith integral solutions oflinear homo- geneous equations. Lemma 1.Let allxl +..·+alnx n=0 ar1X1+...+arnxn=0 beasystem oflinear equations with integer coefficients aij,and n>r.Let A be anumber such thatIaij I<Aforalli,j.Then there exists anintegral, non-trivial solution with IxjI<2(2nA)r/<n-r). Proof We view our system oflinear equationsas alinear equation L(X)=0,where Lisalinear map, L:z<n) z<r>, determined bythematrix of coefficients. IfBisapositive number, wedenote byz<n>(B) the setofvectors X inz<n) such thatIXI<B(where IXIisthemaximum oftheabsolute values ofthecoefficients ofX). Then Lmaps z<n)(B) into Z<r)(nBA). The number of elements inz<n)(B) is>Bnand «2B+l)n. We seek avalue ofBsuch that there will betwo distinct elements X, Yinz<n)(B) having the same image, L(X)=L(Y).Forthis, itwill suffice that Bn>(2nBA)r, and thus itwill suffice that B=(2nA )r/<n-r). Wetake X-Yasthesolution ofourproblem. LetKbeafinite extension ofQ,and letIKbetheintegral closure ofZinK. From Exercise 5ofChapter IX, weknow that IKisafree module over Z,of dimension [K:Q]. We view Kascontained inthecomplex numbers. If APPENDIX 1 THE TRANSCENDENCE OF eAND 'IT869 rxEK,aconjugate ofrxwill betaken tobeanelement (Jrx,where (Jisanembedding ofKin C.Bythesizeofa setofelements ofKwe shall mean themaximum ofthe absolute values ofallconjugates ofthese elements. Bythesizeofavector X =(xl'...,xn)weshall mean thesizeofthe setofits coordinates. Let Wl,...,WM beabasis ofIKover Z.Let rxElK, and write rx=alw l+...+aMwM' LetW'l'...,w:V bethedual basis ofWl,...,WMwith respect tothe trace. Then we can express the(Fourier) coefficients ajof rxasatrace, aj=Tr(rxwj). The trace isasum over theconjugates. Hence the size ofthese coefficients is bounded bythe size ofrx,times afixed constant, depending onthesize ofthe elements wj. Lemma 2. LetKbeafinite extension ofQ. Let rxllX l+...+rxlnxn=0 rxr1X1+...+rxrnxn=0 beasystem oflinear equations with coefficients inIK,and n>r.Let Abea number such thatsize(rxij)<A,foralli,j.Then there exists anon-trivial solution XinIKsuch that size(X)<Cl(C 2nA)r/(n-r), where CbC2are constants depending only onK. Proof. Let Wb. ..,WM beabasis ofIKover Z.Each xjcan bewritten Xj=jlWl +...+jMWM with unknowns j;".Eachrxijcan bewritten rxij=aij 1Wl+...+aijMWM with integers aij;"EZ.Ifwemultiply outtherxijXj,wefind that ourlinear equa- tions with coefficients inIKareequivalent toasystem ofrMlinear equations in thenMunknowns j;",with coefficients inZ,whose size isbounded byCA, where Cisanumber depending onlyonMand thesizeoftheelements W;..,together with theproducts W;..w,inother words where Cdepends onlyonK.Applying Lemma 1,weobtain asolution interms ofthej;",and hence asolution XinIK, whose size satisfies thedesired bound. 870 THE TRANSCENDENCE OF eAND 'IT APPENDIX 1 The next lemma has todowith estimates ofderivatives. Bythe size ofa polynomial with coefficients inK,weshall mean thesizeofitssetofcoefficients. Adenominator for asetofelements ofKwill beanypositive rational integer whose product with every element ofthe setisanalgebraic integer. Wedefine in asimilar wayadenominator for apolynomial with coefficients inK. We abbreviate" denominator" byden. Let P(Tl'...' TN)=Ll1(v)M(v)(T) beapolynomial with complex coefficients, and let Q(Tl'...,TN)=Lp(v)M(v)(T) be apolynomial with real coefficients >O.We say that Qdominates Pif 111(v) I<p(V)forall(v). Itisthen immediately verified that therelation ofdomi- nance ispreserved under addition, multiplication, andtaking partial derivatives with respect tothevariables T1,...,TN. Lemma 3. Let Kbeoffinite degree over Q.Letfl,...,fNbefunctions, holomorphic on aneighborhood ofapointwEC,and assume that D=dldz maps thering K[fl'.. .,fN] intoitself. Assume thatIi(w)EKforalli.Then there. exists anumber C1having thefollowing property. LetP(Tl'.. .,TN)be apolynomial with coefficients inK,ofdegree<r.Ifwe setf=P(fl,...,fN), then wehave, forallpositive integers k, size(Dkf(w»<size(P)rkk! c+r Furthermore, there isadenominator forD".f(w) bounded byden(P)C +r. Proof. There exist polynomials Pi(T l,.. .,TN)with coefficients inKsuch that Dii=Pi(fl'. ..,fN). Let hbethemaximum oftheir degrees. There exists aunique derivation Don K[Tl' ..., TN] such that D7i=Pi(Tl'...' TN). For anypolynomial Pwehave N D(P(T l,.. .,TN»=L(DiP)(Tl'. ..,TN).Pi(T b...,TN)' i=1 where Dl,...,DNarethepartial derivatives. Thepolynomial Pisdominated by size(P)(1 +Tl+...+TN)r, and each Piisdominated bysize(Pi)(l +Tl+...+TN)h. Thus DPisdominated by size(P)C 2r(1+Tl+...+TN)r+h. APPENDIX 1 THE TRANSCENDENCE OF eAND 'IT871 Proceeding inductively,one sees that Dkpisdominated by size(P)C r"k!(1+Tl...+TN)r+kh. Substituting values h(w) for1i,weobtain thedesired bound onDkf(w). The second assertion concerning denominators isproved alsobyatrivial induction. We now come tothemain part oftheproof ofour theorem. Letf,9betwo functions among fl'...,fNwhich arealgebraically independent over K.Let rbeapositive integer divisible by2m. Weshall letrtend toinfinity attheend oftheproof. Let r F=Lbijfi gj i,j=1 havecoefficients bijinK.Let n=r2/2m. We can select thebijnotallequal to0, and such that DkF(w v)=0 for0<k<nand v=1,..., m.Indeed, wehave tosolve asystem ofmnlinear equations inr2=2mn unknowns. Note that mn =1. 2mn-mn Wemultiply these equations byadenominator forthecoefficients. Using the estimate ofLemma 3,and Lemma 2,we can infact take thebijtobealgebraic integers, whose size isbounded by O(r"n! Ci+r)<O(n2") for n 00. Since f,9arealgebraically independent over K,our function Fisnot identicallyzero. We let sbethesmallest integer such that allderivatives ofF uptoorder s-1vanish atallpointsWl,...,wm,but such that DSF does not vanish atone ofthe w,sayw1.Then s>n.Welet y=DSF(Wl) =IO. Then yisanelement ofK,and byLemma 3,ithas adenominator which is bounded byO(C1) for s 00.Let cbethisdenominator. The norm ofcyfrom KtoQisthen anon-zero rational integer. Each conjugate ofcyisbounded by O(S5S). Consequently,weget (1)1<IN(cy) I<O(S5s)[K:Q]- 11yI, 872 THE TRANSCENDENCE OF eAND 'IT APPENDIX 1 whereIyIisthefixed absolute value ofy,which will now beestimated very wellby global arguments. Let f)beanentire function oforder<..:p,such thatf)fand f)gareentire, and f)(Wl) =IO.Then f)2rFisentire. Weconsider theentire function H(z)=(z)2rF(z). n(z-wv)S v= 1 Then H(w l)differs from DSF(Wl) byobvious factors, bounded byCs!. Bythe maximum modulus principle, itsabsolute value isbounded bythemaximum of Honalarge circle ofradius R.Ifwe take Rlarge, then z-Wvhasapproximately the same absolute value asR,andconsequently,onthecircle ofradius R,H(z) isbounded inabsolute value byanexpression oftype s3sc;rRP Rms We select R=Sl/2P. Wethen gettheestimate s4scs Iyl<mS/2:. S We now letrtend toinfinity. Then both nand stend toinfinity. Combining this lastinequality with inequality (1), weobtain the desired bound on m.This concludes theproof. Ofcourse, wemade noeffort tobeespecially careful inthe powers of s occurring intheestimates, and thenumber 10canobviously bedecreased by exercising alittle more care intheestimates. The theorem weproved isonly thesimplest in anextensive theory dealing with problems oftranscendence degree. Insome sense, thetheorem is bestpossible without additional hypotheses. Forinstance, ifP(t) isapolynomial with integer coefficients, then eP(t)willtake thevalue 1atallroots ofP,these being algebraic. Furthermore, thefunctions t t2t"t,e,e,..., e arealgebraically independent, buttake onvalues inQ(e) forallintegral values oft. However, oneexpects rather strong results ofalgebraic independence tohold. Lindemann proved thatifr:J..l,...,r:J..narealgebraic numbers, linearly independent over Q,then eatea",..., arealgebraically independent. APPENDIX 1 THE TRANSCENDENCE OF eAND 'IT873 More generally, Schanuel has made thefollowing conjecture: If(Xl'. ..,(Xn arecomplex numbers, linearly independent over Q,then the transcendence degree of IV IVa1 an 1.J\.1,...,I.J\.n'e,..., e should be>n. From this one would deduce atonce thealgebraic independence ofeand 1t (looking at1,21ti, e,e2ni),and allother independence statements concerning the ordinary exponential function and logarithm which one feels tobetrue, for instance, thestatement that 1tcannot lieinthefield obtained bystarting with the algebraic numbers, adjoining values oftheexponential function, taking algebraic closure, anditerating these twooperations. Such statements have todowith values oftheexponential function lying incertain fields oftranscendence degree <n,and onehopes that by asuitable deepening ofTheorem 1,onewill reach thedesired results. APPENDIX 2 Some SetTheory 1. DENUMERABLE SETS Let nbe apositive integer. Let Jnbethe setconsisting ofallintegers k, 1<k<n.IfSisaset, wesaythat Shas nelements ifthere isabijection between Sand Jn'Such abijection associates with each integer kasabove anelement ofS, sayk ak.Thus wemayuseJnto"count" S.Part ofwhat weassume about the basic facts concerning positive integers isthatifShas nelements, then theinteger nisuniquely determined byS. One also agrees tosaythat asethas 0elements ifthe setisempty. Weshall saythat asetSisdenumerable ifthere exists abijection ofSwith the setofpositive integers Z+.Such abijection isthen said toenumerate the setS. Itisamapping nan which toeach positive integer nassociates anelement ofS,themapping being injective andsurjective. IfDisadenumerable set,andf:S Disabijection ofsome setSwith D, then Sisalso denumerable. Indeed, there isabijection 9:D Z+ ,and hence 90fisabijection ofSwith Z+ . LetTbeaset. Asequence ofelements ofTissimplyamapping ofZ+into T. Ifthe map isgiven bytheassociation n Xn,wealso write the sequenceas {xn}nboralso {xbX2,'. .}.Forsimplicity,wealso write {xn} forthesequence. Thus wethink ofthesequenceasprescribingafirst, second,...,n-th element of T.We usethe same braces forsequencesasforsets, butthecontext willalways make ourmeaning clear. 875 876 SOME SET THEORY APPENDIX 2 Examples. The even positive integers may beviewed asasequence {xn} if weputXn=2nfor n=1,2,....The oddpositive integers may also beviewed asasequence {Yn} ifweput Yn=2n-1for n=1,2,....Ineach case, the sequence gives anenumeration ofthegiven set. Wealso usetheword sequence formappings ofthenatural numbers into aset, thus allowing our sequences tostart from 0instead of1.Ifweneed tospecify whether asequence starts with theO-th term orthefirst term, wewrite {xn}nOor{Xn}n1 according tothedesired case. Unless otherwise specified, however, wealways assume that asequence will start with thefirst term. Note that from asequence {xn}nOwe can define anew sequence byletting Yn=Xn-l for n>1.Then Yl=Xo, Y2=Xl'....Thus there isnoessential difference between the two kinds ofsequences. Given asequence {xn}, wecall Xnthen-th term ofthesequence. Asequence may very well besuch that allitsterms areequal. Forinstance, ifwelet Xn= 1 forall n>1,weobtain the sequence {I,1,1,...}.Thus there isadifference between asequence ofelements inasetT,and asubset ofT.Intheexample just given, the setofallterms ofthesequence consists ofone element, namely the single number 1. Let{xl'X2,...}beasequence inasetS.Byasubsequenceweshall mean a sequence {x n1,xn2,...} such that nl<n2<.... For instance, if{xn} isthe sequence ofpositive integers, Xn=n,thesequence ofeven positive integers {x2n} isasubsequence. Anenumeration ofasetSisofcourse asequence inS. Asetisfinite ifthe setisempty, orifthe sethas nelements for some positive integern.Ifasetisnotfinite, itiscalled infinite. Occasionally,amap ofJninto asetTwill becalled afinite sequence inT. Afinite sequence iswritten asusual, {Xl,...,Xn} or{XJi=l,...,n. When weneed tospecify thedistinction between finite sequences and maps of Z+into T,wecall thelatter infinite sequences. Unless otherwise specified,we shall usetheword sequence tomean infinite sequence. Proposition 1.1. LetDbeaninfinite subset ofZ+. Then Disdenumerable, and infact there isaunique enumeration ofD,say{k., k2,...}such that k1<k2<...<kn<kn+1<... . Proof. Weletklbethesmallest element ofD.Suppose inductively that we have defined kl<...<kn,insuch away that any element kinDwhich isnot equal tok1,...,knis>kn.Wedefine kn+1tobethesmallest element ofDwhich is>kn.Then themapn knisthedesired enumeration ofD. APPENDIX 2 SOME SET THEORY 877 Coronary 1.2. Let Sbe adenumerable setand Daninfinite subset ofS. Then Disdenumerable. Proof Given anenumeration ofS,thesubset Dcorresponds toasubset of Z+inthisenumeration. Using Proposition 1.1, weconclude that wecan enumer- ateD. Proposition 1.3. Every infinite setcontains adenumerable subset. Proof Let Sbeaninfinite set. For every non-empty subset TofS,we select adefinite element aTinT.Wethen proceed byinduction. WeletXlbethe chosen element as.Suppose that wehave chosen Xl'...,Xnhaving theproperty that foreach k=2,...,ntheelement Xkistheselected element inthesubset which isthecomplement of{xl'...,Xk-l}.Weletxn+1betheselected element inthecomplement ofthe set{Xl'...,xn}.Byinduction, wethus obtain an association n Xnforallpositive integers n,and since Xn=IXkforallk<nit follows that ourassociation isinjective, Le.givesanenumeration ofa subset ofS. Proposition 1.4. Let Dbe adenumerable set, andf:D Sasurjective mapping. Then Sisdenumerable orfinite. Proof For each YES, there exists anelementXyEDsuch thatf(x y)=Y because fissurjective. The association y Xvisaninjective mapping ofSinto D,because if y,ZESandXy=Xz then y=f(x y)=f(xz)=z. Letg(y)=XY.The image of9isasubset ofDand Disdenumerable. Since 9 isabijection between Sand itsimage, itfollows that Sisdenumerable orfinite. Proposition 1.5. LetDbeadenumerable set. Then DxD(the setofaUpairs (x,y)with x,yED)isdenumerable. Proof. There isabijection between DxDandZ+XZ+ ,soitwill suffice to prove that Z+XZ+isdenumerable. Consider themapping ofZ+XZ+-+Z+ given by (m,n) 2n3m . Itisinjective, andbyProposition 1.1, our result follows. Proposition 1.6. Let{Dl' D2,. ..}beasequence ofdenumerable sets. Let S betheunion ofallsets Di(i=1,2,. ..).Then Sisdenumerable. 878 SOME SET THEORY APPENDIX 2 Proof For each i=1,2,...weenumerate theelements ofDi,asindicated inthefollowing notation: Dl:{XlbX12,X13,...} D2:{X21,X22,X23,...} Di:{Xil' Xi2, Xi3,...} The mapf:Z+XZ+-+Dgiven by f(i,j)=xij isthen asurjective map ofZ+XZ+onto S.ByProposition 1.4,itfollows that Sisdenumerable. Corollary 1.7. LetFbeanon-emptyfinite setand Dadenumerable set. Then FxDisdenumerable. IfSl'S2'...are asequence ofsets, each ofwhich is finite ordenumerable, then theunion S1US2U...isdenumerable orfinite. Proof. There isaninjection ofFinto Z+and abijection ofDwith Z+.Hence there isaninjection ofFxZ+into Z+XZ+and we canapply Corollary 1.2 andProposition 1.6toprove thefirst statement. One could also define asur- jective map ofZ+XZ+onto FxD.(Cf. Exercises 1and4.)Asforthesecond statement, each finite setiscontained insome denumerable set, sothat thesecond statement follows from Proposition 1.1and 1.6. For convenience, weshall saythat asetiscountable ifitiseither finite or denumerable. 2. ZORN'S LEMMA Inorder todealefficiently with infinitely many setssimultaneously,one needs aspecial property. Tostate it,weneed some more terminology. Let Sbeaset. Anordering (also called partial ordering) ofSisarelation, written x<y,among some pairs ofelements ofS,having thefollowing properties. ORO 1. Wehave x<x. ORO 2.Ifx<yand y<zthen x<z. ORO 3.Ifx<yand y<xthen x=y. APPENDIX 2 SOME SET THEORY 879 Wesometimes write y>xforx<y.Note that wedon't require that therelation x<yory<xhold forevery pair ofelements (x,y)ofS.Some pairs may not be comparable. Iftheordering satisfies thisadditional property, then wesaythat it isatotal ordering. Example 1. Let Gbeagroup. Let Sbethe setofsubgroups. IfH,H'are subgroups ofG,wedefine H<H' ifHisasubgroup ofH'. One verifies immediately that this relation defines an ordering onS.Given twosubgroups H,H'ofG,wedonotnecessarily have H<H'orH'<H. Example 2. LetRbearing, and letSbethe setofleftideals ofR.Wedefine anordering inSinaway similar totheabove, namely ifL,L'areleftideals ofR, wedefine L<L' ifLcL'. Example 3. LetXbeaset,and Sthe setofsubsets ofX.IfY,Zaresubsets ofX,wedefine Y<ZifYisasubset ofZ.This defines anordering onS. Inallthese examples, therelation ofordering issaid tobethat ofinclusion. Inanordered set,ifx<yand x=Iywethen write x<y. LetAbeanordered set,and Basubset. Then wecandefine anordering onB bydefining x<yforx,yEBtohold ifandonly ifx<yinA.Weshall saythat RoistheorderingonBinduced byR,oristherestriction toBofthepartial ordering ofA. Let Sbeanordered set. Byaleast element ofS(or asmallest element) one means anelement aESsuch that a<xforallXES. Similarly, byagreatest element one means anelement bsuch that x<bforallXES. Byamaximal element mofSone means anelement such that ifXES and x>m,then x=m.Note that amaximal element need notbeagreatest element. There may bemany maximal elements inS,whereas ifagreatest element exists, then itisunique (proof ?). Let Sbeanordered set. Weshall saythat Sistotally ordered ifgiven x,YES wehave necessarily x<yory<x. Example 4. The integers Zaretotally ordered bytheusual ordering. So arethereal numbers. Let Sbeanordered set,and Tasubset. Anupper bound ofT(inS)isan element bESsuch that x<bforallxET.Aleast upper bound ofTinSisan upper bound bsuch that, ifcisanother upper bound, then b<c.Weshall say 880 SOME SET THEORY APPENDIX 2 that Sisinductively ordered ifevery non-empty totally ordered subset has an upper bound. Weshall saythat Sisstrictly inductively ordered ifevery non-empty totally ordered subset has aleast upper bound. InExamples 1,2,3,ineach case, the setisstrictly inductively ordered. To prove this, letustake Example 2.LetTbeanon-empty totally ordered subset ofthesetofsubgroups ofG.This means thatifH,H' ET,then HcH'orH'cH. Let Ubetheunion ofallsets inT.Then: 1.Uisasubgroup. Proof: Ifx,yEU,there exist subgroups H,H' ET such that xEHand YEH'.If,say, HcH',then both x,YEH'and hence xYEH'. Hence XYEV. Also, X-1EH', soX-lEV. Hence Visa subgroup. 2.Visanupper bound foreach element ofT.Proof: Every HETiscon- tained inV,soH<VforallHET. 3.Visaleast upper bound for T.Proof: Any subgroup ofGwhich contains allthesubgroups HETmust then contain their union V. Theproof that the sets inExamples 2,3arestrictly inductively ordered is entirely similar. We can now state theproperty mentioned atthebeginning ofthesection. Zorn's Lemma. Let Sbeanon-empty inductively ordered set. Then there exists amaximal element inS. Asanexample ofZorn's lemma, weshall now prove theinfinite version ofa theorem given inChapters 1,7,andXIV, 2,namely: Let Rbeanentire, principal ring and letEbeafree module over R.LetFbea submodule. Then Fisfree. Infact, if{Vi}'el isabasis forE,and F=I{O}, then there exists abasis forFindexed byasubset ofI. Proof. For each subset JofIweletEJbethefreesubmodule ofEgenerated byallVj,jEJ,and weletFJ=EJnF.WeletSbethe setofallpairs (FJ, w) where Jisasubset ofI,and w:J' FJisabasis ofFJindexed byasubset J'ofJ. Wewrite wjinstead ofw(j) forjEJ'.If(FJ,w)and(FK,u)are such pairs,we define (FJ, w)<(FK,u)ifJcK,ifJ'cK', andiftherestriction ofutoJis equal tow.(Inother words, thebasis uforFKisanextension ofthebasis wfor FJ.)This defines anordering onS,anditisimmediately verified that Sisinfact inductively ordered, andnon-empty (saybythefinite case oftheresult). We can therefore apply Zorn's lemma. Let(FJ, w)beamaximal element. Wecontend that J=I(this will prove ourresult). Suppose J=IIand letkEIbut kftJ.Let K =Ju{k}. If EJu{k}nF=FJ, APPENDIX 2 SOME SET THEORY 881 then (FK,w)isabigger pair than (FJ, w)contradicting themaximality assump- tion. Otherwise there exist elements ofFKwhich can bewritten intheform CVk+Y with some yEEJand CER,c=IO.The setofallelements CERsuch that there exists yEEJforwhichCVk+YEFisanideal. Let abeagenerator ofthisideal, and let Wk=aVk+Y be anelement ofF,with YEEJ.IfZEFKthen there exists bERsuch that Z-bWkEEJ.But z-bWkEF,whence z-bWkEFJ.Itfollows atonce that thefamily consisting ofWj(jEJ)and Wkisabasis forFK,thus contradicting the maximalityagain. This proves what wewanted. Zorn's lemma could bejust taken asanaxiom ofsettheory. However, itis notpsychologically completely satisfactoryasanaxiom, because itsstatement istooinvolved, and one does notvisualize easily theexistence ofthemaximal element asserted inthat statement. Weshow how one can prove Zorn's lemma from other properties ofsets which everyone would immediately grant asac- ceptable psychologically. From now ontotheend oftheproof ofTheorem 2.1, weletAbeanon- empty partially ordered andstrictly inductively ordered set. We recall that strictly inductively ordered means that every nonempty totally ordered subset has aleast upper bound. We assume givenamapf:A Asuch that forall xEAwehave x<f(x). Wecould call such amapanincreasing map. Let aEA.LetBbeasubset ofA.Weshall saythat Bisadmissible if: 1.Bcontains a. 2.We havef(B)cB. 3.Whenever Tisanon-empty totally ordered subset ofB,theleast upper bound ofTinAlies inB. Then Bisalsostrictly inductively ordered, bytheinduced ordering ofA.We shall prove: Theorem 2.1. (Bourbaki). Let Abe anon-empty partially ordered and strictly inductively ordered set. Letf:A Abe anincreasing mapping. Then there exists anelement XoEAsuch thatf(xo)=Xo. Proof. Suppose that Awere totally ordered. Byassumption, itwould have aleast upper bound bEA,and then b<f(b)<b, 882 SOME SET THEORY APPENDIX 2 sothat inthis case, our theorem isclear. The whole problem istoreduce the theorem tothat case. Inother words, what weneed tofind isatotally ordered admissible subset ofA. Ifwethrow outofAallelements xEAsuch that xisnot >a,then what remains isobviouslyanadmissible subset. Thus without loss ofgenerality, we may assume that Ahas aleast element a,that isa<xforallxEA. LetMbetheintersection ofalladmissible subsets ofA.Note that Aitself is anadmissible subset, and that alladmissible subsets ofAcontain a,sothat Mis notempty. Furthermore, Misitself anadmissible subset ofA.To seethis, let xEM.Then xisinevery admissible subset, sof(x)isalso inevery admissible subset, and hence f(x)EM. Hence f(M)cM.IfTisatotally ordered non- empty subset ofM,and bistheleast upper bound ofTinA,then bliesinevery admissible subset ofA,and hence liesinM.Itfollows that Misthesmallest admissible subset ofA,and that any admissible subset ofAcontained inMis equal toM. Weshall prove that Mistotally ordered, andthereby prove Theorem 2.1. [First wemake some remarks which don't belong totheproof, butwillhelp intheunderstanding ofthesubsequent lemmas. Since aEM, we seethat f(a)EM,f0f(a)EM,and ingeneral f"(a)EM.Furthermore, a<f(a)<f2(a)<.... Ifwe had anequality somewhere, wewould befinished, sowemay assume that theinequalities hold. LetDobethetotally ordered set{f"(a)}"o.Then Do looks like this: a<f(a) <f2(a) <...<f"(a) <... . Let atbetheleast upper bound ofDo. Then we canform al<f(al) <f2(a l)<... inthe same way toobtain Dl,and wecan continue this process, toobtain Dt,D2,... . Itisclear thatDbD2,...arecontained inM.Ifwehad aprecise way ofex- pressing thefactthat wecanestablish anever-ending string ofsuch denumerable sets, then wewould obtain what wewant. Thepoint isthat weare nowtrying to prove Zorn's lemma, which isthenatural tool forguaranteeing theexistence of such astring. However, given such astring,weobserve that itselements have two properties: Ifcisanelement ofsuch astring and x<c,thenf(x)<c. Furthermore, there isnoelement between candf(c), that isifxisanelement of thestring, then x<corf(c)<x.Weshall now prove two lemmas which show that elements ofMhave these properties.] APPENDIX 2 SOME SET THEORY 883 Let CEM.Weshall saythat cisanextreme point ofMifwhenever xEMand x<c,thenf(x)<c.For each extreme point cEMwelet Mc=setofxEM such that x<corf(c)<x. Note that Mcisnotempty because aisinit. Lemma 2.2. Wehave Mc=Mfor every extreme pointcofM. Proof Itwill suffice toprove that Mcisanadmissible subset. Let xEMc. Ifx<cthenf(x)<csof(x)EMc. Ifx=cthenf(x)=f(c) isagain inMc. Iff(c)<x,then f(c)<x<f(x),soonce more f(x)EMc. Thus wehave proved thatf(M c)cMc. LetTbeatotally ordered subset ofMcand letbbetheleast upper bound of TinM.Ifallelements xETare <c,then b<Cand bEMc. Ifsome xETis suchthatf(c)<x,thenf(c)<x<b,andsobisinM c.This proves ourlemma. Lemma 2.3. Every element ofMisanextreme point. Proof LetEbethe setofextreme points ofM.Then Eisnotempty because aEE.Itwill suffice toprove that Eisanadmissible subset. Wefirst prove that fmapsEintoitself. LetcEE. LetxEMandsupposex <f(c). We must prove thatf(x)<f(c). ByLemma 2.2, M=Mc, and hence wehave x<c,orx=c, orf(c)<x.This lastpossibility cannot occur because x<f(c). Ifx<c then f(x)<C<f(c). Ifx=cthenf(x)=f(c), and hence feE)cE. Next letTbeatotally ordered subset ofE.Let bbetheleast upper bound ofTinM.We must prove thatbEE. Let xEMand x<b.Ifforall cETwe havef(c)<x,then c<f(c)<ximplies that xisanupper bound forT,whence b<x,which isimpossible. Since Mc=Mforall cEE, we must therefore have x<cfor some CET.Ifx<c,thenf(x)<c<b,andifx=c,then c=x<b. Since cisanextreme point andMc=M, wegetf(x)<b.This proves that bEE, that Eisadmissible, and thus proves Lemma 2.3. We now seetrivially that Mistotally ordered. For letx,yEM.Then xisan extreme point ofMbyLemma 2,and YEMxsoy<xor x<f(x)<y, thereby proving that Mistotally ordered. Asremarked previously, this con- cludes theproof ofTheorem 2.1. 884 SOME SET THEORY APPENDIX 2 We shall obtain Zorn's lemma essentially asacorollary ofTheorem 2.1. Wefirst obtain Zorn's lemma inaslightly weaker form. Coronary 2.4. Let Abeanon-empty strictly inductively ordered set. Then A has amaximal element. Proof. Suppose that Adoes not have amaximal element. Then foreach xEAthere exists anelementYxEAsuch that x<Yx'Letf: A Abethemap such thatf(x)=YxforallxEA. Then A,fsatisfy thehypotheses ofTheorem 2.1 andapplying Theorem 2.1yieldsacontradiction. The only difference between Corollary 2.4and Zorn's lemma isthat in Corollary 2.4, we assume that anon-empty totally ordered subset has aleast upper bound, rather than anupper bound. Itis,however, asimple matter to reduce Zorn's lemma totheseemingly weaker form ofCorollary 2.4. We do this inthesecond corollary. Corollary 2.5. (Zorn's lemma). Let Sbeanon-empty inductively ordered set. Then Shas amaximal element. Proof. LetAbethe setofnon-empty totally ordered subsets ofS.Then A isnotempty since any subset ofSwith one element belongs toA.IfX,YEA, wedefine X<Ytomean XcY.Then Aispartially ordered, and isinfact strictly inductively ordered. For letT={X;}ieI beatotally ordered subset ofA. Let z =UXi. ieI Then Zistotally ordered. To seethis, letx,YEZ.Then xEXiand yEXjfor some i,jEI.Since Tistotally ordered, sayXicXj.Then x,yEXjand since Xjistotally ordered, x<yory<x.Thus Zistotally ordered, and isobviously aleast upper bound forTinA.ByCorollary 2.4, weconclude that Ahas a maximal element Xo.This means that X0isamaximal totally ordered subset of S(non-empty). Let mbeanupper bound forX0inS.Then misthedesired maximal element ofS.ForifXES and m<xthen X0u{x} istotally ordered, whence equal toXobythemaximality ofXo. Thus xEXoand x<m.Hence x=m,aswas tobeshown. 3. CARDINAL NUMBERS LetA,Bbesets. We shall saythat thecardinality ofAisthe same asthe cardinality ofB,and write card(A)=card(B) ifthere exists abijection ofAonto B. APPENDIX 2 SOME SET THEORY 885 We saycard(A)<card(B) ifthere exists aninjective mapping (injection) f:A B.We also write card(B)>card(A) inthis case. Itisclear that if card(A)<card(B) andcard(B)<card(C), then card(A)<card(C). This amounts tosaying that acomposite ofinjective mappingsisinjective. Similarly, ifcard(A)=card(B) andcard(B)=card(C) then card(A)=card(C). This amounts tosaying that acomposite ofbijective mappings isbijectIve. Weclearly have card(A)=card(A). Using Zorn's lemma, itiseasy toshow (see Exercise 14)that card(A<card(B) orcard(B)<card(A). Letf:A-+Bbeasurjective mapofasetAonto asetB.Then card(B)<card(A). This iseasily seen, because foreach YEBthere exists anelement xEA, denoted byXy,such thatf(x y)=y.Then theassociation y Xyisaninjective mapping ofBinto A,whence bydefinition, card(B)<card(A). Given twononemptysetsA,Bwehave card(A)<card(B) orcard(B)<card(A). This isasimple application ofZorn's lemma. Weconsider thefamily ofpairs (S,f)where Sisasubset ofAandf: S--+Bisaninjective mapping. From the existence ofamaximal element, theassertion follows atonce. Theorem 3.1. (Schroeder-Bernstein). LetA,Bbesets, and suppose that card(A)<card(B), andcard(B)<card(A). Then card(A)=card(B). Proof. Let f:A-+Band g:B-+A beinjections. WeseparateAinto twodisjoint setsAland A2.WeletAlconsist ofallxEAsuch that, when weliftback xbyasuccession ofinverse maps, x,g-I(X), r-1og-1(x), g-1of-l og-I(X),... then atsome stage wereach anelement ofAwhich cannot belifted back toBby g.WeletA2bethecomplement ofAI'inother words, the setofxEAwhich can belifted back indefinitely, orsuch that wegetstopped inB(i.e. reach anelement ofBwhich has noinverse image inAbyf).Then A =AlUA2.Weshall define abijection hofAonto B. IfxEAI'wedefine h(x)=f(x). IfxEA2,we define h(x)=g-l(X)=unique element YEB such that g(y)=x. Then trivially, hisinjective. We must prove that hissurjective. LetbEB. If,when wetrytoliftback bbyasuccession ofmaps '..Lf-l og-Irf-lo g-Iof-l(b) 886 SOME SET THEORY APPENDIX 2 we canliftback indefinitely,orifwegetstopped inB,then g(b)belongstoA2 andconsequently b=h(g(b)),sobliesintheimage ofh.Ontheother hand, ifwe cannot liftback bindefinitely, and getstoppedinA,thenf-1(b)isdefined (Le., bisintheimage off), andf-l(b) liesinAl.Inthis case, b=H(f-l(b)) isalso intheimage ofh,aswas tobeshown. Next weconsider theorems concerning sums andproducts ofcardinalities. Weshall reduce thestudy ofcardinalities ofproducts ofarbitrary sets tothe denumerable case, using Zorn's lemma. Note first that aninfinite setAalways contains adenumerable set. Indeed, since Aisinfinite, we can first select an element alEA,and thecomplement of{al} isinfinite. Inductively, ifwehave selected distinct elements al,...,aninA,thecomplement of{al'...,an} is infinite, and we can select an+1inthiscomplement. Inthis way, weobtain a sequence ofdistinct elements ofA,giving rise toadenumerable subset ofA. LetAbeaset.Byacovering ofAone means asetrofsubsets ofAsuch that theunion Uc Cer ofalltheelements ofrisequal toA.Weshall saythatrisadisjoint covering if whenever C,C'Er,and C#-C',then theintersection ofCand C'isempty. Lemma 3.2. Let Abeaninfinite set. Then there exists adisjoint covering of Abydenumerable sets. Proof. Let Sbethe setwhose elements arepairs (B,r)consisting ofa subset BofA,and adisjoint covering ofBbydenumerable sets. Then Sisnot empty. Indeed, since Aisinfinite, Acontains adenumerable setD,and thepair (D,{D}) isinS.If(B,r)and(B',r') areelements ofS,wedefine (B,r)<(B',r') tomean that BcB',andrcr'.LetTbeatotally ordered non-empty subset ofS.We may write T={(B i,ri)}ieI for some indexing setI.Let B=UBiand ieIr=Uri. ieI IfC,C'Er,C#-C',then there exists some indices i,jsuch that CEriand C'Erj.Since Tistotally ordered, \vehave, say, (B;,ri)<(Bj,rj). Hence infact, C,C'areboth elements ofrj,and hence C,C'have anempty intersection. Ontheother hand, ifxEB,then xEBifor some i,and hence there issome CErisuch that xEC.Hence risadisjoint covering ofB.Since the APPENDIX 2 SOME SET THEORY 887 elements ofeachriaredenumerable subsets ofA,itfollows thatrisadisjoint covering ofBbydenumerable sets, so(B,r)isinS,and isobviously anupper bound forT.Therefore Sisinductively ordered. Let(M,)be amaximal element ofS,byZorn's lemma. Suppose that M¥A.Ifthecomplement ofMinAisinfinite, then there exists adenumerable setDcontained inthiscomplement. Then (M uD, u{D}) isabigger pair than (M,),contradicting themaximality of(M,). Hence the complement ofMinAisafinite setF.LetDobeanelement of. Let Dl=DouF. Then D1isdenumerable. Let1bethe setconsisting ofallelements of,except Do,together with D1.Then1isadisjoint covering ofAbydenumerable sets, aswas tobeshown. Theorem 3.3. LetAbeaninfinite set,and letDbeadenumerable set. Then card(AxD)=card(A). Proof. Bythelemma, we canwrite A=UDi iel asadisjoint union ofdenumerable sets. Then AxD=U(DiXD). ie1 For each iEI,there isabijection ofDixDon DibyProposition 1.5. Since the sets DixDaredisjoint,wegetinthis wayabijection ofAxDonA,asdesired. Corollary 3.4. IfFisafinite non-empty set, then card(AxF)=card(A). Proof. We have card(A)<card(AxF)<card(AxD)=card(A). We can then useTheorem 3.1togetwhat wewant. Coronary 3.5. LetA,Bbenon-empty sets, Ainfinite, and suppose card(B)<card(A). 888 SOME SET THEORY APPENDIX 2 Then card(A uB)=card(A). Proof. We canwrite AuB=AuCfor some subset CofB,such that C and Aaredisjoint. (We letCbethesetofallelements ofBwhich arenotelements ofA.) Then card(C)<card(A). We can then construct aninjection ofAuC into theproduct Ax{I,2} ofAwith asetconsisting of2elements. Namely,wehave abijection ofAwith Ax{I}intheobvious way, and also aninjection ofCinto Ax{2}. Thus card(A uC)<card(Ax{1,2}). Weconclude theproof byCorollary 3.4and Theorem 3.1. Theorem 3.6. Let Abeaninfinite set. Then card(AxA)=card(A). Proof. Let Sbethe setconsisting ofpairs (B,f)where Bisaninfinite subset ofA,andf:BxBisabijection ofBonto BxB.Then Sisnotempty because if Disadenumerable subset ofA,we canalways find abijection ofDonDxD. If(B,f)and(B',f')areinS,wedefine (B,f)<(B',f')tomean BcB',and the restriction off'toBisequal tof.Then Sispartially ordered, and wecontend that Sisinductively ordered. LetTbeanon-empty totally ordered subset ofS, and sayTconsists ofthepairs (Bi,h)foriinsome indexing setI.Let M =UBi. ieI We shall define abijection g:M MxM.IfxEM,then xliesinsome Bi. We define g(x)=fi(x). This value h(X) isindependent ofthechoice ofBiin which xlies. Indeed, ifxEBjfor somejEI,then say (Bi,h)<(Bj,fj). Byassumption, BiCBj,andfj(x)=hex),so9iswell defined. Toshow 9is surjective, letx,yEMand (x,y)EM xM. Then xEBifor some iEIand yEBjforsomejEI.Again since Tistotally ordered, say(Bi,Ii)<(Bj,fj).Thus BiCBj,and x,YEBj.There exists anelement bEBjsuch that jj(b)=(x,y)EBjxBj. Bydefinition, g(b)=(x,y);so9issurjective. We leave theproof that 9is injective tothereader toconclude theproof that 9isabijection. Wethen see APPENDIX 2 SOME SET THEORY 889 that (M,g)isanupper bound forTinS,and therefore that Sisinductively ordered. Let(M,g)beamaximal element ofS,and letCbethecomplement ofMinA. Ifcard(C)<card(M), then card(M)<card(A)=card(M uC)=card(M) byCorollary 3.5,and hence card(M)=card(A) byBernstein's Theorem. Since card(M)=card(MxM), we aredone with theproof inthis case. If card(M)<card( C), then there exists asubset M1ofChaving the same cardinalityasM.Weconsider (M uMl)x(M uM1) =(MxM)u(M 1xM)u(MxM1)u(M 1xM1). Bytheassumption onMandCorollary 3.5,thelastthree sets inparentheses on theright ofthisequation have the same cardinalityasM.Thus (M uMl)x(M uMl)=(MxM)uM2 where M2isdisjoint from M xM,and hasthe same cardinalityasM. We now define abijection gl:MuM1(M uM1)x(MuM1). Weletgl(X)=g(x) ifxEM,and weletglonM1beanybijection ofM1onM2. Inthis waywehave extended 9toMuM l'and thepair(M uM1,gl)isinS, contradicting themaximality of(M,g).The case card(M)<card(C) therefore cannot occur, and our theorem isproved (using Exercise 14below). Corollary 3.7. IfAisaninfinite set,and A(n) =Ax... xAistheproduct taken ntimes, then card(A(n)=card(A). Proof Induction. Corollary 3.8. IfAI'.. .,Anarenon-emptysets with Aninfinite, and card(A i)<card(An) fori=1,..., n,then card(A 1X... xAn)=card(An). 890 SOME SET THEORY APPENDIX 2 Proof Wehave card(An)<card(A 1x... xAn)<card(AnX...XAn) and we useCorollary 3.7and theSchroeder-Bernstein theorem toconclude the proof. Corollary 3.9. Let Abeaninfinite set,and let <I>bethe setoffinite subsets ofA.Then card(<I»=card(A). Proof. Let <l>nbethe setofsubsets ofAhaving exactlynelements, foreach integern=1,2,... .Wefirst show that card(<I>n)<card(A). IfFisanelement of<l>n, weorder theelements ofFinany way, say F={Xl,...,Xn}. and weassociate with Ftheelement (xl'...,Xn)EA(n), Ft----+(xl'...,Xn). IfGisanother subset ofAhavingnelements, say G={Yl'...,Yn}, and G=IF, then (Xl'.·.,Xn)=1=(yl'...,Yn). Hence our map F (xl'...,Xn) of<l>ninto A(n) isinjective. ByCorollary 3.7, weconclude that card(<I>n)<card(A). Now <I>isthedisjoint union ofthe <l>nfor n=1,2,...anditisanexercise to show that card(<I»<card(A) (cf.Exercise 1).Since card(A)<card (<1», because inparticular, card(<I>l)=card(A), we seethat ourcorollary isproved. Inthenext theorem, weshall seethat given aset,there always exists another setwhose cardinality isbigger. Theorem 3.10. Let Abe aninfinite set, and Tthe setconsisting oftwo elements {O,I}.LetMbethe setofallmaps ofAinto T.Then card(A)<card(M) and card(A) =1=card(M). APPENDIX 2 SOME SET THEORY 891 Proof. For each xEAwelet fx:A {O,I} bethemap such thatfx(x)= 1andfx(Y)=0ifY=Ix.Then xfxisobviously aninjection ofAinto M, sothat card(A)<card(M). Suppose that card(A)=card(M). Let xgx beabijection between Aand M. Wedefine amap h:A {O,I}bytherule h(x)=0ifgx(x)=1, h(x)= 1ifgx(x)=o. Then certainly h=Igxforany x,and thiscontradicts theassumption that x gx isabijection, thereby proving Theorem 3.10. Corollary 3.11. LetAbeaninfinite set,and letSbethe setofallsubsets ofA. Then card(A)<card(S) andcard(A) =Icard(S). Proof Weleave itasanexercise. [Hint: IfBisanon-empty subset ofA, usethecharacteristic function lpBsuch that QJB(X)=1if xEB, QJB(X)=0if xfJB. What can you sayabout theassociation B(fJB?] 4. WELL-ORDERING Anordered setAissaid tobewell-ordered ifitistotally ordered, andifevery non-empty subset Bhas aleast element, that is,anelement aEBsuch that a<xforallxEB. Example 1.The setofpositive integers Z+iswell-ordered. Any finite set can bewell-ordered, and adenumerable setDcan bewell-ordered: Anybijection ofDwith Z+willgive rise toawell-ordering ofD. Example 2. Let Sbeawell-ordered setanq letbbeanelement ofsome set, bftS.LetA=Su{b}.Wedefine x<bforallXES. Then Aistotally ordered, and isinfact well-ordered. 892 SOME SET THEORY APPENDIX 2 Proof. LetBbeanon-empty subset ofA.IfBconsists ofbalone, then bisa least element ofB.Otherwise, Bcontains some element aEA.Then BnAisnot empty, and hence has aleast element, which isobviously also aleast element for B. Theorem 4.1. Every non-empty set can bewell-ordered. Proof LetAbeanon-empty set. Let Sbethe setofallpairs (X,w),where Xisasubset ofAand wisawell-ordering ofX.Note that Sisnotempty because anysingle element ofAgives rise tosuch apair. If(X,w)and(X',w') aresuch pairs,wedefine (X,w)<:(X',w')ifXCX',iftheordering induced onXby w'isequal tow,andifXistheinitial segment ofX'.Itisobvious that this defines anorderingonS,and wecontend that Sisinductively ordered. Let {(Xi' Wi)} be atotally ordered non-empty subset ofS.LetX=UXi.Ifa,bEX, tlien a,blieinsome Xi'and wedefine a<:binXifa<:bwith respecttothe ordering Wi.This isindependent ofthechoice ofi(immediate from theassumption oftotal ordering). Infact, Xiswell ordered, forifYisanon-empty subset of X,then there issome element yEYwhich lies insome Xj.Let cbe aleast element ofXjnY.One verifies atonce that cisaleast element ofY.We can therefore apply Zorn's lemma. Let(X,w)beamaximal element inS.IfX A, then, using Example 2,we can define awell-orderingon abigger subset than X,contradicting themaximality assumption. This proves Theorem 4.1. Note. Theorem 4.1isanimmediate andstraightforward consequence of Zorn's lemma. Usually inmathematics, Zorn's lemma isthemost efficient tool when dealing with infinite processes. EXERCISES 1.Prove thestatement made intheproof ofCorollary 3.9. 2.IfAisaninfinite set,and {J)nisthe setofsubsets ofAhaving exactlynelements, show that card(A)<card({J)n) for n>1. 3.LetAibeinfinite setsfori=1,2,...and assume that card( Ai)<card( A) for some setA,and alli.Show that card(91Ai)<card(A). APPENDIX 2 SOME SET THEORY 893 4.LetKbe asubfield ofthecomplex numbers. Show that foreach integer n>1,the cardinality ofthe setofextensions ofKofdegreeninCis<:card(K). 5.LetKbeaninfinite field, and Eanalgebraic extension ofK.Show that card( E)=card( K). 6.Finish theproof oftheCorollary 3.11. 7.IfA,Baresets, denote byM(A, B)the setofallmaps ofAinto B.IfB,B'are setswith the same cardinality, show that M(A, B)andM(A, B')have the same cardinality. If A,A'have the same cardinality, show that M(A, B)and M(A', B)have the same cardinality. 8.LetAbeaninfinite setand abbreviate card(A) bya.IfBisaninfinite set, abbreviate card(B) by/3.Define a/3tobecard(A xB).LetB'beasetdisjoint from Asuch that card(B)=card(B'). Define a+/3tobecard(A uB'). Denote byBAthe setofallmaps ofAintoB,and denote card(BA) by/3a .LetCbeaninfinite setandabbreviate card( C) by)'. Prove thefollowing statements: (a) rx({3+y)=rx{3+rxy. (b) rx{3={3rx. (c)rxP+y=rxPrxy. 9.Let Kbe aninfinite field. Prove that there exists analgebraically closed field Ka containing Kasasubfield, andalgebraic over K.[Hint: LetQbeasetofcardinality strictly greater than thecardinality ofK,andcontaining K.Consider the set8ofall pairs (E,lp)where Eisasubset ofQsuch that KcE,and lpdenotes alawofaddition andmultiplication onEwhich makes Einto afield such that Kisasubfield, and Eis algebraic over K.Define apartial ordering on8inanobvious way; show that 8is inductively ordered, and that amaximal element isalgebraic over Kandalgebraically closed. You will need Exercise 5inthelaststep.] 10.LetKbeaninfinite field. Show that thefield ofrational functions K(t) hasthe same cardinality asK. 11.LetJnbethe setofintegers {I,...,n}. LetZ+bethe setofpositive integers. Show that thefollowing sets have the same cardinality: (a)The setofallmaps M(Z+, In). (b)The setofallmaps M(Z+ ,J2). (c)The setofallreal numbers xsuch that 0<x<1. (d)The setofallreal numbers. 12. 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Math. 224 (1966) pp.209-220 G.SHIMURA, Introduction tothearithmetic theory ofAutomorphic Functions, Iwanami Shoten and Princeton University Press, 1971 M.SHUB, Global Stability ofDynamical Systems, Springer-Verlag, New York 1987 J.SILVERMAN, Wieferich's criterion and theabcconjecture, J.Number Theory 30(1988) pp.226-237 N.SNYDER, Analternate proof ofMason's theorem, Elemente derMath. 55(2000) pp.93-94 R.SOLOMON, Abrief history oftheclassification ofthefinite simple groups, Bull. AMS Vol. 38No.3 (2001) pp.315-322 [Sou 90] [SteT 86] [Sto 81] [Sw69] [Sw83] [SwD73] [TaW 95] [Uch 77] [Uch79] [Uch 81] [vdW 29] [vdW 30] [Wil 95] [Win 91] [Wit 35] [Wit 36] [Wit 37] [Za95]BIBLIOGRAPHY 901 C.SoULE, Geometrie d'Arakelov ettheorie des nombres transcendants, Preprint, 1990 C.L. STEWART and R.TUDEMAN, On theOesterle-Masser Conjecture, Mon. Math. 102(1986) pp.251-257 W. STOTHERS, Polynomial identites andhauptmoduln, Quart. J.Math. 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Amsterdam 31(1929) pp.749-770 B.L.VAN DER WAERDEN, Modern Algebra, Springer-Verlag, 1930 A.WILES, Modular ellipticcurves and Fermat's lasttheorem, Annals of Math. 141(1995) pp.443-551 K.WINBERG, OnGalois groups ofp-closed algebraic number fields with restricted ramification, I,J.reine angew. Math. 400(1989) pp.185-202 andII,ibid. 416(1991) pp.187-194 E.WITT, Der Existenzsatz fur abelsche Funktionenkorper, J.reine angew Math. 173(1936) pp.43-51 E.WITT, Konstruktion vongaloisschen Korpern derderCharakteristic p mitvorgegebener Gruppe derOrdnung pi, J.reine angew. Math. 174 (1936) pp.237-245 E.WITT, Zyklische Korper undAlgebren derCharakteristik pvom Grad pn. Struktur diskret bewerteter perfekter Korper mit vollkommenen Restklassenkorper der Charakteristik p,J.reine angew. Math. 176 (1937) pp. 126-140 U.ZANNIER, OnDavenport's bound forthedegree off3-g2and Rie- mann's existence theorem, Acta Arithm. LXXI.2 (1995) pp. 107-137 INDEX abcconjecture, 195 abelian, 4 category, 133 extension, 266, 278 group, 4,42,79 Kummer theory, 293 tower, 18 absolute value, 465 absolutely semisimple, 659 abstract nonsense, 759 abut, 815 action ofagroup, 25 acyclic, 795 Adams operations, 726, 782 additive category, 133 additive functor, 625, 790 additive polynomial, 308 adic completion, 163, 206 expansion, 190 topology, 162, 206 adjoint, 533, 581 adjoint functor, 629 affine space, 383 algebra, 121, 629, 749 algebraic closure, 178, 231, 272 element, 223 extension, 224 group, 549 integer, 371 set, 379 space, 383, 386 algebraically closed, 272 independent, 102, 308, 356 almost all, 5 alternating algebra, 733 form, 511, 526, 530, 571, 598 group, 31,32,722 matrix, 530, 587 multilinear map, 511, 731product, 733, 780 annihilator, 417 anti-dual, 532 anti-linear, 562 anti-module, 532 approximation theorem, 467 Aramata's theorem, 701 archimedean ordering, 450 Artin conjectures, 256, 301 theorems, 264, 283, 290, 429 artinian, 439, 443, 661 Artin-Rees theorem, 429 Artin-Schreier theorem, 290 associated graded ring, 428, 430 group andfield, 301 ideal ofalgebraic set, 381 linear map, 507 matrix ofbilinear map, 528 object, 814 prime, 418 associative, 3 asymptotic Fermat, 196 automorphism, 10,54 inner, 26 ofaform, 525, 533 Banach space, 475 balanced, 660 base change, 625 basis, 135, 140 Bateman-Horn conjecture, 323 belong group andfield, 263 ideal andalgebraic set, 381 prime andprimary ideal, 421 Bernoulli numbers, 218 polynomials, 219 bifunctor, 806 bijective, ix bilinear form, 146, 522 903 904 INDEX bilinear map, 48, 121, 144 binomial polynomial, 434 Blichfeldt theorem, 702 blocks, 555 Borel subgroup, 537 boundaries, 767 bounded complex,762 Bourbaki theorems onsets, 881 ontraces andsemisimplicity,650 bracket product,121 Brauer's theorems, 701, 709 Bruhat decomposition,539 Burnside theorems onsimple modules, 648 ontensor representations,726 butterfly lemma, 20cocycle GLn, 549 Hilbert's theorem, 90, 288 Sah's theorem, 303 coefficient function, 681 coefficients oflinear combination, 129 ofmatnx, 503 ofpolynomial, 98, 101 coerasable, 805 cofinal, 52 cohomology, 288, 302, 303, 549, 764 ofgroups,826 cokernel, 119, 133 column operation,154 rank, 506 vector, 503 commutative, 4 diagram, ix group,4 ring, 83,84, 86 commutator, 20,69, 75 commutator subgroup, 20, 75 ofSLn,539, 541 commute, 29 compact Krull topology, 329 spec ofaring, 411 complete family, 837 field, 469 ring and local ring, 206 completely reducible, 554 completion, 52,469, 486 complex, 445, 761, 765 complex numbers, 272 component, 503, 507 ofamatrix, 503 composition ofmappings,85 compositum offields, 226 conjugacy class, 673 conjugate elements ofagroup,26 ofafield, 243 conjugate embeddings, 243, 476 fields, 243, 477 subgroups, 26,28, 35 conjugation, 26,552, 570, 662 connected, 411 connected sum, 6 connection, 755e-dimension, 772 cancellation law, 40 canonical map, 14, 16 cardinal number, 885 Cartan subgroup, 712 Casimir, 628, 639 category, 53 Cauchy family, 52 sequence, 51,162,206,469 Cayley-Hamilton theorem, 561 center ofagroup, 26, 29 ofaring, 84 central element, 714 centralizer, 14 chain condition, 407 character, 282, 327, 667, 668 independence, 283, 676 characteristic, 90 characteristic polynomial, 256, 434, 561 oftensor product, 569 Chevalley's theorem, 214 Chinese remainder theorem, 94 class formula, 29 class function, 673 class number, 674 Clifford algebra, 749, 757 closed complex, 765 subgroup, 329 under lawofcomposition,6 coboundary, 302 INDEX 905 <:onstant polynomial, 175 constant term, 100 content, 181 conttagredient, 665 conttavariant functor, 62 convergence, 206 convolution, 85, 116 coordinates, 408 coproduct, 59, 80 ofcommutative rings, 630 ofgroups, 70, 72 ofmodules, 128 cocrespondence, 76 coset, 12 representative, 12 countable, 878 covariant functor, 62 Cramer's rule, 513 cubic extension, 270 cuspidal, 318 cycle inhomology, 767 inpermutations, 30 cyclic endomorphism, 96 extension, 266, 288 group, 8,23,96, 830 module, 147, 149 tower, 18 cyclotomic field, 277-282, 314, 323 polynomials, 279dependent absolute values, 465 deRham complex, 748 derivation, 214, 368, 746, 754 over asubfield, 369 universal, 746 derivative, 178 derived functor, 791 descending chain condition, 408, 439, 443, 661 determinant, 513 ideal, 738, 739 ofcohomology, 738 oflinear map, 513, 520 ofmodule, 735 ofWitt group, 595 diagonal element, 504 diagonalizable, 568 diagonalized form, 576 difference equations, 256 differential, 747, 762, 814 dihedral group, 78, 723 dimension ofcharacter, 670 ofmodule, 146, 507 oftranscendental extension, 355 ofvector space, 141 dimension inhomology, 806, 811, 823 shifting, 805 direct limit, 160, 170, 639 product, 9,127 sum, 36,130, 165 directed family, 51, 160 discrete valuation ring, 487 discriminant, 193, 204, 325 distinguished extensions offields, 227, 242 ofrings, 335, 291 distinguished polynomials, 209 distributivity, 83 divide, 111, 116 divisible, 50 division ring, 84,642 Dolbeault complex, 764 dominate (polynomials), 870 double coset, 75, 693 doubly transitive, 80 dual basis, 142, 287 group, 46, 145 module, 142, 145, 523, 737 representation, 665Davenport theorem, 195 decomposable, 439 decomposition field, 341 group, 341 Dedekind determinant, 548 ring, 88,116, 168, 353 defined, 710, 769 definite form, 593 degree ofextension, 224 ofmorphism, 765 ofpolynomial, 100, 190 ofvariety, 438 Weierstrass, 208 Deligne-Secre theorem, 319 density theorem, 647 ..denumerable set, 875 906 INDEX effective character, 668, 685 eigenvalue, 562 eigenvector, 562, 582-585 Eisenstein criterion, 183 elementary divisors, 153, 168,521,547 group, 705 matrix, 540 symmetric polynomials, 190, 217 elimination, 391 ideal, 392 embedding, 11, 120 offields, 229 ofrings, 91 endomorphism, 10,24,54 ofcyclic groups, 96 enough injectives, 787 T-exacts, 810 entire, 91 functions, 87 epimorphism, 120 equivalent norms, 470 places, 349 valuations, 480 erasable, 800 euclidean algorithm, 173, 207 Euler characteristic, 769 Euler-Grothendieck group,771 Euler phifunction, 94 Euler- Poincare characteristic, 769, 824 map, 156, 433, 435, 770 evaluation, 98, 101 even permutation,31 exact, 15, 120 for afunctor, 619 sequence ofcomplexes, 767 expansion ofdeterminant, 515 exponent ofanelement, 23, 149 ofafield extension, 293 ofagroup, 23 ofamodule, 149 exponential, 497 Ext, 791,808,810,831,857 extension ofbase, 623 ofderivations, 375 offields, 223 ofhomomorphisms, 347, 378 ofmodules, 831exterior algebra, 733 product,733 extreme point,883 factor group,14 module, 119, 141 ring, 89 factorial, Ill, 115, 175, 209 faithful, 28,334, 649, 664 faithfully flat, 638 Fermat theorem, 195, 319 fiber product, 61, 81 field, 93 ofdefinition ofarepresentation,710 filtered complex,817 filtration, 156, 172, 426, 814, 817 finite complex,762 dimension, 141, 772, 823 extension, 223 field, 244 free resolution, 840 homological dimension, 772, 823 module, 129 resolution, 763 sequence,877 set, 877 type,129 under aplace,349 finitely generated algebra,121 extension, 226 group, 66 module, 129 ring, 90 finitely presented,171 Fitting ideal, 738-745 Fitting lemma, 440 five lemma, 169 fixed field, 261 point, 28,34, 80 flat, 612, 808 for amodule, 616 forgetful functor, 62 form multilinear, 450, 466 polynomial,384 formal power series, 205 Fourier coefficients, 679 INDEX 907 fractional ideal, 88 fractions, 107 free abelian group, 38, 39 extension, 362 generators, 137 group, 66, 82 module, 135 module generated byaset, 137 resolution, 763 Frey polynomial, 198 Frobenius element, 180, 246, 316, 346 reciprocity, 686, 689 functionals, 142 functor, 62 fundamental group, 63graded algebra, 172, 631 module, 427,751,765 morphism, 765, 766 object, 814 ring, 631 Gram-Schmidt orthogonalization, 579, 599 Grassman algebra, 733 greatestcommon divisor, 111 Grothendieck algebra andring, 778-782 group, 40, 139 power series, 218 spectral sequence, 819 group,7 algebra, 104, 121 automorphism,10 extensions, 827 homomorphism, 10 object, 65 ring, 85,104, 126Gor(G,k)-module, 664, 779 G-homomorphism, 779 G-object, 55 G-regular, 829 G-set, 27, 55 Galois cohomology, 288, 302 extension, 261 group, 252, 262, 269 theory, 262 Gauss lemma, 181, 209, 495 Gauss sum, 277 g.c.d., III Gelfand-Mazur theorem, 471 Gelfand-Naimark theorem, 406 Gelfond-Schneider, 868 generate and generators for agroup, 9,23, 68 for anideal, 87 for amodule, 660 for aring, 90 generating function orpower series, 211 generators andrelations, 68 generic forms, 390, 3924 hyperplane,374 pfaffian, 589 point, 383, 408 polynomial, 272, 345 ghost components, 330 GL2,300, 317,537,715 GLn,19,521,543,546,547 global sections, 792 Goursat's lemma, 75Hall conjecture,197 harmonic polynomials, 354, 550 Hasse zeta function, 255 height, 167 Herbrand quotient, 79 Hermite-Lindemann, 867 hermitian form, 533, 571, 579 linear map, 534 matrix, 535 Hilbert Nullstellensatz, 380, 551 polynomial,433 -Serre theorem, 431 syzygy theorem, 862 theorem onpolynomial rings, 185 theorem 90, 288 -Zariski theorem, 409 homogeneous, 410, 427, 631 algebraic space, 385 ideal, 385, 436, 733 integral closure, 409 point, 385 polynomial, 103, 107, 190, 384, 436 quadratic map, 575 homology, 445, 767 isomorphism, 767, 836 homomorphisms incategories, 765 homomorphism ofcomplex, 445, 765 908 INDEX homomorphism (continued) ofgroups,10 ofinverse systems, 163 ofmodules, 119, 122 ofmonoid, 10 ofrepresentations, 125 ofrings, 88 homotopies ofcomplexes, 787 Horrock's theorem, 847 Howe's proof, 258 hyperbolic enlargement, 593 pair, 586, 590 plane, 586, 590 space, 590 hyperplane, 542 section, 374, 410map, ix module, 782, 830 resolution, 788, 801, 819 inner automorphism, 26 inseparable degree, 249 extension, 247 integers mod n,94 integral, 334, 351, 352, 409 closure, 336, 409 domain, 91 equation, 334 extension, 340 homomorphism, 337 map, 357 root test, 185 valued polynomials, 216, 435 integrally closed, 337 integrality criterion, 352, 409 invariant bases, 550 submodule, 665 invariant oflinear map, 557, 560 ofmatrix, 557 ofmodule, 153, 557, 563 ofsubmodule, 153, 154 inverse, ix,7 inverse limit, 50,51,161,163,169 ofGalois groups, 313, 328 inverse matrix, 518 invertible, 84 Irr(z,k,x), 224 irreducible algebraic set,382, 408 character, 669, 696 element, III module, 554 polynomial, 175, 183 polynomial ofafield element, 224 irrelevant prime, 436 isolated prime, 422 isometry, 572 isomorphism, 10,54 ofrepresentations, 56, 667 isotropy group, 27 Iss'sa-Hironaka theorem, 498Ideal, 86 class group, 88, 126 idempotent, 443 image,11 indecomposable, 440 independent absolute values, 465 characters, 283, 676 elements ofmodule, 151 extensions, 362 variables, 102, 103 index, 12 induced character, 686 homomorphism, 16 module, 688 ordering, 879 representation, 688 inductively ordered, 880 inertia form, 393 group, 344 infinite cyclic group, 8,23 cyclic module, 147 extension, 223, 235 Galois extensions, 313 period, 8,23 set, 876 under aplace, 349 infinitely large, 450 small, 450 injectiveJacobson density, 647 radical, 658 Jordan-Holder, 22, 156 Jordan canonical form, 559 INDEX 909 K-family, 771 K-theory, 139, 771-782 kernel ofbilinear map, 48, 144, 522, 572 ofhomomorphism, 11, 133 Kolchin's theorem, 661 Koszul complex, 853 Krull theorem, 429 topology, 329 Krull-Remak-Schmidt, 441 Kummer extensions abelian, 294-296, 332 non-abelian, 297, 304, 326norm, 478 parameter,487 ring, 110, 425, 441 uniformization, 498 localization, 110 locally nilpotent, 418 logarithm, 497, 597 logarithmic derivative, 214, 375 Mackey's theorems, 694 MacLane's criterion, 364 mapping cylinder, 838 Maschke's theorem, 666 Mason-Stothers theorem, 194, 220 matrix, 503 ofbilinear map, 528 over non-commutative ring, 641 maximal abelian extension, 269 archimedean, 450 element, 879 ideal, 92 metric linear map, 573 minimal polynomial, 556, 572 Mittag-Leffler condition, 164 modular forms, 318, 319 module, 117 over principal ring, 146, 521 modulo anideal, 90 Moebius inversion, 116, 254 monic, 175 monoid, 3 algebra, 106, 126 homomorphism,10 monomial, 101 monomorphism, 120 Morita's theorem, 660 morphism, 53 ofcomplex, 765 offunctor, 65,625, 800 orrepresentation, 125 multilinear map, 511, 521, 602 multiple root, 178, 247 multiplicative function, 116 subgroup ofafield, 177 subset, 107 multiplicity ofcharacter, 670 ofroot, 178 ofsimple module, 644 Nakayama's lemma, 424, 661 natural transformation, 65L-functions, 727 lambda operation, 217 lambda-ring, 218, 780 Langlands conjectures, 316, 319 lattice, 662 lawofcomposition,3 Lazard's theorem, 639 leading coefficient, 100 least common multiple,113 element, 879 upper bound, 879 left coset, 12 derived functor, 791 exact, 790 ideal, 86 module, 117 length ofcomplex, 765 offiltration, 433 ofmodule, 433, 644 Liealgebra, 548 lieabove prime, 338 valuation ring, 350 lifting, 227 linear combination, 129 dependence, 130 independence, 129, 150, 283 map, 119 polynomial, 100 linearly disjoint, 360 local degree, 477 homomorphism, 444 910 INDEX negative, 449 definite, 578 Newton approximation,493 nilpotent, 416, 559, 569 Noether normalization, 357 Noetherian, 186,210,408-409,415,427 graded ring, 427 module, 413 non-commutative variables, 633 non-degenerate, 522, 572 non-singular, 523, 529 norm, 284, 578, 637 on avector space, 469 on afinitely generated abelian group, 166 normal basis theorem, 312 endomorphism, 597 extension, 238 subgroup,14 tower, 18 normalizer, 14 Northcott theorems, 864 null sequence, 52 space, 586 nullstellensatz, 380, 383 occur, 102, 176 oddpermutation,31 one-dimensional character, 671 representation, 671 open complex, 761 open set, 406 operate on amodule, 664 on anobject, 55 on aset,25,76 orbit, 28 decomposition formula, 29 order ofagroup,12 atp,113, 488 atavaluation, 488 ofazero, 488 ordering, 449,480, 878 ordinary tensor product, 630 orthogonal basis, 572-585 element, 48, 144, 572 group, 535 map, 535 sum, 572orthogonality relations, 677 orthogonalization, 579 orthonormal, 577 over amap, 229 p-adic integers, 51,162, 169, 488 numbers, 488 p-class, 706 p-conjugate,706 p-divisible,50 p-elementary,705 p-group, 33 p-regular,705 p-singular,705 p-subgroup,33 pairing, 48 parallelogram law, 598 partial fractions, 187 partition,79 function, 211 perfect, 252 period, 23, 148 periodicity ofClifford algebra, 758 permutation, 8,30 perpendicular, 48, 144, 522 Pfaffian, 589 Pic orPicard group, 88, 126 place, 349, 482 Poincare series, 211, 431 point ofalgebraic set, 383 inafield, 408 polar decomposition, 58 polarization identity, 580 pole, 488 polynomial,97 algebra, 97,633 function, 98 invariants, 557 irreducible, 175, 183 Noetherian, 185 Pontrjagin dual, 145 positive,449 definite, 578, 583 power map,10 power series, 205 factorial, 209 Noetherian, 210 pnmary decomposition, 422 ideal, 421 module, 421 INDEX 911 radical ofanideal, 388, 417ofaring, 661 ofaninteger, 195 Ramanujan power series, 212 ramification index, 483 rank, 42,46 ofamatrix, 506 rational conjugacy class, 276, 326, 725 element, 714 function, 110 real, 451 closed, 451 closure, 452 place, 462 zero, 457 reduced decomposition, 422, 443 polynomial,177 reduction criterion, 185 map, 99, 102 modulo anideal, 446, 623 mod p,623 refinement ofatower, 18 regular character, 675, 699 extension, 366 module, 699, 829 representation, 675, 829 sequence, 850 relations, 68 relative invariant, 171, 327 relatively prime,113 representation, 55,124, 126 functor, 64 ofagroup, 55,317,664 ofaring, 553 space, 667 residue class, 91 degree, 422, 483 ring, 91 resolution, 763, 798 resultant, 200, 398,410 system, 403 variety, 393 Ribet, 319 Rieffel's theorem, 655 Riemann surface, 275 Riemann-Roch, 212, 218, 220, 258 right coset, 12,75 derived functor, 791 exact functor, 791, 798prime element, 113 field, 90 ideal, 92 ring, 90 primitive element, 243, 244 group, 80 operation, 79 polynomials, 181, 182 power series, 209 root, 301 root ofunity, 277, 278 principal homomorphism, 418 ideal, 86, 88 module, 554, 556 representation, 554 ring, 86,146, 521 product incategory, 58 ofgroups, 9 ofmodules, 127 ofrings, 91 profinite,51 projection, 388 projective module, 137, 168, 848, 850 resolution, 763 space, 386 proper, ix congruence, 492 pull-back, 61 purely inseparable element, 249 extension, 250 push-out, 62, 81 quadratic extension, 269 form, 575 map, 574 symbol, 281 quadratically closed, 462 quatemions, 9,545, 723, 758 Quillen-Suslin theorem, 848 quotient field, 110 ring, 107 912 INDEX right (continued) ideal, 66 module, 117 rigid, 275 rigidity theorem, 276 ring, 83 homomorphism, 88 offractions, 107 root, 175 ofunity, 177, 276 roW operation, 154 'rank, S06 vector, 503simple character, 669 group, 20 module, 156, 554, 643 ring, 653, 655 root, 247 simplicity of5Ln, 539, 542 size ofamatrix, 503 skew symmetric, 526 5L2' 69,537, 539, 546 generators andrelations, 69,70, 537 5Ln, 521, 539, 541, 547 snake lemma, 158, 169, 614-621 Snyder's proof, 220 solvable extension, 291, 314 group, 18,293, 314 byradicals, 292 spec ofaring, 405, 410 special linear group, 14,52,59,69,541, 546, 547 specializing,101 specialization,384 spectral sequence, 815-825 theorem, 581, 583, 585 split exact sequence, 132 splitting field, 235 square matrix, 504 group, 9,77, 270 root ofoperator, 584 stably free, 840 dimension, 840 stably isomorphic, 841 stalk, 161 standard complex, 764 alternating matrix, 587 Steinberg theorem, 726 Stewart- Tijdeman, 196 strictly inductively ordered, 881 stripping functor, 62 Sturm's theorem, 454 subgroup, 9 submodule, 118 submonoid, 6 subobject, 134 subring, 84 subsequence, 876 subspace, 141 substituting, 98, 10153and54'722 scalar product, 571 Schanuel conjecture, 873 lemma, 841 Schreier's theorem, 22 Schroeder-Bernstein theorem, 885 Schur Galois groups, 274 lemma, 643 Schwarz inequality, 578, 580 section, 64, 792 self-adjoint, 581 semidirect product, 15,76 semilinear, 532 seminorm, 166, 475 semipositive, 583, 597 semisimple endomorphism, 569, 661 module, 554, 647, 659 representation, 554, 712 ring, 651 separable closure, 243 degree, 239 element, 240 extension, 241, 658 polynomial, 241 separably generated, 363 separating transcendence basis, 363 sequence, 875 Serre's conjecture, 848 theorem, 844 sesquilinear form, 532 Shafarevich conjecture, 314 sheaf, 792 sign ofapermutation, 31, 77 INDEX 913 super algebra, 632 commutator, 757 product, 631, 751 tensor product, 632, 751 supersolvable, 702 support, 419 surjective, ix Sylow group, 33 Sylvester's theorem, 577 symmetric algebra, 635 endomorphism, 525, 585, 597 form, 525, 571 group, 29,269, 272-274 matrix, 530 multilinear map, 635 polynomial, 190, 217 product, 635, 781, 861 symplectic, 535 basis, 599 syzygy theorem, 862 Szpiro conjecture, 198transcendence basis, 356 degree, 355 ofe,867 transcendental, 99 transitive, 28, 79 translation, 26, 227 transpose ofbifunctor, 808 oflinear map, 524 ofmatrix, 505 transposition,13 transvection, 542 trigonometric degree, 115 polynomial, 114, 115 trivial character, 282 operation, 664 representation, 664 subgroup, 9 valuation, 465 two-sided ideal, 86, 655 type ofabelian group, 43 ofmodule, 149 Taniyama-Shimura conjecture, 316, 319 Tate group, 50, 163, 169 limit, 598 Taylor series, 213 tensor, 581, 628 algebra, 633 exact, 612 product, 602, 725 product ofcomplexes, 832, 851 product representation, 725, 799 Tits construction offree group,81 tor(fortorsion), 42,47, 149 Tor, 622, 791 dimension, 622 Tomheim proof, 471 torsion free, 45, 147 module, 147, 149 total complex, 815 degree, 103 totally ordered, 879 tower offields, 225 ofgroups, 18 trace ofelement, 284, 666 oflinear map, 511, 570 ofmatrix, 505, 511unimodular, 846 extension property, 849 unipotent, 714 unique factorization, Ill, 116 uniquely divisible, 575 unit, 84 element, 3,83 ideal, 87 unitary, 535, 583 universal, 37 ddta-functor, 800 derivation, 746 universally attracting, 57 repelling, 57 upper bound, 879 upper diagonal group, 19 valuation, 465 valuation ring, 348, 481 determined byordering, 450, 452 value group, 480 Vandermonde determinant, 257-259, 516 vanishing ideal, 38 variable, 99, 104 variation ofsigns, 454 914 INDEX variety, 382 vector space, 118, 139 volume, 735Witt group, 594, 599 theorem, 591 vector, 330, 492 Witt-Grothendieck group, 595 Warning's theorem, 214 Wedderburn's theorem, 649 Weiersttass degree, 208 polynomial,208 preparation theorem, 208 weight, 191 well-behaved, 410, 478 well-defined, x well-ordering, 891 Weyl group, 570Zariski-Matsusaka theorem, 372 Zariski topology, 407 Zassenhaus lemma, 20 zero divisor, 91 element, 3 ofideal, 390, 405 ofpolynomial, 102, 175, 379, 390 zeta function, 211, 212, 255 Zorn's lemma, 880, 884 Graduate Texts inMathematics TAKEUTIlZARING. Introduction to 35 ALExANDERlWERMER. Several Complex Axiomatic SetTheory 2nd ed. Variables and Banach Algebras. 3rded. 2 OXTOBY. Measure andCategory. 2nd ed. 36 KELLEY!NAMIOKA etal.Linear 3 SCHAEFER. Topological Vector Spaces. Topological Spaces. 2nd ed. 37 MONIC Mathematical Logic. 4 HILTON/ST AMMBACH ACourse in 38 GRAUERTlFRrrzsCHE. Several Complex Homological Algebra. 2nd ed. Variables. 5 MAc LANE. Categories fortheWorking 39 ARVESON. AnInvitation toC*-Algebras. Mathematician. 2nd ed. 40 KEMENY/SNEWKNAPP. Denumerable 6 HUGHESIPIPER. Projective Planes. Markov Chains. 2nd ed. 7 SERRE. ACourse inArithmetic. 41 APOSTOL. Modular Functions andDirichlet 8 TAKEUTI/ZARING. Axiomatic SetTheory. Series inNumber Theory 9 HUMPHREYs. Introduction toLieAlgebras 2nd ed. andRepresentation Theory.42 SERRE Linear Representations ofFinite 10 COHEN ACourse inSimple Homotopy Groups. Theory.43 GILLMAN/JERISON. Rings ofContinuous II CONWAY. Functions ofOneComplex Functions. Variable I.2nd ed. 44 KENDIG. Elementary Algebraic Geometry. 12 BEALS. Advanced Mathematical Analysis. 45 LoE-VE. Probability TheoryI.4thed. 13 ANDERSON/fuLLER. Rings andCategories 46 LoE-VE. Probability Theory II.4thed. ofModules 2nd ed. 47 MOISE. Geometric Topology in 14 GOLUBITSKY/GUILLEMIN. Stable Mappings Dimensions 2and 3. andTheir Singularities.48 SACHS/WU. General Relativity for 15 BERBERIAN. Lectures inFunctional Mathematicians. Analysis andOperator Theory. 49 GRUENBERG/WEIR. Linear Geometry. 16 WINTER. The Structure ofFields. 2nd ed. 17 ROSENBLATI. Random Processes. 2nd ed. 50 EDWARDS. Fermat's Last Theorem. 18 HALMos. Measure Theory. 51 KLINGENBERG ACourse inDifferential 19 HALMos. AHilbert Space Problem Book. Geometry. 2nd ed. 52 HARTSHORNE. Algebraic Geometry. 20 HUSEMOLLER. Fibre Bundles. 3rded. 53 MANIN. ACourse inMathematical Logic. 21 HUMPHREYs. Linear Algebraic Groups. 54 GRAVERIW ATKINS. Combinatorics with 22 BARNES/MACK. AnAlgebraic Introduction EmphasisontheTheory ofGraphs. toMathematical Logic. 55 BROWNIPEARCY. Introduction toOperator 23 GREUB. Linear Algebra. 4thed. TheoryI:Elements ofFunctional 24 HOLMEs. Geometric Functional Analysis Analysis. and ItsApplications.56 MASSEY. Algebraic Topology: An 25 HEWITT/STROMBERG. Real and Abstract Introduction. Analysis.57 CROWELrlFOX. Introduction toKnot 26 MANES. Algebraic Theories. Theory. 27 KELLEY. General Topology.58 KOBLITZ. p-adic Numbers, p-adic 28 ZARISKIISAMUEL. Commutative Algebra. Analysis, andZeta-Functions. 2nd ed. V01.I. 59 LANG. Cyclotomic Fields. 29 ZARISKIISAMUEL. Commutative Algebra. 60 ARNOLD. Mathematical Methods in Vol. II. Classical Mechanics. 2nd ed. 30 JACOBSON. Lectures inAbstract AlgebraI. 61 WHITEHEAD. Elements ofHomotopy Basic Concepts. Theory. 31 JACOBSON. Lectures inAbstract Algebra II. 62 KARGAPOLOV /MERLZJAKOV. Fundamentals Linear Algebra. oftheTheory ofGroups. 32 JACOBSON Lectures inAbstract Algebra 63 BOLLOBAS. Graph Theory. III.Theory ofFields and Galois Theory. 64 EDWARDS. Fourier Series. Vol. I.2nd ed. 33 HIRSCH. Differential Topology. 65 WELLS. Differential AnalysisonComplex 34 SPITZER. Principles ofRandom Walk. Manifolds. 2nd ed. 2nd ed. 66 WATERHOUSE. Introduction toAffine 100 BERG/CHRISTENSENlREssEL. Harmonic Group Schemes. AnalysisonSemigroups: Theory of 67 SERRE. Local Fields. Positive Definite and Related Functions. 68 WEIDMANN. Linear Operators inHilbert 101 EDWARDS. Galois Theory. Spaces. 102 VARADARAJAN. LieGroups, LieAlgebras 69 LANG. Cyclotomic Fields II. andTheir Representations. 70 MASSEY. Singular Homology Theory. 103 LANG. Complex Analysis. 3rd ed. 71 FARKAS/KRA. Riemann Surfaces. 2nd ed. 104 DUBROVIN/FOMENKO/NOVIKOV. Modem 72 STILLWELL. Classical Topology and Geometry-Methods andApplications. Combinatorial Group Theory. 2nd ed. Part II. 73 HUNGERFORD. Algebra. 105 LANG. S(R). 74 DAVENPORT. Multiplicative Number 106 SILVERMAN. The Arithmetic ofElliptic Theory. 3rd ed. Curves. 75 HocHSCHILD. Basic Theory ofAlgebraic 107 OLVER. Applications ofLieGroupsto Groups andLieAlgebras. Differential Equations. 2nd ed. 76 lITAKA. Algebraic Geometry. 108 RANGE. Holomorphic Functions and 77 HECKE. Lectures ontheTheory of Integral Representations inSeveral Algebraic Numbers. Complex Variables. 78 BURRIS/SANKAPPANAVAR. ACourse in 109 LEHTo. Univalent Functions and Universal Algebra. Teichmiiller Spaces. 79 WALTERS. AnIntroduction toErgodic 110 LANG. Algebraic Number Theory. Theory. III HUSEMOLLER. Elliptic Curves. 80 ROBINSON. ACourse intheTheory of 112 LANG. Elliptic Functions. Groups. 2nd ed. 113 KARATZAS/SHREVE. Brownian Motion and 81 FORSTER. Lectures onRiemann Surfaces. Stochastic Calculus. 2nd ed. 82 Borrffu. Differential Forms inAlgebraic 114 KOBLm. ACourse inNumber Theory and Topology. Cryptography. 2nd ed. 83 WASHINGTON. Introduction toCyclotomic 115 BERGERIGOSTIAUX. Differential Geometry: Fields. 2nd ed. Manifolds, Curves, and Surfaces. 84 IRELAND/ROSEN. AClassical Introduction 116 KELLEY/SRINIV ASAN. Measure and toModem Number Theory. 2nd ed. Integral. Vol. I. 85 EDWARDS. Fourier Series. Vol. II.2nd ed. 117 SERRE. Algebraic Groups and Class Fields. 86 VAN LINT. Introduction toCoding Theory. 118 PEDERSEN. Analysis Now. 2nd ed. 119 ROTMAN. AnIntroduction toAlgebraic 87 BROWN. Cohomology ofGroups. Topology. 88 PIERCE. Associative Algebras. 120 ZIEMER. Weakly Differentiable Functions: 89 LANG. Introduction toAlgebraic and Sobolev Spaces and Functions ofBounded Abelian Functions. 2nd ed. Variation. 90 BR0NDSTED. AnIntroduction toConvex 121 LANG. Cyclotomic Fields Iand II. Polytopes. Combined 2nd ed. 91 BEARDON. OntheGeometry ofDiscrete 122 REMMERT. Theory ofComplex Functions. Groups. Readings inMathematics 92 DIESTEL. Sequencesand Series inBanach 123 EBBINGHAUSIHERMES etal.Numbers. Spaces. Readings inMathematics 93 DUBROVIN/FoMENKO/NOVIKOV. Modem 124 DUBROVIN/FoMENKo/NOVIKOV. Modem Geometry-Methods andApplications. Geometry-Methods andApplications. Part I.2nd ed. Part III. 94 WARNER. Foundations ofDifferentiable 125 BERENSTEINIGA Y.Complex Variables: Manifolds andLieGroups. AnIntroduction. 95 SHIRYAEV. Probability. 2nd ed. 126 BOREL. Linear Algebraic Groups. 2nd ed. 96 CONWAY. ACourse inFunctional 127 MASSEY. ABasic Course inAlgebraic Analysis. 2nd ed. Topology. 97 KOBLITZ. Introduction toElliptic Curves 128 RAUCH. Partial Differential Equations. andModular Forms. 2nd ed. 129 FuLTON/HARRIS. Representation Theory: A 98 BROCKERITOM DIECK. Representations of First Course. Compact LieGroups. Readings inMathematics 99 GRovE/BENSON Finite Reflection Groups. 130 DODSON/POSTON. Tensor Geometry. 2nd ed. 131 LAM. AFirst Course inNoncommutative 163 DIXON/MORTIMER. Permutation Groups. Rings. 2nd ed. 164 NATHANSON. Additive Number Theory: 132 BEARDON. Iteration ofRational Functions. The Classical Bases. 133 HARRIs. Algebraic Geometry: AFirst 165 NATHANSON. Additive Number Theory: Course. Inverse Problems and theGeometry of 134 ROMAN. Coding andInformation Theory. Sumsets. 135 ROMAN. Advanced Linear Algebra. 166 SHARPE. Differential Geometry: Cartan's 136 ADKINslWEINTRAUB. Algebra: An Generalization ofKlein's Erlangen Approach viaModule Theory. Program. 137 AxLERIBoURDoNIRAMEY. Harmonic 167 MORANDI. Field andGalois Theory. Function Theory. 2nd ed. 168 EWALD. Combinatorial Convexity and 138 COHEN. ACourse inComputational Algebraic Geometry. Algebraic Number Theory. 169 BHATIA. Matrix Analysis. 139 BREDON. Topology andGeometry. 170 BREOON. Sheaf Theory. 2nd ed. 140 AUBIN. Optima andEquilibria. An 171 PETERSEN. Riemannian Geometry. Introduction toNonlinear Analysis.172 REMMERT. Classical Topics inComplex 141 BECKERIWEISPFENNING/KREDEL. Grabner Function Theory. Bases. AComputational Approach to 173 DIESTEL. Graph Theory. 2nd ed. Commutati veAlgebra. 174 BRIDGES. Foundations ofReal and 142 LANG. Real and Functional Analysis.Abstract Analysis. 3rd ed. 175 LICKORISH. AnIntroduction toKnot 143 DOOB. Measure Theory. Theory. 144 DENNIslFARB. Noncommutative 176 LEE. Riemannian Manifolds. Algebra.177 NEWMAN. Analytic Number Theory. 145 VICK. Homology Theory. An 178 CLARKFlLEDY AEV/STERN/WOLENSKI. Introduction toAlgebraic Topology.Nonsmooth Analysis andControl 2nd ed. Theory. 146 BRIDGES. Computability: A 179 DOUGLAS. Banach Algebra Techniques in Mathematical Sketchbook. Operator Theory. 2nd ed. 147 ROSENBERG. Algebraic K-Theory 180 SRIVASTAVA. ACourse onBorel Sets. and ItsApplications.181 KREss. Numerical Analysis. 148 ROTMAN. AnIntroduction tothe 182 WALTER. Ordinary Differential Theory ofGroups. 4thed. Equations. 149 RATCLIFFE. Foundations of 183 MEGGINSON. AnIntroduction toBanach Hyperbolic Manifolds. Space Theory. 150 EISENBUD. Commutative Algebra 184 BOLLOBAS. Modem Graph Theory. with aView Toward Algebraic 185 Cox/LITILEIO'SHEA. Using Algebraic Geometry. Geometry. 151 SILYERMAN. Advanced Topics in 186 RAMAKRISHNANN ALENZA. Fourier theArithmetic ofElliptic Curves. AnalysisonNumber Fields. 152 ZIEGLER. Lectures onPolytopes. 187 HARRIslMORRISON. Moduli ofCurves. 153 fuLTON. Algebraic Topology: A 188 GOLDBLA 17.Lectures ontheHyperrea1s: First Course. AnIntroduction toNonstandard Analysis. 154 BROWNIPEARCY. AnIntroduction to 189 LAM. Lectures onModules andRings. Analysis. 190 ESMONDFlMURTY. Problems inAlgebraic 155 KASSEL. Quantum Groups. Number Theory. 156 KECHRIS. Classical Descriptive Set 191 LANG. Fundamentals ofDifferential Theory. Geometry. 157 MALLIA VIN' Integration and 192 HIRSCH/LACOMBE. Elements of Probability. Functional Analysis. 158 ROMAN. Field Theory. 193 COHEN. Advanced Topics in 159 CONW AY.Functions ofOne Computational Number Theory. Complex Variable II. 194 ENGEIlNAGEL. One-Parameter Semi groups 160 LANG. Differential andRiemannian forLinear Evolution Equations. Manifolds. 195 NATHANSON. Elementary Methods in 161 BORWEIN/ERDEL VI.Polynomials and Number Theory. Polynomial Inequalities. 196 OSBORNE. Basic Homological Algebra. 162 ALPERIN/BELL. Groups and 197 EISENBUD/HARRIS. TheGeometry of Representations. Schemes 198 ROBERT. ACourse inp-adic Analysis. 205 FELIXIHALPERlNffHOMAS. Rational 199 HEDENMALMIKoRENBLUMlZHU. Theory Homotopy Theory. 2nd ed. ofBergman Spaces. 206 MURTY. Problems inAnalytic Number 200 BAO/CHERN/SHEN. AnIntroduction to Theory. Riemann-Finsler Geometry. Readings inMathematics 201 HINDRY/SILVERMAN. Diophantine 207 GODSIIlROYLE. Algebraic Graph Theory. Geometry: AnIntroduction. 208 CHENEY. Analysis forApplied 202 LEE. Introduction toTopological Mathematics. Manifolds. 209 ARVESON. AShort Course onSpectral 203 SAGAN. TheSymmetric Group: Theory. Representations, Combinatorial 210 ROSEN. Number Theory inFunction Fields Algorithms, andSymmetric Functions. 211 LANG. Algebra, 3rded. 204 EsCOFIER. Galois Theory. iit, i|