6
$\begingroup$

I was looking at a proof of the following theorem...

Let $S/\mathbb{C}$ be a smooth projective surface, let $C$ be a non-singular irreducible curve on $S$. Then for all $L \in \text{Pic}\,S$, the intersection number $\langle \mathcal{O}_S(C), L \rangle$ is equal to $\text{deg}(L|_C)$.

Proof: we have an exact sequence $0 \to \mathcal{O}_S(-C) \to \mathcal{O}_S \to \mathcal{O}_C \to 0$ which remains exact upon tensoring with $L^{-1}$, hence giving an exact sequence $0 \to L^{-1}(-C) \to L^{-1} \to L^{-1} \otimes \mathcal{O}_C \to 0$. By additivity of the Euler-Poincaré characteristic, we get $\chi(\mathcal{O}_S) - \chi(\mathcal{O}_S(-C)) = \chi(\mathcal{O}_C)$ and $\chi(L^{-1}) - \chi(L^{-1}(-C)) = \chi(L|_C^{-1})$. This allows us to write the intersection number as $\langle \mathcal{O}_S(C) , L \rangle = \chi(\mathcal{O}_C) - \chi(L|_C^{-1}) = -\text{deg}(L|_C^{-1}) = \text{deg}(L|_C)$, by Riemann-Roch.

Question

Where did we use the fact that $C$ is non-singular? It has to be essential, but I don't see why...

References:

http://math.stanford.edu/~vakil/02-245/sclass5A.pdf

http://math.stanford.edu/~vakil/02-245/sclass6A.pdf

  • 0
    The smoothness assumption is not used. – 2012-03-14

1 Answers 1

1

The statement should work even when $C$ is not smooth, but the degree of $L|_C$ is not even defined for singular $C$ in Vakil's lecture notes. In fact, his notes only ever cover the smooth case - probably because it is more intuitive and he can save a lot of time by not generalizing. If you want to generalize, you will have to use Liu's definition 7.3.1 (p.275), and then the proof should just work in the same way.