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Let $X\longrightarrow \mathbb P^1$ be a rational ellipic surface. In the Paper "pencils of cubic curves and rational elliptic surfaces" (by C.T.C.Wall) it is stated that C.K < 0, where C is a curve not contained in a fiber and K ist a canonical divisor on X. I cannot see this, can anyone help me?

M. E.
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The canonical class of a rational elliptic surface is the negative of the fibre class. Now consider any irreducible curve $C \subset X$, and its image under $X \to \mathbb{P}^1$. This image is either a point, in which case $C$ is contained in a fibre, or all of $\mathbb{P}^1$, in which case $C$ must intersect the fibre class positively.

Rhys
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  • You say "The canonical class of a rational elliptic surface is the negative of the fibre class". Though I don't really see this yet, I'm sure I will be able to calculate this (probably with the adjunction formula, somehow). Still I wanted to ask if there is a "nice" (or more or less easy) way to see this? – M. E. Feb 26 '13 at 11:48
  • An easy way to see it is to note that since $X$ is obtained by blowing up $\mathbb{P}^2$ at the intersections of two cubics, it can be represented as a degree (1,3) hypersurface in $\mathbb{P}^1\times\mathbb{P}^2$. Now use adjunction. – Rhys Feb 26 '13 at 11:53
  • Hmmm, for my purpose this is possibly the worst way to prove it :). In fact I want to show that every rational elliptic surface is obtained that way. (That was the reason for the Question in the first place). Thanks anyway, your answer was quite helpful. – M. E. Feb 26 '13 at 11:59