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If $C$ is a curve of genus $g$, I'm trying to prove the dimension of the divisor $(2g-1)P$ associated to this curve is $g$.

I'm using the Riemann-Roch theorem which says:

Let $W$ be a canonical divisor on $C$. Then for any divisor $D$: $$l(D)=deg(D)+1-g+l(W-D)$$

Using this theorem I get $l((2g-1)P)=g+l(W-(2g-1)P)$. So why $l(W-(2g-1)P)=0$?

Remark: I'm studying Fulton's Algebraic Curves book.

Thanks in advance

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Remember that the canonical divisor has degree $2g - 2$, and that if $\deg D < 0$ then $l(D) = 0$. The reason is that it's impossible to find a non-zero rational function $f$ satisfying $D + \operatorname{div}(f) \geq 0$ since $\operatorname{div}(f)$ always has degree $0$.

[Looking at the book, this is given as Proposition 8.2.3 (2).]

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  • $\begingroup$ I'm sorry but I don't know what is bundle and global sections yet, my level is the Fulton's algebraic curves book. $\endgroup$ – user42912 Nov 10 '14 at 0:21
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    $\begingroup$ It's just a difference of language. I'll rewrite. I haven't read Fulton's book but I think I can guess what he's saying. $\endgroup$ – Hoot Nov 10 '14 at 0:22
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    $\begingroup$ thank you very much! the book is free available online in this link: math.lsa.umich.edu/~wfulton/CurveBook.pdf, the page he speaks about Riemann-Roch theorem is 108, any help is very welcome. $\endgroup$ – user42912 Nov 10 '14 at 0:35

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