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?

  • $\begingroup$ @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

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.


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