1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $A-P\sim E$.

$\endgroup$
4
  • $\begingroup$ 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 $D$). $\endgroup$ – Seven Apr 25 '16 at 19:41
  • $\begingroup$ Thanks, that was very helpful. $\endgroup$ – Seven Apr 25 '16 at 20:00
  • $\begingroup$ 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) $\endgroup$ – Seven Apr 25 '16 at 20:15
  • 2
    $\begingroup$ If you just assume $A$ is effective, then the only reasonable argument to prove the same is to use Riemann-Roch. $\endgroup$ – Mohan Apr 25 '16 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.