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Let $X$ be an integral scheme over a field. Let $G$ be a finite group acting on $X$ faithfully. Assume the quotient stack $[X/G]$ is separated (e.g., when $G$ acts on $X$ properly). Then $[X/G]$ is a separated Deligne-Mumford (DM) stack and there is a coarse moduli space $$\pi:[X/G] \to X/G.$$ Is $\pi$ always a birational morphism of DM stacks?

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    $\begingroup$ What if $X$ has multiple components, with different elements of $G$ acting trivially on each? $\endgroup$
    – Will Sawin
    Jun 26 at 15:47
  • $\begingroup$ @WillSawin Thanks for the comment. I edited the question to avoid that case. $\endgroup$ Jun 26 at 16:07
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    $\begingroup$ I think we're still in trouble if we just glue the components together. (Say two $\mathbb P^1$s joined at a point, and the group generated by two involutions of the separate $\mathbb P^1$s that each fix the other one, and thus that point.) Maybe you want to assume $X$ is irreducible? $\endgroup$
    – Will Sawin
    Jun 26 at 16:53
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    $\begingroup$ @WillSawin That is still not enough. Consider the action of $\mu_\ell$ on $\text{Spec}\ k[\epsilon]/\langle \epsilon^{\ell} \rangle$ by $\zeta\bullet \epsilon = \zeta\epsilon$. This is a "faithful" action, but the map from the stack to the coarse moduli space is not birational. The correct hypothesis is that for every generic point of $X$, the induced action of $G$ on the residue field of the point is faithful. $\endgroup$ Jun 26 at 17:28
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    $\begingroup$ Yes, I think. For each nontrivial element $g\in G$, the fixed points form a closed set, which must not contain the whole space as then $g$ would act trivially. The complementary open set thus contains the generic point. The intersection of these open sets over all nontrivial $g\in G$ forms an open set $U$ which is $G$-invariant. Restricted to $U$, the action of $G$ is free. Thus, the image of $U$ in the stack $X/G$ is the quotient $U/G$, which is an algebraic space, and so the image of $U/G$ in the coarse moduli space of $X/G$ is again isomorphic to $U/G$. $\endgroup$
    – Will Sawin
    Jun 27 at 14:07

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Yes. For each nontrivial element $g\in G$, the fixed points form a closed set, which must not contain the whole space as then $g$ would act trivially (by reducedness). The complementary open set thus contains the generic point.

The (nonempty, by irreducibility) intersection of these open sets over all nontrivial $g\in G$ forms an open set $U$ which is $G$-invariant. Restricted to $U$, the action of $G$ is free. Thus, the image of $U$ in the stack $[X/G]$ is the quotient $U/G$, which is an algebraic space, and so the image of $U/G$ in the algebraic space $X/G$ is again isomorphic to $U/G$.

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