Let X be an algebraic space such that there exists an etale surjective map f: X -> Y, where Y is a scheme. Is it true that this implies X is also a scheme?
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The point must be to avoid separatedness hypotheses on $f$. (D. Knutson proved algebraic spaces locally quasi-finite and separated over schemes are schemes; he may have had noetherian hypotheses, in which case those hypotheses are removed in the appendix to Champs Algebriques.)
Anyway, in the absence of separatedness the answer is negative. In the Introduction to Knutson's book there's a procedure beginning with smooth algebraic variety $Y$ of dimension $> 1$ over a field and "replacing" a smooth hypersurface in $Y$ with an irreducible etale double cover while remaining smooth and irreducible. This resulting $X$ is an algebraic space etale over $Y$, and the rank-jumping of $f:X \rightarrow Y$ is really weird: it jumps up along a closed set rather than down, contrary to Zariski's Main Theorem (in the formulation of EGA) for locally quasi-finite flat maps of schemes. Hence, $X$ is not a scheme (though the failure to be a scheme can surely be seen in a zillion other ways).
The preceding does actually have an "application". It yields a geometrically irreducible smooth algebraic space $X$ over $\mathbf{Q}$ of dimension 2 such that $X_{\mathbf{C}}$ admits an analytification in the sense of complex-analytic spaces but $X_k$ does not admit an analytification in the sense of rigid-analytic spaces or Berkovich $k$-analytic spaces for any non-archimedean field $k$ of characteristic 0. (This is explained in detail in Example 3.1.1 of my paper "Non-archimedean analytification of algebraic spaces" with M. Temkin.)
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1$\begingroup$ Jim, even in the category of complex-analytic spaces there are etale equivalence relations admitting no quotient as such (even equivalence relations arising from analytifying etale scheme charts for algebraic spaces of finite type over $\mathbf{C}$). So do you consider the complex-analytic spaces to be an inadequate category? $\endgroup$– BCnrdCommented Oct 26, 2010 at 12:39
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1$\begingroup$ OK, that's interesting. Can you give me an example or a reference? $\endgroup$– JBorgerCommented Oct 26, 2010 at 22:26
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1$\begingroup$ Jim, see Theorem 2.2.5 of my paper with Temkin for a proof that "diagonal being immersion" is a necessary condition for analytifiability (= existence of quotient for analytified etale equivalence relation) of a quasi-separated algebraic space locally of finite type over $\mathbf{C}$ (or over any non-archimedean field; same whether rigid-analytic or Berkovich). That this diagonal condition is actually sufficient over $\mathbf{C}$ for existence of the quotient is an instructive exercise. At the end of my answer I give a reference for a counterexample to that over any non-archimedean field. $\endgroup$– BCnrdCommented Oct 26, 2010 at 22:42
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1$\begingroup$ Thanks. After looking at an example that isn't locally separated, my inclination would be to say that complex-analytic spaces are an inadequate category. I would think that you'd want your category to have a simple universal property. Does the category of usual complex-analytic spaces have one? If not, I'd expect that you'd run into trouble when defining functors to and from your category (which actually might be happening here). Surjective local isomorphisms surely have all the descent properties you'd ever want. Why not adjoin the quotients by such equivalence relations for convenience? $\endgroup$– JBorgerCommented Oct 27, 2010 at 0:14
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1$\begingroup$ Jim, so you're proposing to make a kind of complex-analytic version of the theory of algebraic spaces. Yes, perhaps a reasonable thing, though it would be appropriate to first undertake a study of the work of Binenger et al. who carried over Artin's methods to the complex-analytic category to see what they did. I think it is nonetheless interesting to get affirmative existence results for etale quotients under some mild separatedness hypotheses, and one probably needs some genuinely compelling examples to justify serious effort towards (or interest in!) working with a bigger category. $\endgroup$– BCnrdCommented Oct 27, 2010 at 0:40