The generic fibre is a genus 1 curve Let $S/\mathbb{C}$ be a smooth projective surface and let $C/\mathbb{C}$ be a smooth projective curve such that there exists a surjective morphism $\pi: S \to C$. Are the following conditions equivalent and if so, why?


*

*The generic fibre $S_\eta \to \text{Spec}(\mathbb{C}(C))$ is a (smooth) genus 1 curve.

*For all but finitely many closed points $P$ of $C$, the fibre $S_P$ is an elliptic curve.

 A: Yes if you suppose $O_C\to \pi_*O_S$ is an isomorphism. By the generic smoothness, we know that all fibers are smooth except for a proper closed subset of (so finitely many points in) $C$. Moreover, as $\pi$ is flat, the fibers have the same arithmetic genus. The hypothesis on $\pi_*O_S$ implies that all fibers are connected (Zariski's connectedness theorem). So for smooth fibers, arithmetic genus equal to geometric genus. 
(1) implies (2): if the generic fiber has genus 1, then all smooth fibers are curves of genus $1$ over $\mathbb C$, so they are elliptic curves. 
(2) implies (1): if one fiber is elliptic, then the generic fiber has genus $1$. 
Without the hypothesis on $\pi_*O_S$, the implication from (1) to (2) is not true: consider a non trivial cover $C' \to C$, a surface $S$ fibered over $C'$ and compose $S\to C'\to C$. Then most of the fibers of $S\to C$ are not connected and therefore are not elliptic curves.
Edit the counterexample may not be one, it depends on what you call a genus $1$ curve. If it is geometrically connected, then (1) implies the condition on $\pi_*O_S$ and hence (2). On the other hand, the method of the "counterexample" shows that (2) implies (1) (and $O_C\simeq\pi_*O_S$). 
