# Quotient of a rational variety by a finite group

Let $X$ be a rational variety and let $G$ be a finite group acting on $X$. Let us consider the diagonal action of $G$ over the product $X^{h} = X\times...\times X$,

$$G\times(X\times...\times X)\rightarrow X\times...\times X,\quad (g,x,...,x)\mapsto (gx,...,gx).$$

Let $Y= X^{h}/G$ be the quotient. Are there hypothesis under which $Y$ is rational ?

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I don't see the point of the $h$: we start with any rational variety $X$ and a finite group $G$ acting on it. Then you produce $X^h$ which is...a rational variety acted upon by a finite group $G$. –  Pete L. Clark Dec 10 '13 at 23:48
@Pete: While Saltman's example shows that $X/G$ may not be rational, one may wonder if $X^h/G$ for some values of $h$, for example for large $h$. (I have not thought to the problem, though.) –  ACL Dec 11 '13 at 8:26
@ACL: Do you mean that the OP wants to fix $X$ and ask for which $h$ is $X^h/$ rational? That would make sense... –  Pete L. Clark Dec 11 '13 at 14:00
@Pete: I don't know what he meant. But that's a way to fix the question. (For $h=1$, the action is not really diagonal.) –  ACL Dec 11 '13 at 22:03

As mentioned above, I don't see the point of the $h$: your general case is an instance of the $h=1$ case, since $X^h$ is still a rational variety acted upon by a finite group $G$.

The first thing to say is that the answer is not always affirmative. Indeed the situation when $X = \mathbb{A}^{\# G}$ and the $G$-action corresponds to the regular representation of $G$ on a vector space over a field $K$ is called Noether's Problem, because Emmy Noether pointed out that an affirmative solution over $K = \mathbb{Q}$ would imply that every finite group is a Galois group over $\mathbb{Q}$.

Unfortunately (or fortunately, depending upon your perspective) Noether's problem has a negative solution in general. The first counterexample was given by Swan in 1969: here $G$ is cyclic of order $47$ and $K = \mathbb{Q}$. The case of faithful (but not necessarily regular) representations of any commutative group over any fielf was handled by Lenstra, obtaining in particular the cyclic group of order $8$ over $\mathbb{Q}$ as a minimal counterexample. In particular Lenstra's work gives many instances where the quotient is rational, e.g. for any commutative group over any algebraically closed field. The first negative example over $K = \mathbb{C}$ was given by Saltman in 1984. There is an excellent survey paper by R.G. Swan (unfortunately the link I gave requires institutional access).

In my opinion the fact that there are many instances where Noether's problem has an affirmative answer may be underappreciated, in particular as a tool for realizing some finite groups as Galois groups over $\mathbb{Q}$ (or any Hilbertian field of characteristic $0$). In fact I wrote a mostly expository note about the case of finite pseudoreflection groups as Galois groups over Hilbertian fields: in characteristic $0$ one has the theorem of Shephard-Todd-Chevalley: the invariant ring $K[t_1,\ldots,t_n]^G$ of a finite group acting faithfully and linearly in characteristic $0$ is a polynomial ring if and only $G$ is a pseudoreflection group (and the representation is a pseudoreflection representation): this conclusion is much stronger than the needed rationality. In particular one gets all Weyl groups (of simple Lie algebras) as Galois groups over $\mathbb{Q}$ -- I found a paper by a Russian mathematician which works harder to prove this result for all but one of the Weyl groups. One also gets all Coxeter groups over a real abelian number field, and so forth: I feel like someone should go through the Chevalley-Todd classification, record the character field of each of the pseudoreflection representations and compare it to the smallest known abelian number field over which each of these groups has otherwise been shown to be a Galois group (usually by methods of Belyi-Thompson-Malle...): as far as I can see one might get some of these groups as Galois groups over smaller fields than was otherwise known.

The Shephard-Todd-Chevalley Theorem is not true in characteristic $p > 0$, but rather recent work has shown that the invariant field of a pseudoreflection representation is nevertheless rational, which is still enough to realize these groups as Galois groups over $\mathbb{F}_q(t)$, where the $q$ depends on the character field of the representation once again. Here I strongly suspect that patiently going through the (more complicated) classification of pseudoreflection groups will yield some new Galois groups. Unfortunately I have not gotten around to trying these calculations / tabulations myself and I'm not sure whether my paper is really worth much without them, so this project is in a bit of limbo. (If anyone is interested in doing this, please contact me!)

This is just an example to show that of course there will be some hypotheses in which your question has an affirmative answer. If $K = \mathbb{C}$ then it follows from the classification of surfaces that the case $\operatorname{dim} X = 2$ will have an affirmative answer. This is not true in positive characteristic or over (e.g.) $\mathbb{Q}$, but I'm not sure whether the known examples come from group actions. Someone else here will!

To get a more useful answer you should say a bit more about what hypotheses you have in mind.

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No. There are counter-examples (the first one is due to Saltman) even for $h=1$, $X$ a complex vector space and $G$ acting linearly.

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What about the case $dim(X) = 2$ ? –  MorFel1921 Dec 10 '13 at 22:12