3
$\begingroup$

In This notse, Vakil gives a proof of the theorem stated in the title. (page 5). In the proof he made use of a section s of $\mathcal{O}_C(kp)$ that has only one zero of order k at p. However such a section doesn't seem to be exist. In fact, any such section s(considered as element of K(x)) must have poles at some other points, and therefor can't be a section of $\mathcal{O}_C(kp)$ since $div(s) + kp \ngtr 0$ .

I am correct?

$\endgroup$
  • $\begingroup$ @QiaochuYuan, I don't think such a section would give rise to a hyperplane that meet the image of X at only one point.(Although it can be used to give another proof of the theorem). $\endgroup$ – Saberization May 28 '15 at 20:42
2
$\begingroup$

The reason of misunderstanding is follows. There are two languages to think about $\Gamma ( \mathcal{O}(kp) )$:

  • Functions $f$ on surface that $div(f) + kp >0$

If you talk this language, then for you mentioned function is just $1$.

  • Sections of line bundle

Then to understand what is order of zero at point $p$ you have take a neighbourhood of $p$ where your sheaf is isomorphic to $\mathcal{O}$. After applying this isomorphism to a section, you will see what is called order of zero. Vakil speaks in the second language.

$\endgroup$

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.