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 $\Bbb{P}^1$.

Thus $F=E/G$ is an elliptic curve (and $\Bbb{P}^1/G=\Bbb{P}^1$)

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 ($F$, in fact).

An exercise on Beauville's book (chap. IX) says that if $S$ is a ruled surface over an elliptic curve $E$, and $S$ is an elliptic surface, then $S$ is isomorphic to the above example. Any hint for attacking this ? Thank you.

Edit: This question has been answered here.


1 Answer 1


This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Francesco Polizzi below.

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 $S=(E \times \mathbb{P}^1)/G$.


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