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I have the following problem:

Let $S$ be a compact Riemann Surface of genus $g$, and let $p\in S$. Show that there is a meromorphic function $F$ on $S$ with pole of order $n$ at $p$, if $n\ge 2g$.

I made a rough draft of the solution:

Solution: Let $p \in S$ and consider the divisor $D=np$. For a canonical divisor $K$ on a compact Riemann surface of genus $g$, we have

$deg K=2g-2$,

and so $deg (K-D)=degK-degD$.

Like $D=np \Rightarrow degD=n \Rightarrow deg(K-D)=2g-n-2.$ But $n\ge 2g \Rightarrow 0\ge2g-n \Rightarrow deg(K-d)=2g-n-2 <0$, and thus, $h^{0}(K-D)=0$. The riemann-roch teorem therefore yields

$h^{0}(D)=deg(D) -g+1=n-g+1=n+1-g.$

But $n\ge 2g \Rightarrow n+1>2g>g \Rightarrow n+1-g>0$. Therefore, $h^{0}(D)>0$.

Therefore, we can find a nonconstant meromorphic function $F$ with $D+(F) ≥ 0$, i.e. with at most a pole of order $n$ at $p$, and as $F$ is nonconstant, it must have a pole somewhere, and so it does have a pole of order $n$ at $p$.

I need some help to make sure that it is correct and also for writing. The final part seems unclear (confused), in my view.

Please be welcome to any correction!

I noticed that in case the genre be $g=0$, this exercise shows that: $S$ is equivalent to the Riemann sphere.

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To show that there exists a meromorphic function with a pole of order precisely $n$ at $p$, you should show that $h^0(np) > h^0((n-1)p)$, i.e. that there exists a section $\mathcal O_S(np)$ which fails to be a section of $\mathcal O_S((n-1)p)$. You can do this using the method in your post. Does this make sense?

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  • $\begingroup$ This no make sense for me :..( And my solution, does this make sense? $\endgroup$
    – Manoel
    Feb 26, 2017 at 5:11
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    $\begingroup$ Your method of calculation is fine, but you really need to compute both $h^0(np)$ and $h^0((n-1)p)$. Let me explain like this. $h^0(np)$ is the dimension of the space of meromorphic functions there are with at most an order $n$ pole at $p$. $h^0((n-1)p)$ is the dimension of the space of meromorphic functions with at most an order $n-1$ pole at $p$. So $h^0(np) - h^0((n-1)p)$ is the dimension of the space of meromorphic functions with at most an order $n$ pole at $p$ quotiented by the space of meromorphic functions with at most an order $n-1$ pole. $\endgroup$
    – Kenny Wong
    Feb 26, 2017 at 5:25
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    $\begingroup$ Thus, if $h^0(np) - h^0((n-1)p) \geq 1$, then there must exist a meromorphic function with a pole at $p$ of order precisely equal to $n$. $\endgroup$
    – Kenny Wong
    Feb 26, 2017 at 5:25
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    $\begingroup$ I got to understand what you said! Thank you for your attention! $\endgroup$
    – Manoel
    Mar 1, 2017 at 1:30

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