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Let $X$ be an algebraic variety over $\mathbb C$. Let $X^{an}\to Y$ be a finite etale morphism with $Y$ a complex analytic space.

I read somewhere that $Y$ algebraizes, ie, $Y=V^{an}$ for some algebraic space $V$ over $\mathbb C$. (Edit: I previously wrote $Y$ algebraic variety. For my purposes, I just want $Y$ to be an algebraic space.)

Why is this, and what is an "easy" proof for this? (Consequently, the morphism $X^{an}\to Y = V^{an}$ also algebraizes by Riemann's existence theorem. Actually, this is not right. It algebraizes in the sense that it comes from an algebraic morphism $W\to V$, but $W$ might not be isomorphic to $X$. )

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    $\begingroup$ You need $Y$ quasi-projective. Otherwise, consider Hironaka's examples (Hartshorne Appendix B.3) of $X$ an algebraic, nonprojective, $3$-fold and $Y$ a non-algebraic $3$-fold. If you make the right choices, there is a $2$-fold unbranched cover $X \to Y$. $\endgroup$
    – David E Speyer
    Feb 16, 2017 at 18:33
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    $\begingroup$ Hironaka's example exists as an algebraic space, though not as a scheme, so it does have meaningful algebro-geometric structure. Indeed, his analytic constructions are compact Hausdorff Moishezon spaces, and Artin proved that analytification is an equivalence from the category of proper algebraic spaces over $\mathbf{C}$ to the category of compact Hausdorff Moishezon spaces. $\endgroup$
    – nfdc23
    Feb 16, 2017 at 21:08
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    $\begingroup$ The parenthetical at the end is an incorrect argument, because the Riemann Existence Theorem doesn't ensure in the non-proper case that the unique compatible algebraization of $X^{\rm{an}}$ to a finite etale cover of $V$ is necessarily $X$; it could be another algebraization of the same analytic space. One gets examples of this via the footnote near the end of Chapter 1 of Mumford's book on abelian varieties. If you are requiring that $V$ be somehow "compatible" with $X$ then that is not always possible. Probably you are forgetting some conditions from what you read; where did you read it? $\endgroup$
    – nfdc23
    Feb 16, 2017 at 21:11
  • $\begingroup$ @nfdc23 Thank you for your comment. I corrected my last parenthetical statement. I actually don't remember where I read it. It's just something that I vaguely recall reading somewhere, but I might actually be misremembering. $\endgroup$
    – Jean-Paul
    Feb 17, 2017 at 22:36

1 Answer 1

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Up to passing to the Galois closure $\bar{X} \to X$ and using Riemann Extension Theorem (ensuring that $\bar{X}$ is algebraic) we may assume that $X \to Y$ is a Galois cover, induced by the action of a finite group $G$. Then, quoting M. Roth answer to this MathOverflow question

In the case of a quotient of a scheme $X$ by a finite group $G$, a necessary and sufficient condition for the quotient scheme $X/G$ to exist is that the orbit of every point of $X$ be contained in an affine open subset of $X$. This is proved in SGA I, Exposé V, Proposition 1.8.

The condition is in particular satisfied when $X$ is a projective scheme, because the orbit is formed by a finite number of points and we can take the affine subset of $X$ given by the complement of a hyperplane section avoiding all of them (note that the property of being projective is preserved in the Galois closure, since the pullback of an ample divisor by a finite map is again ample).

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    $\begingroup$ But we don't know that G acts algebraically, right? A priori it only acts by holomorphic automorphism's. $\endgroup$
    – Dan Petersen
    Feb 17, 2017 at 7:33
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    $\begingroup$ Any element $g \in G$ gives a biholomorphism $\varphi_g \colon X \to X$. If $X$ is projective, this should be a projective automorphism by GAGA, right? So the action of $G$ on $X$ is actually an algebraic one. $\endgroup$
    – Francesco Polizzi
    Feb 17, 2017 at 7:52
  • $\begingroup$ I agree that there is no problem in the projective case. $\endgroup$
    – Dan Petersen
    Feb 17, 2017 at 8:06

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