failure of Xn with non-singular matrices

Failure of Xn with non-singular matrices PhL 2.6.13 This was an interesting idea for Chapter 10 that blew up, but record it here anyway. ___________________________________________________________________________________ As our next step, we define Xm to be the restriction of Xm such that all non-zero matrices must have a non-vanishing determinant. Since Xm has pm elements, it seems clear that Xm is going to have some smaller number of elements since we require det(X) ≠ 0 for any X in Xm. A 2x2 matrix might have the form and then det(X) = 0 means xt-yz = 0 or f(x,y,z,t) = 0. For real continuous variables, g(x,y,z) = 0 defines a 2D surface in 3D space. Similarly, the determinant condition f(x,y,z,t) = 0 defines a 3D surface in a 4D space. Points on this surface have det(X) = 0. In general the condition det(X) = 0 specifies an m2-1 dimensional surface in the m2 dimensional space of the matrix elements. Since the elements of an Xm matrix are in Zp, the equation det(X) = 0 defines some kind of "digital surface" which touches a certain number of points and the matrices corresponding to those points must be excluded from Xm . We shall not attempt to calculate the number of matrices in Xm but we know it is some number smaller than pm but probably not too much smaller than pm. For the case p = 3 and m = 3, there are 39 = 19,683 matrices. Of these, exactly 6,891 have det(X) = 0, including the zero matrix, so about 35% of the matrices of Xm are not present in Xm. Fact: The ring Xm of mxm non-singular matrices having elements in Zp is a field. [ wrong! ] Test: Since Xm is I claim a field, and since it has a finite number of elements, it must be some Galois field! That means there must be some prime p and integer m such that pm = 19,683 – 6,891 + 1 = 12,793 Remember that the 6891 includes the all-zero matrix 0, so I want to add it back in. But here is how Maple factors that number 12,973 = 11 * 1163 so something is wrong in the machine!!! Proof: The only item missing from the ring-with-identity Xm is the multiplicative inverse. By requiring det ≠ 0 for matrices in Xm, we assure that an inverse for every non-zero element Xm exists. We have closure because if A = BC then det(A) = det(B)det(C) so if B and C are non-singular, A is also non-singular. But how about additive closure!! If A = B + C then we get det(A) = det(B+C). But suppose I select C = - B . The both det(C) and det(B) ≠ 0, but det(B+C) = 0 so things are NOT closed, the whole thing goes blewie! Fine by me. That explains why the numbers above don't work, and it was amusing to do this Maple calculation.