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.



Surfaces ruled over elliptic curves
**EDIT** We show that the answer to the OP's question is *yes*. Thanks to Will Sawin for his comments.


I use the notation of [Hartshorne, *Algebraic Geometry*, Chapter V Section 2].


Since $S$ is a ruled surface, there exists a section $C\_0$ of minimal self-intersection; set $C\_0^2 = -e$. If we write $S=\mathbb{P}(\mathcal{E})$, with $\mathcal{E}$ normalized, then $e= - \deg \mathcal{E}$.


Moreover, since the base of the ruling is an elliptic curve $E$, then $e \in \{-1, 0, 1, 2, 3, 4, \ldots \}$.


We can exclude the case $e >0$. Indeed, any divisor $D \in \textrm{Pic}(S)$ can be written as $a C\_0 + b f$, where $f$ is the fibre of the ruling. Now the elliptic fibre of the morphism over $\mathbb{P}^1$ must satisfy $(aC\_0+bf)^2=0$, that is $b = \frac{1}{2}ae$. But if $e >0$ then an effective divisor must also satisfy $b \geq ae$ (Hartshorne, Proposition 2.20 p. 382) and this is a contradiction.


Then the only possibility is $e \in \{-1, 0 \}$.


The case $e=-1$ corresponds to the second symmetric product $\textrm{Sym}^2(E)$; in this case $\mathcal{E}$ is the unique indecomposable rank two vector bundle on $E$ with $\deg \mathcal{E} =1$. Then the linear system $|4C\_0-2f|$ is a pencil of elliptic curves. One can also write $S$ in the desired for (see Wil Sawin's first comment). 


The case $\mathcal{E}=0$ corresponds either to the trivial bundle (so $S$ is a product), or to $\mathcal{O}\_E \oplus \mathcal{L}$, with $\deg \mathcal{L}=0$ or to $\mathbb{P}(\mathcal{E})$, where $\mathcal{E}$ is the unique indecomposible rank two vector bundle on $E$ such that $\deg \mathcal{E}=0$.


Let us consider first the case $\mathcal{O}\_E \oplus \mathcal{L}$, with $\deg \mathcal{L}=0$. Then the curves in the elliptic pencil should correspond to elements of $h^0(aC\_0)$, that is to sections of $\textrm{Sym}^a (\mathcal{O}\_E \oplus \mathcal{L})$. Since we must have two independent sections, we obtain that $\mathcal{L}$ is a torsion bundle. In this case the elliptic fibration do exist, and one can easily write $S$ in the desired form (see Will Sawin's second comment).


Finally, let us exclude the case $\mathbb{P}(\mathcal{E})$, with $\mathcal{E}$ indecomposable of degree $0$. Again, the curves in the elliptic pencil must correspond to elements of $h^0(aC\_0)$, that is to sections of $\textrm{Sym}^a \mathcal{E}$. But from [Atiyah, *Vector bundles on an elliptic curve*, Theorem 9] one sees that $\textrm{Sym}^a \mathcal{E}$ is again indecomposable of degree $0$, and that $h^0(\textrm{Sym}^a \mathcal{E})=1$. Thus one does not have two independent sections, and the elliptic pencil cannot exist.


Summing up, we obtain



> 
> Let $p \colon S \to \mathbb{P}^1$ be an elliptic surface and assume that $S$ is ruled over an elliptic curve $E$. Then $S=\mathbb{P}(\mathcal{E})$, where either 
> 
> 
> $\bullet$ $\mathcal{E}$ is the unique indecomposable rank $2$ vector bundle of degree $1$ on $E$, or
> 
> 
> $\bullet$ $\mathcal{E}= \mathcal{O}\_E \oplus \mathcal{L}$, where $\mathcal{L}$ is a (possibly trivial) torsion line bundle. 
> 
> 
> In both cases, we can also write $S$ in the desired form $S=(E \times \mathbb{P}^1)/G$. 
> 
> 
> 


