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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.

Gabriel Soranzo
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1 Answers1

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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.

Tabes Bridges
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  • Thanks. I find another solution here: $C.H=\deg_C \mathcal{O}X(-H){|C}=\deg_C \mathcal{O}{\mathbb{P}^n}(-H){|C}=\deg_C(H\cap C)=\deg C$ the last equality is not very clear for me but it comes from the generalized Bezout theorem in I.7.7. – Gabriel Soranzo Apr 24 '21 at 08:22
  • A question/remark: it seems that this argument with Riemann-Roch works only with non-singular curves. In the context of section V.1 in Hartshorne curves mean effective divisors. Maybe the second solution I mention above work in this case? – Gabriel Soranzo Apr 24 '21 at 20:14
  • Look at exercise IV.1.9(a) in Hartshorne. You still have $\chi(D) = \deg(D) + 1 - p_a$ with $p_a$ the arithmetic genus of $C$, so the same argument works. – Tabes Bridges Apr 24 '21 at 22:51
  • For this version of R.R. to work $H\cap C$ should be in the regular locus of $C$. I guess that the idea is then to take $H'\sim H$ with $H'\cap C$ are all regulars point of $C$ with Bertini theorem. – Gabriel Soranzo Apr 26 '21 at 09:05
  • Yes; since $H$ is very ample there is no issue in avoiding any given finite set (such as the singular locus of a curve) for purposes of computing intersection numbers. – Tabes Bridges Apr 26 '21 at 09:27