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?
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.