Let $X$ be a compact Riemann surface and $P\in X$ a point. Show there exists a meromorphic function $f:X\to\mathbb{C}$ such that $f$ has a pole of order $1$ at $P$.
We are currently learning about the Riemann-Roch theorem, so I have in particular the following theorem that I believe will help solve the problem.
For a divisor $D$ on $X$ let $L(D)=\{f:X\to\mathbb{C}\mbox{ meromorphic}:\mbox{div}(f)+D\geq0\}$ and $\ell(D)=\mbox{dim}L(D)$ and let $\theta(D)=\mbox{deg}(D)-\ell(D)$. Then $\ell(D)\leq\mbox{deg}(D)+1$ when $\mbox{deg}(D)\geq0$ and there exists a global upper bound $M$ such that $\theta(D)\leq M$ for all divisors $D$ on $X$.
Here is what I got so far.
From this we get $\ell(D)\geq\mbox{deg}(D)-M$, so in particular $\ell((M+2)\cdot P)\geq2$, so there exists a meromorphic function $f:X\to\mathbb{C}$ that only has a pole at $P$.
We also find for distinct points $P_1,…,P_{M+2}\in X$ that $\ell(\sum1\cdot P_i)\geq2$, so there exists a meromorphic function $f:X\to\mathbb{C}$ that only has simple poles and only at the specified points and has at least one such pole. So $f$ has a simple pole at at least one of the specified points. So there can be at most $M+1$ distinct points for which there exists no meromorphic function $f:X\to\mathbb{C}$ with a simple pole at that point.
Best Answer
From Riemann-Roch one gets the existence of an $a$ such that $\ell(nP)=n-a$ for all sufficiently large $n$. This means that $\ell((n+1)P)= \ell(nP)+1$ for all sufficiently large $n$. Then there is a function $f$ in $L((n+1)P)$ but not in $L(nP)$. This has a pole of order $n+1$ at $P$. There is also $g$ with a pole of order $n+2$ at $P$. Then $g/f$ has a simple pole at $P$.