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I have a few doubts about Ruled Surfaces.

Edit:

  1. We're working on ground field $\Bbb{C}$
  2. The term ruled stands here for birationally ruled

Let $S$ be an algebraic smooth surface and $C$ a smooth projective curve.

If there is a morphism $p:S\rightarrow C$ with each fiber birational to $\Bbb{P}^1$ let's say that $S$ is a rational fibration. By the Noether-Enriques theorem each rational fibration is a ruled surface.

Claim: If $S\neq\Bbb{P}^2$ then $S$ is a rational fibration if and only if $S$ is ruled.

Assume this is true. Then given a ruled surface $S\neq\Bbb{P}^2$ we consider $S$ as a rational fibration and there can be at most a finite number of fibers which are birational but not isomorphic to $\Bbb{P}^1$. So they are singular irreducible curves (right?). Hence by a finite sequence of blow-ups we get a geometrically ruled surface, i.e. a rational fibration with each fiber isomorphic to the line. Is this argument correct?

Do you think that claim is true?

Edit: Since Asal's answer suggests the claim is false. Let me re-question: suppose $S\neq\Bbb{P}^2$ is birationally ruled. Can we show there exists a morphism $p:S\rightarrow C$ whose generic fiber is isomorphic to $\Bbb{P}^1$ ?

Heitor Fontana
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  • I think I'm missing something here. You say "By the Noether-Enriques theorem each rational fibration is a ruled surface" but then you ask "Can we say that in a rational fibration the generic fiber is isomorphic to P^1 ?" Aren't you answering your own question? –  May 10 '13 at 17:15
  • Sorry @AsalBeagDubh. I'll try to be more clear: if $p:S\rightarrow C$ is a rational fibration we know that $p^{-1}(x)$ is a curve birational to $\Bbb{P}^1$. Can we say that for $x$ generic this curve is also isomorphic to $\Bbb{P}^1$ ? – Heitor Fontana May 10 '13 at 17:22
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    Well, the Noether-Enriques theorem still seems to tell you the answer. However, there's a much simpler reason: if $f:X \rightarrow Y$ is any dominant morphism of nonsingular varieties (over an alg.closed field of char. 0), then the general fibre of $f$ is nonsingular. And a nonsingular curve birational to $\mathbf{P}^1$ is itself $\mathbf{P}^1$. Does that answer your question? –  May 10 '13 at 17:31
  • @AsalBeagDubh. Yes this does. But what is the underlying reason of that? does it follow from Bertini's theorem? – Heitor Fontana May 10 '13 at 17:35
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    Yes, this is what's called "Bertini's second theorem" in Shafarevich's book, II.6.2 Theorem 2. –  May 10 '13 at 17:36
  • A small nitpick: to apply Noether-Enriques theorem, $p$ should be smooth over a point $x\in C$ such that $p^{-1}(x)\cong\mathbb P^1.$ And then, if $p$ is locally smooth, then the fibre over $x$ is smooth, and it is birational to $\mathbb P^1,$ hence isomorphic to $\mathbb P^1,$ which implies that $S$ is ruled. But Noether-Enriques not only gives ruled-ness, it also tells us that over a Zariski open subset of $C$ the map $p$ is just the projection $U\times\mathbb P^1\to U.$ Thus, the general fibre is clearly $\mathbb P^1.$ – Andrew May 10 '13 at 17:51
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    Andrew: I think the crux of this question is whether we're talking about an algebraically closed field of characteristic zero, or not. If we are (as I assumed in my answer), then as I explained, Bertini's theorem guarantees that $p$ is generically smooth. If not, then certainly you need an extra hypothesis. In any case, N-E seems like overkill just to say that the generic fibre is $\mathbf{P}^1$. –  May 10 '13 at 17:59
  • Dear @AsalBeagDubh, I don't disagree with you that Noether-Enriques is not needed to imply the general fibre is $\mathbb P^1,$ but I did want to mention that (1) there seemed to be a missing hypothesis in the OP's application of N-E (which we can patch by second Bertini), and (2) even though the OP mentioned N-E, he didn't seem to use all of it! – Andrew May 10 '13 at 18:20
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    Dear @Andrew, yes, I don't disagree with you either! :) I think we are both working toward the same goal of clarifying the question as far as possible. Thanks for your helpful comments! –  May 10 '13 at 18:26

1 Answers1

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Let me try to answer the modified question. (I assume we are working over an algebraically closed field of characteristic zero.)

The answer to your question depends crucially on what is meant by ruled. (I think this was a source of confusion in the comments.) There are two different defintions I know of, neither one standard:

  • $S$ is geometrically ruled (what Hartshorne calls ruled) if $p: S \rightarrow C$ has every fibre isomorphic to $\mathbf{P}^1$;
  • $S$ is birationally ruled (what some other people called ruled) if $S$ is birational to a product $C \times \mathbf{P}^1$.

Now Noether--Enriques tells you that a rational fibration is birationally ruled (at least once you know there are some smooth fibres, which you get from Bertini's second theorem) but you might worry that there are some singular fibres that stop it from being geometrically ruled. However, this can't actually happen, for the following reason. By your definition of rational fibration, every fibre of $p$ is an irreducible reduced rational curve, but any such curve which is not isomorphic to $\mathbf{P}^1$ must be singular, hence have arithmetic genus greater than 0; however, this cannot happen for a flat family such as $p: S \rightarrow C$. (See Hartshorne for justifications of all these assertions.)

So a rational fibration is the same thing as a geometrically ruled surface. Just to be sure that these are really different from the birationally ruled surfaces, let's give an example: take $p: S \rightarrow C$ a geometrically ruled surface, and let $\pi: S' \rightarrow S$ be the blowup of any point on $S$. Then $p \circ \pi: S' \rightarrow C$ is a birationally ruled surface, but one of its fibres is a reducible curve, so it is not geometrically ruled.

Actually, this last example also highlights a problem with your idea in the final paragraph: blowing up points unfortunately can't turn a fibration with singular fibres into one with smooth fibres, because it introduces extra components into some of the fibres.

I hope this helps.

Edit: Finally I think I can answer the intended question. Let $S$ be a birationally ruled surface, not isomorphic to $\mathbf{P}^2$. Contract $(-1)$-curves on $S$ until we reach a minimal surface $S_m$. So we have a morphism $p: S \rightarrow S_m$. Now according to Enriques' classification, there are three possibilities for $S_m$:

  1. It is a geometrically ruled surface $\pi: S_m \rightarrow C$ over a curve of genus $>0$;

  2. It is a Hirzebruch surace $\pi: S_m \rightarrow \mathbf{P}^1$;

  3. It is $\mathbf{P}^2$.

The first two types are geometrically ruled, so $\pi \circ p$ gives a morphism to a curve with generic fibre $\mathbf{P}^1$. (We get $S$ from $S_m$ by blowing up a finite number of points, so that doesn't affect the generic fibre.)

The only problem is if we arrive at $\mathbf{P}^2$. But in that case we just stop contracting curves at the penultimate step, to get a morphism $p: S \rightarrow \Sigma_1$ where $\Sigma_1$ is the blowup of $\mathbf{P}^2$ in one point. Now $\Sigma_1$ is also a geometrically ruled surface, so we can argue as in the other cases.

And if that doesn't answer the question, I quit! :)

  • Thank you. This does help. What I meant with ruled was birationally ruled. Then, since you say that a rational fibration is isomorphic to a geometrically ruled surface, let me change my question: if $S\neq\Bbb{P}^2$ is birationally ruled can we show that there is a morphism $p:S\rightarrow C$ with generic fiber isomorphic to $\Bbb{P}^1$ ? – Heitor Fontana May 15 '13 at 08:50
  • By the way, I was wondering: in your example you show that $p\circ\pi$ is not geometrically ruling for $S'$, but how can we be sure that there doesn't exist some other gemetrically ruling $q:S'\rightarrow B$ ? – Heitor Fontana May 15 '13 at 09:10
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    @Heitor, to answer your second question: as long as the "birational ruling" (i.e. the morphism with some singular fibres) has at least one fibre which is a reducible curve, then it cannot be geometrically ruled. For a geometrically ruled surface has Picard number 2, while a fibration with reducible fibre must have Picard number at least 3. (To be honest, I am not sure what can happen with a birational ruling that only has non-reduced, but not reducible, fibres.) Again, I'll leave you to find the proofs of those claims. –  May 15 '13 at 14:30