Let $H$ be a very ample divisor on a surface $X$ and let $C$ be any curve in $X$. The divisor $H$ gives an embedding $X\to \mathbb{P}^n$ which gives sense in the degree of the curve $C$: it is the highest coefficient of the Hilbert polynomial (ie $n\mapsto \chi(\mathcal{O}_C(n))$) of $C$ embedded in $\mathbb{P}^n$. The exercice V.1.2 in R. Hartshorne book ask to prove that $\deg(C)=C.H$ where $C.H$ is the intersection number of $H$ and $C$ in the surface $X$.
I can't see how to prove that.
An idea: as the Hilbert polynomial of $C$ in $\mathbb{P}^n$ is in fact linear then $\deg(C)=\chi(\mathcal{O}_C(1))-\chi(\mathcal{O}_C)$ but with $i:C\to X$ we have that
$$ \mathcal{O}_C(1)=i^*\mathcal{O}_X(1)=i^*\mathcal{O}_X(H)=\mathcal{L}(H)\otimes_{\mathcal{O}_X}\mathcal{O}_C $$
so that
$$ \deg(C)=\chi(\mathcal{L}(H)\otimes\mathcal{O}_C)-\chi(\mathcal{O}_C) $$
and it is proved before that
$$ C.H=\deg_C(\mathcal{L}(H)\otimes\mathcal{O}_C) $$
where here $\deg_C$ is the degree of the associated divisor, so that (if what I say is correct) one has to prove that
$$ \chi(\mathcal{L}(H)\otimes\mathcal{O}_C)-\chi(\mathcal{O}_C)=\deg_C(\mathcal{L}(H)\otimes\mathcal{O}_C) $$
but how? I'm missing some ingredient.
Best Answer
Compute $\chi(\mathcal L(H) \otimes \mathcal O_C)$ and $\chi(\mathcal O_C)$ using Riemann-Roch on $C$. We get $$ \chi(\mathcal L(H) \otimes \mathcal O_C) = \deg_C(\mathcal L(H) \otimes \mathcal O_C) - g(C) + 1 $$ $$ \chi(\mathcal O_C) = 0 - g(C) + 1. $$
Subtracting the second line from the first give you the desired equality.