# Show that a meromorphic function on a compact Riemann surface can have a simple pole anywhere.

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.

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 $$P$$.