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This question is about an argument on page 358 of Hartshorne, proof of Lemma 1.3. Consider curves $C$ and $D$ on a surface $X$ meeting transversely. There is an exact sequence $0\to \mathcal{L}(-D)\otimes\mathcal{O}_C\to\mathcal{O}_C\to\mathcal{O}_{C\cap D}\to 0$. And then it concludes that $\mathcal{L}(D)\otimes\mathcal{O}_C$ is the invertible sheaf correspond to the divisor $C\cap D$. My question is how does this implication follow? Maybe I am missing some obvious point....

Thanks, Minimax

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The exact sequence you've given follows from the fact (Proposition II.6.18) that $\mathcal I_D=\mathcal O_X(-D),$ after tensoring the sequence $$0\to\mathcal I_D\to\mathcal O_X\to\mathcal O_D\to 0$$ by $\mathcal O_C.$ By exactness of the tensored sequence, we know that $$\mathcal I_{C\cap D}=\mathcal O_X(-D)\otimes\mathcal O_C =: \mathcal O_C(-D).$$

The main point is that $\mathcal O_C(-D)=\mathcal O_C(-C\cap D).$ Then applying Proposition II.6.18 a second time, on $C\cap D\subseteq C,$ implies $\mathcal O_C(C\cap D)=\mathcal I_{C\cap D}^\vee=\mathcal O_C(D)=\mathcal O_X(D)\otimes\mathcal O_C.$

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But why is tensor with $\mathcal{O}_C$ preserve left exactness? –  minimax Jan 8 '13 at 6:39
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Dear @minimax, I think we can argue using Exercise II.1.2c. Namely, a sequence of sheaves and morphisms is exact iff it is exact on stalks. Localizing at any point $p\in C\cap D,$ we find that the stalk $(\mathcal O_X(-D)\otimes\mathcal O_C)_p$ is an $\mathcal O_{C,p}$-module generated by $f_p,$ where $f$ is some local defining equation for $D.$ We can also write $\mathcal O_{C\cap D,p}=\mathcal O_{C,p}/(f_p)$ since we consider the scheme-theoretic intersection. Thus, locally the sequence is exact. –  Andrew Jan 8 '13 at 13:57
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Dear @Andrew, yes, I agree that using the stalk version of exactness is the correct ingredient in this situation. The algebraic content is then this: for $A$ is regular local ring of dimension 2 and $f, g$ a regular sequence generating the maximal ideal, then the followng is a short exact sequence $0\to A/(f)\to A/(f)\to A/(f,g)\to 0$, where the first map is multiplication by $g$. –  minimax Jan 9 '13 at 0:45
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