Every divisor $D$ on $X$ is dominated by a divisor linearly equivalent to $mA$ I am reading the proof of Riemann-Roch theorem from Shafarevich's Basic Algebraic Geometry 1 (3rd edition), but I'm stuck on a Lemma on pg 215. It says

(II) Every divisor $D$ on $X$ is dominated by a divisor linearly equivalent to $mA$ for some integer $m$.

Here $X$ is a non-singular projective curve and $A$ is the divisor of poles of some $f \in k(X)$ (though I believe that's not important, it seems that the only important thing is that $A$ must be effective). 
The book dismisses the proof as an easy verification but I have not been able to see it. Any help will be appreciated.

There are many ways of proving this. I assume that when you say `dominated' you mean $mA-D$ is effective and $f$ is non-constant. Clearly, suffices to prove this for a single point $P$ as $D$. If $P$ is in the support of $A$, then $A-P$ is effective and we are done. So, assume not. So, $f$ is regular at $P$ and change $f$ to $f-f(P)$ and then $A$ is unchanged, and the new $f$ vanishes at $P$. So, $\mathrm {div} f^{-1}=A-D$ where $D$ is the divisor of zeros of $f$. But, by choice, $D=E+P$ where $E\geq 0$ and so $A-P\sim E$.
