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Expository notes by Keith Conrad, not Phil's own work, on exterior powers. They cover symmetric, skew-symmetric and alternating multilinear functions, and the exterior power as the universal alternating quotient of the tensor power. Later sections treat free modules, induced linear maps, determinants, linear independence, the wedge product and the exterior algebra.

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EXTERIOR POWERS KEITH CONRAD 1.Introduction LetRbe a commutative ring. Unless indicated otherwise, all modules are R-modules and all tensor products are taken over R, so we abbreviate Rto . A bilinear function out ofM1M2turns into a linear function out of the tensor product M1 M2. In a similar way, a multilinear function out of M1Mkturns into a linear function out of thek-fold tensor product M1 Mk. We will concern ourselves with the case when the component modules are all the same: M1==Mk=M. The tensor power M k universally linearizes all the multilinear functions on Mk. A function on Mkcan have its variables permuted to give new functions on Mk. When looking at permutations of the variables, two important types of functions on Mkoccur: symmetric and alternating. These will be dened in Section 2. We will introduce in Section 3 the module which universally linearizes the alternating multilinear functions on Mk: the exterior power k(M). It is a certain quotient module of M k. The special case of exterior powers of nite free modules will be examined in Section 4. Exterior powers will be extended from modules to linear maps in Section 5. Applications of exterior powers to determinants are in Section 6 and to linear independence are in Section 7. Section 8 will introduce a product k(M)`(N)!k+`(M) Finally, in Section 9 we will combine all the exterior powers of a xed module into a noncommutative ring called the exterior algebra of the module. The exterior power construction is important in geometry, where it provides the language for discussing dierential forms on manifolds. (A dierential form on a manifold is related to exterior powers of the dual space of the tangent space of a manifold at each of its points.) Exterior powers also arise in representation theory, as one of several ways of creating new representations of a group from a given representation of the group. In linear algebra, exterior powers provide an algebraic mechanism for detecting linear relations among vectors and for studying the \geometry" of the subspaces of a vector space. 2.Symmetric and alternating functions For any function f:Mk!Nand any2Sk, we get a new function Mk!Nby permuting the variables in faccording to : (m1;:::;mk)7!f(m(1);:::;m(k))2N: (Warning: if we regard this new function on Mkas the eect of onf, and write it as (f)(m1;:::;mk), then1(2f) equals (21)f,not(12)f, so we don't have a left action of Skon the functions Mk!Nbut a right action. We won't be using group actions, so don't worry about this.) 1 2 KEITH CONRAD Denition 2.1. We callf:Mk!Nsymmetric if f(m(1);:::;m(k)) =f(m1;:::;mk) for all2Sk. We callfskew-symmetric if (2.1) f(m(1);:::;m(k)) = (sign)f(m1;:::;mk) for all2Sk. Fork2,fis called alternating if f(m1;:::;mk) = 0 whenever mi=mjfor somei6=j: Symmetric functions are unchanged by any permutation of the variables, while skew- symmetric functions are unchanged by even permutations and change sign under odd per- mutations. For example, the value of a skew-symmetric function changes by a sign if we permute any two of the variables. Alternating functions might not seem so intuitive. When Ris a eld of characteristic 0, like the real numbers, we will see that alternating and skew-symmetric multilinear functions are the same thing (Theorem 2.10). Example 2.2. The function M n(R)Mn(R)!Rby (A;B)7!Tr(AB) is symmetric. Example 2.3. The function R2R2!Rgiven by (a c ;b d )7!adbcis skew-symmetric and alternating. Example 2.4. The cross product R3R3!R3is skew-symmetric and alternating. Example 2.5. The function CC!Rgiven by (z;w)7!Im(zw) is skew-symmetric and alternating. Example 2.6. LetRcontain Z=2Z, so1 = 1 inR. The multiplication map RR!R is symmetric and skew-symmetric, but notalternating. In Denition 2.1, the variables are indexed in the order from 1 to n. Let's show the properties of symmetry, skew-symmetry, and alternating don't depend on this particular ordering. Theorem 2.7. Fix a listing of the numbers from 1tokasi1;i2;:::;ik. If a function f:Mk!Nis symmetric then f(m(i1);:::;m(ik)) =f(mi1;:::;mik) for all2Sk. Iffis skew-symmetric then f(m(i1);:::;m(ik)) = (sign)f(mi1;:::;mik) for all2Sk. Fork2, iffis alternating then f(mi1;:::;mik) = 0 whenevermis=mitfor someis6=it: Proof. We will discuss the skew-symmetric case, leaving the other two cases to the reader. Lete2Skbe the permutation where e(1) =i1;:::;e(k) =ik. Iffis skew-symmetric, f(mi1;:::;mik) =f(me(1);:::;m e(k)) = (signe)f(m1;:::;mk); EXTERIOR POWERS 3 so for any2Sk f(m(i1);:::;m(ik)) =f(m(e)(1);:::;m (e)(k)) = sign(e)f(m1;:::;mk) = sign() sign(e)f(m1;:::;mk) = sign()f(mi1;:::;mik): The next two theorems explain the connection between alternating multilinear functions and skew-symmetric multilinear functions, which is suggested by the above examples. Theorem 2.8. Fork2, a multilinear function f:Mk!Nwhich is alternating is skew-symmetric. Proof. We rst do the case k= 2, because the basic idea is already evident there. When f:M2!Nis alternating, f(m;m ) = 0 for all minM. So for all mandm0inM, f(m+m0;m+m0) = 0: Expanding by linearity in each component, f(m;m ) +f(m;m0) +f(m0;m) +f(m0;m0) = 0: The rst and last terms are 0, so f(m;m0) =f(m0;m). This means fis skew-symmetric. For the general case when k2, we want to show (2.1) for all 2Sk. Notice rst that iffsatises (2.1) for the permutations 1and2inSkthen f(m(12)(1);:::;m (12)(k)) =f(m1(2(1));:::;m1(2(k))) = (sign1)f(m2(1);:::;m2(k)) = (sign1)(sign2)f(m1;:::;mk) = sign(12)f(m1;:::;mk): Hence to verify (2.1) for all 2Skit suces to verify (2.1) as runs over a generating set ofSk. We will check (2.1) when runs over the generating set of transpositions f(i i+ 1) : 1ik1g: That is, for ( m1;:::;mk)2Mk, we want to show any alternating multilinear function f:Mk!Nsatises (2.2)f(:::;mi1;mi;mi+1;mi+2;:::) =f(:::;mi1;mi+1;mi;mi+2;:::); where we only interchange the places of miandmi+1. Fix all components for fexcept those in positions iandi+ 1, reducing us to a function of two variables: choose m1;:::;mi1;mi+2;:::;mk2M(an empty condition if k= 2) and letg(x;y) =f(m1;:::;mi1;x;y;mi+2;:::). Thengis bilinear and alternating. Therefore by thek= 2 casegis skew-symmetric: g(x;y) =g(y;x). This implies (2.2), so we're done. Corollary 2.9. Fork2, a function f:Mk!Nis skew-symmetric if and only if it satises f(m1;:::;mi+1;mi;:::;mk) =f(m1;:::;mi;mi+1;:::;mk) for1ik1, andfis alternating if and only if f(m1;:::;mi;mi+1;:::;mk) = 0 4 KEITH CONRAD whenevermi=mi+1for1ik1. Proof. The second paragraph of the proof of Theorem 2.8 applies to all functions, not just multilinear functions, so the condition equivalent to skew-symmetry follows. If we now suppose fvanishes at any k-tuple with a pair of adjacent equal coordinates, then what we just proved shows fis skew-symmetric. Therefore the value of fat any k-tuple with a pair of equal coordinates is up to sign its value at a k-tuple with a pair of adjacent equal coordinates, and that value is 0 by hypothesis. Example 2.6 shows the converse of Theorem 2.8 can fail: a multilinear function can be skew-symmetric and not alternating. But if 2 is a unit in R(e.g.,R=R) then the converse of Theorem 2.8 does hold: Theorem 2.10. Letk2. If22Rthen a multilinear function f:Mk!Nwhich is skew-symmetric is alternating. Proof. We will show f(m1;m2;:::;mk) = 0 when m1=m2. The argument when mi=mj for other distinct pairs iandjis the same. By skew-symmetry, f(m2;m1;m3;:::;mk) =f(m1;m2;m3;:::;mk): Therefore f(m;m;m 3;:::;mk) =f(m;m;m 3;:::;mk); so 2f(m;m;m 3;:::;mk) = 0: Since 2 is in R,f(m;m;m 3;:::;mk) = 0. Strictly speaking, the assumption that 2 2Rcould be weakened to 2 not being a zero divisor inR. (This is stronger than saying Rdoesn't have characteristic 2, since R=Z=(4) doesn't have characteristic 2 but the proof doesn't work for such R.) Over the real and complex numbers the terms \alternating" and \skew-symmetric" can be used interchangeably for multilinear functions. For instance in [5], where only real vector spaces are used, multilinear functions satisfying the skew-symmetric property (2.1) are called alternating. At any point in our discussion where there is a noticeable dierence between being alternating and being skew-symmetric, it is just a technicality, so don't worry too much about it. 3.Exterior powers of modules LetMandNbeR-modules and k1. Whenf:Mk!Nis multilinear and g:N!P is linear, the composite gfis multilinear. Moreover, if fis symmetric, skew-symmetric, or alternating then gfhas the corresponding property too. So we can create new (symmetric, skew-symmetric, or alternating) multilinear maps from old ones by composing with a linear map. Thekth tensor power Mk !M k, sending (m1;:::;mk) tom1 mk, is a particular example of a multilinear map out of Mk, and every other example comes from this one: given any multilinear map f:Mk!Nthere is a unique linear map ef:M k!Nwhose composite with Mk !M kisf. That is, there is a unique linear map ef:M k!N EXTERIOR POWERS 5 making the diagram Mk f//N M kef== commute, which means ef(m1 mk) =f(m1;:::;mk). We now focus our attention on multilinear fwhich are alternating. When k2 and f:Mk!Nis alternating, fvanishes on any k-tuple with a pair of equal coordinates, so ef:M k!Nvanishes on any tensor m1 mkwithmi=mjfor somei6=j. Thus the submodule J kspanned by these special tensors is automatically in the kernel of ef. The quotient of M kby the submodule J kwill be our main object of interest. Denition 3.1. For anR-moduleMand an integer k2, thekthexterior power ofM, denoted k(M), is theR-moduleM k=Jkwhere Jkis the submodule of M kspanned by allm1 mkwithmi=mjfor somei6=j. For anym1;:::;mk2M, the coset of m1 mkin k(M) is denoted m1^^mk. For completeness, set 0(M) =Rand 1(M) =M(so J 0= J1=f0g). We could write k R(M) to place the ring Rin our notation. But since we will never be changing the ring, we suppress this extra decoration. The general element of k(M) will be denoted !or(as we write tfor a general tensor). Since M kis spanned by the tensorsm1 mk, the quotient module M k=Jk= k(M) is spanned by their images m1^^mk. That is, any !2k(M) is a nite R-linear combination !=X ri1;:::;ikmi1^^mik; where the coecients ri1;:::;ikare inRand themi's are inM. We callm1^m2^^mkanelementary wedge product and read it as \ m1wedgem2::: wedgemk." Another name for elementary wedge products is decomposable elements. (More synonyms: simple, pure, monomial). Since r(m1^m2^^mk) = (rm1)^m2^^mk, every element of k(M) is a sum (not just a linear combination) of elementary wedge products. A linear { or even additive { map out of k(M) is completely determined by its values on elementary wedge products because they additively span k(M). More general wedge products will be met in Section 8. The modules k(M) were introduced by Grassmann (for M=Rn), who called expres- sions likem1^m2outer products. Now we use the label \exterior" instead. Perhaps it would be better to call k(M) an alternating power instead of an exterior power, but it's too late to change the terminology. Example 3.2. SupposeMis spanned by the two elements xandy:M=Rx+Ry. (This doesn't mean xandyare a basis over R,e.g.,R=Z[p5] andM= (2;1 +p5).) We will show 2(M) is spanned by the single element x^y. The tensor square M 2is spanned by all terms m m0wheremandm0are inM. Writem=ax+byandm0=cx+dy. Then inM 2 m m0= (ax+by) (cx+dy) =ac(x x) +ad(x y) +bc(y x) +bd(y y): 6 KEITH CONRAD The tensors x xandy yare in J 2, so in 2(M) bothx^xandy^yvanish. Therefore m^m0=ad(x^y) +bc(y^x): Moreover, the tensor ( x+y) (x+y) is in J 2, so x y+y x= (x+y) (y+x)x xy y2J2: Therefore in 2(M) we havex^y+y^x= 0, so m^m0=ad(x^y) +bc(x^y) = (adbc)(x^y); which means 2(M) is spanned by the single element x^y. It could happen that x^y= 0 (so 2(M) could be zero), or even if x^y6= 0 it could happen that r(x^y) = 0 for some nonzeror2R. It all depends on the nature of the R-linear relations between xandyin M. By comparison to 2(M),M 2is spanned by x x,x y,y x, andy y, and without further information we have no reason to collapse this spanning set (usually x y6=y x, for instance). So when Mhas a 2-element spanning set, a spanning set for M 2is typically larger than 2 while a spanning set for 2(M) is denitely smaller. Fork2, the standard map Mk !M kis multilinear and the reduction map M k!M k=Jk= k(M) is linear, so the composite map ^:Mk !M k!k(M) is multilinear. That is, the function (3.1) ( m1;:::;mk)7!m1^^mk fromMkto k(M) is multilinear in the mi's. For example, m1^^cmi^^mk=c(m1^^mi^^mk); m 1^^ 0^^mk= 0: Working in k(M) forcesm1^^mk= 0 ifmi=mjfor somei6=j. (Think about what working modulo J kmeans for the tensors m1 mkwhenmi=mjfor some i6=j.) Therefore (3.1) is an example of an alternating multilinear map out of Mk. Now we show it is a universal example: all others pass through it using linear maps out of k(M). Theorem 3.3. LetMbe anR-module and k2. For anyR-moduleNand alternating multilinear map f:Mk!N, there is a unique linear map ef: k(M)!Nsuch that the diagram Mk ^ f//N k(M)ef<< commutes, i.e., ef(m1^^mk) =f(m1;:::;mk). This theorem makes no sense when k= 0 ork= 1 since there are no alternating multilinear functions in those cases. Proof. Sincefis multilinear, it induces a linear map M k!Nwhose behavior on elemen- tary tensors is m1 mk7!f(m1;:::;mk): EXTERIOR POWERS 7 Mk f//N M k== (We are going to avoid giving this map M k!Na specic notation, since it is just an intermediate device in this proof.) Because fis alternating, it vanishes at any k-tuple (m1;:::;mk) wheremi=mjfor somei6=j. Thus the linear map which finduces from M ktoNvanishes at any elementary tensor m1 mkwheremi=mjfor somei6=j. Hence this linear map out of M kvanishes on the submodule J kofM k, so we get an induced linear map efout ofM k=Jk= k(M). Specically, ef: k(M)!Nis given by m1^^mk7!f(m1;:::;mk): Mk f//N M k ;; k(M)efEE Since the elements m1^^mkspan k(M), a linear map out of k(M) is uniquely determined by its eect on these elements. Thus, having constructed a linear map out of k(M) whose eect on any m1^^mkis the same as the eect of fon (m1;:::;mk), it is the unique such linear map. We will call the particular alternating multilinear map Mk^! k(M) given by (m1;:::;mk)7!m1^^mk thecanonical map . Remark 3.4. We could have constructed k(M) as the quotient of a huge free module on the set Mk, bypassing the use of M k. Since the canonical map Mk^! k(M) is multilinear, we get a linear map M k!k(M) and can recover k(M) as a quotient of M kanyway. Corollary 3.5. Suppose is anR-module and there is an alternating multilinear map :Mk!with the same universal mapping property as the canonical map Mk^! k(M): for every R-moduleNand alternating multilinear map f:Mk!Nthere is a unique linear map !Nmaking the diagram Mk f////N == 8 KEITH CONRAD commute. Then there is a unique R-linear map : !k(M)such that the diagram Mk ~~^ ## //k(M) commutes, and is an isomorphism. Proof. This is the usual argument that an object equipped with a map satisfying a universal mapping property is determined up to a unique isomorphism: set two such objects and maps against each other to get maps between the objects in both directions whose composites in both orders have to be the identity maps on the two objects by the usual argument. Since the canonical map Mk^! k(M) is alternating multilinear, by Theorem 2.8 it is skew-symmetric: (3.2) m(1)^^m(k)= (sign)m1^^mk for every2Sk. In particular, an elementary wedge product m1^^mkin k(M) is determined up to an overall sign by the terms miappearing in it ( e.g.,m^m0^m00= m0^m^m00). Example 3.6. Returning to Example 3.2 and working directly in 2(M) from the start, we have (3.3) ( ax+by)^(cx+dy) =ac(x^x) +ad(x^y) +bc(y^x) +bd(y^y) by multilinearity. Since (3.1) is alternating, x^xandy^yvanish. By (3.2), y^x=x^y. Feeding this into (3.3) gives (ax+by)^(cx+dy) =ad(x^y)bc(x^y) = (adbc)(x^y); so 2(M) is spanned by x^ywhenMis spanned by xandy. That was faster than Example 3.2! Example 3.7. SupposeMis spanned by three elements e1;e2, ande3. We will show e1^e2, e1^e3, ande2^e3span 2(M). We know by the denition of 2(M) that 2(M) is spanned by allm^m0, so it suces to show every m^m0is a linear combination of e1^e2,e1^e3, ande2^e3. Writing m=ae1+be2+ce3; m0=a0e1+b0e2+c0e3; the multilinearity and the alternating property ( ei^ei= 0 andei^ej=ej^eifori6=j) imply m^m0= (ae1+be2+ce3)^(a0e1+b0e2+c0e3) =ae1^(b0e2+c0e3) +be2^(a0e1+c0e3) +ce3^(a0e1+b0e2) =ab0(e1^e2) +ac0(e1^e3) +ba0(e2^e1) +bc0(e2^e3) + ca0(e3^e1) +cb0(e3^e2) = (ab0ba0)(e1^e2) + (ac0ca0)(e1^e3) + (bc0cb0)(e2^e3): If we write this formula with the rst and third terms exchanged as (3.4)m^m0= (bc0cb0)(e2^e3) + (ac0ca0)(e1^e3) + (ab0ba0)(e1^e2); EXTERIOR POWERS 9 it looks quite close to the cross product on R3: (a;b;c )(a0;b0;c0) = (bc0cb0;(ac0ca0);ab0ba0): (There is a way of making e3^e1the more natural wedge product than e1^e3in 2(R3), so (3.4) would then match the coordinates of the cross-product everywhere. This uses the Hodge-star operator. We don't discuss that here.) In a tensor power of a module, every tensor is a sum of elementary tensors but most elements are not themselves elementary tensors. The same thing happens with exterior powers: a general element of k(M) is a sum of elementary wedge products, but is notitself of this form. Example 3.8. LetMbe spanned by e1,e2,e3, ande4. The exterior square 2(M) is spanned by the pairs e1^e2; e1^e3; e1^e4; e2^e3; e2^e4; e3^e4: WhenMis free andfe1;e2;e3;e4gis a basis of M, the sume1^e2+e3^e4in 2(M) is notan elementary wedge product: it can't be expressed in the form m^m0. We'll see why (in most cases) in Example 8.9. On the other hand, the sum e1^e2+ 3(e1^e3) + 3(e1^e4) + 2(e2^e3) + 2(e2^e4) in 2(M) doesn't look like an elementary wedge product but it is! It equals (e1+e2+e3+e4)^(e2+ 3e3+ 3e4): Check equality by expanding this out using multilinearity and the relations ei^ei= 0 and ei^ej=ej^eifori6=j. A linear map out of k(M) is completely determined by its values on the elementary wedge products m1^^mk, since they span the module. But elementary wedge products, like elementary tensors, are not linearly independent, so verifying there is a linear map out of k(M) with some prescribed behavior on all the elementary wedge products has to be done carefully. Proceed by rst introducing a function on Mkwhich is multilinear and alternating whose value at ( m1;:::;mk) is what you want the value to be at m1^^mk, and then it automatically factors through k(M) as a linear map with the desired value at m1^^mk. This is like creating homomorphisms out of a quotient group G=N by rst making a homomorphism out of Gwith the desired values and then checking Nis in the kernel. Example 3.9. There is a unique linear map 2(M)!M 2such that m1^m27!m1 m2m2 m1: To construct such a map, start by letting f:M2!M 2byf(m1;m2) =m1 m2m2 m1. This is bilinear and f(m;m ) = 0, sofis alternating and thus there is a linear map 2(M)! M 2sending any elementary wedge product m1^m2tof(m1;m2) =m1 m2m2 m1. Since a linear map out of 2(M) is determined by its values on elementary wedge products, fis the only linear map with the given values on all m1^m2. Here are some basic questions about exterior powers. Questions (1) What does it mean to say m1^^mk= 0? (2) What does it mean to say k(M) = 0? 10 KEITH CONRAD (3) What does it mean to say m1^^mk=m0 1^^m0 k? Answers (1) Saying m1^^mkequals 0 means every alternating multilinear map out of Mk vanishes at ( m1;:::;mk). Indeed, since every alternating multilinear map out of Mkinduces a linear map out of k(M) which sends m1^^mkto the same place as (m1;:::;mk), ifm1^^mk= 0 then the linear map out of k(M) must send m1^^mkto 0 (linear maps send 0 to 0) so the original alternating multilinear map we started with out of Mkhas to equal 0 at ( m1;:::;mk). Conversely, if every alternating multilinear map out of Mksends (m1;:::;mk) to 0 thenm1^^mk= 0 because the canonical map Mk^! k(M) is a particular example of an alternating multilinear map out of Mkand it sends ( m1;:::;mk) tom1^^mk. Thus you can prove a particular elementary wedge product m1^^mkis not0 by nding an alternating multilinear map on Mkwhich is not equal to 0 at (m1;:::;mk). (2) To say k(M) = 0 means every alternating multilinear map on Mkis identically 0. To show k(M)6= 0, nd an example of an alternating multilinear map on Mk which is not identically 0. (3) The condition m1^^mk=m0 1^^m0 kmeans every alternating multilinear map onMktakes the same values at ( m1;:::;mk) and at (m0 1;:::;m0 k). Remark 3.10. Unlike the tensor product, which can be dened between dierent R- modules, there is no \exterior product" of two unrelated R-modules. This is because the concept of exterior power is bound up with the idea of alternating multilinear functions, and permuting variables in a multivariable function only makes sense when the function has its variables coming from the same module. 4.Spanning sets for exterior powers Let's look more closely at spanning sets of an exterior power module. If Mis nitely generated (not necessarily free!), with spanning set x1;:::;xd, then any tensor power M k is nitely generated as an R-module by the dktensorsxi1 xikwhere 1i1;:::;ikd: at rst we know M kis spanned by all the elementary tensors m1 mk, but write each mias anR-linear combination of x1;:::;xdand then expand out using the multilinearity of to express every elementary tensor in M kas anR-linear combination of the tensors xi1 xik. (As a general rule this spanning set for M kcan't be reduced further: when Mis free andx1;:::;xdis a basis then the dkelementary tensors xi1 xikare a basis ofM k.) Since k(M) is a quotient module of M k, it is spanned as an R-module by the dkelementary wedge products xi1^^xikwhere 1i1;:::;ikd. Thus exterior powers of a nitely generated R-module are nitely generated. Theorem 4.1. IfMhas ad-element spanning set then k(M) =f0gfork>d . For example, 2(R) = 0, and more generally k(Rd) = 0 fork>d . Proof. Letx1;:::;xdspanM. Whenk>d , anyxi1^^xikcontains two equal terms, so it is zero. Thus k(M) is spanned by 0, so it is 0. There is a lot of redundancy in the dkelementary wedge products xi1^^xikcoming from a spanning set fx1;:::;xdgofM. For instance, by the alternating property such an elementary wedge product vanishes if two terms in it are equal. Therefore we can discard EXTERIOR POWERS 11 from our spanning set for k(M) thosexi1^^xikwhere anxiappears twice and we are still left with a spanning set. Moreover, by (3.2) two elementary wedge products containing the same factors in dierent order are equal up to sign, so our spanning set for k(M) as anR-module can be reduced further to the elements xi1^^xikwhere the indices are strictly increasing: 1 i1<<ikd. The number of such k-tuples of indices isd k . So k(M) has a spanning set of sized k , which may or may not be reducible further. We will now prove that there can be no further reduction when Mis free and x1;:::;xdis a basis ofM. Theorem 4.2. IfMis free then k(M)is free, provided krank(M)ifMhas a nite rank and with no constraint on kifMhas innite rank. Explicitly, if M6= 0 is nite free with basise1;:::;ed, then for 1kdthed k elementary wedge products ei1^^eikwhere 1i1<<ikd are a basis of k(M). In particular, k(M)is free of rankd k for0kdandk(M) = 0 fork>d . IfMhas an innite basis feigi2Iand we put a well-ordering on the index set I, then for any k1a basis of k(M)isfei1^^eikgi1<i2<<ik. Theorem 4.2 is the rst nontrivial result about exterior powers, as it tells us a situation where exterior powers are guaranteed to be nonzero, and in fact be \as big as possible." Read the proof closely, as otherwise you may feel somewhat uneasy about exactly why exterior powers of free modules are free. Proof. Fork= 0 there is nothing to show. Take k1. The idea in the proof is to embed k(M) as a submodule of M kand exploit what we know already about M kfor freeM. The embedding we will write down may look like it comes out of nowhere, but it is very common in dierential geometry and we make some remarks about this after the proof. The basic idea is to turn elementary wedge products into skew-symmetric tensors by an averaging process. The function Mk!M kgiven by (m1;:::;mk)7!X 2Sk(sign)m(1) m(k) is multilinear since each summand contains each mionce. This function is also alternating. To prove this, by Corollary 2.9 it suces to check the function vanishes at k-tuples with adjacent equal coordinates. If mi=mi+1then for each 2Skthe terms in the sum at and(i i+ 1) are negatives of each other. Now the universal mapping property of the exterior power tells us there is an R-linear map k;M: k(M)!M ksuch that (4.1) k;M(m1^^mk) =X 2Sk(sign)m(1) m(k): The casek= 2, by the way, is Example 3.9. Althoughk;Mexists for all M, injectivity of k;Mis not a general property. Our proof of injectivity for free Mwill use a basis. We'll write the proof with a nite basis, and the reader can make the minor changes to see the same argument works if Mhas an innite basis. LetMhave basise1;:::;ed. Injectivity of k;M: k(M)!M kis clear fork>d , so we may take 1kd. We may even take k2, as the theorem is obvious for k= 1. Since 12 KEITH CONRAD theei's spanMas anR-module, the elementary wedge products (4.2) ei1^^eikwhere 1i1<<ikd span k(M). (Here we need that the indices are totally ordered.) We know already that M khas a basis ei1 eikwhere 1i1;:::;ikd; where no inequalities are imposed on the indices. Suppose!2k(M) satisesk;M(!) = 0. Write !=X 1i1<<ikdci1;:::;ikei1^^eik: withci1;:::;ik2R. Then the condition k;M(!) = 0 becomes X 1i1<<ikdci1;:::;ikX 2Sk(sign)ei(1) ei(k)= 0; which is the same asX 2SkX I(sign)cIe(I)= 0; whereIruns over all strictly increasing k-tuples (i1;:::;ik) from 1 to d, withcIande(I) having an obvious meaning in this context. The vectors fe(I)g;Iare a basis ofM k, so allcIare 0. This proves k;Mis injective and it also shows (4.2) is a linearly independent subset of k(M), so it is a basis (spans and is linearly independent). Exterior powers are closely connected to determinants, and most proofs of Theorem 4.2 for nite free Muse the determinant. What we used in lieu of theorems about determinants is our knowledge of bases of tensor powers of a free module. For aesthetic reasons, we want to come back later and prove properties of the determinant using exterior powers, so we did not use the determinant directly in the proof of Theorem 4.2. However, the linear map k;Mlooks a lot like a determinant. WhenVis a nite-dimensional vector space, it is free so k;V: k(V),!V kgiven by (4.1) is an embedding. It means we can think of k(V) as a subspace of the tensor powerV kinstead of as a quotient space. This viewpoint is widely used in dierential geometry, where vector spaces are dened over RorCand the image of k(V) inV kis the subspace of skew-symmetric tensors. The embedding k;Vhas an unfortunate scaling problem: when we embed k(V) intoV kwithk;Vand then reduce V kback to k(V) with the canonical map ^, the composite map k(V)k;V!V k^! k(V) isnotthe identity map on k(V), but is multiplication by k!. We can verify this by checking it on elementary wedge products: v1^^vk7!X 2Sk(sign)v(1) v(k) 7!X 2Sk(sign)v(1)^^v(k) =X 2Sk(sign)(sign)v1^^vk =k!v1^^vk: EXTERIOR POWERS 13 This suggests a better embedding of k(V) intoV kis1 k!k;V, which is given by the formula v1^^vk7!1 k!X 2Sk(sign)v(1) v(k): The composite map k(V)(1=k!)k;V! V k^! k(V) is the identity, but this rescaled embedding only makes sense if k!6= 0 in the scalar eld. That is ne for real and complex vector spaces (as in dierential geometry), but it is not a universal method. So either you can take your embedding k(V),!V kusingk;Vfor all vector spaces and make k(V)!V k!k(V) be multiplication by k!, or you can have an embedding k(V),! V kthat only makes sense when k!6= 0. Either way, this mismatch between k(V) as a quotient space of V k(correct denition) and as a subspace of V k(incorrect but widely used denition) leads to a lot of excess factorials in formulas when exterior powers are regarded as subspaces of tensor powers instead of as quotient spaces of them. There are other approaches to the proof of Theorem 4.2 when Mis nite free. In [1, pp. 90{91] an explicit nite free R-module is constructed which has the same universal mapping property as k(M), so k(M) has to be nite free by Corollary 3.5. The other aspects of Theorem 4.2 (the rank of k(M) and an explicit basis) can be read o from the proof in [1]. In [4, pp. 747{751], Theorem 4.2 for nite free Mis proved using what is called there the Grassmann algebra of M(which we will meet later under the label exterior algebra ofM). WhenMis free of rank dandkd, we will call the basis fei1^^eik: 1i1<< ikdgof k(M) the corresponding basis from the choice of basis e1;:::;edofM. Example 4.3. IfMis free of rank dwith basise1;:::;edthen 2(M) is free of rankd 2 with corresponding basis fei^ej: 1i<jdg. Remark 4.4. WhenMis a freeR-module, its tensor powers M kare freeR-modules. While k(M) is a quotient module of M k, it does not follow from this alone that k(M) must be free: the quotient of a free module is not generally free (consider R=I whereIis a proper nonzero ideal). Work was really needed to show exterior powers of free modules are free modules. WhenMis free of rank d, k(M)6= 0 whenkdand k(M) = 0 when k >d . This is why we call d(M) the top exterior power . It is free of rank 1; a basis of d(M) ise1^^ed ife1;:::;edis a basis of M. Although d(M) is isomorphic to Ras anR-module, it is not naturally isomorphic: there is no canonical isomorphism between them. WhenMhas ad-element spanning set with dminimally chosen, and Mis not free, it might happen that d(M) = 0. For example, consider a non-principal ideal I:=Rx+Ryin Rwith two generators. The module 2(I) is spanned as an R-module by x^y.1Sometimes 2(I) is zero and sometimes it is nonzero. Example 4.5. LetR=Z[p5] andI= (2;1 +p5). Set!:= 2^(1 +p5)22(I). We will show 2 != 0 and 3!= 0, so!= 0 (just subtract) and thus 2(I) = 0: 2!= 2^2(1 +p 5) = (1 +p 5)(2^2) = 0; 3!= 6^(1 +p 5) = (1p 5)((1 +p 5)^(1 +p 5)) = 0: 1It is important to realize x^yhere means an elementary wedge product in 2(I),notin 2(R); the latter exterior square is 0 all the time. 14 KEITH CONRAD Example 4.6. LetR=A[X;Y ] be the polynomial ring in two variables over a nonzero commutative ring A. LetI= (X;Y ) inR. We will show X^Yin 2(I) is nonzero by writing down a linear map out of 2(I) whose value on X^Yis nonzero. DeneB:I2!Ato be the determinant on degree one coecients: B(aX+bY+;cX+dY+) =adbc: Regard the target module Aas anR-module through scaling by the constant term of a polynomial: f(X;Y )a=f(0;0)a. (That is, we basically treat AasR=I=A[X;Y ]=(X;Y ).) ThenBisR-bilinear, and it is alternating too. Since B(X;Y ) = 1,Binduces a linear map L: 2(I)!AwhereL(X^Y) =B(X;Y ) = 1, so 2(I)6= 0. Remark 4.7. An analogue of Example 4.5 in the real quadratic ring Z[p 5] has a dierent result. For J= (2;1 +p 5), 2^(1 +p 5) in 2(J) is not 0, so 2(J)6= 0. Constructing an alternating bilinear map out of JJthat is not identically 0 can be done using the ideas in Example 4.6, and details are left to the reader (Hint: Find a basis for Jas aZ-module.) The moral from these two examples is that for nonfree M, the highest kfor which k(M)6= 0 need not be the size of a minimal spanning set for the module. It only gives an upper bound, when Mis nitely generated.2 An important distinction to remember between tensor and exterior powers is that exterior powers are not recursively dened. Whereas M (k+1)=M RM k, we can't say that k+1(M) is a product of Mand k(M); later on (Section 9) we will introduce the exterior algebra, in which something like this does make sense. To appreciate the lack of a recursive denition of exterior powers, consider the following problem. When Mhas ad-element spanning set, k(M) = 0 forkd+ 1 by Theorem 4.1. Treating Mas 1(M), we pose a generalization: if i(M) for somei >1 has ad-element spanning set, is k(M) = 0 for kd+i? The next theorem settles the d= 0 case in the armative. Theorem 4.8. Ifi(M) = 0 for somei1then j(M) = 0 for allji. Proof. It suces to show i(M) = 0)i+1(M) = 0, as then we are done by induction. To prove i+1(M) = 0 we show all ( i+ 1)-fold elementary wedge products (4.3) m1^^mi^mi+1 are 0. First we give a fake proof. Since i(M) = 0,m1^^mi= 0, so (4.3) equals 0 ^mi+1= 0. What makes this absurd, at our present level of understanding, is that there is no sense (yet) in which the notation ^is \associative," as the notation ^is really just a placeholder to tell us where things go. We can't treat the piece m1^^miin (4.3) as its own elementary wedge product having any kind of relation to (4.3). This is like calculus, where students are warned that the separate parts of d y=dxdo not have an independent meaning, although later they may learn otherwise, as we too will learn otherwise about ^in Section 9. Now we give a real proof, which in fact contains the germ of the idea in Section 9 to make ^into a genuine operation and not just a placeholder. For each elementary wedge product (4.3), which belongs to i+1(M), we will create a linear map i(M)!i+1(M) with (4.3) in its image. Then since i(M) = 0, so a linear map out of i(M) has image 0, (4.3) is 0. 2Since ( Q=Z) Z(Q=Z) = 0, 2(Q=Z) = 0 where we regard Q=Zas aZ-module, so the vanishing of k(M) for some kdoes not force Mto be nitely generated. EXTERIOR POWERS 15 Consider the function Mi!i+1(M) given by (x1;:::;xi)7!x1^^xi^mi+1: This is multilinear and alternating, so by the universal mapping property of exterior powers there is a linear map i(M)!i+1(M) where x1^^xi7!x1^^xi^mi+1: The left side is 0 for all choices of x1;:::;xiinM, so the right side is 0 for all such choices too. In particular, (4.3) is 0. Here is a dierent proof. To say i(M) = 0 means any alternating multilinear function out ofMiis identically 0. If ':Mi+1!Nis an alternating multilinear function, and (m1;:::;mi;mi+1)2Mi+1, consider the function '(x1;:::;xi;mi+1) inx1;:::;xi. It is alternating multilinear in ivariables from M. Therefore '(x1;:::;xi;mi+1) = 0 for all x1;:::;xiinM, so'(m1;:::;mi;mi+1) = 0. Returning to the general question, where i(M) has ad-element spanning set, asking if k(M) = 0 forkd+iis the same as asking if d+i(M) = 0 by Theorem 4.8. The answer is \yes" although more technique is needed for that than we will develop here. 5.Exterior powers of linear maps Having constructed exterior powers of modules, we extend the construction to linear maps between modules. First recall any linear map ':M!Nbetween two R-modules induces a linear map ' k:M k!N kon thekth tensor powers, for any positive integer k, which has the eect ' k(m1 mk) ='(m1) '(mk) on elementary tensors. Theorem 5.1. Let':M!Nbe a linear map of R-modules. Then for each k2there is a unique linear map ^k('): k(M)!k(N)with the eect m1^^mk7!'(m1)^^'(mk) on all elementary wedge products. For a second linear map :N!P,^k( ') = ^k( )^k('). Moreover,^k(idM) = idk(M). Proof. There is at most one such linear map k(M)!k(N) since the elementary wedge products span k(M). To show there is such a linear map, start by backing up and dening a functionf:Mk!k(N) by f(m1;:::;mk) ='(m1)^^'(mk): This is a multilinear map which is alternating, so by the universal mapping property of the kth exterior power there is a linear map k(M)!k(N) with the eect m1^^mk7!f(m1;:::;mk) ='(m1)^^'(mk): Mk ^ f//k(N) k(M):: This proves the existence of the linear map ^k(') we are seeking. 16 KEITH CONRAD To show^k( ') =^k( )^k('), it suces since both sides are linear to check that both sides have the same value on each elementary wedge product in k(M). At any elementary wedge product m1^^mk, the left side and right side have the common value ('(m1))^^ ('(mk)). That^k(idM) = idk(M)is easy:^k(idM) is linear and xes everym1^^mkand these span k(M), so^k(idM) xes everything. Theorem 5.1 is also true for k= 0 andk= 1 by setting^0(') = idRand^1(') ='. Recall 0(M) =Rand 1(M) =M. That the passage from 'to^k(') respects composition and sends the identity map on a module to the identity map on its kth exterior power is called functoriality of thekth exterior power. Armed with bases for exterior powers of nite free modules, we can write down matrices for exterior powers of linear maps between them. When MandNare nite free of respective ranksmandn, a choice of bases of MandNturns any linear map ':M!Ninto an nmmatrix. Using the corresponding bases on k(M) and k(N), we can write^k(') as ann k m k matrix. If we want to look at an example, we need to keep mandnsmall or we will face very large matrices. Example 5.2. IfMandNare free of rank 5 and ':M!Nis linear, then^2(') is represented by a 10 10 matrix since5 2 = 10. Example 5.3. LetL:R3!R3be the linear map given by the matrix (5.1)0 @0 2 0 1 1 1 0 3 21 A: We will compute the matrix for ^2(L): 2(R3)!2(R3) with respect to the basis e1^ e2;e1^e3;e2^e3, where the ei's are the standard basis of R3. Going in order, ^2(L)(e1^e2) =L(e1)^L(e2) =e2^(2e1+e2+ 3e3) =2(e1^e2) + 3(e2^e3); ^2(L)(e1^e3) =L(e1)^L(e3) =e2^(e2+ 2e3) = 2(e2^e3); and ^2(L)(e2^e3) =L(e2)^L(e3) = (2e1+e2+ 3e3)^(e2+ 2e3) = 2(e1^e2) + 4(e1^e3)e2^e3: Therefore the matrix for ^2(L) relative to this ordered basis is 0 @2 0 2 0 0 4 3 211 A: Theorem 5.4. Let':M!Nbe linear. If 'is anR-module isomorphism then ^k(')is anR-module isomorphism for every k. If'is surjective then every ^k(')is surjective. EXTERIOR POWERS 17 Proof. It is clear for k= 0 andk= 1. Letk2. Suppose 'is an isomorphism of R- modules, with inverse :N!M. Then' = idNand '= idM, so by Theorem 5.1 we have^k(')^k( ) =^k(idN) = idk(N)and similarly^k( )^k(') = idk(M). If'is surjective, then ^k(') is surjective because k(N) is spanned by the elementary wedge products n1^^nkfor allni2Nand these are in the image of ^k(') explicitly: writingni='(mi), the elementary wedge product of the ni's in k(N) is'(m1^^mk). Since the image of the linear map ^k(') is a submodule of k(N) which contains a spanning set for k(N), the image is all of k(N). As with tensor products of linear maps, it is false that^k(') has to be injective if 'is injective. Example 5.5. LetR=A[X;Y ] withAa nonzero commutative ring and let I= (X;Y ). In our discussion of tensor products, it was seen that the inclusion map i:I!Ris injective while its induced R-linear map i 2:I 2!R 2=Ris not injective. Therefore it should come as no surprise that the map ^2(i): 2(I)!2(R) also is not injective. Indeed, 2(R) = 0 and we saw in Example 4.6 that 2(I)6= 0 because there is an R-linear map L: 2(I)!AwhereL(X^Y) = 1. (In 2(I) we have X^Y6= 0 while in 2(R) we have X^Y=XY(1^1) = 0. There is nothing inconsistent about this, even though IR, because the natural map ^2(i): 2(I)!2(R) is not injective. The X^Y's in 2(I) and 2(R) lie in dierent modules, and although ^2(i) sendsX^Yin the rst module to X^Yin the second, linear maps can send a nonzero element to 0 and that is what is happening.) We now show the linear map L: 2(I)!Ais actually an isomorphism of R-modules. In 2(I) =fr(X^Y) :r2Rg,X^Yis killed by multiplication by XandYsinceX(X^Y) = X^XY=Y(X^X) = 0 and likewise for Y(X^Y). Sof(X;Y )(X^Y) =f(0;0)(X^Y) in 2(I), which means every element of 2(I) has the form a(X^Y) for somea2A. Thus the function L0:A!2(I) given byL0(a) =a(X^Y) isR-linear and is an inverse to L. The isomorphism 2(I)=Ageneralizes to the polynomial ring R=A[X1;:::;Xn] for anyn2: the ideal I= (X1;:::;Xn) inRhas n(I)=A=R=I, so the inclusion i:I!R is injective but^n(i): n(I)!n(R) = 0 is not injective. Although exterior powers don't preserve injectivity of linear maps in general, there are some cases when they do. This is the topic of the rest of this section. A number of ideas here and later are taken from [2]. Theorem 5.6. Suppose':M!Nis injective and the image '(M)Nis a direct summand: N='(M)Pfor some submodule PofN. Then^k(')is injective for all k0andk(M)is isomorphic to a direct summand of k(N). Proof. The result is trivial for k= 0 andk= 1. Suppose k2. We will use the splitting criteria for short exact sequences of modules. Since N='(M) P, we have a linear map :NMwhich undoes the eect of ': let ('(m) +p) =m. Then ('(m)) =mfor allm2M, so (5.2) '= idM: (The composite in the other direction, ' , is denitely not id NunlessP= 0, but this does not matter.) Applying ^kto (5.2) gives us linear maps ^k('): k(M)!k(N) and ^k( ): k(N)!k(M) with ^k( )^k(') =^k( ') =^k(idM) = idk(M) 18 KEITH CONRAD by functoriality (Theorem 5.1). In particular, if ^k(')(!) = 0 for some !2k(M), then applying^k( ) to both sides gives us !=^k( )(0) = 0, so^k(') has kernel 0 and thus is injective. For the short exact sequence 0 !k(M)^k(')! k(N)!k(N)=k(M)!0, the fact that^k( ) is a left inverse to ^k(') implies by the splitting criteria for short exact sequences that k(N)=k(M)(k(N)=k(M)), so k(M) is isomorphic to a direct summand of k(N). Example 5.7. For anyR-modulesMandM0, the inclusion i:M!MM0where i(m) = (m;0) has image M0, which is a direct summand of MM0, so the induced linear map k(M)!k(MM0) sendingm1^^mkto (m1;0)^^ (mk;0) is one-to-one. Remark 5.8. It is instructive to check that the hypothesis of Theorem 5.6 does not apply to the inclusion i: (X;Y )!A[X;Y ] in Example 5.5 (which must be so because ^2(i) is not injective). We can write A[X;Y ] =A(X;Y ), so (X;Y ) is a direct summand of A[X;Y ] as abelian groups, or even as A-modules, but this is not a direct sum of A[X;Y ]-modules: Ais not an ideal in A[X;Y ]. Corollary 5.9. LetKbe a eld and VandWbeK-vector spaces. If the K-linear map ':V!Wis injective then^k('): k(V)!k(W)is injective for all k0. Proof. The subspace '(V)Wis a direct summand of W: pick a basis of '(V) overK, extend it to a basis of the whole space W, and letPbe the span of the new part of this full basis:W='(V)P. Thus the hypothesis of Theorem 5.6 applies to this situation. When working with linear maps of vector spaces (not necessarily nite-dimensional), we have shown ':V!Winjective =) ^k(') injective for all k(Corollary 5 :9); ':V!Wsurjective =) ^k(') surjective for all k(Theorem 5 :4); ':V!Wan isomorphism = ) ^k(') an isomorphism for all k: Replacing vector spaces with modules, the second and third properties are true but the rst one may fail (Example 5.5 with R=A[X;Y ] andk= 2). The rst property does remain true for free modules, however. Theorem 5.10. SupposeMandNare freeR-modules. If a linear map ':M!Nis injective then^k('): k(M)!k(N)is injective for all k. Notice the free hypothesis! We can't use Corollary 5.9 here (if Ris not a eld), as the image of'need not be a direct summand of N. That is, a submodule of a free module is often not a direct summand (unless Ris a eld). We are not assuming MandNhave nite bases. Proof. The diagram k(M)k;M// ^k(') M k ' k k(N)k;N//N k EXTERIOR POWERS 19 commutes, where the top and bottom maps come from Theorem 4.2. The explicit eect in this diagram on an elementary wedge product in k(M) is given by m1^^mkk;M//_ ^k(') 2Sk(sign)m(1) m(k)_ ' k '(m1)^^'(mk)k;N//2Sk(sign)'(m(1)) '(m(k)) Sincek;Mandk;Nare injective by Theorem 4.2, and ' kis injective from our development of the tensor product (any tensor power of an injective linear map of atmodules is injective, and free modules are at), ^k(') has to be injective from commutativity of the diagram. Here is a nice application of Theorem 5.10 and the nonvanishing of k(Rn) forkn (but notk>n ). Corollary 5.11. If':Rm!Rnis a linear map, then surjectivity of 'impliesmnand injectivity of 'impliesmn. Proof. First suppose 'is onto. Taking nth exterior powers, ^n(') : n(Rm)!n(Rn) is onto by Theorem 5.4. Since n(Rn)6= 0, n(Rm)6= 0, sonm. Now suppose 'is one-to-one. Taking mth exterior powers, ^m('): m(Rm)!m(Rn) is one-to-one by Theorem 5.10, so the nonvanishing of m(Rm) implies m(Rn)6= 0, so mn. The proof of Corollary 5.11 is short, but if we unravel it we see that the injectivity part of Corollary 5.11 is a deeper result than the surjectivity, because exterior powers of linear maps don't preserve injectivity in general (Example 5.5). There will be another interesting application of Theorem 5.10 in Section 7 (Theorem 7.4). Corollary 5.12. IfMis a free module and fm1;:::;msgis a nite linearly independent subset then for any ksthes k elementary wedge products (5.3) mi1^^mikwhere 1i1<<iks are linearly independent in k(M). Proof. We have an embedding Rs,!MbyPs i=1riei7!Ps i=1rimi. SinceRsandMare free, thekth exterior power of this linear map is an embedding k(Rs),!k(M) which sends the basis ei1^^eik of k(Rs), where 1i1<<iks, to the elementary wedge products in (5.3), so they are linearly independent in k(M). Remark 5.13. In a vector space, any linearly independent subset extends to a basis. The corresponding result in Rnis generally false. In fact Z2already provides counterexamples: the vector (2 ;2) is linearly independent by itself but can't belong to a basis because a basis vector in Z2must have relatively prime coordinates. Since any linearly independent subset ofRnhas at most nterms in it, by Corollary 5.11, it is natural to ask if every maximal linearly independent subset of Rnhasnvectors in it (which need not be a basis). This is true in Z2,e.g., (2;2) is part of the linearly independent subset f(2;2);(1;0)g. However, it is nottrue in general that every maximal linearly independent subset of Rn hasnvectors in it. There are rings Rsuch thatR2contains a vector vsuch thatfvgis 20 KEITH CONRAD linearly independent (meaning it's torsion-free) but there is no linearly independent subset fv;wginR2. An example, due to David Speyer, is the following: let Rbe the ring of functions C2f(0;0)g!Cwhich coincide with a polynomial function at all but nitely many points. (The nitely many exceptional points can vary.) Letting zandwbe the coordinate functions on C2, inR2the vector ( z;w) is linearly independent and is not part of any larger linearly independent subset. 6.Determinants Now we put exterior powers to work in the development of the determinant, whose properties have up until now played no role except for a 2 2 determinant in Example 5.5. LetMbe a freeR-module of rank d1. The top exterior power d(M) is a freeR- module of rank 1, so any linear map d(M)!d(M) is scaling by an element of R. For a linear map ':M!M, the induced linear map ^d('): d(M)!d(M) is scaling by what element of R? Theorem 6.1. IfMis a freeR-module of rank d1and':M!Mis a linear map, its top exterior power ^d('): d(M)!d(M)is multiplication by det'2R. Proof. We want to show ^d(')(!) = (det')!for all!2d(M). It suces to check this when!is a basis of d(M). Lete1;:::;edbe a basis for M, soe1^^edis a basis for d(M). We will show ^d(')(e1^^ed) = (det')(e1^^ed): By denition, ^d(')(e1^^ed) ='(e1)^^'(ed): Let'(ej) =Pd i=1aijei. Then (aij) is the matrix representation for 'in the ordered basis e1;:::;edand ^d(')(e1^^ed) =dX i=1ai1ei^^dX i=1aidei =dX i1=1ai11ei1^^dX id=1aiddeid; where we introduce dierent labels for the summation indices because we are about to combine terms using multilinearity: ^d(')(e1^^ed) =dX i1;:::;id=1ai11aiddei1^^eid: In this sum, terms with equal indices can be dropped (the wedge product vanishes) so all we are left with is a sum over d-tuples of distinct indices. Since ddistinct integers from 1 todmust be 1;2;:::;d in some rearrangement, we can write i1=(1);:::;id=(d) as runs overSd: ^d(')(e1^^ed) =dX 2Sda(1)1a(d)d(e(1)^^e(d)): EXTERIOR POWERS 21 By (3.2), this becomes ^d(')(e1^^ed) =dX 2Sd(sign)a(1)1a(d)d(e1^^ed): Thus^d(') acts on the 1-element basis e1^^edof d(M) as multiplication by the number we recognize as det(( aij)>) = det(aij) = det('), so it acts on every element of d(M) as scaling by det( '). Since we did not use determinants before, we could dene the determinant of a linear operator'on a (nonzero) nite free module Mto be the scalar by which 'acts on the top exterior power of M. Then the proof of Theorem 6.1 shows det 'can be computed from any matrix representation of 'by the usual formula, and it shows this formula is independent of the choice of matrix representation for 'since our construction of exterior powers was coordinate-free. Since ^k(idM) = idk(M), the determinant of the identity map is 1. Here is a slick proof that the determinant is multiplicative: Corollary 6.2. IfMis a nonzero nite free R-module and 'and are linear maps M!M, then det( ') = det( ) det('). Proof. LetMhave rankd1. By Theorem 5.1, ^d( ') =^d( )^d('). Both sides are linear maps d(M)!d(M) and d(M) is free of rank 1. The left side is multiplication by det( '). The right side is multiplication by det( ') followed by multiplication by det( ), which is multiplication by det( ) det('). Thus, by checking both sides on a one-element basis of d(M), we obtain det( ') = det( ) det('). Continuing a purely logical development (not assuming prior knowledge of determinants, that is), at this point we could introduce the characteristic polynomial and prove the Cayley- Hamilton theorem. One of the corollaries of the Cayley-Hamilton theorem is that GL d(R) = fA2Md(R) : detA2Rg. That is used in the next result, which characterizes bases in Rdusing d(R). Corollary 6.3. LetMbe a freeR-module of rank d1. Forx1;:::;xd2M,fx1;:::;xdg is a basis of Mif and only if x1^^xdis a basis of d(M). Proof. We know by Theorem 4.2 that if fx1;:::;xdgis a basis of Mthenx1^^xdis a basis of d(M). We now want to go the other way: if x1^^xdis a basis of d(M) we showfx1;:::;xdgis a basis of M. SinceMis free of rank dit has some basis, say fe1;:::;edg. Write the xj's in terms of this basis:xj=Pn i=1aijej, whereaij2R. That means the linear map A:M!Mgiven byA(ej) =xjfor alljhas matrix representation ( aij) in the basisfe1;:::;edg. Therefore x1^^xd=Ae1^^Aed=^d(A)(e1^^ed) = (detA)(e1^^ed): Sincex1^^xdande1^^edare both bases of d(M) (the rst by hypothesis and the second by Theorem 4.2), the scalar by which they dier must be a unit: det A2R. ThereforeA2GLd(R), so thexj's must be a basis of RdbecauseA(ej) =xjand theej's are a basis. While the top exterior power of a linear operator on a nite free module is multiplication by its determinant, the lower-order exterior powers of a linear map between nite free modules have matrix representations whose entries are determinants. Let ':M!Nbe 22 KEITH CONRAD linear, with MandNnite free of positive ranks mandn, respectively. Take kmand knsince otherwise k(M) or k(N) is 0. Pick bases e1;:::;emforMandf1;:::;fnfor N. The corresponding basis of k(M) is allej1^^ejkwhere 1j1<<jkm; Similarly, the corresponding basis for k(N) is allfi1^^fikwhere 1i1<<ikn. The matrix for^k(') relative to these bases of k(M) and k(N) is therefore naturally indexed by pairs of increasing k-tuples. Theorem 6.4. With notation as above, let [']denotes the nmmatrix for'relative to the choice of bases for MandN. Relative to the corresponding bases on k(M)andk(N), the matrix entry for ^k(')in row position (i1;:::;ik)and column position (j1;:::;jk)is the determinant of the kkmatrix built from rows i1;:::;ikand columns j1;:::;jkof[']. Proof. The matrix entry in question is the coecient of fi1^^fikin the expansion of ^k(')(ej1^^ejk) ='(ej1)^^'(ejk). Details are left to the reader. In short, this says the kth exterior power of a linear map 'has matrix entries which are kkdeterminants of submatrices of the matrix of '. Example 6.5. In Example 5.3 we computed the second exterior power of the linear map R3!R3given by the 33 matrixLin (5.1). The result is a matrix of size3 2 3 2 = 33 matrix whose rows and columns are associated to basis pairs ei^ei0andej^ej0, where the ordering of the basis was e1^e2;e1^e3, ande2^e3. For instance, the upper right entry in the matrix for^2(L) is in its rst row and third column, so it is the e1^e2-coecient of ^2(L)(e2^e3). This matrix entry has row position (1 ;2) (index for the rst basis vector) and column position (2 ;3) (index for the third basis vector). The 2 2 submatrix of L using rows 1 and 2 and columns 2 and 3 is (2 0 1 1), whose determinant is 2, which matches the upper right entry in the matrix at the end of Example 5.3. 7.Exterior powers and linear independence This section discusses the connection between elementary wedge products and linear independence in a free module. We will start o with vector spaces, which are easier to handle. If we are given nvectors in Rn, there are two ways to determine if they are linearly independent using the nnmatrix with the vectors as the columns. The rst way is to row reduce the matrix and see if you get the identity matrix. The second way is to compute the determinant of the matrix and see if you get a nonzero value. How can we decide if a set of k<n vectors in Rnis linearly dependent? Again there are two ways, each generalizing one of the previous two methods in terms of the nkmatrix having the vectors as the columns. The rst way is to row reduce the matrix to see if a kksubmatrix is the identity. The second way is to compute the determinants of all kksubmatrices and see if any of them is not 0. Whereas the rst way (row reduction) provides a set of steps that always keeps you going in the right direction, the second way involves one determinant computation after the other, and that will take longer to complete (particularly if all the determinants turn out to be 0!). Using exterior powers, we can carry out all these determinant computations at once. That is the algorithmic content of the next theorem for nite-dimensional vector spaces, if you keep in mind how determinants are related to coecients in wedge product expansions. EXTERIOR POWERS 23 Theorem 7.1. LetVbe a vector space. The vectors v1;:::;vkinVare linearly independent if and only if v1^^vk6= 0ink(V). Proof. The casek= 1 is trivial, so take k2. First assume fv1;:::;vkgis a linearly independent set. This set extends to a basis of V(we are working over a eld!), so v1^^vk is part of a basis of k(V) by Theorem 4.2. In particular, v1^^vk6= 0 in k(V). Now supposefv1;:::;vkgis linearly dependent, so one viis a linear combination of the others. Whether or not v1^^vkis nonzero in k(V) is independent of the order of the factors, since permuting the terms only changes the elementary wedge product by a sign, so we may suppose vkis a linear combination of the rest: vk=c1v1++ck1vk1: Then in k(V), v1^^vk1^vk=v1^^vk1^ k1X i=1civi! : Expanding the right side gives a sum of k1 wedge products, each containing a repeated vector, so every term vanishes. In this theorem Vcan be an innite-dimensional vector space since we never required nite-dimensionality in the proof. At one point we invoked Theorem 4.2, which was proved for all free modules, not just free modules with a nite basis. Corollary 7.2. IfVis a vector space and v1;:::;vkare linearly independent in V, then an element v2Vis a linear combination of v1;:::;vkif and only if v1^^vk^v= 0in k+1(V). Proof. Because the vi's are linearly independent, vis a linear combination of them if and only iffv1;:::;vk;vgis linearly dependent, and that is equivalent to their wedge product vanishing by Theorem 7.1. Example 7.3. InR3, does the vector v= (4;1;1) lie in the span of v1= (2;1;3) and v2= (1;2;4)? Sincev1andv2are linearly independent, vis in their span if and only if v1^v2^vvanishes in 3(R3). Lete1;e2;e3be the standard basis of R3. Then we compute v1^v2= (2e1+e2+ 3e3)^(e1+ 2e2+ 4e3) = 4(e1^e2) + 8(e1^e3)e1^e2+ 4(e2^e3) 3(e1^e3)6(e2^e3) = 3(e1^e2) + 5(e1^e3)2(e2^e3) so v1^v2^v= (3(e1^e2) + 5(e1^e3)2(e2^e3))^(4e1e2+e3) = 3(e1^e2^e3)5(e1^e3^e2)8(e2^e3^e1) = 3(e1^e2^e3) + 5(e1^e2^e3)8(e1^e2^e3) = 0: Since this vanishes, (4 ;1;1) is in the span of (2 ;1;3) and (1;2;4). This method does not explicitly represent (4 ;1;1) as a linear combination in the span of (2;1;3) and (1;2;4). (Here is one: (4 ;1;1) = 3(2;1;3)2(1;2;4).) On the other hand, if we are only concerned with an existence question (is it in the span, not how is it in the 24 KEITH CONRAD span) then this procedure works just as the computation of a determinant does to detect invertibility of a matrix without providing a formula for the inverse matrix. There is a generalization of Theorem 7.1 to describe linear independence in a free R- moduleMrather than in a vector space. Given m1;:::;mkinM, their linear independence is equivalent to a property of m1^^mkin k(M), but the property is notthe nonvanishing of this elementary wedge product: Theorem 7.4. LetMbe a freeR-module. Elements m1;:::;mkinMare linearly indepen- dent inMif and only if m1^^mkink(M)is torsion-free: for r2R,r(m1^^mk) = 0 only whenr= 0. WhenRis a eld, we recover Theorem 7.1. What makes Theorem 7.4 more subtle than Theorem 7.1 is that a linearly independent set in a free module usually does not extend to a basis, so we can't blindly adapt the proof of Theorem 7.1 to the case of free modules. Proof. The casek= 1 is trivial by the denition of a linearly independent set in a module, so we can take k2. Assumem1;:::;mkis a linearly independent set in M. By Corollary 5.12, the elementary wedge product m1^^mkis a linearly independent one-element subset in k(M), so m1^^mkhas noR-torsion. If themi's are linearly dependent, say r1m1++rkmk= 0 withri2Rnot all 0, we may re-index and assume r16= 0. Then 0 = (r1m1++rkmk)^m2^^mk=r1(m1^m2^^mk); som1^m2^^mkhasR-torsion. Corollary 7.5. A system of dequations in dunknowns a11x1++a1dxd= 0 a21x1++a2dxd= 0 ... ad1x1++addxd= 0 in a commutative ring Rhas a nonzero solution x1;:::;xd2Rif and only if det(aij)is a zero divisor in R. Proof. We rewrite the theorem in terms of matrices: for A2Md(R), the equation Av= 0 has a nonzero solution v2Rdif and only if det Ais a zero divisor in R. We consider the negated property, that the only solution of Av= 0 isv= 0. This is equivalent to the columns of Abeing linearly independent. The columns are Ae1;:::;Aed, and their elementary wedge product is Ae1^^Aed= (detA)e1^^ed2d(Rd): Sincee1^^edis a basis of d(Rd) as anR-module, (det A)e1^^edis torsion-free if and only if the only solution of rdetA= 0 isr= 0, which means det Aisnota zero divisor. ThusAv= 0 has a nonzero solution if and only if det Ais a zero divisor. We can turn Theorem 7.4 into a characterization of linear independence of kvectors v1;:::;vkinRdwhenkd.3The test is that v1^^vkis torsion-free in k(Rd). If we 3We may as well let kd, since a linearly independent subset of Rdhas at most dterms by Corollary 5.11. EXTERIOR POWERS 25 letA2Mdk(R) be the matrix whose columns are v1;:::;vk, linear independence of the columns is the same thing as injectivity of Aas a linear map Rk!Rd. So we can now say, forkd, when a matrix in M dk(R) is injective as a linear transformation: if and only if the elementary wedge product of its columns is torsion-free in k(Rd). What does that mean concretely? Writing vj=Pd i=1aijeiusing the standard basis e1;:::;edofRd, v1^^vk=X 1i1<<ikda1i1a1ik......... aki1akikei1^^eik: This is torsion-free precisely when the coecients are not all killed by a common nonzero element of R. Therefore we get the rule: for kd,A2Mdk(R) is injective as a linear mapRk!Rdif and only if the determinants of its kksubmatrices have no common nonzero annihilator in R. Example 7.6. Let A=0 @2 2 1 5 1 21 A2M32(Z=6Z): Its 22 submatrices have determinants 8 ;3;2, which equal 2 ;3;2 inZ=6Z. Although none of these determinants is a unit in Z=6Z, which would be an easy way to see injectivity, Ais still injective because 2 and 3 have no common nonzero annihilator in Z=6Z(that is, 2r= 0 and 3r= 0 only for r= 0). In the terminology of linear independence, we showed the two columns of Aare linearly independent in ( Z=6Z)3. (This does notmean neither column is a scalar multiple of the other, but it means the stronger assertion that no linear combination of the columns is 0 except for the trivial combination with both coecients equal to 0.) We can also check this in 2((Z=6Z)3): the two columns of Aare 2e1+e2+e3and 2e1+ 5e2+ 2e3, and (2e1+e2+e3)^(2e1+ 5e2+ 2e3) = 2e1^e2+ 2e1^e3+ 3e2^e3; which is torsion-free since 2 r= 0 and 3r= 0 in Z=6Zonly forr= 0. That the coecients in the elementary wedge product match the determinants of the 22 submatrices illustrates that the rules about elementary wedge product computations really do encode everything about determinants of submatrices. I learned the following neat use of Corollary 7.5 from [2, pp. 6{7]. Theorem 7.7. LetMbe anR-module admitting a linear injection Rd,!Mand a linear surjectionRdM. ThenM=Rd. Proof. The hypotheses say Mhas ad-element spanning set and a d-element linearly inde- pendent subset. Call the former x1;:::;xdand the latter e1;:::;ed, so we can write M=Rx1++Rxd; ei=dX j=1aijxj: Thexj's spanM. We want to show they are linearly independent, so they form a basis and Mis free of rank d. 26 KEITH CONRAD The expression of the ei's in terms of the xj's can be written as the vector-matrix equation (7.1)0 B@e1 ... ed1 CA= (aij)0 B@x1 ... xd1 CA; where we treat ( aij)2Md(R) as a matrix acting on the d-tuples inMd. We show by contradiction that := det( aij) is not a zero divisor in R. If is a zero divisor then the columns of ( aij) are linearly dependent by Corollary 7.5. A square matrix and its transpose have the same determinant, so the rows of ( aij) are also linearly dependent. That means there are c1;:::;cd2Rnot all 0 such that (c1;:::;cd)(aij) = (0;:::; 0): Using this, we multiply both sides of (7.1) on the left by ( c1;:::;cd) to get (c1;:::;cd)0 B@e1 ... ed1 CA= 0: That saysc1e1++cded= 0, which contradicts linear independence of the ei's. So is not a zero divisor. Suppose now that a1x1++adxd= 0. We want to show every aiis 0. Multiply both sides of (7.1) on the left by the cofactor matrix for ( aij) (that's the matrix you multiply by to get the scalar diagonal matrix with the determinant along the main diagonal): cof(aij)0 B@e1 ... ed1 CA= 0 B@x1 ... xd1 CA: Now multiply both sides of this equation on the left by ( a1;:::;ad): (a1;:::;ad) cof(aij)0 B@e1 ... ed1 CA= (a1x1++adxd) = 0: The product ( a1;:::;ad) cof(aij) is a (row) vector, say ( b1;:::;bd). Thenb1e1++bded= 0, so everybiis 0 by linear independence of the ei's. Therefore ( a1;:::;ad) cof(aij) = (0;:::; 0). Multiply both sides of this on the right by (aij) to get (a1;:::;ad) = (0;:::; 0), so ai= 0 for alli. Since is not a zero divisor, every aiis 0 and we are done. 8.The wedge product By concatenating elementary wedge products, we introduce a multiplication operation between dierent exterior powers of a module. Lemma 8.1. Iff:MM|{z} k timesN1N`is multilinear and is alternating in the M's, there is a unique multilinear map ef: k(M)N1N`!Nsuch that ef(m1^^mk;n1;:::;n`) =f(m1;:::;mk;n1;:::;n`): EXTERIOR POWERS 27 Proof. Uniqueness of effollows from multilinearity. To prove existence of ef, x a choice of n12N1;:::;n`2N`. Denefn1;:::;n`:Mk!Nby fn1;:::;n`(m1;:::;mk) =f(m1;:::;mk;n1;:::;n`): Sincefis multilinear, fn1;:::;n`is multilinear. Since fis alternating in the mi's,fn1;:::;n`is an alternating map. Therefore there is a unique linear map efn1;:::;n`: k(M)!Nsuch that efn1;:::;n`(m1^^mk) =fn1;:::;n`(m1;:::;mk) =f(m1;:::;mk;n1;:::;n`): on elementary wedge products. Deneef: k(M)N1N`!Nbyef(!;n 1;:::;n`) =fn1;:::;n`(!). Since each fn1;:::;n`is linear,efis linear in its rst component. To check efis linear in one of its other coordinates, we carry it out for n1(all the rest are similar). We want to verify that ef(!;n 1+n0 1;n2;:::;n`) =ef(!;n 1;n2;:::;n`) +ef(!;n0 1;n2;:::;n`) and ef(!;rn 1;n2;:::;n`) =ref(!;n 1;n2;:::;n`): Both sides of both equations are additive in !, so it suces to check the identity when !=m1^^mkis an elementary wedge product. In that case the two equations turn into linearity of fin itsN1-component, and that is just a special case of the multilinearity off. Theorem 8.2. LetMbe anR-module. For positive integers kand`, there is a unique R-bilinear map k(M)`(M)!k+`(M)satisfying the rule (m1^^mk;m0 1^^m0 `)7!m1^^mk^m0 1^^m0 ` on pairs of elementary wedge products. Proof. Since the elementary wedge products span each exterior power module, and a bilinear map is determined by its values on pairs coming from spanning sets, there is at most one bilinear map with the prescribed behavior. As usual in this game, what needs proof is the existence of such a map. Start by backing up and considering the function f:MkM`!k+`(M) by f(m1;:::;mk;m0 1;:::;m0 `) =m1^^mk^m0 1^^m0 `: This is multilinear and alternating in the rst k-coordinates, so by Lemma 8.1 there is a multilinear map ef: k(M)M`!k+`(M) such that ef(m1^^mk;m0 1;:::;m0 `) =m1^^mk^m0 1^^m0 `: and it is alternating in its last `coordinates. Therefore, again by Lemma 8.1, there is a bilinear map B: k(M)`(M)!k+`(M) such that B(!;m0 1^m0 `) =ef(!;m0 1;:::;m0 `); so B(m1^^mk;m0 1^m0 `) =ef(m1^^mk;m0 1;:::;m0 `) =m1^^mk^m0 1^^m0 `: We have produced a bilinear map on the kth and`th exterior powers with the desired value on pairs of elementary wedge products. 28 KEITH CONRAD The operation constructed in Theorem 8.2, sending k(M)`(M) to k+`(M), is denoted^and is called the wedge product . We place the operation in between the elements it acts on, just like other multiplication functions in mathematics. So for any !2k(M) and2`(M) we have an element !^2k+`(M), and this operation is bilinear in ! and. The formula in Theorem 8.2 on two elementary wedge products looks like this: (m1^^mk)^(m0 1^^m0 `) =m1^mk^m0 1^^m0 `: Notice we have given a newmeaning to the notation ^and to the terminology \wedge prod- uct," which we have until now used in a purely formal way always in the phrase \elementary wedge product." This new operational meaning of the wedge product is consistent with the old formal one, e.g.the (new) wedge product operation ^: 1(M)1(M)!2(M) sends (m;m0) tom^m0(old notation). This is very much like the denition of R[T] as formal nite sumsP iaiTiwhere, after the ring operations are dened, the symbol Tiis recognized as thei-fold product of the element T. We have dened a wedge product k(M)`(M)!k+`(M) whenkand`are positive. What if one of them is 0? Recall (by denition) 0(M) =R. Theorem 8.2 extends to the casek= 0 or`= 0 if we let the maps 0(M)`(M)!`(M) and k(M)0(M)! 0(M) be scalar multiplication: r^=rand!^r=r!. Now we can think about the old notation m1^^mkin an operational way: it is the result of applying the wedge product operation ktimes with elements from the module 1(M) =M. Since we are now able to speak about a wedge product !1^^!kwhere the !i's lie in exterior power modules ki(M), the elementary wedge products m1^^mk, where the factors are in M, are merely special cases of wedge products. Actually, to speak of !1^^!kunambiguously we need ^to be associative! So let's check that. Theorem 8.3. The wedge product is associative: if !2a(M),2b(M), and2c(M) then (!^)^=!^(^)ina+b+c(M). Proof. This is easy if a,b, orcis zero (then one of !,, andis inR, and wedging with R is just scalar multiplication), so we can assume a,b, andcare all positive. Letfandgbe the functions from a(M)b(M)c(M) to a+b+c(M) given by both choices of parentheses: f(!;; ) = (!^)^; g (!;; ) =!^(^): (Note!^2a+b(M) and^2b+c(M).) Since the wedge product on two exterior power modules is bilinear, fandgare both trilinear functions. Therefore to show f=git suces to verify equality on triples of elementary wedge products, since they are spanning sets of a(M), b(M), and c(M). On such elements the equality is obvious by the denition of the wedge product operation, so we are done. The following theorem puts associativity of the wedge product to work. Theorem 8.4. LetMbe anR-module. If some i(M)is nitely generated, then j(M) is nitely generated for all ji. EXTERIOR POWERS 29 The important point here is that we are not assuming Mis nitely generated, only that some exterior power is nitely generated. Because exterior powers are not dened recursively, this theorem is not a tautology.4 Proof. It suces to show that if i(M) is nitely generated then i+1(M) is nitely gener- ated. Our argument is a simplication of [3, Lemma 2.1]. The module i(M) has a nite spanning set, which we can take to be a set of elementary wedge products. Let x1;:::;xp2Mbe the terms appearing in those elementary wedge products, so i(M) is spanned by the i-fold wedges of x1;:::;xp. We will show i+1(M) is spanned by the ( i+ 1)-fold wedges of x1;:::;xp. It suces to show the span of these wedges contains all the elementary wedge products in i+1(M). Choose (i+ 1)-fold elementary wedge product, say y1^^yi^yi+1= (y1^^yi)^yi+1: Sincey1^^yiis in i(M), it is anR-linear combination of i-fold wedges of x1;:::;xp. Thereforey1^^yi^yi+1is anR-linear combination of expressions (xj1^^xji)^yi+1=xj1^(xj2^^xji^yi+1); where the equation uses associativity of the wedge product. Since xj2^^xji^yi+1is in i(M), it is anR-linear combination of i-fold wedges of x1;:::;xp. Theorem 8.5. IfMis spanned as an R-module by x1;:::;xdthen for 1kdevery element of k(M)is a sum !1^x1+!2^x2++!d^xd for some!i2k1(M). We don't consider k>d in this theorem since k(M) = 0 for such kby Theorem 4.1. Proof. The result is clear when k= 1 since 0(M) =Rby denition, so we can assume k2. From the beginning of Section 4, k(M) is spanned as an R-module by the k-fold elementary wedge products xi1^^xikwhere 1i1<< ikd. Using the wedge product multiplication k1(M)M!k(M), we can write xi1^^xik=^xik; where=xi1^^xik12k1(M). Therefore every element of k(M) is anR-linear combination r1(1^x1) ++rd(d^xd), whereri2Randi2k1(M). Since ri(i^xi) = (rii)^xiwe can set!i=riiand we're done. Here is an analogue of Theorem 4.8 for linear maps. Theorem 8.6. Let':M!Nbe a linear map of R-modules. If^i(') = 0 for someithen ^j(') = 0 for allji. Theorem 4.8 is the special case when N=Mand'= idM(why?). 4The theorem doesn't go backwards down to M,i.e., if some i(M) is nitely generated this does not imply Mis nitely generated. For example, 2(Q=Z) = 0 as a Z-module, but Q=Zis not nitely generated as aZ-module. 30 KEITH CONRAD Proof. It suces so show ^i+1(') = 0. Since^i(') = 0 for any m1;:::;miinMwe have 0 =^i(')(m1^^mi) ='(m1)^^'(mi) in i(N). For anym1;:::;mi+1inM, in i+1(N) we have ^i+1(')(m1^^mi^mi+1) ='(m1)^^'(mi)^'(mi+1) = ('(m1)^^'(mi))^'(mi+1) = 0^'(mi+1) = 0: Such terms span the image of ^i+1('), so^i+1(') = 0. In an elementary wedge product m1^^mk, where the factors are in M= 1(M), transposing two of them introduces a sign change. What is the sign-change rule for trans- posing factors in a wedge product !1^^!k? By associativity, we just need to understand how a single wedge product !^changes when the factors are reversed. Theorem 8.7. For!2k(M)and2`(M),!^= (1)k`^!. Proof. This is trivial if k= 0 or`= 0 (in which case the wedge product is simply scaling and (1)k`= 1), so we can take kand`positive. Both sides of the desired equation are bilinear functions of !and, so to verify equality for all!andit suces to do so on spanning sets of the modules. Thus we can take !=m1^^mkand=m0 1^^m0 `, so we want to show (8.1) m1^^mk^m0 1^^m0 `= (1)k`(m0 1^^m0 `^m1^^mk): Starting with the expression on the left, we successively move m0 1;m0 2;:::;m0 `to the front. First move m0 1past each of the mi's, which is a total of kswaps, so m1^^mk^m0 1^^m0 `= (1)km0 1^m1^^mk^m0 2^^m0 `: Now movem0 2past everymi, introducing another set of ksign changes: m1^^mk^m0 1^^m0 `= (1)2km0 1^m0 2^m1^^mk^m0 3^^m0 `: Repeat until m0 `has been moved past every mi. In all, there are k`swaps, so the overall sign at the end is ( 1)k`. Theorem 8.8. For oddkand!2k(M),!^!= 0. Proof. There is a quick proof using Theorem 8.7 when 2 2R:!^!= (1)k2(!^!) = (!^!), so 2(!^!) = 0, so!^!= 0. To handle the general case when 2 may not be a unit, write !=X i1;:::;ikci1;:::;ikmi1^^mik=X IcI!I; whereI= (i1;:::;ik) is a multi-index of kintegers and !I=!(i1;:::;ik)is an abbreviation formi1^^mik. Then, by the bilinearity of the wedge product, (8.2) !^!=X IcImI^X JcJ!J=X I;JcIcJ!I^!J; EXTERIOR POWERS 31 whereIandJrun over the same set of multi-indices. Note each !I^!Jis an elementary wedge product with 2 kfactors. The multi-indices IandJcould be equal. In that case !I^!J= 0 since it is a 2 k-fold elementary wedge product with repeated factors. When I6=Jthe double sum in (8.2) contains cIcJ!I^!J+cJcI!J^!I=cIcJ(!I^!J+!J^!I): Since!Iand!Jare in k(M),!J^!I= (1)k2(!I^!J) =(!I^!J), socIcJ(!I^!J+ !J^!I) = 0. What about Theorem 8.8 when kis even? If !=m1^^mkis an elementary wedge product in k(M) then!^!vanishes since it is an elementary wedge product with a repeated factor from M. But it is not generally true that !^!= 0 for all!2k(M). Example 8.9. Fork2, letMbe nite free with linearly independent subset e1;:::;e 2k. Set!=e1^^ek+ek+1^^e2k2k(M). This is a sum of two elementary wedge products, and !^!= 2e1^^ek^ek+1^^e2k22k(M): By Corollary 5.12 and Theorem 7.4, e1^^ek^ek+1^^e2kis torsion-free in 2k(M), so when 26= 0 inRwe have!^!6= 0. Elementary wedge products always \square" to 0, so!isnotan elementary wedge product when 2 6= 0 inR. To get practice computing in an exterior power module, we look at the equation v^!= 0 in k+1(V), whereVis a vector space and !2k(V). Theorem 8.10. LetVbe a vector space. For nonzero !2k(V), dim(fv2V:v^!= 0g)k; with equality if and only if !is an elementary wedge product. Proof. The result is obvious if k= 0, so we can suppose k1. Letv1;:::;vdbe linearly independent vectors in Vthat each wedge !to 0. We want to show dk. SinceVmight be innite-dimensional, we rst create a suitable nite-dimensional subspace Win which we can work. (If you want to assume Vis nite-dimensional, set W=Vand skip the rest of this paragraph.) Let Wbe the span of v1;:::;vdand the nonzero vectors appearing in some xed representation of !as a nite sum of elementary wedge products in k(V). Then Wis nite-dimensional. There's a natural embedding W ,!Vand we get an embedding `(W),!`(V) in a natural way for all `by Corollary 5.9. If we view each viinWand! in k(W) then the condition vi^!= 0 in k+1(V) impliesvi^!= 0 in k+1(W). Setn= dimW, so obviously dn. Ifknthen obviously dk, so we can suppose kn1. Extendfv1;:::;vdgto a basisfv1;:::;vngofW. Using this basis we can write !in k(W) as a nite sum of linearly independent elementary wedge products: !=X 1i1<<iknci1;:::;ikvi1^^vik 32 KEITH CONRAD and some coecient is not 0. Fix ibetween 1 and d, and compute vi^!using this formula: 0 =vi^! =X 1i1<<iknci1;:::;ikvi^vi1^^vik =X 1i1<<ikn i1;:::;ik6=ici1;:::;ikvi^vi1^^vik: This equation is taking place in k+1(W), where the ( k+ 1)-fold wedges vi^vi1^^vik fori62fi1;:::;ikgare linearly independent. Therefore the coecients here are all 0: i62fi1;:::;ikg)ci1;:::;ik= 0: Hereiwas any number from 1 to d, so ci1;:::;ik6= 0)f1;:::;dgfi1;:::;ikg: There is at least one nonzero coecient, so we must have f1;:::;dgfi1;:::;ikgfor some k-tuple of indices. Counting the two sets, dk. Ifd=kthenf1;:::;kg=fi1;:::;ikg, which allows just one nonzero term and != c1;:::;kv1^^vk, which is an elementary wedge product. Conversely, if !is a nonzero elementary wedge product in k(V) thenfv2V:!^v= 0ghas dimension kby Corollary 7.2 and associativity of the wedge product. Theorem 8.11. LetVbe a vector space and k1. For nonzero v2Vand!2k(V), v^!= 0if and only if !=v^for some2k1(V). Proof. By associativity, if !=v^thenv^!=v^(v^) = (v^v)^= 0. The point of the theorem is that the converse direction holds: if v^!= 0 then we can write !=v^ for some. As in the proof of the previous theorem, we can reduce to the nite-dimensional case (details left to the reader), so we'll just take Vto be a nite-dimensional vector space. Set n= dimV1. Since!2k(V), ifk>n then!= 0 and we can trivially write !=v^0. So we may suppose kn. Ifk= 0 then!20(V) is a scalar and wedging with !is scalar multiplication, so the conditions v6= 0 andv^!= 0 imply!= 0. Thus again !=v^0. Now suppose 1kn. Extendvto a basis of V, sayv1;:::;vnwherev=v1. Ifk=n the condition v^!= 0 is automatic since n+1(V) = 0. And it is also automatic that ! is \divisible" by vifk=n: n(V) has basis v1^^vn, so!=c(v1^^vn) for some scalarc. Then!=v^where=cv2^^vn. We now assume 1 kn1 (son2). Using the basis of k(V) coming from our chosen basis of Vthat includes vas the rst member v1, (8.3) !=X 1i1<<iknci1;:::;ikvi1^^vik with scalar coecients. Since v1=v,v^v1= 0 so (8.4) v^!=X 2i1<<iknci1;:::;ikv^vi1^^vik: The elementary wedge products on the right are part of a basis of k+1(V), so fromv^!= 0 we see all the coecients in (8.4) vanish. Thus in (8.3) the only nonzero terms are among EXTERIOR POWERS 33 those with i1= 1, so we can pull out v1=v: !=v^X 1<i2<<iknc1;i2;:::;ikvi2^^vik: Letbe the large sum here, so !=v^. 9.The exterior algebra Since wedge products move elements into higher-degree exterior powers, we can view ^as multiplication in a noncommutative ring by taking the direct sum of all the exterior powers of a module. Denition 9.1. For anR-moduleM, its exterior algebra is the direct sum (M) =M k0k(M) =RM2(M)3(M); provided with the multiplication rule given by the wedge product from Theorem 8.2, ex- tended distributively to the whole direct sum. Each k(M) is only an R-module, but their direct sum ( M) is anR-algebra: it has a multiplication which commutes with scaling by Rand has identity (1 ;0;0;:::). NoticeR is a subring of ( M), embedded as r7!(r;0;0;:::). When k(M) is viewed in the exterior algebra ( M), it is called a homogeneous part and its elements are said to be the homogeneous terms of degreekin (M). For instance, whenm0,m1,m2,m3, andm4are inMthe summ0+m1^m2+m3^m4in (M) has homogeneous parts m0andm1^m2+m3^m4. This is the same terminology used for homogeneous multivariable polynomials, e.g.,X2XY+Y3has homogeneous parts X2XY(of degree 2) and Y3(of degree 3). The wedge product on the exterior algebra ( M) is bilinear, associative, and distributive, but not commutative (unless 1 = 1 inR). The replacement for commutativity in Theorem 8.7 does notgeneralize to a rule between all elements of ( M). However, if !2k(M) and kis even then !commutes with every element of ( M), so more generally the submoduleL kevenk(M) lies in the center of ( M). A typical element of ( M) is a sequence ( !k)k0with!k= 0 fork0, and we write it as a formal sumP k0!kwhile keeping the direct sum aspect in mind. In this notation, the wedge product of two elements of ( M) is X k0!k^X `0`=X p00 @X k+`=p!k^`1 A; where the inner sum on the right is actual addition in p(M) and the outer sum is purely formal (corresponding to the direct sum decomposition dening ( M)). This extension of the wedge product from operations k(M)`(M)!k+`(M) on dierent exterior power modules to a single operation ( M)(M)!(M) is analogous to the way the multiplication rule ( aTi)(bTj) =abTi+jon monomials, which is associative, can be extended to the usual multiplication between any two polynomials in R[T] =L i0RTi. WhenM is a nitely generated R-module with ngenerators, every k(M) is nitely generated as an R-module and k(M) = 0 fork > n so (M) =Ln k=0k(M) is nitely generated as an R-module. Because k(M) is spanned as an R-module by the elementary wedge products, 34 KEITH CONRAD and an elementary wedge product is a wedge product of elements of M, (M) is generated as anR-algebra (not as an R-module!) by M. Theorem 9.2. WhenMis a freeR-module of rank d, its exterior algebra is a free R-module of rank 2d. Proof. Each exterior power module k(M) is free with rankd k for 0kdand vanishes fork>d , so their direct sum ( M) is free with rank dX k=0d k = 2d: Concretely, when Mis free with basis e1;:::;ed, we can think of ( M) as anR-algebra generated by the ei's subject to the relations e2 i= 0 andeiej=ejeifori6=j. The construction of ( M) was basis-free, but this explicit description when there is a basis is helpful when doing computations. Example 9.3. LetVbe a real vector space of dimension 3 with basis e1;e2;e3. Then 0(V) =R, 1(V) =V=Re1Re2Re3, 2(V) =R(e1^e2)R(e1^e3)R(e2^e3), and 3(V) =R(e1^e2^e3). The exterior algebra ( V) is the direct sum of these vector spaces and we can count the dimension as 1 + 3 + 3 + 1 = 8. To get practice computing in an exterior algebra, we ask which elements of the exterior algebra of a vector space wedge a given vector to 0. Theorem 9.4. LetVbe a vector space over the eld K. For nonzero v2Vand!2(V), v^!= 0if and only if !=v^for some2(V). Proof. The reduction to the case of nite-dimensional Vproceeds as in the reduction step of the proof of Theorem 8.11. By associativity, if !=v^thenv^!=v^(v^) = (v^v)^= 0. The point of the theorem is that the converse direction holds. First we reduce to the nite-dimensional case. Let Wbe the span of vand all the nonzero elementary wedge products in an expression for !. SinceW ,!V, k(W),!k(V), so (W),!(V). From these embeddings, it suces to prove the theorem with Vreplaced by W, so we may assumeVis nite-dimensional. Write!=Pn k=0!kwhere!k2k(V) is the degree kpart of!. (If you think about direct sums as sequences, != (!0;:::;!n).) Then v^!=nX k=0v^!k: Sincev^!k2k+1(V), the terms in the sum are in dierent homogeneous parts of ( V), so the vanishing of v^!impliesv^!k= 0 for each k. By Theorem 8.11, !0= 0 and fork1 we have !k=v^k1for somek12k1(V). Thus!=v^where =Pn k=1k12(V). While we created the exterior algebra ( M) as a direct sum of R-modules with a snazzy multiplicative structure, it can be characterized on its own terms among R-algebras by a universal mapping property. Since each k(M) is spanned as an R-module by the ele- mentary wedge products m1^^mk, (M) is generated as an R-algebra (using wedge EXTERIOR POWERS 35 multiplication) by M. Moreover, m^m= 0 in (M) for allm2M. We now turn this into a universal mapping property. Theorem 9.5. LetAbe anyR-algebra and suppose there is an R-linear map L:M!A such thatL(m)2= 0for allm2M. Then there is a unique extension of Lto anR-algebra mapeL: (M)!A. That is, there is a unique R-algebra map eLmaking the diagram (M) eL ""M? OO L//A commute. This theorem is saying that any R-linear map fromMto anR-algebraAsuch that the image elements square to 0 can always be extended uniquely to an R-algebra map from (M) toA. Such a universal property determines ( M) up toR-algebra isomorphism by the usual argument. Proof. SinceMgenerates ( M) as anR-algebra, there is at most one R-algebra map (M)!Awhose values on Mare given by L. The whole problem is to construct such a map. For anymandm0inM,L(m)2= 0,L(m0)2= 0, andL(m+m0)2= 0. Expanding (L(m) +L(m0))2and removing the squared terms leaves 0 =L(m)L(m0) +L(m0)L(m); which shows L(m)L(m0) =L(m0)L(m). Therefore any product of L-values onMcan be permuted at the cost of an overall sign change. This implies L(m1)L(m2)L(mk) = 0 if twomi's are equal, since we can permute the terms to bring them together and then use the vanishing of L(mi)2. This will be used later. If there is going to be an R-algebra map f: (M)!AextendingL, then on an elemen- tary wedge product we must have f(m1^^mk) =f(m1)f(mk) =L(m1)L(mk): since^is the multiplication in ( M). To show there is such a map, we start on the level of the k(M)'s. Fork0, letMk!Aby (m1;:::;mk)7!L(m1)L(mk). This is multilinear since LisR-linear. It is alternating because L(m1)L(mk) = 0 when two mi's are equal. Hence we obtain an R-linear map fk: k(M)!Asatisfying fk(m1^^mk) =L(m1)L(mk) for all elementary wedge products m1^^mk. Denef: (M)!Aby letting it be fkon k(M) and extending to the direct sum ( M) =L k0k(M) by additivity. This functionfisR-linear and it is left to the reader to show fis multiplicative. Notice the individual k(M)'s don't appear in the statement of Theorem 9.5. This theorem describes an intrinsic feature of the full exterior algebra as an R-algebra. 36 KEITH CONRAD References [1] W. C. Brown, \A Second Course in Linear Algebra," J. Wiley & Sons, New York, 1988. [2] H. Flanders, Tensor and Exterior Powers , J. Algebra 7, 1{24 (1967). [3] R. Gardner, Modules Admitting Determinants , Linear and Multilinear Algebra 3(1975/76), 209{214. [4] J. Rotman, \Advanced Modern Algebra," Prentice-Hall, Upper Saddle River, NJ, 2002. [5] F. Warner, \Foundations of Dierentiable Manifolds and Lie Groups," Springer-Verlag, New York, 1983.