# Intersection divisor on a smooth projective curve

Let $$D$$ be a very ample divisor on an algebraic curve, inducing a holomorphic embedding $$\phi_D:X\to \Bbb P^n$$. The image (which we also call $$X$$) is a smooth projective curve. Fix a degree $$k$$ and a homogeneous polynomial $$F_0$$ of degree $$k$$ in $$n+1$$ variables such that $$F_0$$ is not identically zero on $$X$$. Consider the intersection divisor $$\text{div}(F_0)$$ on $$X$$. Since the hyperplane divisors on $$X$$ are exactly the divisors in the linear system $$|D|$$, we see that $$\text{div}(F_0)\sim kD$$ since $$F_0$$ has degree $$k$$.

This is a paragraph in p.204 of Miranda's book Algebraic Curves and Riemann Surfaces. I can see that the hyperplane divisors on $$X$$ are exactly the divisors in $$|D|$$, but how does this imply that $$\text{div}(F_0)\sim kD$$?

I can't recall exactly how Miranda defines these things, so just to be clear: you understand that hyperplane divisors arise as intersections with literal hyperplanes in $$\mathbb P^n$$ (i.e. zero loci of linear equations in $$n+1$$ variables), and that $$\operatorname{div}(F_0)$$ will be the intersection of the image of $$X$$ with a degree $$k$$ hypersurface, right?
Once you know that, the basic idea is just that all hypersurfaces of a fixed degree in a fixed projective space are linearly equivalent. We can degenerate the hypersurface cut out by $$F_0$$ to a union of $$k$$ hyperplanes or even a single hyperplane with multiplicity $$k$$ via the linear equivalence $$sF_0 + t F_1 = 0$$, $$(s:t) \in \mathbb P^1$$, where $$F_1$$ is the product of $$k$$ distinct linear equations (for the union of $$k$$ hyperplanes) or a single linear equation raised to the $$k$$-th power (for the hyperplane of multiplicity $$k$$).