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

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.

share|cite|improve this question
Are gentlewomen allowed to write? – Matthieu Romagny May 19 '13 at 11:23

1 Answer 1

up vote 6 down vote accepted

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 $D= \sum_{i=1}^k r_i F_i$.

share|cite|improve this answer

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.