$\newcommand{\MX}{\mathcal{M}(X)}\newcommand{\OR}{\text{ord}}$
Let $X$ be a Riemann surface. Then $\mathcal{M}(X)$ is a field of meromorphic functions on $X$.
I saw that we can define a divisor for any meromorphic function $f\in\MX$ by taking for any $a\in Y\subset X$ open: $$\OR_a:\MX\to\Bbb{Z}\cup\{\infty\},\quad \OR_a(f)=\begin{cases}0,&\text{ if } f \text{ is nonzero and holomorphic at }a\\k&,\text{ if f has a zero of order }k\text{ at a}\\-k,&\text{ if f has a pole of order k at a}\\ \infty,&\text{ if f is identically zero at a}\end{cases}$$
Where we then have for any $f\in\MX\backslash\{0\}$ that $\OR_{-}(f):X\to\Bbb{Z}$ is a divisor.
I think that this makes $\OR_a$ a valuation. If I could write any $f,g\in\MX$ as polynomials, then this would be obvious, since $\OR_a(fg)$ would just be the sum of the zeros of the numerator, minus the sum of zeros of the denominator.
Is this a valuation? How do I show it rigorously?
Yes, it is a valuation (in fact it s a discrete valuation).
Just show that $\operatorname{ord}_p(f+g)\geq\min(\operatorname{ord}_p f,\operatorname{ord}_pg)$ and $\operatorname{ord}_p(fg)=\operatorname{ord}_p f+\operatorname{ord}_p g$, which of course is just taking a coordinate neighbourhood of $p$ and write out what $\operatorname{ord}_p f$ tells you about the form $f$ takes on that neighbourhood (transfer what you know about $\mathbb{D}$ to this setting).