MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove this? Let G be a finite group acting on a connected normal affine variety.Then the stabilizer at x is trivial iff Then the orbit space morphism q : X---->X // G is E'tale at x.

share|cite|improve this question
I will assume that you are working over an algebraically closed field. Let $\mathcal{O}$ be the completion of $X$ at $x$, and let $\mathcal{O}'$ be the completion of $X/G$ at the image of $x$. Then the fact is that $\mathcal{O}'$ is the ring of invariants in $\mathcal{O}$ under the stabilizer of $x$. In particular, $\mathcal{O}'=\mathcal{O}$ precisely when the stabilizer is trivial. – Keerthi Madapusi Pera Dec 21 '11 at 17:45
I suppose we should also assume the action is faithful for this to be true. – Jack Huizenga Dec 21 '11 at 18:14
Yes, of course. – Keerthi Madapusi Pera Dec 21 '11 at 18:36

see SGA1 Exposé V Corollaire 2.3

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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