# Intersection divisor linearly equivalent to a multiple of a very ample divisor?

Towards the bottom of pg. 204 of Rick Miranda's Algebraic Curves and Riemann Surfaces, he starts describing equations for smooth projective curves and says that a general intersection divisor is equivalent to a multiple of a very ample divisor.

In particular, $D$ is a very ample divisor on a Riemann surface $X$, so $\phi_D : X \to \mathbb{P}^n$ is a holomorphic embedding. Then for a fixed homogeneous polynomial $F_0$ of degree $k$ which is not zero on all of $\phi_D(X)$, the intersection divisor $\mathrm{div}(F_0)$ is linearly equivalent to $kD$.

Why are these equivalent? He mentions that the complete linear system $|D|$ is the set of hyperplane divisors, but I don't really see why that helps.

• How I would think about it: div(F_0) on $\Bbb{P}^n$ is linearly equivalent to k times a hyperplane; a hyperplane section of X is equivalent to D; so div(F_0) on X is equivalent to kD. – Eoin Aug 10 '18 at 21:12