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Mar 22, 2010 at 4:20 comment added Victor Miller Ooops,forgot to throw in $(0,0,1,2)$ as a generator.
Mar 22, 2010 at 4:18 comment added Victor Miller Continuation: and thus finitely generated. For $n=2$ we get as generators $(1,0,0,0),(0,1,0,1),(0,0,2,0),(0,2,1,0),(0,4,0,0),(0,0,0,4)$. For each such $\alpha=(a,b,c,d)$, set $z_{\alpha} = \prod_j y_j^{\alpha_j}$. Note that some of the the $z_{\alpha}$ are congruent modulo $\pi$, the unique prime above 2. So take sums and differences of those, dividing by $\pi$ and reduce mod $\pi$, plus the reduction of all of the others. Do those reductions generated the invariant ring in characteristic 2 ?
Mar 22, 2010 at 4:10 comment added Victor Miller I was also thinking that the symmetrized square free monomials might generate. Here's another approach that may be fruitful. Over $\mathbb{Q}$, let $K_n$ the field obtained by adjoining by a primitive $2^n$th root of unity $\zeta_n$. The prime 2 is totally ramified in $K_n$. Set $y_j = \sum_{k=0}^{2^n-1} \zeta_n^{jk} x_k$. Then the action of $C_{2^n}$ on the $y_j$ is to multiply them by a suitable root of unity. Thus, one wants the set of non-negative integers $a_j$ such $\prod y_j^{a_j}$ is invariant, which amounts to them being an additive submonoid of an integral lattice ..continued..
Mar 21, 2010 at 19:37 comment added Torsten Ekedahl Note that the invariant ring is in general not Cohen-Macaulay so it is not free as a module over the symmetric polynomials. I think that excludes the possibility of a basis permuted by the group.
Mar 21, 2010 at 19:23 history edited Wilberd van der Kallen CC BY-SA 2.5
added 141 characters in body
Mar 21, 2010 at 18:51 history edited Wilberd van der Kallen CC BY-SA 2.5
is harder than I thought
Mar 21, 2010 at 16:38 history answered Wilberd van der Kallen CC BY-SA 2.5