MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 2 deleted 967 characters in body

I don't know if it works, but my stupid idea to find an example is the following: take a (relative) curve $f:C\to S$ where $S=Spec(R)$, $R=dvr$. Take $C$ integral and normal (or even regular, it's better) $f$ flat and projective with smooth generic fibre that we suppose geometrically integral. We also take a section $x\in C(S)$ so that everything is nice. Now consider the generic fibre $C_K$ and its schematic closure $D$ on $\mathbf{Pic}_C^0$ (which is a separated $S$-scheme in this case). Then $D$ is $S$-flat, integral, with smooth generic fibre and closed subscheme of $\mathbf{Pic}_C^0$, then separated. If necessary we can desingularize $D$ in order to obtain $D'$ regular and $S$-flat with generic fibre isomorphic to $C_K$ but $D'$ is not necessarily projective

Message deleted because false.I would like to say that $D'$ and $C$ have the same $\mathbf{Pic}_C^0$ and the same "Picard algebraic space"...can this make sense? Can the reduction to special fibre solve the problem asked at the beginning of this forum?
show/hide this revision's text 1

I don't know if it works, but my stupid idea to find an example is the following: take a (relative) curve $f:C\to S$ where $S=Spec(R)$, $R=dvr$. Take $C$ integral and normal (or even regular, it's better) $f$ flat and projective with smooth generic fibre that we suppose geometrically integral. We also take a section $x\in C(S)$ so that everything is nice. Now consider the generic fibre $C_K$ and its schematic closure $D$ on $\mathbf{Pic}_C^0$ (which is a separated $S$-scheme in this case). Then $D$ is $S$-flat, integral, with smooth generic fibre and closed subscheme of $\mathbf{Pic}_C^0$, then separated. If necessary we can desingularize $D$ in order to obtain $D'$ regular and $S$-flat with generic fibre isomorphic to $C_K$ but $D'$ is not necessarily projective. I would like to say that $D'$ and $C$ have the same $\mathbf{Pic}_C^0$ and the same "Picard algebraic space"...can this make sense? Can the reduction to special fibre solve the problem asked at the beginning of this forum?