I'd like to show that if $X$ is a (nonsingular, projective, algebraic) surface, $H$ a very ample divisor on $X$, and $C$ an effective divisor (curve) on $X$, then the intersection number $C.H$ equals the degree of $C$ in the projective embedding $X \subset \mathbb{P}^n$ defined by $H$ (where degree in this case is defined as the leading coefficient of the Hilbert polynomial).

I see that I can assume $C$ is irreducible, and moreover understand how to solve the problem in the case that $C$ is nonsingular using the Riemann-Roch theorem for $C$: if $P$ is the Hilbert polynomial of $C$ in $\mathbb{P}^n$, one can compute $$ P(0) = \chi_C(\mathcal{O}_C) = 1 - g_C, \quad P(1) = \chi_C(\mathcal{O}_C(1)) = 1 - g_C + \deg_C(\mathcal{O}_X(H)|_C), $$ yielding leading coefficient equal to $C.H$.

However, I've not been able to do anything in the case where $C$ is a possibly singular irreducible curve on $X$. I've tried applying Riemann-Roch for singular curves (Hartshorne IV Exercise 1.9), but run into trouble relating degrees of effective divisors to intersection numbers in the singular case. How can I get around this?

  • 3
    $\begingroup$ How about the following: pick a hyperplane which misses all the singular points, and then use that all hyperplanes are linearly equivalent and therefore must give the same intersection numbers? $\endgroup$
    – KReiser
    Commented Jan 5, 2022 at 23:57
  • $\begingroup$ @KReiser Got it, thanks! $\endgroup$
    – Legendre
    Commented Jan 7, 2022 at 21:29
  • $\begingroup$ Glad to help. To be honest, I was kind of spitballing - intersection theory was one of the things my education sort of lacked in, but if this answered your question, I'd be happy to write it as an answer below. $\endgroup$
    – KReiser
    Commented Jan 7, 2022 at 21:39
  • $\begingroup$ @KReiser I've posted what I think works as an answer below -- does it seem correct to you? Of course if you post your comment as an answer below I'd be happy to accept it. $\endgroup$
    – Legendre
    Commented Jan 8, 2022 at 2:55
  • 1
    $\begingroup$ Looks good to me (though, you know, caveat emptor), and it has more details than I would have been able to provide. +1, thanks for recording your work. $\endgroup$
    – KReiser
    Commented Jan 8, 2022 at 2:57

1 Answer 1


By writing C as a sum of prime divisors, it suffices to assume that $C$ is irreducible. Let $i: X \to \mathbb{P}^n$ be the embedding determined by $H$, and let $j: C \to X$ be the embedding of $C$ in $X$. Since $C$ is a curve, it has finitely many singular points, so we may choose a hyperplane $H' \subset \mathbb{P}^n$ not passing through any singular points of the image of $C$ in $\mathbb{P}^n$.

Let $P(z) = az + b$ be the Hilbert polynomial of $C \subset \mathbb{P}^n$, so that $a = \deg_{\mathbb{P}^n}(C)$. By Riemann-Roch for singular curves, we have $$ b = P(0) = \chi(\mathcal{O}_C) = 1 - p_a(C), \\ a + b = P(1) = \chi((i \circ j)^* \mathcal{O}_{\mathbb{P}^n}(1)) = \chi((i \circ j)^* \mathcal{O}_{\mathbb{P}^n}(H')) = 1 - p_a(C) + \deg(H' \cap C). $$ Thus $a = \deg_{\mathbb{P}^n}(C)$ is the degree of $H' \cap C$ considered as a divisor on $C$. But this degree is just the sum of the points of intersection of the curves $(H' \cap X)$ and $C$ counted with multiplicity, which we know (e.g. by Hartshorne V Prop. 1.4), equals the intersection number $C.i^*H'$. Now since any two hyperplanes in $\mathbb{P}^n$ are linearly equivalent, we have $C.i^*H' = C.H$, and we are done.


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