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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 $\text{div}(F_0)\sim kD$?

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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 $k$).

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