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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.


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


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You apply Riemann-Roch for non-singular irreducible curves. There's a more general Riemann-Roch for singular curves. I think you can find it in chapter 7 of Liu's book on alg. geometry. – seporhau Mar 11 '12 at 12:24
Yes, but doesn't Corollary 3.18 on page 279 of Liu give exactly the formula I want, i.e. $\chi(\mathcal{O}_X(D)) = \deg D + \chi(\mathcal{O}_X)$, also without the assumption that $X$ is non-singular? – Evariste Mar 11 '12 at 15:19
You mean thm 3.17, but that's surely a typo. I think you're right. Maybe the problem lies in the definition. Is the intersection number of $C$ with $L$ well-defined if $C$ is singular in Vakil's notes? Does he use the dvr associated to $C$? – seporhau Mar 11 '12 at 16:07
Yes, I meant theorem 3.17, sorry. Vakil defines intersection numbers on page 3 of - as far as I can see he doesn't need non-singularity there, but I'm probably missing something? Thanks for helping... – Evariste Mar 11 '12 at 16:11
The smoothness assumption is not used. – user18119 Mar 14 '12 at 21:17
up vote 1 down vote accepted

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.

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