Is an algebraic space over a DVR, whose special fibre (and all its infinitesimal neighborhood) and generic fibre are schemes, actually a scheme?



Is an algebraic space over a DVR, whose special fibre and generic fibre are schemes, actually a scheme?
The answer is negative. Let me give a counter-example by modifying the example of a singular complex algebraic surface that is not a scheme given by Knutson in [Algebraic Spaces, p.21-22].


Let me work over $\mathbb{C}$. Consider the pencil in $\mathbb{P}^2$ generated by two smooth cubic curves $C$ and $C'$ intersecting transversally in nine points $P\_1,\dots, P\_9$. Blowing up these nine points, we get a morphism $X\to\mathbb{P}^1$ whose fiber over $0$ is the elliptic curve $X\_0=C$. Choose, as usual, an inflection point $O\in C$ as the origin of the group law on $C$.


Let $\hat{X}$ be the blow-up of $X$ in a tenth point $Q\in X\_0=C$ (chosen generic, so that no multiple of $Q$ is in the subgroup of $C$ generated by $P\_1,\dots, P\_9$ : here, we use the uncountability of the base field). At this point, the strict transform of $X\_0$ in $\hat{X}$ has negative self-intersection and may be contracted to a point $y$ in a surface $Y\to\mathbb{P}^1$. Let $T$ be the local ring of $\mathbb{P}^1$ at $0$. I claim that $Y\_T\to T$ is the counter-example we are looking for.


First, the fibers of $Y\_T\to T$ (and their infinitesimal neighbourhoods) are schemes because they are one-dimensional [Algebraic Spaces V 4.9]. However, if $Y\_T$ were a scheme, it would be possible to find a curve $D$ in $Y$ intersecting the tenth exceptional divisor, but not containing $y$. Its strict transform in $\mathbb{P}^2$ would be a plane curve meeting $C$ only at $P\_1,\dots, P\_9, Q$. By the choice of $Q$, this plane curve would meet $C$ only at $P\_1,\dots, P\_9$. This contradicts the fact that $D$ intersects the tenth exceptional divisor.


