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I have a question about the definition of an elliptic surface. One defines an elliptic surface $S$ over a base curve $C$ (over some field $k$) as a surjective morphism $f: S \to C$ such that almost all fibres are smooth genus 1 curves, which is moreover relatively minimal with respect to the elliptic fibration - i.e. no fibre contains a smooth rational curve with self-intersection $-1$ ("exceptional curve of the first kind").

Often one also includes the condition that there is at least one singular fibre.

I've never understood why one really wants to include this last condition. For example, there exist "elliptic surfaces without singular fibre" which are NOT isomorphic to a trivial surface, i.e. a product of $C$ with another curve. (I'm not sure whether this can happen for $C = \mathbb{P}^1$ though.)

Could someone shed some light on this matter?

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Isn't the Hopf surface is an example of an elliptic surface over $\mathbb P^1$ which is not a product and has no singular fibers? –  Gunnar Magnusson Apr 4 '12 at 12:59

2 Answers 2

up vote 6 down vote accepted

In most papers on elliptic surfaces (at least among those I am aware of) the condition "there is at least one singular fiber" is to exclude elliptic surfaces that are a product or to enforce similar properties e.g., the Mordell-Weil group is finitely generated.

There are many examples of elliptic surfaces that are not a product and have no singular fiber. For a fixed elliptic curve $E$ and fixed base curve $C$, all elliptic surfaces with a section, having no singular fiber and such that the $j$-invariant of the general fiber equals $j(E)$ are parametrized by either $Pic(C)[2]$ ($j(E)\neq0,1728$), $Pic(C)[6]$ ($j(E)=0$) or $Pic(C)[4]$ ($j(E)=1728$).

A sketch of the construction: A non-trivial torsion line bundle yields an unramified Galois cover $D \to C$ with group $Z/2Z$, $Z/3Z$, $Z/4Z$ or $Z/6Z$. One can extend to action of $G$ to $E$ such that it has finitely many fixed points on $E$. Then $G$ has no fixed points on $D \times E$ and $S=(D\times E)/G$ is smooth, and the projection $D\times E\to D$ induces an elliptic fibration $S\to C=D/G$ without singular fibers. (This is a generalization of the construction of the so-called bielliptic surfaces.)

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Thanks for your answer! –  Wanderer Apr 4 '12 at 19:44

When the base curve is rational, the condition "there is at least one singular fiber" implies that the elliptic surface is regular ($q(S)=0$) meaning that the Picard variety $Pic^0(S)$ is trivial.

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