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.



Basics of minimal Elliptic Surfaces [following Beauville]
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 $D= \sum\_{i=1}^k r\_i F\_i$.


