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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.

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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.

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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

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