# Intersection number of very ample divisor and curve on a surface equals the degree

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?

• 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? Commented Jan 5, 2022 at 23:57
• @KReiser Got it, thanks! Commented Jan 7, 2022 at 21:29
• 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. Commented Jan 7, 2022 at 21:39
• @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. Commented Jan 8, 2022 at 2:55
• 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. Commented Jan 8, 2022 at 2:57

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.