Let $f : X \to Y$ be a finite, surjective étale morphism of algebraic spaces (say, of finite type over some noetherian scheme). Assume that $X$ is a scheme. Does this imply that $Y$ is a scheme? Is $Y$ the quotient of $X$ by the relation $R = \{(x,y) \in X \times X \ : \ f(x) = f(y) \} \subseteq X \times X$?

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    $\begingroup$ The answer is no. A version of Hironaka’s examples of a proper nonprojective variety (see e.g. the appendix in Hartshorne’s Algebraic Geometry) has a fixed point free involution whose orbit does not admit an affine neighborhood. It follows that the quotient is not a scheme. $\endgroup$ – Piotr Achinger Aug 30 at 17:55
  • $\begingroup$ Is it known whether projectivity of X ensures that Y is a scheme? $\endgroup$ – Mellon Aug 30 at 18:36
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    $\begingroup$ Yes, projectivity is sufficient: the graph of the morphine is a flat family of closed subschemes of $X$ parameter used by $Y$. This determines a closed immersion of $Y$ into the Hilbert scheme of $X$. $\endgroup$ – Jason Starr Aug 30 at 18:55
  • $\begingroup$ "morphine" --> "morphism", "parameter used" --> "parameterized" (autocorrect error) $\endgroup$ – Jason Starr Aug 30 at 19:28

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