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.
Best Answer
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