# Elliptic surface with $\kappa = 1$

I have two "rookie questions" about elliptic surfaces:

1. Let $S$ be an elliptic surface over $\mathbb{C}$, i.e. a smooth, projective algebraic surface equipped with a morphism $f: S \to C$ to a curve $C$ such that the generic fibre is an elliptic curve over $\mathbb{C}(C)$. The Kodaira dimension of an elliptic surface is at most $1$ (but can be $0$ or $-\infty$). If one asks that $g(C) \geq 2$, does this force the Kodaira dimension to be equal to $1$, or can it still be $0$ or $-\infty$?

2. A minimal elliptic surface is usually defined to be an elliptic surface which does not contain any vertical $(-1)$-curves, i.e. $(-1)$-curves contained in the fibres of the morphism $f: S \to C$. However there can be horizontal $(-1)$-curves. Does contracting such a curve always give another elliptic surface? If the Kodaira dimension is $1$ then this is certainly the case since it is a birational invariant. How do the fibres change in this case? I mean, they still have to be elliptic curves, but I don't have a very clear picture of what the relationship between the "new" and the "old" fibres is. I guess $C$ will not change, but I just can't imagine how the picture looks.

• Answer to 1 is given by Francesco. Concerning the new question 2 two things can be said. First, if the base is a curve of genus $>0$ there can not be a section that is a rational curve (this is a simple exercise). Second, the answer to the beginning of your second question is NO. Indeed, the simplest such elliptic surface is obtained by blowing up $\mathbb CP^2$ in $9$ points - the base points of an elliptic pencil. If you blow down these curves you get back $\mathbb CP^2$, which is not elliptic. Commented Apr 16, 2012 at 23:11

If $g(C) \geq 2$ then $\textrm{kod}(S)=1$.
The same holds also if $g(C)=1$ and $f$ is not locally trivial.