During a lesson of the course on Riemann Surfaces our lecturer made the following remark, saying this could be proved as an exercise:
The space $\mathcal{M}(X)$ of meromorphic functions on a compact Riemann surface, if not empty, is not finite dimensional (as a vector space over $\mathbb{C}$).
This is a bit confusing: this was stated before Riemann-Roch, whence the "if not empty". Anyway, holomorphic functions are always also meromorhic, so this is in any case a problem. Besides, the space of meromorphic functions which are not holomorphic is not a vector space, so what is meant here? I have adjusted the statement as follows (still in a "pre-Riemann-Roch" setting):
Let $X$ be a compact Riemann surface. If there exists a function $f \in \mathcal{M}(X)-\mathcal{O}(X)$, then $\mathcal{M}(X)$ is infinite dimensional. Does this hold true? How can this be proved (without Riemann-Roch)?
I have tought to go by induction on number of generators. I am able to prove directly that it cannot be generated by two elements: suppose these two elements are called $g$ and $h$, one of the two, say $g$, is a constant, then if $f$ (the same as above) has a pole of order $n$ at a point $p$, also $h$ must have a pole of order $n$ at $p$, now also $h^2$ is meromorphic but cannot be written as a linear combination of $g$ and $h$, since $\mathrm{ord}_p h^2 = 2n$.
Now I'd like to prove that if $\mathcal{M}(X)$ is generated by $n$ functions, it is also generated by $n-1$ functions; but how?
Alternatively: every "polynomial in $f$", that is any expression of the form $a_0+ a_1 f + \ldots + a_n f^n$, where powers are respect to point-wise multiplication, is again in $\mathcal{M}(X)$, if this map $\mathbb{C}[z] \to \mathcal{M}(X)$ is injective, we are done. But is it?
Best Answer
As long as you have a meromorphic function $f$ on a compact Riemann surface, that is not constant, then $f$ will have a pole; of order $m$ at a point $P$. Then the $f^n$ are linearly independent over $\Bbb C$ since $f^n$ has a pole of order $mn$ at $P$, and functions with poles of different orders at $P$ are linearly independent.