The question in the title can be reformulated as follows. Let $f : Y \to X$ be a monomorphism of algebraic spaces where $X$ is a scheme. Is it true that $Y$ is a scheme?

If $f$ is locally of finite type, then the answer is yes. Namely, any locally quasi-finite, separated morphism of algebraic spaces is representable, see for example Lemma Tag 0418.

If $f$ is flat, then the answer is yes (David Rydh).

In general, we can immediately reduce to the case that $X = \text{Spec}(A)$ and that $Y$ has an \'etale covering by an affine scheme $\text{Spec}(B)$. Then the two maps $B \to B \otimes_A B$ are \'etale. Moreover, we may assume that $A$ is the equalizer of these two ring maps $B \to B \otimes_A B$. If we could show that this forces $A \to B$ to be flat, then we would win using Rydh's result...

  • $\begingroup$ The fact that $B \to B \otimes_A B$ is flat seems relevant. $\endgroup$ – Will Sawin Jan 21 '14 at 2:22

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