TITLE: Invariant polynomials under a group action (hidden GIT)
QUESTION [30 upvotes]: Let's say I start with the polynomial ring in $n$ variables $R = \mathbb{Z}[x_1,...,x_n]$ (in the case at hand I had $\mathbb{C}$ in place of $\mathbb{Z}$).
Now the symmetric group $\mathfrak{S}_n$ acts by permutation on the indeterminates.
The subring of invariant polynomials $R^{\mathfrak{S}_n}$ has a nice description (by generators and relations) in terms of symmetric functions.
What if I only consider the action of the cyclic group $Z_n$? 
Does anyone know if the ring $R^{Z_n}$ admits a nice presentation?
(in the case at hand I had $\mathbb{C}[x_1,x_2,x_3]$ and the action of the cyclic group $Z_3$. Maybe in this case we can use some formula (assuming there is one) for groups splitting as semidirect products?)

REPLY [12 votes]: Even without knowing an explicit set of generators, you can compute the Hilbert series with very little work as follows. In general, suppose a finite group $G$ acts on a vector space $V$ over a field $k$ of characteristic not divisible by $|G|$ via an action map $\rho : G \to \text{GL}(V)$. Then $G$ acts on the symmetric algebra $S(V^{\ast})$ (a coordinate-independent description of the polynomial functions on $V$), and the ring of functions on the quotient $V/G$ is the invariant subalgebra
$$S(V^{\ast})^G \cong \bigoplus_{n \ge 0} S^n(V^{\ast})^G.$$
The Hilbert series of this subalgebra can be computed using Molien's formula, which gives
$$\sum_{n \ge 0} t^n \dim S^n(V^{\ast})^G = \frac{1}{|G|} \sum_{g \in G} \frac{1}{\det (1 - t \rho(g))}.$$
In particular the Hilbert series is invariant under extension of scalars (so e.g. the answer doesn't depend on whether we take $k = \mathbb{Q}$ or $k = \mathbb{C}$), which is not completely obvious. 
In this case $G = \mathbb{Z}_n$ and $V$ is the regular representation. If an element $g \in G$ has order $d \mid n$, its action in the regular representation consists of $\frac{n}{d}$ cycles of length $d$, so the determinant above is $(1 - t^d)^{n/d}$. Altogether this gives the Hilbert series
$$\frac{1}{n} \sum_{d \mid n} \frac{\varphi(d)}{(1 - t^d)^{n/d}}.$$