# Is this true that algebraic spaces etale and surjective over a scheme is a scheme ?

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.)
Brian, regarding your application, what goes wrong? Suppose you write $X$ as the quotient of an affine scheme $A$ by an etale equivalence relation $B$, and we write $B$ as a quotient of an affine scheme $C$ by an etale equivalence relation $D$, with $D$ affine. The most obvious way of making the analytification of $X$ would by the following steps. 1) Analytify $C$ and $D$. 2) Presumably this is an etale equivalence relation. 3) Let $B'$ be the quotient. 4) Presumably $B'$ is an etale equivalence relation on the analytification of $A$. 5) Let $X'$ be the quotient. What goes wrong? – JBorger Oct 26 '10 at 6:01
Jim, since $B$ is a scheme, it has an analytification, and that is $B'$. We cannot just "let" something be the quotient. We have to prove a quotient (according to reasonable and useful definition; e.g., does indeed make $B^{\rm{an}}$ serve as $B'$) actually exists. For item 5, it doesn't exist. (Surprised? I sure was!) If you look up the reference at the end of my answer you'll see the reason that it does not exist. The dichotomy with the complex-analytic case is that there's no "Gelfand-Mazur theorem" over non-arch. fields, even alg. closed ones. – BCnrd Oct 26 '10 at 6:30
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? – BCnrd Oct 26 '10 at 12:39