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Let $L$ be a line bundle on a smooth connected complete complex algebraic surface $X$. Assume that $L$ has enough sections i.e. that $H^0(L,X)$ has dimension $> 1$. A nonzero section $s$ of $L$ will cut out some divisor $C_s \subset X$.

Need it be the case that, for generic $s$, the curve $C_s$ is irreducible?

The example I have in mind is that of an elliptic fibration on a K3 surface $X$. Here, we take $L$ to be a line bundle whose $H^0$ induces a flat morphism $\pi: X \longrightarrow \mathbf{P}^1$ where a general fiber is a smooth connected curve of genus 1.

For such an $L$, we can consider its tensor square $L' = L \otimes L$. There are obvious sections $s \otimes s' \in H^0(L',X)$ coming from the product structure, whose divisors are disjoint unions of two fibers $C_s$ and $C_s'$ of $\pi$. But need $L'$ have other sections where the divisor is a single, connected elliptic curve?

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  • $\begingroup$ You should not expect this. In the example that you consider the projection map from $C_s$ to the base has degree $0$. So its image will always be $0$-dimensional, i.e. $C_s$ will always be a union of fibres. $\endgroup$
    – Dmitri Panov
    Commented Jul 12, 2018 at 22:58
  • $\begingroup$ How does this rule out the possibility that $C_s$ might be irreducible? It is a union of fibers, yes. But perhaps the union is one fiber, and that fiber is connected? $\endgroup$
    – Kim
    Commented Jul 12, 2018 at 23:18

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Not in general. Take a high genus hyperellitpic curve $C$ with a degree 2 map $f:C\to\mathbb{P}^1$. Take your surface to be $C\times D$ for some smooth curve $D$ and let $L$ be the pull back of $\mathcal{O}_{\mathbb{P}^1}(1)$ by the obvious map. Then $H^0(L)=2$ which is globally generated and every section vanishes along a reducible curve.

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  • $\begingroup$ Thanks for the example. Do you think it may have a chance of holding for the elliptic fibration case above? $\endgroup$
    – Kim
    Commented Jul 13, 2018 at 1:35
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    $\begingroup$ @Kim No, all the sections of $L^2$ comes from $\mathbb{P}^1$ and so no divisor in this system would be irreducible, since any section of $\mathcal{O}_{\mathbb{P}^1}(2)$ vanishes at two points, counted with multiplicity. $\endgroup$
    – Mohan
    Commented Jul 13, 2018 at 3:01

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