# 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.

## 1 Answer

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 $A-P\sim E$.

• By dominated I mean there exists $D^{\prime}$ such that $D^{\prime} - D$ is effective and $D^{\prime} \equiv mA$. (I think you also meant the same thing in your second line but were picturing $mA$ itself to dominate $D$). – Seven Apr 25 '16 at 19:41
• Thanks, that was very helpful. – Seven Apr 25 '16 at 20:00
• I assume from this answer that $A$ being the divisor of poles of $f$ is crucial. Is the lemma true when $A$ is just effective? If so, does it yield to an elementary proof like above? (General curiosity) – Seven Apr 25 '16 at 20:15
• If you just assume $A$ is effective, then the only reasonable argument to prove the same is to use Riemann-Roch. – Mohan Apr 25 '16 at 20:49