Let $X$ be a normal, projective surface (I'm happy to work over $\mathbb{C}$) and let $p: X \to S$ be a flat morphism to a smooth projective curve $S$. Suppose that for a dense open subset of $s \in S$, the fibers $X_s$ are isomorphic to $\mathbb{P}^1$. Is there anything we can say about the special fibers? I'm trying to understand the proof of Mori's "bend-and-break" lemma (e.g., Proposition 7.3 in Debarre's notes), where it is claimed that the special fiber $X_t$ has the property that its reduction is always a union of smooth rational curves glued together.

The arithmetic genus is constant in a flat family, so $X_t$ has arithmetic genus zero: that is, the Euler characteristic of the structure sheaf is equal to one. Moreover, since $X$ is normal of dimension two, it is Cohen-Macaulay, and each fiber is therefore Cohen-Macaulay as well. However, I don't see why the special fiber couldn't be, say, an elliptic curve but with a generically nonreduced structure sheaf to push the Euler characteristic up to 1. What am I missing?

  • $\begingroup$ If the special curve is generically nonreduced, then won't we get too many singular points on the normal surface? $\endgroup$ – Andrew Mar 18 '13 at 14:45
  • $\begingroup$ @Andrew: I don't think this is true in general. Consider for instance the map $\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$ where the map is projection followed by squaring; the fiber over zero is generically nonreduced (it's the dual numbers times $\mathbb{P}^1$). $\endgroup$ – Akhil Mathew Mar 18 '13 at 14:47
  • $\begingroup$ Ah, I see, that's a good example. $\endgroup$ – Andrew Mar 18 '13 at 14:53
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    $\begingroup$ A proof of what you want which is a bit more straightforward than David Speyer's is in Koll'ar, Rational Curves on Algebraic Varieties, Proposition II.2.2. The basic idea seems to be to take an arbitrary point $s \in S$, take $T$ to be the spectrum of its local ring, and resolve the birational map $\mathbf{P}^1_T \dashrightarrow X$. $\endgroup$ – user64687 Mar 18 '13 at 18:22
  • $\begingroup$ @AsalBeagDubh: In Kollar, rational curves can be singular. Also, in general there is no non-constant map from $\mathbb P^1_T$ to $X$. $\endgroup$ – user18119 Mar 18 '13 at 22:44

Let $F=X_t$ be a closed fiber of $X\to S$. Let $F_r=F_{\mathrm{red}}$.

Let $k$ be the ground field, supposed to be algebraically closed. The crucial point is to show that $H^0(F, O_F)=k$. Once we prove this, as in the notes you are reading, we have $H^1(F, O_F)=0$ by the invariance of the arithmetic genus in a flat family; the surjective homomorphism $O_F\to O_{F_{r}}$ implies that $H^1(F, O_F)\to H^1(F_{r}, O_{F_r})$ is surjective because $\dim F=1$, hence $F_r$ has arithmetical genus $0$. Then it is easy to show that $F_r$ is a transversal (not necessarily semi-stable) union of projective lines.

The property $H^0(F, O_F)=k$ is called cohomological flatness in dimension $0$. This property is studied in Raynaud's fondamental paper "Spécialisation du foncteur de Picard", IHES 1970. The definition is given in 1.4. Then in 6.1.4 he defines the condition (N) (I don't know why 'N') which is satisfied here by $F=X_t$. In 6.4.2, he shows that $X\to S$ is cohomologically flat if $k$ has characteristic $0$. This proves what you want.

Some remarks:

  1. If $X$ is smooth over $k=\mathbb C$, there is a direct proof of cohomological flatness in Xiao's Lectures notes (somewhere in the begining) and it is probably well known. Unfortunately, one can not reduce the case $X$ normal to the case $X$ smooth just by desingularization.

  2. If $X$ is regular and if the generic fiber of $X\to S$ is also a $\mathbb P^1$ (it is a smooth conic in general if $S$ is just is a regular one-dimensional scheme), then $X$ is obtained by successive blowing ups of closed points starting from $\mathbb P^1_S$, so all irreducible components of $F$ are projective lines.

  3. In general, it is not true (even when $X$ is smooth over $k$ of positive characteristic and $F$ is irreducible) that $H^0(F, O_F)=k$. So in the note you are reading, I think it misses some arguments (it is said that $F$ has no embedded point, so $H^0(F, O_F)⁼k$), unless something in characteristic $0$ is implicit there.

  • $\begingroup$ Thanks for this answer. I haven't gotten too far yet in Raynaud's paper, but on point 2, unless I've misunderstood, a smooth conic over the function field of a curve is always a $\mathbb{P}^1$ by Tsen's theorem. Could you spell out the argument in this case? $\endgroup$ – Akhil Mathew Mar 18 '13 at 22:54
  • $\begingroup$ You are right, the Brauer group of $k(S)$ is trivial. I was thinking over a general regular scheme of dimension $1$. When the generic fiber has a rational point, it specializes to a smooth point of $F$ (if $X$ is regular), the component of $F$ passing through this point then has multiplicity $1$. Then see, for instance, "Algebraic geometry and arithmetic curves", Cor. 9.1.24, and 9.1.32 for the assertion on the specialization of a rational point of the generic fiber. $\endgroup$ – user18119 Mar 18 '13 at 23:01
  • $\begingroup$ Thanks. But in the non-smooth case, it seems a different argument becomes necessary. I'll keep looking at Raynaud's paper. $\endgroup$ – Akhil Mathew Mar 19 '13 at 0:40

From an advanced perspective, this follows from a variant of semistable reduction which Harris and Morrison call Nodal Reduction. I quote from their textbook Moduli of Curves, Proposition 3.49.

Let $B$ be a smooth curve, $0$ a point of $B$ and $B^{\ast} = B \setminus \{ 0 \}$. Let $X \to B^{\ast}$ be a flat family of nodal curves of genus $g$, $\psi:X \to Z$ any morphism to a projective scheme $Z$, ...

Then there exists a branched cover $B' \to B$ and a [projective] family $X' \to B'$ of nodal curves extending the fiber product $X \times_{B^{\ast}} B'$ with the following properties:

The total space $X'$ is smooth.

The morphism $X \times_{B^{\ast}} B' \to X \overset{\psi}{\longrightarrow} Z$ extends to a morphism on all of $X'$. ...

The word "projective" in square brackets in the second line is my addition but was clearly intended, since otherwise we don't have to fill in the central fiber at all. The ellipses conceal conditions concerning marked points, which we won't need.

Let $\bar{X} \to B$ be a flat projective family, whose fibers over $B^{\ast}$ are reduced and nodal. I will take $Z=\bar{X}$ with $\psi$ the inclusion $X \hookrightarrow \bar{X}$. So I get to conclude that there is a branched cover $B' \to B$, and a nodal family $X'$ over $B'$, so that there is a map $\phi: X' \to \bar{X}$. This theorem is stronger than stable and semi-stable reduction in that the new family maps to the old family, but weaker in that we have less control over the fibers of $X'$; they are reduced curves with nodal singularities, but there may be rational components with only one node.

In particular, in your setting, we have a map $\phi: X'_0 \to \bar{X}_0$. Since every component of $X'_0$ is genus $0$, if $\bar{X}_0$ has higher genus, the map $\phi$ is constant. So $\phi(X')$ only meets the central fiber of $\bar{X}$ at only one point. This is absurd, the next paragraph gives one way to turn it into a contradiction.

But $X' \to B'$ is projective, hence proper, and $B' \to B$ is finite, hence proper, so $X' \to B$ is proper. So $X' \times_B \bar{X} \to \bar{X}$ is a closed map. But $X' \times_B \bar{X} = X'$ and the map $X' \times_B \bar{X} \to \bar{X}$ is just $\phi: X' \to \bar{X}$, so we deduce that $\phi$ is a closed map and $\phi(X')$ is closed. As $\phi(X')$ contains all of $X \times_{B'} B^{\ast}$, the map $\phi$ must be surjective, contradiction.

  • $\begingroup$ I expect there is a more elementary answer; I look forward to learning it! $\endgroup$ – David E Speyer Mar 18 '13 at 16:03
  • $\begingroup$ It can (and this does) happen that $X_0$ is singular and $X'_0\to X_0$ is the normalization map. In this case, you just proved that $X_0$ is rational, not smooth. $\endgroup$ – user18119 Mar 18 '13 at 21:37
  • $\begingroup$ Thanks for this answer. I wasn't aware of this version of semistable reduction. $\endgroup$ – Akhil Mathew Mar 18 '13 at 22:42
  • $\begingroup$ @QiL'8 You are, of course, correct. Did I write something that suggested otherwise? $\endgroup$ – David E Speyer Mar 19 '13 at 0:54
  • $\begingroup$ Dear David, I am a little confused. In your initial setting, are the fibers of $X\to B$ reduced ? $\endgroup$ – user18119 Mar 19 '13 at 6:18

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