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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 $X_s$. 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$. 2 typo In a more positive direction, if the special fiber$X_s$is not uniruled, then$X_s$is birational to$Y_s$. This can be found in a paper of Ulf Persson (memoirs of AMS 1977) perhapes perhaps over complex numbers. But the idea works over excellent rings (e.g. localization of finitely generated$\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$. Therefore$\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$. 1 In a more positive direction, if the special fiber$X_s$is not uniruled, then$X_s$is birational to$Y_s$. This can be found in a paper of Ulf Persson (memoirs of AMS 1977) perhapes over complex numbers. But the idea works over excellent rings (e.g. localization of finitely generated$\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$. Therefore$\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\$.