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Feb 17, 2017 at 22:36 comment added Jean-Paul @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.
Feb 17, 2017 at 22:34 history edited Jean-Paul CC BY-SA 3.0
I took into account the comments.
Feb 17, 2017 at 7:25 answer added Francesco Polizzi timeline score: 6
Feb 16, 2017 at 21:11 comment added nfdc23 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?
Feb 16, 2017 at 21:08 comment added nfdc23 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.
Feb 16, 2017 at 18:33 comment added David E Speyer 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 $X \to Y$.
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Feb 16, 2017 at 18:10 history asked Jean-Paul CC BY-SA 3.0