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 <tan_khengfeung@pacific.net.sg> 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  <adler@hera.wku.edu> 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^<A^2> = 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