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...