TITLE: Surfaces ruled over elliptic curves
QUESTION [6 upvotes]: Ground field $\Bbb{C}$. Algebraic category. Elliptic surfaces are those surfaces endowed with a morphism onto some smooth curve, with generic fiber an elliptic curve.
Suppose $E$ is an elliptic curve and consider the ruled surface
$$ S=\frac{E\times\Bbb{P}^1}{G} $$
where $G$ is a group of translations of $E$, acting on $\Bbb{P}^1$. Then $S$ is an elliptic surface, for the projection on the second factor induces a morphism $S\rightarrow\Bbb{P}^1$ whose fibers are elliptic curves ($F=E/G$, in fact).
Now, let $p\colon S\rightarrow C$ be an elliptic surface and suppose $S$ is ruled over an elliptic curve $E$. 
Is $S$ isomorphic to the above example? Any hint for attacking this? Thank you.

REPLY [5 votes]: One more attempt. 
All the fibers of $p\colon S\to C$ dominate $E$, so   they are all isogenous to $E$ and  by Kodaira's classification of elliptic fibers they all have smooth support. Hence $p$ is a so-called "quasi-bundle'' and by a result of Serrano [F.Serrano, 
Isotrivial fibred surfaces
Ann. Mat. Pura Appl. (4) 171 (1996), 63–81]  is of the form $(F\times  B)/G$, where:

$F$ is the elliptic fiber
$B$ is a smooth curve
$G$ is a finite group that acts faithfully on $F$,  on $B$ and diagonally on the product $F\times B$
the action of $G$ on $F\times B$ is free. 

The surface $S$ has  exactly two morphims to a curve, $p_1\colon S\to F/G$ and $p_2\colon S\to B/G$. The general fiber of $p_1$ is $B$ and the general fiber of $p_2$ is $F$.
Since $S$ is ruled over an elliptic curve $E$,  we have $B=\mathbb P^1$ and $F/G=E$. So $G$ acts on $F$ by translation and $S$  is as required.
A final remark: since $G$ acts  faithfully also on $\mathbb P^1$, the possibilities for $G$  are $G=\mathbb Z_2\times \mathbb Z_2$ and $G=\mathbb Z_m$, $m\ge 2$.