# Showing that meromorphic functions form a field (without using Laurent series)

Thinking of meromorphic functions on a Riemann surface $X$ as holomorphic maps $X → \hat{ℂ}$, where $\hat{ℂ}$ is the Riemann sphere, how can one show that the set of meromorphic functions $\mathcal{M}(X)$ forms a field? I want to avoid Laurent series.

I tried to define addition and multiplication in the following way: \begin{align} f + g&\colon X → \hat{ℂ},\; z ↦ \lim_{w → z} (f(w) + g(w))\\ f · g&\colon X → \hat{ℂ},\; z ↦ \lim_{w → z} (f(w)·g(w)) \end{align} I have to show that this is well-defined, then the assertion follows easily. By the identity theorem, the poles of $f$ and $g$ are discrete and closed in $X$, so one can indeed look at the sums and products of $f$ and $g$ in a punctured neighbourhood of $z$.

So I have to establish that those limits exist. I also know $\lvert h (w)\rvert → ∞ ⇔ h (w) → ∞$ which I then can apply to $h = f + g$ and $h = f · g$. This is where I get stuck.

I’m also open to different ways of proving that meromorphic functions form a field. For example one can prov it for the unit disk first and then lift it to $X$ by the use of charts. I’m also interested in plausible arguments to convince me, that Laurent series or lifting are the way to go.

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Since everything happens locally, it is enough to do it on a disc. Can you prove all this there? –  Mariano Suárez-Alvarez Mar 20 '13 at 17:15
@MarianoSuárez-Alvarez I can, but I’m not content with it. It’s a little bit messy in my opninion. –  k.stm Mar 20 '13 at 17:40
@K.Stm.: In that case it might be best for you to write out the argument you know and point to the parts that you find less than fully satisfactory. –  Pete L. Clark Mar 20 '13 at 17:45

## 2 Answers

One way is to start with a slightly different definition of a meromorphic function on $X$: namely, it is a holomorphic function $f$ on an open subset $Y$ of $X$ such that for each $x_0 \in X \setminus Y$ there is an open coordinate neighborhood* $U$ of $x_0$ such that $U \setminus \{x_0\} \subset Y$ and there is some positive integer $k$ such that $x^k f$ extends to be holomorphic on $U$. This gives a meromorphic function on your sense in an evident way, but it is easier to add, subtract, multiply and invert meromorphic functions as defined above.

*: Here I mean that there is a biholomorphic map from $U$ to the open unit disk in the complex plane which carries $x_0$ to the origin. We write $x$ for the pulback of the identity function $z \mapsto z$ under this map, so that $x$ has a simple zero at $x_0$.

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Thanks for the suggestion. This may indeed be easier. But then you have to keep track of your poles and erase unnecessary ones after adding and multiplying, or else you have no negatives or reciprocals. –  k.stm Mar 20 '13 at 17:30
Identify two of Pete's functions when they coincide in the intersection of their domains, and work with that. –  Mariano Suárez-Alvarez Mar 20 '13 at 17:32
@Mariano: right, there is an implicit equivalence relation here. –  Pete L. Clark Mar 20 '13 at 17:44

Use another definition of meromorphic:

A function is meromorphic if it is analytical except for an isolated set of points. This way things are quite easy now.

E.g. $f,g$ meromorphic with exceptional sets $F,G$ which consist of isolated points. $\Rightarrow f\cdot g$ meromorphic, since $f\cdot g$ is analytical except on the set $F\cup G$ which also consists of isolated points, since we did no infinite union. Etc. ...

So do not care so much about the function values/limits, work with the poles.

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Your definition is incorrect, as shown by the non-meromorphic function $e ^{1/z}$ –  Georges Elencwajg Dec 19 '13 at 18:22
Apart from that, it ignores the premise that I wanted to show this using the given definition. So thanks for your answer, but it is not useful at all. –  k.stm Dec 21 '13 at 8:29