# Invariant Laurent polynomials under cyclic group action

Start with the cyclic group $G:=\mathbb{Z}/p$ of prime order $p$ and and an integer lattice $P:=\mathbb{Z}^p$. Let $G$ act on $P$ by cyclic permutation of coordinates. There is an induced action on the group ring $\mathbb{Z}[P]$ of the lattice $P$ which is isomorphic to the ring $\mathbb{Z}[\mathbf{x^{\pm 1}}]$ of Laurent polynomials in $p$ variables $\mathbf{x}=(x_1,\dots,x_p)$ over $\mathbb{Z}$.

How do I find a (hopefully) finite/minimal set of invariants which generates the ring of $G$-invariants $\mathbb{Z}[\mathbf{x^{\pm 1}}]^G$?

Since $\mathbb{Z}[\mathbf{x^{\pm 1}}]^G=\mathbb{Z}[\mathbf{x}]^G[\sigma_p^{-1}]$ where $\sigma_p=x_1\dots x_p$, the question about Invariant Polynomials under a Group Action (hidden GIT) is of some help but not much is said there about what happens over $\mathbb{Z}$.

From the Lie theory point of view this is related to the weight lattice of the root system $A_{p-1}$ where $G$ is a subgroup of the Weyl group $S_p$ and the fundamental domain I'm considering is some sort of "larger Weyl chamber".

• If we take the group ring with coefficients from the ring of $p$-th cyclotomic field, then the action can be diagonalized, and invariants are readily found: then to solve your original question one has to carry out 'Galois Descent'. This is done in H W Lenstra's paper in Inventiones 1974 (in the context of Noether's problem) – P Vanchinathan Jan 22 '15 at 3:05

Even over $\mathbb{Q}$, the invariants (polynomial invariants, not rational invariants, which are easier) of the cyclic group are messy. I don't see a reason why localizing by $\sigma_p$ should make things easier. Now doing the computation over $\mathbb{Z}$ adds to the difficulty since computing over $\mathbb{Z}$ means that all anomalies of the modular situation will show up.
To give evidence to my first claim, here are the numbers $r$ of minimal generators of the invariant rings $\mathbb{Q}[x_1,\ldots,x_n]^{C_n}$ for some $n$, computed by MAGMA: $n = 3$: $r = 4$; $n = 4$: $r = 7$; $n = 5$: $r = 15$, $n = 7$: $r = 48$; $n = 9$: $r = 119$; $n = 11$: $r = 348$.