On the bottom of page 204 in Rick Miranda's book 'Algebraic curves and Riemann Surfaces' he writes "Since the hyperplane divisors on $X$ are exactly the divisors in the linear system |D|, we see that div$(F_0) \sim kD$ since $F_0$ has degree $k$."
The context is that $X$ is a compact Riemann surface embedded in $\mathbb{P}^n$ via the map $ x \mapsto [f_0(x) : \dots : f_n(x) ] $ where $f_0, \dots f_n$ is a basis for $L(D)$, and $F_0$ is a homogeneous polynomial on $\mathbb{P}^n$ of degree $k$.
Does anyone understand why div$(F_0) \sim kD$? It is not clear to me what he means.
The divisors of any two nonzero homogeneous polynomials of the same degre on $\Bbb P^n$ are linearly equivalent since their difference is the divisor associated to the rational function obtained by taking the quotient of these two polynomials. So $(F_0)\sim (h^k)$ where $h$ is any linear form, and as $(h^k)=k\cdot(h)$ we have that $(F_0)\sim kH$ for $H$ a hyperplane divisor on $\Bbb P^n$. Now we restrict everything to your curve and maintain the equivalence (showing this last claim is probably a lemma or an exercise somewhere in your book, I've never used Miranda).