Let $S, S'$ be flat schemes over a DVR. Their generic fibers are isomorphic and their special fibers are isomorphic as well. Does that imply $S$ and $S'$ isomorphic? If not, what can go wrong?
Thanks in advance!
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2$\begingroup$ On a more down-to-earth level, it seems like the following should work: take two projective varieties that are birational but not isomorphic. Take affine opens, and embed them in affine space so that they have the same points with (say) first coordinate 0. Consider projection to the first coordinate as a map to $\mathbb{A}^1$, making them schemes over $\mathbb{C}[x]$, and localize at (x). Then the generic fibers are the same since they're birational, and the special fibers are the same since they have the same points over 0. $\endgroup$– Kevin CastoCommented Nov 27, 2016 at 5:49
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1$\begingroup$ @KevinCasto: how do you make sure that this localisation doesn't accidentally make them isomorphic? $\endgroup$– R. van Dobben de BruynCommented Nov 27, 2016 at 7:15
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6$\begingroup$ What is the motivation for this question? Let $A$ be a dvr with fraction field $F$, and $K/F$ a finite separable extension of degree $d > 1$ with no non-trivial autmorphism, and suppose the maximal ideal of $A$ is totally split in the integral closure $B$ of $A$ in $K$. (Algebraic curves give many such.) Let $S$ and $S'$ be complements of distinct closed points in ${\rm{Spec}}(B)$. Their generic fibers are uniquely $F$-isomorphic since ${\rm{Aut}}(K/F)=1$, so $S$ and $S'$ are not $A$-isomorphic (missing different closed points from ${\rm{Spec}}(B)$) but their special fibers are isomorphic. $\endgroup$– nfdc23Commented Nov 27, 2016 at 7:33
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2$\begingroup$ @nfdc23: This is an answer, right? Why posting it as a comment? $\endgroup$– HeinrichDCommented Nov 27, 2016 at 8:15
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2$\begingroup$ Well, your comment precisely answers the question as given. When people come to mathoverflow sites they tend to read the answers first, and it is not so good if valuable answers are hidden in the comments. Luckily, good comments get upvoted many times so that people also consider reading them. Also notice that there is no possibility to easily find again comments hidden in answers. Comments have smaller font size and cannot be edited later on. It doesn't really matter if the answer is optimal or not, usually every answer is valuable for many people. $\endgroup$– HeinrichDCommented Nov 27, 2016 at 18:12
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3 Answers
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Here is a counterexample (vaguely inspired by Kevin Casto's approach, and very similar to nfdc23's example). The moduli space $\mathcal M_g$ of smooth genus $g \geq 2$ curves is irreducible, and has a closed subset corresponding to curves admitting a nontrivial automorphism (this is a strict subset if $g \geq 3$). Thus we can form a family whose general member has no automorphisms, but a special member is, say, hyperelliptic.
That is, we can produce a family $\mathscr C \to \operatorname{Spec} \mathbb C[x]_{(x)}$ whose generic fibre $C$ has no automorphisms and whose special fibre $C_0$ has a hyperelliptic involution $\sigma \colon C_0 \to C_0$ that does not lift to an automorphism of $C$. Let $p \in C_0$ be a point not fixed by $\sigma$, and consider the flat (but non-proper) families $\mathscr C \setminus\{p\}$ and $\mathscr C\setminus\{\sigma(p)\}$ over $\operatorname{Spec}\mathbb C[x]_{(x)}$.
Now the generic fibres are both $C$, and the special fibres are $C_0 \setminus\{p\}$ and $C_0\setminus\{\sigma(p)\}$, which are isomorphic through $\sigma$. But a global isomorphism would induce an automorphism of $C$, which therefore has to be the identity on $C$. The only extension of the identity to a map $\mathscr C \setminus \{p\} \to \mathscr C$ is the inclusion, which does not land in $\mathscr C \setminus\{\sigma(p)\}$. (Informally: '$p$ cannot go to $\sigma(p)$'.)
Remark. I do not have an example of a proper family. In fact, all the counterexamples suggested so far are non-proper.
answered Nov 27, 2016 at 7:42
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3$\begingroup$ It's possible to do this also with genus 1 curves and subtracting 2 points. Say, the family $y^2=x^3-x-t$ with the point at $\infty$ and a point $(1,0)$ removed from one copy and the point at $\infty$ and $(-1,0)$ removed from the author. None of the automorphisms that send one pair of points to the other pair lift, because the differences of the two pairs are two different $2$-torsion points on the Jacobian, and the only liftable automorphisms of the Jacobian, the identity and minus the identity, fix all two-torsion points. $\endgroup$ Commented Nov 27, 2016 at 8:34
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1$\begingroup$ This gives simple explicit equations, and it can be used to make proper non-smooth examples, by gluing the points instead. $\endgroup$ Commented Nov 27, 2016 at 8:35
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Let $\mathcal{X}\rightarrow \textrm{Spec}\,\mathbb{C}[[t]]$ be a flat family of K3 surfaces such that the central fiber $X_0$ has a $-2$ curve $C$. Then one can perform a flop along $C$ inside the threefold $\mathcal{X}$, thus producing another family $\mathcal{X}'\rightarrow \textrm{Spec}\,\mathbb{C}[[t]]$. These families are not isomorphic even though they are fiberwise isomorphic (i.e. the general fibers and special fibers are individually isomorphic). This is related to the fact that the moduli stack of smooth K3 surfaces is not separated. To produce a separated moduli space, one must contract ADE configurations of $-2$ curves, of which the above example is the simplest.
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2$\begingroup$ Specifically you need the normal bundle of $C$, which lies in an exact sequence $\mathcal O_C(-2) \to N_C \to \mathcal O_C$, to be $\mathcal O_C(-1) + \mathcal O_C(-1)$ and not $\mathcal O_C(-2) + \mathcal O_C$ to perform the flop, right? $\endgroup$ Commented Nov 27, 2016 at 8:58
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1$\begingroup$ True. Equivalently, the family must be chosen so that $C$ does not admit a first order thickening over $\textrm{Spec}\,\mathbb{C}[[t]]$, i.e. should be transverse to the locus of K3s with a $-2$-curve. $\endgroup$ Commented Nov 27, 2016 at 9:08
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If we are all pitching in our favorite counterexamples, consider a smooth, projective morphism $\pi:S\to \text{Spec}(R)$ over a DVR whose generic fiber is the Hirzebruch surface $\mathbb{P}^1\times \mathbb{P}^1$ and whose special fiber is the Hirzebruch surfaces $\Sigma_2 = \mathbb{P}_{\mathbb{P}^1}(\mathcal{O}(-1)\oplus \mathcal{O}(+1))$. For such a family, there is an integer invariant; roughly the order of vanishing at the closed point of the DVR of the "modulus" of the moduli space of Hirzebruch surfaces. More precisely, it is the length of the torsion sheaf $R^1\pi_* \textit{Hom}_{\mathcal{O}_S}(\Omega_\pi,\mathcal{O}_S)$. This integer invariant can take on any positive integer value. Thus, there exist families $S$, $S'$ as above such that the invariant is $1$, respectively $2$.