Question

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?

Answer 1

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.)