Basics of minimal Elliptic Surfaces [following Beauville]

I am reading Beauville's chapter IX on Elliptic surfaces.

Let $S$ be a minimal elliptic surface with $\kappa=1$ and $p:S\rightarrow C$ be the elliptic fibration.

We know $K^2=0$. Suppose the $m$-canonical system is non-empty and let $D\in \lvert m K \rvert$.

Then $D.F=0$, where $F$ is a fiber of $p$, by the genus formula. So the components of $D$ are contained in the fibers. I don't get the following:

Since $K^2=0$ we have $D=\sum r_i F_i \quad$ with $r_i\in\Bbb{Q}$ and $F_i$ fibers.

(he says this comes from Proposition VII.4, but this is probably a misprint, since that proposition has nothing to do with this situation. I thought it could follow from VIII.3 or VIII.4, but it doesn't seem any obvious to me -actually it seems that we'd get $D=r F$, which he does not use).

Sigh... I don't see why this holds. Any hint gentlemen? Thank you.

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Are gentlewomen allowed to write? – Matthieu Romagny May 19 '13 at 11:23

You are right, this follows from VIII.4.

The point is that you do not know if $D \in |mK|$ is connected.

If it is, then by VIII.4 and $D^2=0$ one deduces $D=rF$ with $r \in \mathbb{Q}$.

Otherwise, write $$D=D_1+D_2 + \ldots +D_k,$$ where the $D_i$ are the connected components. Then $D_i D_j=0$ for $i \neq j$, hence $D^2=0$ implies $\sum_{i=1}^k D_i^2 =0$.

Now the connectedness of $D_i$ implies that is contained in a single fibre, hence $D_i^2 \leq 0$; this gives $D_i^2=0$ for all $i$.

Therefore there exists $r_i \in \mathbb{Q}$ and a fibre $F_i$ such that $D_i= r_iF_i$, that is $D= \sum_{i=1}^k r_i F_i$.

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