# Exact sequence in Beauville's “Complex Algebraic Surfaces”

On page 3 of Beauville's book (Lemma I.5) he takes two curves $C$ and $C'$ in a surface $S$ an takes global sections $s\in H^0(S,\mathcal{O}_S(C))$ and $s'\in H^0(S,\mathcal{O}_S(C'))$. In a recent post, I was explained that you can take $s$ and $s'$ to be 1. Beauville then writes the sequence

$0\longrightarrow\mathcal{O}_S(-C-C') \xrightarrow{(s',-s)}\mathcal{O}_S(-C)\oplus \mathcal{O}_S(-C')\xrightarrow{(s,s')}\mathcal{O}_S\longrightarrow\mathcal{O}_{C\cap C'}\longrightarrow 0.$

I assume that if $\mathcal{O}_S(-C)$ (resp. $\mathcal{O}_S(-C')$) is generated on an open set $U_\alpha$ by $f_\alpha$ (resp. $f_\alpha'$), then it makes sense that the first map in the sequence takes $f_\alpha f_\alpha'$ to $(f_\alpha,-f_\alpha')$, but I don't see how this is consistent with the notation $(s',-s)$, especially since $s$ and $s'$ can be taken to be 1!

Where am I wrong here or not understanding something correctly?

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There is a typo in the displayed exact sequence: the first morphism is $(s',-s)$ and not $(s',-s')$. –  Georges Elencwajg Jun 29 '12 at 9:46
Corrected, thanks. –  rfauffar Jun 29 '12 at 9:53

Let me just explain the morphism of sheaves $\mathcal{O}_S(-C-C') \xrightarrow{s'}\mathcal{O}_S(-C)$, the rest being similar.

On $U_\alpha$ the divisor $(-C-C')\mid U_\alpha$ is given by the holomorphic function $f_\alpha f'_\alpha$ and the section $s'\in \Gamma (U _\alpha ,\mathcal{O}_S(C'))$ is given by a meromorphic function function $\frac {S'_\alpha}{f'_\alpha }$ where $S'_\alpha$ is holomorphic on $U_\alpha$.
The morphism $\mathcal{O}_S(-C-C') \xrightarrow{s'}\mathcal{O}_S(-C)$ then sends the section $\frac {h_\alpha}{f_\alpha f'_\alpha}\in \Gamma (U_\alpha, \mathcal{O}_S(-C-C') )$ to the section $\frac {h_\alpha S'_\alpha}{f_\alpha }\in \Gamma (U_\alpha, \mathcal{O}_S(-C) )$.

In order to reassure you, let us take for an example the canonical section $s'_0=1=\frac {f'_\alpha}{ f'_\alpha}\in \Gamma(U_\alpha,\mathcal O_C')$ i.e. $S'_\alpha=f'_ \alpha$, about which you worry in your question.
The morphism $\mathcal{O}_S(-C-C') \xrightarrow{s'_0}\mathcal{O}_S(-C)$ is then given on $U_\alpha$ by $$\Gamma (U_\alpha, \mathcal{O}_S(-C-C') )\to \Gamma (U_\alpha, \mathcal{O}_S(-C) ):\frac {h_\alpha}{f_\alpha f'_\alpha}\mapsto \frac {h_\alpha f'_\alpha}{f_\alpha }$$

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Ok perfect! I completely understood; thank you very much!! –  rfauffar Jun 29 '12 at 12:33
I'm glad about that, Robert! –  Georges Elencwajg Jun 29 '12 at 12:40
@GeorgesElencwajg, is it true that there is some mixing up of denominators and numerators going on? If $\mathcal{O}_S(-C)$ is given by $f_\alpha$, shouldn't it be $S'_\alpha \cdot f'_\alpha$ instead of $\frac{S'_\alpha}{f'_\alpha}$? Also concerning the second alinea, if $C$ and $C'$ are effective curves, how can the divisor $-C-C'$ be given by a holomorphic function? Shouldn't this be a function with no zeros and only poles? I might be mixing it up though, there is a lot of inverting going on in this invertible sheaf business.. Pretty fascinating though.. –  Joachim Aug 16 '12 at 17:28
@GeorgesElencwajg, after more thought i think i know what you mean. If you assume $C,C'$ to be locally given by regluar functions $f_\alpha, f'_\alpha$, all you need to do is whenever you talk about sections of $\mathcal{O}_C$, to change the $1/f_\alpha$'s into $f_\alpha$'s. Also the "multiplication" by $s'$ now doesnt make much sense: you multiply by both numerator and denominator. There's an obvious fix to that as well. I'd feel bad editing your post, since it is an accepted answer and you normally answer my questions, so i wont. =) Your answer helped me out in any case, so thanks for that! –  Joachim Aug 16 '12 at 17:39