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.
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$\begingroup$ 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. $\endgroup$– Keerthi MadapusiCommented Dec 21, 2011 at 17:45
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$\begingroup$ I suppose we should also assume the action is faithful for this to be true. $\endgroup$– Jack HuizengaCommented Dec 21, 2011 at 18:14
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$\begingroup$ Yes, of course. $\endgroup$– Keerthi MadapusiCommented Dec 21, 2011 at 18:36
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