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The quotient of the affine space $\mathbb{A}^n$ by the symmetric group $Sym_n$ is again an affine space of the same dimension, and invariants are given by elementary symmetric polynomials.

What about a $n$-cycle? More precisely, what are the generators of the invariants of the action of $\mathbb{Z}/n\mathbb{Z}$ on $\mathbb{A}^n$ given by $(x_1,\dots,x_n)\mapsto (x_2,x_3,\dots,x_n,x_1)$ ?

It is probably classical, but I was not able to find it.

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If the field contains $n$-th roots of 1 and the characteristic does not divide $n$, then up to a linear change of coordinates the action is $(x_1,\dots x_n)\mapsto (x_1, \zeta x_2,\dots \zeta^{n-1}x_n)$. – rita Jan 16 '13 at 17:25
...$\zeta$ is a primitive $n$-th root of 1, of course. – rita Jan 16 '13 at 17:26
Yes, I agree. Does it help to give the answer? PS: By the way, I am in fact more interested in the case where the field is $\mathbb{Q}$. But an answer over $\mathbb{C}$ good be a good start. – Jérémy Blanc Jan 16 '13 at 17:45
@Jérémy: With respect to rita's change of coordinates, which monomials in $x_1,\dots,x_n$ are invariant? This answers your question. By the way, if you prove that the invariant ring is $\textit{not}$ a polynomial ring after base change from $\mathbb{Q}$ to $\mathbb{C}$, then that also implies that the invariant ring over $\mathbb{Q}$ is not a polynomial ring. – Jason Starr Jan 16 '13 at 18:32
Also, there is a general theorem, the Chevalley-Shephard-Todd theorem, that answers many similar questions. – Jason Starr Jan 16 '13 at 20:08
up vote 5 down vote accepted

Rita's answer is perfectly correct. However, Jérémy posed his question over a (not necessarily algebrically closed) characteristic $0$ field $k=\mathbb{Q}$, so here is another approach. Consider the subring $S = k[s_1,\dots,s_n]$ generated by the elementary symmetric polynomials, i.e., the invariant ring under the entire symmetric group, not just the cyclic group. The entire polynomial ring $R = k[x_1,\dots,x_n]$ is an $S$-algebra that is, in fact, a free $S$-module of rank $n!$. The fact that this is a free $S$-module follows automatically since $S$ is regular and $R$ is Cohen-Macaulay; however, there are also perfectly explicit choices of $S$-basis, e.g., the monomials $\underline{x}^{\underline{e}} =x_1^{e_1}x_2^{e_2}\cdots x_n^{e_n}$ indexed by all exponents $(e_1,\dots,e_n)\in \mathbb{Z}_{\geq 0}^n$ with $e_i < i$ for every $i$. The $k$-algebra homomorphism $$\phi:R\to R, \ \ \phi(x_1,\dots,x_{n-1},x_n) = (x_2,\dots,x_n,x_1),$$ is an $S$-algebra homomorphism. The ring of $\phi$-invariants is the kernel of the $S$-module homomorphism $\text{Id}_R-\phi:R\to R$, which is the same as the image of the "averaging homomorphism" / Maschke homomorphism / Reynolds operator $$E=\frac{1}{n} ( \text{Id} + \phi + \phi^2 + \dots + \phi^{n-1} ).$$ In particular, the subring of $\phi$-invariants is a direct summand of $R$ as an $S$-module. Of course one set of generators is the set of images $E(\underline{x}^{\underline{e}})$ as above; you can compute a smaller set of $S$-generators for the kernel using a term order on $R$ as an $S$-module and using Gröbner basis methods. Any collection of $S$-generators give $k$-algebra generators once you add in the $k$-algebra generators of $S$, i.e., the elementary symmetric polynomials.

Regarding the question when the ring of invariants is a polynomial ring, or even a smooth $k$-algebra, this fails for all $n\geq 3$. As explained in the comments, this can be checked after base change to $\overline{k}$, where it becomes an immediate corollary of the Chevalley-Shephard-Todd theorem (or just direct computation). However, as you can see, the invariant ring is Cohen-Macaulay; it is a free $S$-module (special case of the Hochster-Roberts Theorem).

$\textbf{Edit}$. It seems quite plausible that, as a free $S$-module, the ring of invariants has a basis consisting of $E(\underline{x}^{\underline{e}})$ for those $(n-1)!$ exponents as above satisfying the additional condition that $e_n=0$.

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This problem (over an algebraically closed field) has been studied by quite a few people. Rather than repeat myself I'll point you to my answer (and the references there) to this subsequent posing of your question cyclically symmetric functions

Working over the rationals introduces an extra level of difficulty.

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