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Let $K$ be an algebraic closure of $\mathbb{F}_2$. The cyclic group $C_{2^n}$ acts on $K[x_0, \dots, x_{2^n-1}]$ by cyclically permuting the $x_i$: $a : x_i \rightarrow x_{i + a \bmod 2^n}$. Is there a nice description of the ring of invariants of $C_{2^n}$ acting this way on $K[x]$? Things are quite easy when the characteristic $ \ne 2$, but look quite a bit more intricate here since, e.g. the group ring $K[C_{2^n}]$ is not semi-simple.

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What would be really nice if there was a recursive description. If $R_n$ is the ring of invariants (given explicitly with generators and syzygies) for $C_{2^n}$, then looking at $R_n \times R_n$ in $R_{n+1}$ there is an action (involving the cocycle giving the group extension) of $C_{2^{n+1}}$, which could be used to "clean up" the invariants. –  Victor Miller Mar 20 '10 at 17:19
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A reasonable amount of work seems to have been done (for any indecomposable representation not just a permutation representation). A starting point is (as well as the references to it in SciMath): MR0499459 (81b:14024) Almkvist, Gert; Fossum, Robert Decomposition of exterior and symmetric powers of indecomposable $Z/pZ$-modules in characteristic $p$ and relations to invariants. Séminaire d'Algèbre Paul Dubreil, 30ème année (Paris, 1976--1977), pp. 1--111, Lecture Notes in Math., 641, Springer, Berlin, 1978. –  Torsten Ekedahl Mar 21 '10 at 10:58
    
@Torsten: Thanks! I just starting looking at the Almkvist-Fossum paper, and it looks like it has just what I need. –  Victor Miller Mar 22 '10 at 4:48

4 Answers 4

It seems to me that the characteristic of the field does not play a big role in this question: here is a sketch of an argument.

Note first of all, that all the invariants are linear combinations of "symmetrized monomials": if m is a monomial in the polynomial ring, then form the sum of all the translates of m by the elements of your group. This means that every invariant in the polynomial ring comes from an invariant polynomial with coefficients in the prime field $\mathbb{F}_2$ of K and that invariants with coefficients in $\mathbb{F}_2$ are the reduction of invariants with integer coefficients. Thus we have translated the question over characteristic two to a question over the integers: it suffices to find generators and relations for the ring of invariants of your group over the integers to find generators and relations in any ring.

Over the integers I do not know what the answer is, but if you do know what the answer is over any field of characteristic different from two, maybe you can now fill in the argument. Thinking briefly about the set of generators, it seems like you might simply need the "symmetrized square-free monomials", with relations that are a bit tedious to write down, but that maybe can be nicely interpreted.

EDIT: The square-free monomials are not enough, but it seems that you do not have to look much further to describe explicitly a finite set of generators for the group algebra of a finite cyclic group over the integers. Indeed, let S be the set of monomials m for which there exists an integer r such that the exponents of m are the integers $\{0,1,\ldots,r\}$. Then the product of all the variables of the polynomial ring together with the symmetrizations of all the monomials in S seems to generate the ring of invariants. "Symmetrize a monomial m" means sum over the cosets of the stabilizer in the cyclic group of m. I have not thought about relations, but there will be plenty!

If you really need to make explicit the fact that the ring of invariants is not Cohen-Macaulay, maybe you can, but maybe you do not need to do that...

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The characteristic certainly plays a role in the answer. For all characteristics but two the invariant ring is Cohen-Macaulay but in characteristic two it isn't. It is of course true that you might find a (more or less) characteristic free presentation of the invariant ring where this fact is not apparent. –  Torsten Ekedahl Mar 22 '10 at 5:17

Indeed the "symmetrized square-free monomials" seem to generate. (Order lexicographically and look what the highest term in a product looks like. Now use that to concoct rewriting rules.)

[Oops! This is less obvious than it seemed. The symmetrizations are not with respect to the full symmetric group. In fact it fails for the cyclic group of order four, where the square-free case is not enough to generate all invariants of degree three.]

They also seem to be independent, as the transcendence degree matches.

[Oops! Also wrong. It would contradict the Chevalley–Shephard–Todd theorem. There may be many orbits of our cyclic group in the set of square free monomials of a given degree.]

One may wish to check the degree of the full ring as a module over the predicted subring. For example, K[x,y] as a module over K[x+y,xy] has basis 1, x, but why?

[Because of the minimal polynomial (T-x)(T-y) over that subring. But this reasoning is less helpful for larger degree. Nevertheless one may wish to look at our full ring as a (free) module over the polynomial ring in the elementary symmetric functions. Is there a basis of that module that is permuted by our cyclic group? And one really wants the ring structure, not just the vector space.]

Wilberd

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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. –  Torsten Ekedahl Mar 21 '10 at 19:37
    
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.. –  Victor Miller Mar 22 '10 at 4:10
    
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 ? –  Victor Miller Mar 22 '10 at 4:18
    
Ooops,forgot to throw in $(0,0,1,2)$ as a generator. –  Victor Miller Mar 22 '10 at 4:20

The paper [H.E.A. Campbell, J. Harris and D.L. Wehlau, Internal duality for resolutions of rings, J. of Algebra, 215 (1999) 1--33.] considers this question (or a closely related one) when n=3. Also note that since the group acts via permutations, the answer is (essentially) the same for all fields of characteristic 2 so it suffices to work over ${\mathbb F}_2$.

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I found an answer to my question (well almost, I'd still like a more explicit description) in the following paper:

MR894515 (88k:13004) 13B05 (11T99 20J06) Landweber, Peter S. (1-RTG); Stong, Robert E. (1-VA) The depth of rings of invariants over finite fields. Number theory (New York, 1984–1985), 259–274, Lecture Notes in Math., 1240, Springer, Berlin, 1987.

In it they prove (their Theorem 5) that if $V$ is a finite dimensional vector space over a field $k$ of characteristic $p$, and $G$ is a finite group acting linearly on $V$ that if the subspace of covariants $V_G = V/\{ gv - v : v \in V, g \in G\}$ has codimension 1 in $V$ ( which is the case for my question ), then the ring of invariants $k[V]^{G}$ is a polynomial ring.

There's another paper: "Invariants of some Abelian $p$-groups in characteristic $p$" by Mara D. Neusel in Proc. AMS v. 125, no 7, pp. 1921-1931, showing that a similar result hold when codim$(V^G) = 2$ or codim$(V_G) = 2$, where $V^G = \{ v : v = gv \text{ for all } g \in G\}$.

Unfortunately neither applies in my case.

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This looks to be in contradiction with the result that says that the invariant ring is not Cohen-Macaulay as a polynomial ring is certainly Cohen-Macaulay. Also according to the Math Review the condition is that the subspace $\{gv-v\}$ be $1$-dimensional not of codimension $1$. (The space of covariants is the quotient of $V$ by that space.) –  Torsten Ekedahl Apr 1 '10 at 18:21

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