On 10/27/08, Eugene Salamin <gene_salamin@yahoo.com> wrote:
I rather like the discussion of alternate bilinear forms and Pfaffians in Nathan Jacobson's "Basic Algebra", vol. 1. ... Thus det B is a square, but this is as an element of the field.
A remarkably roundabout description of what is essentially just common-or-garden Jordan-style elimination!
However, the matrix whose determinant we wish to show is a square is one in which B[i,j] = x[i,j]. Let R be the ring Z[x[i,j]] of polynomials with integer coefficients in the (2n)(2n-1)/2 indeterminates x[i,j], i < j. Let F = Q(x[i,j]) be the fraction field of R. We want to show that det B is a square in R, but so far we only know that det B is a square in F, i.e. there are polynomials f and g in R such that det B = (f / g)^2. ...
Good point. An sledgehammer which cracks this --- at the cost of some combinatorial reasoning --- is to demonstrate the Pfaffian polynomial explicitly, in the same way as the determinant familiarly in n! terms. The expansion formula (paraphrased from Wikipedia is: Pf(B) = \sum_p parity(p) \product_i B[p(2i), p(2i+1)], where B[i,j] denote matrix elements, parity(p) is the usual permutation parity (+/-)1, and p ranges over permutations of [0..2n-1] satisfying constraints p(2i) < p(2i+1), p(2i) < p(2i+2). The number of terms is 1.3.5...(2n-1), denoted there (2n-1)!! It's a pity that the square of this does not obviously equal det(B) --- this train of thought just leads back to the Parameswaran proof. WFL