If the quotient of an algebraic space $X$ by a finite group $G$ is a scheme, is $X$ already a scheme? Here $G$ is just a finite group, but I'd like to know the answer when $X$ is defined over Spec(Z).
Is this at least true if $G$ acts freely?
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$\begingroup$ This follows from the easy direction of Chevalley's theorem for algebraic spaces. For the hard direction, you can consult Knutson's "Algebraic Spaces" or the more recent proof by David Rydh in the not-necessarily-Noetherian setting. $\endgroup$– Jason StarrMay 4, 2015 at 20:43
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$\begingroup$ I googled "David Rydh Chevalley theorem." Is the Chevalley theorem you mean, "let $X$ be an affine scheme, let $Y$ be an algebraic space, and let $f:X \rightarrow Y$ be an integral and surjective morphism. Then $Y$ is affine." ? Excuse my ignorance but I don't know how to apply this. $\endgroup$– cooldude99May 4, 2015 at 21:06
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3$\begingroup$ Chevalley's theorem is the statement, roughly, that $Y$ is affine if and only if $X$ is affine. The easy direction is: if $Y$ is affine, then also $X$ is affine. This is the direction you are interested in: every scheme is covered by open affines, you can consider the inverse image of these open affines in $X$, and then you apply the easy direction of Chevalley's Theorem. The hard direction is the one that Rydh generalizes: if $X$ is affine, then also $Y$ is affine. $\endgroup$– Jason StarrMay 4, 2015 at 21:17
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1$\begingroup$ I'm voting to close this question as off-topic because it has been answered in the comments $\endgroup$– Dima PasechnikMay 5, 2015 at 14:59
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