TITLE: Is an algebraic space over a DVR, whose special fibre and generic fibre are schemes, actually a scheme?
QUESTION [7 upvotes]: Is an algebraic space over a DVR, whose special fibre (and all its infinitesimal neighborhood) and generic fibre are schemes, actually a scheme?

REPLY [10 votes]: Olivier's example is perfect, but let me just point out that counterexamples are easier to construct if you allow nonseparated spaces.
For example, start with two DVRs $R\hookrightarrow R'$ (with spectra $X'\to X$) such that $R'$ is finite free of rank $2$ over $R$, and the fraction field extension $K\hookrightarrow K'$ is separable. 
The constant group scheme $(\mathbb{Z}/2\mathbb{Z})_X$ acts naturally on $X'$. Let $G$ be the open subgroup scheme obtained by removing the  nontrivial point over the closed point of $X$. Put $Y:=X'/G$. The morphism $Y\to X$ is an isomorphism on the generic points, but has the same closed fiber as $X'$.
If $R\hookrightarrow R'$ is split unramified, then $Y$ is the familiar ``$X$ with the closed point doubled'' wich is a scheme. Otherwise, the closed fiber of $Y\to X$ is a one-point scheme, which has no affine (or even separated) Zariski neighborhood in $Y$. In particular, $Y$ is not a scheme.
You can check that $Y$ is locally separated (in Artin's sense: the diagonal map is an immersion) iff $R\hookrightarrow R'$ is unramified.