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Let $\mathbb{k}$ be an algebraically closed field of characteristic 0. Denote $S=\mathrm{Spec}\:\mathbb{k}[t]$, $U=\mathrm{Spec}\:\mathbb{k}[t, t^{-1}]$, $Z=\mathrm{Spec}\:\mathbb{k}[t]/(t)$.

What is the minimum integer $n$ such that there exist smooth projective morphisms of relative dimension $n$ $X\rightarrow S$, $Y\rightarrow S$ such that there is no $Z$-isomorphism $X_{Z}\rightarrow Y_{Z}$ but there is an $S$-morphism $X\rightarrow Y$ inducing a $U$-isomorphism $X_{U}\rightarrow Y_{U}$?

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    $\begingroup$ I don't think this is possible. The morphism $X_Z\rightarrow Y_Z$ would contract some subvariety, hence it would not induce an isomorphism on cohomology, while your hypotheses imply that it does. $\endgroup$
    – abx
    May 28 '20 at 16:26
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    $\begingroup$ @abx how does your cohomology argument work? Is it for coherent cohomology of the structure sheaf? $\endgroup$
    – user145520
    May 28 '20 at 17:35
  • $\begingroup$ No, for cohomology with rational (if you are over $\Bbb{C}$) or $\ell$-adic coefficients. $\endgroup$
    – abx
    May 28 '20 at 18:56
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I believe that this never happens. The reason is as follows. The morphism $f:X\to Y$ is a small birational morphism (it contracts some divisors on the special fiber) and so by standard arguments $Y$ is not $\mathbb Q$-factorial which is a contradiction as $Y$ is smooth.

To see that $Y$ is not $\mathbb Q$-factorial, let $A$ be a relatively ample divisor, then if $f_*A$ is $\mathbb Q$-Cartier and so $A=f^*f_*A$ (as $f$ is small), but $f^*f_*A$ can not be relatively ample.

Far reaching generalizations of this argument are in https://arxiv.org/pdf/0901.0389.pdf. Counterexamples when $Y$ has mild singularities are in Remark 4.4 of that paper (dim 2) and in Section 4 of Wilson's paper (dim 3).

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  • $\begingroup$ You can also get this out of the purity theorem in SGA 2 (this gives a way to prove a version of the result over a mixed characteristic DVR as well as over $k[t]_{\langle t \rangle}$). $\endgroup$ Jun 6 '20 at 4:16
  • $\begingroup$ @JasonStarr could you elaborate? $\endgroup$
    – user145520
    Jun 6 '20 at 16:29

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