MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello all, I am attempting to understand the proof of Lemma III.8 of Beauville's Complex Algebraic Surfaces:

Let $S$ be a minimal surface, $C$ a smooth curve, $p:S\rightarrow C$ a morphism with generic fibre isomorphic to $\mathbb{P}^1$. Then $S$ is geometrically ruled by $p$.

The proof begins by selecting an arbitrary fibre $F$ of $p$, and observing that $F^2=0$ and $F.K=-2$, where $K$ is the canonical divisor of $S$.

My question is how do we know that $F.K=-2$? The tools that come to mind for deducing this all involve knowing the genus of $F$, which we are trying to calculate.

Thank you for any responses.

share|cite|improve this question
The general fibre, say $G$, and $F$ are algebraically equivalent. Therefore $F\cdot D= G\cdot D$ for any divisor $D$, in particular for $D=K$. Since $G\cong \mathbb{P}^1$, we can use the adjunction formula $G\cdot K=G^2+G\cdot K=2(0)-2$. – Donu Arapura Jan 1 '13 at 22:38
Thank you for the response. To see that $F$ and $G$ are algebraically equivalent, do we just observe that they are fibres over two points, which are topologically equivalent, and so the Chern classes of $F$ and $G$ agree? – Matt Grimes Jan 1 '13 at 23:47
$F$ and $G$ are algebraically equivalent by definition, but you can also use the argument you mention to show that $F\cdot D=G\cdot D$. – Donu Arapura Jan 2 '13 at 1:37
up vote -1 down vote accepted

You use that S is a minimal surface. Thus relatively minimal. Thus, it has no minus one curves. Also, the arithmetic genus is constant in the fibers. The latter is a fairly general fact about fibered surfaces.

share|cite|improve this answer
Ah! So then knowing that the self-intersection number of a fibre is $0$ and Riemann-Roch tell us that $K.F=K.G$ for a general fibre $G$? Thank you for the response. – Matt Grimes Jan 1 '13 at 23:56
The fact that $S$ is minimal is irrelevant here, and Riemann-Roch is not used, either. The point, as explained in Donu's comment above, is that the intersection number is independent of the fiber. – rita Jan 2 '13 at 7:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.