TITLE: Does isomorphic generic fibre imply isomorphic special fibre for smooth morphisms?
QUESTION [14 upvotes]: Let $X$ and $Y$ be regular integral Noetherian schemes. Assume that $X$ and $Y$ are smooth and proper over a base scheme $S=Spec R$, where $R$ is a discrete valuation ring.
If $X$ and $Y$ have isomorphic generic fibres, is it also the case that their special fibres are isomorphic?
Remarks:

The answer is yes when $X$ and $Y$ are abelian schemes (this follows from the theory of the Néron model). In general though it is not the case that morphisms between the generic fibres extend to the special fibre.
I am also particularly interested in case where $R$ is the localisation of $\mathbb{Z}$ at some prime, and hence the generic fibres are smooth proper varieties over $\mathbb{Q}$.

REPLY [7 votes]: In a more positive direction, if the special fiber $X_s$ is not uniruled ruled, then $X_s$ is birational to $Y_s$. 
This can be found in a paper of Ulf Persson (memoirs of AMS 1977) perhaps over complex numbers. But the idea works over excellent rings (e.g. localization of finitely generated $\mathbb Z$-algebras): consider the graph $\Gamma$ of the birational map from $X$ to $Y$ and normalize it. By a theorem of Abhyankar, any irreducible component of the exceptional locus of $\Gamma\to Y$ is ruled, hence doesn't dominate $X_s$ (because they would be birational). Therefore $\Gamma\to X$ and $\Gamma\to Y$ are isomorphic over the generic points of $X_s$ and $Y_s$. So $X_s$ is birational to $Y_s$.