If $X$ is a K3 surface and $X\to C$ is an elliptic fibration, then $C\simeq \Bbb{P}^1$

Question

Sometime ago I heard someone say in a talk that if an algebraic surface $X$ is K3 and has an elliptic fibration $\pi:X\to C$ (where $C$ is a curve), then $C$ must be rational.

The speaker just said this is a simple consequence of the cohomology of $X$.

By definition of K3 surface, $H^1(X,\mathcal{O}_X)=0$ and that's all we know about the cohomology. Is that enough to conclude that $H^1(C,\mathcal{O}_C)=0$? Is that enough to conclude $C\simeq \Bbb{P}^1$ (birational equivalence)?

Maybe this is really trivial, but I'm just not very familiar with sheaf cohomology.

Answer 1

In characteristic zero, use the fact that $H^1(O_X)=0$ if and only if $H^0(\Omega_X)=0$ and the same for $C$. Now if $C$ has a nonzero differential form, you can pull it back to $X$; since we're in characteristic zero the pullback form is still nonzero, a contradiction.