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

Solution 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.