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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$\begingroup$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 $x_0$.
$\endgroup$ 3 $\begingroup$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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