# Let $S \subset \mathbb{P}^3$ be a quartic containing a line $l$, $H$ a hyperplane section of $S$, then $|H-l|$ is a pencil of elliptic curves.

Reading the proof of Proposition VIII.15 in Beauville's "Complex algebraic surfaces", I got stuck with the following fact he is using:

Let $S \subset \mathbb{P}^3$ be a quartic containing a line $l$, $H$ a hyperplane section of $S$, then $|H-l|$ is a pencil of elliptic curves. Why is this?

He also uses the fact that if $Q \subset \mathbb{P}^4$ is a quadric with an ordinary double point, and $V \subset \mathbb{P}^4$ is a cubic such that $Q \cap V$ is a smooth surface, then one of the two pencils of planes on $Q$ cuts out on $V$ a pencil of elliptic curves. Again, why would this be true? Thanks for your help.

We have an exact sequence $0\mathcal{O}_S(H-l)\to \mathcal{O}_S(H)\to \mathcal{O}_l(H)\to 0$ and taking cohomologies, and noting that $H^0$ surjects and $H^0(\mathcal{O}_S(H)$ has dimension 4 and $H^0(\mathcal{O}_l(H))$ has dimension 2, we see that $H-l$ is a pencil. Genus of $H-l$ is 1 follows from adjunction, since $(H-l)^2=0$.
• Or more simply, the family of planes in $\mathbb P^3$ containing a line forms a pencil; as the plane varies, it cuts the quartic in a degree-4 plane curve, one component of which is a line. The residual curve is a plane cubic (generically a smooth elliptic curve, at least in characteristic 0, by Bézout). Mar 14 '16 at 17:10
• So "$H^0$" surjects comes from the fact that $H^1(\mathcal{O}_S(H-l))=0$? Why is that? Mar 14 '16 at 18:32
• No, the composite $H^0(\mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(H))\to H^0(\mathcal{O}_S(H))\to H^0\mathcal{O}_l(H))$ is surjective, since $l$ in a line and thus the second map is surjective. Mar 14 '16 at 18:44