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?)


7 Answers 7


The actions of $S_n$ and $\mathbb Z_n$ differ in the sense that in the first case the quotient is smooth (it is again $\mathbb C^n$) while in the second case it is singular. This is why in the fist case we have a nice presentation, but in the second not really. For example, the number of generators of the quotient can not be less than the dimension of Zariski tangent space to the singularity at zero of $\mathbb C^n/\mathbb Z_n$.

Still in principle the presentation can be provided by toric geometry (https://dacox.people.amherst.edu/toric.html) because the quotient is the toric singularity. For example, in your case of $\mathbb C^3/\mathbb Z_3$ let us change the coordinates so that $\mathbb Z_3$ is acting as $w_0\to w_0$, $w_1\to \mu w_1$, $w_2\to \mu^2 w_2$ (here $\mu^3=1$). Then you can write the minimal set of four generators:

$w_0, w_1^3, w_2^3, w_1w_2$, and one obvious relation $(w_1^3w_2^3)=(w_1w_2)^3$

The case $\mathbb C^n/\mathbb Z_n$ for $n>3$ will be more involved, but the idea is the same roughly. First you chose the coordinates on $\mathbb C^n$ $w for which the action is diagonal. Then pick the minimal set of monomials (in these new coordinates) that are invariant under the action, and generate the whole set of invariant monomials (of positive degree).

Consider one more case $n=4$, and chose the coordinates $w_i$, so that $Z_4$ is acting as $w_i\to \mu^iw_i$, $\mu^4=1$. The number of generators is $7$ this time:

$w_0, w_1^4, w_3^4, w_2^2, w_1w_3, w_1^2w_2, w_3^2w_2$

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    $\begingroup$ Note however that diagonalisation is not possible over $\mathbb Z$ (or even $\mathbb Q$) which make things trickier. However, my guess is that the easiest way is to first do it over a splitting field and then descend (at least to $\mathbb Q$, over $\mathbb Z$ there should be even more complications). $\endgroup$ Jul 28, 2011 at 5:39
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    $\begingroup$ Link in the post seems to be dead, here is a Wayback Machine snapshot. $\endgroup$ Aug 7, 2019 at 13:48
  • $\begingroup$ The link is fixed for now. $\endgroup$ Oct 22, 2020 at 15:21

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

$$\frac{1}{n} \sum_{d \mid n} \frac{\varphi(d)}{(1 - t^d)^{n/d}}.$$


The minimal number of invariants needed to generate ${\mathbb Z}[x_1,...,x_n]^{{\mathbb Z}_n}$ has been considered by a number of authors including Erdos, Dixmier and Kac. (see the references in [John C. Harris and David L. Wehlau Non-Negative Integer Linear Congruences, Indagationes Mathematicae 17 No. 1 (2006) 37-44]. It is easily seen to be bounded below by the number of partitions of n, ${\mathcal P}(n)$. Dixmier produced a number of papers giving the asymptotic behavior of this number as a function of $n$. The results in the above Harris-Wehlau paper are completed by using the main result in [Pingzhi Yuan, On the index of minimal zero-sum sequences over finite cyclic groups, Journal of Combinatorial Theory, Series A 114 (2007) 1545–1551]. These two papers combine to show that the number of homogeneous generators of this ring of invariants of degree $k$ is exactly $\phi(n){\mathcal P}(n-k)$ if $k \geq \lfloor n/2\rfloor + 2$ (here $\phi$ is Euler's totient function). Surprisingly (at least to me) much less is known about the number of generators in lower degrees.


By the general theory of invariants of finite groups, there exist $p=(n-1)!$ homogeneous polynomials $u_1, \dots, u_p$ (which can be chosen to be monomials) such that every element $f$ of $R^{Z_n}$ can be uniquely written $f = u_1 g_1 + \cdots+ u_p g_p$, where $g_1,\dots,g_p$ are symmetric functions. I don't know whether an explicit description of $u_1,\dots,u_p$ is known for arbitrary $n$.

A reference is http://math.mit.edu/~rstan/pubs/pubfiles/38.pdf.


I recommend that you read the second chapter of this book https://www.springer.com/gp/book/9783211774168, entitled Algorithms in Invariant Theory. In particular it shows that you you can use Groebner bases to calculate generators for the ring of invariants under the action of a finite group.


Reposting VA's wonderful answer (with a trivial correction), since someone else just asked me this question:

This is just to add 1% to Dmitri's 99% complete answer. Change the coordinates to $w_0,\dots, w_{n-1}$ defined by the formula

$$ w_i = x_0 + \mu^i x_1 + \mu^{2i} x_2 + \dots, $$

where $\mu$ is a primitive $n$-th root of identity. Then the ring of invariants is the subring of monomials

$$ w_0^{k_0}\dots w_{n-1}^{k_{n-1}} \quad \text{such that}\quad n \mid k_1 + 2k_2 + \dots + \left(n-1\right) k_{n-1}$$

and a set of generators can be obtained by taking minimal such monomials (i.e. not divisible by smaller such monomials). And relations between these generators are of the form (monomial in $w_i$) = (another monomial in $w_i$). That's a pretty easy presentation by any standard.

P.S. This works over $\mathbb C$ or any ring containing $1/n$ and $\mu$.

Note that the number of minimal monomials (in the sense described above) is Sequence A096337 in the OEIS.

A comment by Victor Miller (slightly edited):

The above construction works when the group, $G$, is any abelian group. The reason is that all irreducible representations are $1$-dimensional. More specifically, for all characters $\chi \in \widehat{G}$, set $w_{\chi} = \sum_{g \in G} \chi(g) x_{i \cdot g}$. Then $w_{\chi}^g = \chi(g) w_{\chi}$. So, as above, take monomials in the $w_\chi$, choosing the powers so that they are invariant.


First about $S_n$: when it acts by permutating the variable there is a nice description of invariants, as the transpositions will be represented by reflections. Chevalley's theorem states the invariants will be a polynomial ring in such cases. But for other representations it won't be. So even for cyclic groups acting by permuting the variables, it fails to be a polynomial ring.

Slightly different from the question, but hope it is relevant: Instead of polynomial ring with integer coefficients going to rational function field with rational coefficients the questions has been considered by Emmy Noether. This is relevant to Inverse Galois Problem.

R. Swan (1969) answered that the field of invariant rational functions will not be purely transcendental for cyclic permutation group on 47 variables. HW Lenstra (1974) considered all transitive permutation action of abelian groups on the variables and answered when it would be purely transcendental. From his theorem cyclic group of order 8 will fail to give purely transcendental invariant field.


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