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Say I have an algebraic surface $S$ (smooth, projective over $\mathbb{C}$) and a morphism $f: S \to B$ to some curve $B$ (smooth, projective over $\mathbb{C}$) defined by the complete linear system corresponding to some divisor $D$ on $S$. We assume this linear system to be base point free.

Is it true that $D$ will be a sum of fibres of $f$ and if so, why?


In fact I'm just trying to understand the proof of Proposition IX.2.(b) in Beauville's book on algebraic surfaces. The proof goes as follows:

(Here $S$ is a smooth projective surface over $\mathbb{C}$ with $\kappa(S) = 1$. Moreover, $M$ is some effective divisor such that $|M|$ is base point free - corresponding to the mobile part of $|nK_S|$ for some $n$ such that the $n$-th plurigenus is at least $2$ - and $K \cdot M = 0$ where $K$ is the canonical divisor.)

"(...) It follows that $|M|$ defines a morphism $f$ from $S$ to $\mathbb{P}^N$ whose image is a curve $C$. Consider the Stein factorization $f: S \to B \to C \subseteq \mathbb{P}^N$, where $p: S \to B$ has connected fibres. Let $F$ be a fibre of $p$. Since $M$ is a sum of fibres of $p$ and $K \cdot M = 0$, we must have $K \cdot F = 0$. (...)"

There are three statements which are not clear to me:

*Why should $M$ be a sum of fibres of $p$?

*Why does $K \cdot M = 0$ imply that $K \cdot F = 0$ for ANY fibre $F$?

*Why on earth should the curve $B$ coming out of this construction be smooth? I mean, all we did is starting with a smooth, projective surface $S$ with $\kappa(S) = 1$, taking the mobile part of the linear system $|nK_S|$ for some $n$ such that the $n$-th plurigenus is $\geq 2$ (which is base point free), giving a curve $C$ which is the image of the corresponding pluricanonical map, and then taking the curve $B$ coming out of the Stein factorization. I see no reason why $B$ should be smooth, I'm really lost here.


On a related note, in the proof of Proposition IX.3:

"(...) Let $S$ be a minimal elliptic surface. Since $K_S \cdot F_b = 0$ for all $b \in B$, the maps contract the fibres $F_b$. It follows that the image of $\varphi_{nK}$ has dimension $\leq 1$. (...)"

Why is this true?

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2 Answers

This looks like very simple. By construction $D \sim f^*E$ for some ample divisor $E$ on $B$.

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So $D$ linearly equivalent to a sum of vertical divisors. –  user18119 Jun 9 '12 at 20:12
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Here's an attempt to clarify the three statements.

Why should $M$ be a sum of fibres of $p$?

$f=\varphi_M$ is the morphism associated to $|M|$ and $f$ factorizes with $p$ plus an embedding of the curve, so geometrically you have to think of the fibers of $p$ as fibers of $f=\varphi_M$. Now, the condition $M^2=0$ geometrically means that the curves in $|M|$ are vertical: indeed a fiber of $p$ (read: of $f$) will be an element in $|M|$, and has self-intersection zero. In other words the components of $M$ are contained in some fibers of $p$. In symbols this means we can write $$ M=\sum r_F F $$ where $r_F\geq 0$ and $F$ are the fibers of $p$.

Why does $K \cdot M = 0$ imply that $K \cdot F = 0$ for ANY fibre $F$?

Observe $K\cdot F\geq 0$ for any fiber since $K$ is nef. Thus $$ 0=K\cdot M=\sum r_F K\cdot F $$ which gives $K\cdot F=0$.

Why on earth should the curve $B$ coming out of this construction be smooth?

Yes, it could also be singular. If that's the case then just replace $B$ with its desingularization. You still have your morphism $p:S\rightarrow B$ onto a smooth curve, with the desired properties.

Hope that helps :)

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