4
$\begingroup$

Let $X$ be a smooth, quasi-projective variety, $G$ be a finite group which acts freely and properly on $X$. Denote by $\alpha:X \to X/G$ the quotient. Is $\alpha$ generically etale?

Also, as I am new to this topic (of group action on varieties) can someone suggest a good reference for the topic? I am mainly interested in finite group actions.

$\endgroup$
  • $\begingroup$ The morphism $X\to X/G$ is finite. Therefore, it is generically etale if and only if the extension of function fields $K(X/G) \subset K(X)$ is separable. This is the case if the base field (which you didn't specify) is of characteristic zero, because fields of characteristic zero are perfect. $\endgroup$ – Ariyan Javanpeykar Apr 17 '16 at 17:58
  • $\begingroup$ @AriyanJavanpeykar Thanks a lot for the answer. Could you please suggest some text to study this topic as well? $\endgroup$ – Ron Apr 17 '16 at 18:06
  • $\begingroup$ You could have a look at Liu's book on Algebraic Geometry, and specifically Chapter 4 in which he discusses flat morphisms, etale morphisms and smooth morphisms. Of course, the same is done in Hartshorne's book. $\endgroup$ – Ariyan Javanpeykar Apr 17 '16 at 19:39
  • 3
    $\begingroup$ If a quotient exists as a variety or a scheme, then the fibres of $\alpha$ will be the same as $G$-orbits, and all contained in an open affine of $X$. But there exist group actions like in your setting where this condition fails, and hence no scheme quotient exists. Thus you must clarify what you mean by "the quotient". Another comment is that since you assume the action is free, if a quotient exists then $\alpha$ will in fact be \'etale everywhere. $\endgroup$ – Matthieu Romagny Apr 17 '16 at 19:53
  • 1
    $\begingroup$ Dear Ron, the best-known counterexample is due to Hironaka. It appears here and there in the literature; a good source is wikipedia's page for "Hironaka's example". $\endgroup$ – Matthieu Romagny Apr 18 '16 at 7:00
7
$\begingroup$

Let $X$ be a normal variety over an algebraically closed field of characteristic 0 with a finite group $G$ acting effectively. Since $G$ is finite it is reductive and a geometric quotient $X/G$ exists. The quotient map $X\to X/G$ is étale at a point $x\in X$ if and only if the stabilizer at $x$ is trivial. Since the equations $gx-x$ are algebraic and define the locus where the action fails to be free, the complement is Zariski open and shows that the quotient map is always generically étale.

See Section 4.3 here.

$\endgroup$
  • 1
    $\begingroup$ I believe you need to impose that $X$ be quasi-projective. $\endgroup$ – ACL Apr 18 '16 at 14:27
  • 1
    $\begingroup$ You don't need quasi-projectiveness. Being of finite type is enough: given any (irreducible) finite-type $X$ with $G$ action, let $Y\subset X$ be a closed divisor such that the complement $X\setminus Y$ is affine. Then the complement $X\setminus G\cdot Y$ is an affine scheme, and Sean's answer applies. $\endgroup$ – Dmitry Vaintrob Apr 18 '16 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.