# Abelian Elliptic Surfaces

By abelian surface we mean a 2-dimensional algebraic complex torus. Thus $$S=\Bbb{C}^2/\Gamma$$ where $\Gamma$ is a rank $4$ lattice in $\Bbb{C}^2$ and such that $S$ is algebraic. It has trivial canonical bundle $K=0$, so all plurigenera $P_n$ are equal to $1$. Also, topologically $S$ is a $4$-torus so one finds $q(S)=2$.

By elliptic surface we mean that there exists a fibration $S\rightarrow C$ on a smooth curve $C$, such that the generic fiber is an elliptic curve.

In examples (ix.4) on Beauville's book he claims: an abelian surface $S$ is elliptic if and only if there are two elliptic curves $E,F$ and an exact sequence $$0\rightarrow E\rightarrow S\rightarrow F\rightarrow 0$$

Any hint to understand why is this true? Thank you.

Suppose an abelian surface $S$ is elliptic, so there is a fibration $S\rightarrow C$ s.t. the generic fiber is an elliptic curve. We pick a generic fiber $f$ which is an elliptic curve, then we have the closed embedding $f\rightarrow S$.
By Proposition V.12, any morphism between two complex tori is a composition of a translation and a group homomorphism, so there is a translation $E$ of $f$ such that $E$ is a subgroup of $S$. Since $S$ is an abelian surface, $S/E$ is a one-dimensional abelian variety, hence an elliptic curve $F$, which proves the only if part.
Remark: Actually you can get more. Since any curve is movable (in the sense of algebraically equivalence) by translation, the intersection number of any two fibers is zero implies that every fiber is a translation of $f$, so the elliptic fibration $S\rightarrow C$ must isomorphic to $S\rightarrow F$.