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?


  1. 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.

  2. 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 $\mathbb{Q}$.


3 Answers 3


Here is an example showing the answer is no:

Start with $Z =\mathbb{P}^2_R$, $R$ an arbitrary dvr. Let $P$ be a section of $Z \to Spec(R)$ and let $W$ be the blowup of $Z$ along the image of the section (so both fibres are $\mathbb{P}^2$ with a point blown up). Let $Q$ be a section of $W \to Spec(R)$ whose image does not intersect the exceptional divisor of the first blow up and let $Q'$ be a section whose image intersects the exceptional divisor only in the special fibre. Let $X$ be the blow up of $W$ along $Q$ and $Y$ the blow up of $W$ along $Q'$.

The generic fibres of $X$ and $Y$ are isomorphic since they are both just $\mathbb{P}^2$ over some field blown up in two distinct (rational) points and any two points on $\mathbb{P}^2$ over a field are "the same". That the special fibres are not isomorphic can be seen by considering curves of self intersection $-2$:

The special fibre of $X$ has none since we have blown up two distinct points, so the only curves with negative self intersection are three $(-1)$-curves i.e. the two exceptional divisors and the strict transform of the line joining the two points. The special fibre of $Y$ does have one; this is the strict transform in the second blowup of the exceptional divisor of the first blowup. (The second blow up changes the $(-1)$-curve into a $(-2)$-curve since we blow up a point on it.)

  • $\begingroup$ Excellent answer thanks! I think I have come across this example before in a different context, don't know why I didn't think of it. $\endgroup$ Sep 14, 2011 at 14:57

It has already been pointed out that the answer is no and there is a good example to show why. On the other hand it may be worth noting that, as in the example, essentially the only way this fails is if the special fibers have ruled components.

More precisely, a theorem of Matsusaka and Mumford says that if $X$ and $Y$ are projective over the base and there are ample relative divisors (one on each) that correspond to each other on the general fiber via the isomorphism, and the special fibers do not have ruled components, then they are also isomorphic.

For the precise statement see Theorem 2 of Matsusaka's and Mumford's paper.

The philosophical point of this question/statement is that whether this holds for a class of varieties/schemes is essentially the same as whether their moduli problem is separated, that is, if there exists a moduli space/stack/etc for them, then that moduli space is separated.

Matsusaka-Mumford's theorem implies that moduli spaces of smooth polarized varieties (at least over an algebraically closed field) are separated. If one wants to deal with more moduli spaces (say stable curves/varieties) then one needs more modern tools such as Kollár-Shepherd-Barron's results and others.

A good reference for related questions is Viehweg's book on moduli.

  • $\begingroup$ Thanks for this extra answer, it is most useful to know. $\endgroup$ Sep 14, 2011 at 19:35
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    $\begingroup$ To be precise, in Matsusaka-Mumford theorem, the relative divisors must be ample. $\endgroup$
    – Qing Liu
    Sep 15, 2011 at 8:08

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 $Y_s$.

  • 3
    $\begingroup$ This already occurs in the Matsusaka-Mumford paper mentioned by Sándor Kovács. $\endgroup$
    – naf
    Sep 15, 2011 at 9:55
  • $\begingroup$ Yes you are right, this is their Theorem 1. Thanks ! $\endgroup$
    – Qing Liu
    Sep 15, 2011 at 11:31

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