From: rusin@math.niu.edu (Dave Rusin) Newsgroups: sci.math.research Subject: Re: Are they isomorphic? Date: 1 Nov 1998 06:38:29 GMT Summary: yes Kheng Feung Tan writes: >Let W=F_2[x_1,x_2,x_3] be the polynomial ring in 3 variables over the >field with 2 elements, F_2. [...] >Consider the matrix - - > | 1 1 1 | > A = | 0 1 1 | > | 0 0 1 | > - -. >[...] We have 4 invariants: > >f_1 = x_1 >f_2 = x_2(x_1 + x_2) >f_3 = x_2^3 + x_1^2x_3 + x_1x_3^2 + x_1^2x_2 >f_4 = x_3^4 + (x_1^2 +x_1x_2 + x_2^2)x_3^2 + (x_1^2x_2 + x_1x_2^2)x_3; > >and a relation among them: > >f_1^2f_4 + f_3^2 + f_2^3 + f_1f_2f_3 = 0. > >We are wondering if the following is true: >Let F_2[x_1,x_2,x_3]^H denote the ring of invariants under the group H. >Consider the polynomial ring P = F_2[f_1,f_2,f_3,f_4]. >Let I be the ideal generated by f_1^2f_4 + f_3^2 + f_2^3 + f_1f_2f_3 in >P. >Then F_2[x_1,x_2,x_3]^H is isomorphic to P/I. This is true. These are pretty standard computations in invariant theory; I believe this can be easily treated with current versions of Magma and Macaulay, and those of us who compute mod-2 cohomology of group extensions see this quite often (see. e.g. Math Comp 53 (1989) 359-385.) To carry the computations out by hand, it helps to find a subring of W^H which is a polynomial ring and over which W is a free module of finite rank. You have already found W0 = F_2[f1, f2, f4], which will suffice: W is recovered from W0 by adjoining roots of a monic quadratic and a monic quartic, so is the tensor product of the free modules on generators {1, x2} and {1, x3, x3^2, x3^3}. If we order the basis of W over W0 to be [1, x2, x3, x2*x3, x3^2, x2*x3^2, x3^3, x2*x3^3] then the action of A on W is represented by a certain 8x8 matrix M with coefficients in W0. The invariants in W are the kernel of M-I. If we tensor with the quotient field of W0 we may compute the kernel by row-reduction; Maple reports it to be the span of {[1, 0, 0, 0, 0, 0, 0, 0], [0, f2/f1, f1, 0, 1, 0, 0, 0]}; clearly the only vectors in this subspace with W0-integral coordinates are those in the span of 1 = [1, 0, 0, 0, 0, 0, 0, 0] and f3 = [f1*f2, f2, f1^2, 0, f1, 0, 0, 0]. So the invariant subring is given as a W0-module as W^H = W0 . 1 + W0 . f3, a rank-2 free-module. Its description as a W0-algebra is then completed by describing the product structure, which in this case requires only that we compute f3^2 = w1 . 1 + w2 . f3 ; you have already found f3^2 = (f1^2f4 + f2^3).1 + (f1f2).f3 . Then W^H = W0 [ f3 ] / (minimal polynomial of f3) = F_2[f1,f2,f3,f4]/I, as desired. I would also like to comment on the suggestions of Allen Adler who wrote: >(1) Try to compute the ring of invariants of A^2, which might be > easier. A will then induce an automorphism of order 2 on that > ring, which might (or might not) be easier to work with. The invariant rings of subgroups may well be more complicated than the invariants of the full group. In this case, W^ = F_2[x1, x2, x3(x3+x1)] is a pure-polynomial ring, but the generators are no longer all of degree 1 (which means A no longer acts by a matrix of scalars). >(2) I think the ring of invariants for GL_n(K) acting on K[x_1,...,x_n] > is known in general when K is a finite field. I vaguely recall a couple > of articles about it in Comptes Rendus. Your group GL_2 would > be denoted GL_3(F_2) and its ring R of invariants is therefore known. > Your polynomial ring W is then a module over R and you might be able, > with enough computing power, to write down the generators and relations > of that module. Having done so, you can then try to write down the > submodule fixed by A. Indeed, the invariants are the Dickson invariants, obtained as the nonzero coefficients (those of X^(|K|^n - |K|^i) ) of the polynomial Prod( X - v , v in span of x1, ..., xn ); the invariant subring is simply a polynomial algebra. As Adler suggests this makes it possible to compute invariant subrings for any subgroup G of GL_n(K) using essentially the same method I outlined above. But the computational complexity is likely to be prohibitive if [GL_n(K) : G] is large. It pays to find a smaller group than GL_n(K) whose invariant ring is a polynomial algebra, as I did above. There are several nice summaries of the properties of the Dickson invariants, among them Clarence Wilkerson's "Primer", Contemp. Math., AMS v. 19 (1983) 421-434. dave