Let $X$ be a scheme proper over $\mathbb{Z}$, $G$ a finite group acting on $X$. Suppose that the quotient scheme $Y := X/G$ is well-defined.
Should $Y$ be then proper over $\mathbb{Z}$ as well?
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3By Stacks 01W6 this is equivalent to $Y$ separated. I know examples of nonseparated quotients when quotienting by a $\Bbb C^\times$ action but I can't remember right now whether this can happen or not when taking the quotient by a finite group. – KReiser Mar 19 '20 at 05:04
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1I don't think this can go wrong when a finite group acts. For example if $X$ is a Hausdorff topological space and $G$ is a finite group acting on $X$, then the quotient space $X/G$ is automatically Hausdorff. – Pol van Hoften Mar 19 '20 at 10:36