Divisor of meromorphic functions on Riemann surfaces Let $C$ be the Riemann surface $y^2=x^3+1$ defined over $\mathbb{C}$, then I want to calculate the divisor of the meromorphic function of $g=\frac{x^2}{y}$. In the class of Riemann surface, we know the divisor of a meromorphic function is defined by $\sum\text{ord}_p(g)·p$. We need to choose a coordinate atlas $U\rightarrow \varphi(U)\subset C$ such that $p\in U$ and we define the order of $p$ is the least $n$ such that $a_n\neq0$, where $a_n$ is the coefficient of the Laurent expansion of $g\circ\varphi^{-1}: \varphi(U)\rightarrow\mathbb{C}$.
But when we regard $C$ as an algebraic curve, for example this question, when we calculate the order of $g$ at $(0,1)$, people always prove $x$ is a uniformizer of the localization $\mathbb{C}[x,y]/(y^2-x^3-1)$ at (x,y-1) first, and then regard $x$ as a meromorphic function of order $1$.
So my question is that, if we know $x-a$ is a uniformizer of the localization of $\mathbb{C}[x,y]/(f(x,y))$ at a point $(x-a,y-b)$ where $f(a,b)=0$, then in the sense of Riemann surface, how do we see the order of $x-a$ is $1$ by choosing a coordinate atlas and Laurent expansion?
 A: I try to explain what I think is correct and a bit about the relation between the algebraic and analytic views. My arguments, here, are a bit incomplete.I will try to complete them ASAP. I also hope I understand your question correctly.

Suppose $f$ is a non-singular irreducible polynomial in two variables,$x,y$. Then $f=0$ is a Riemann surface. For this Riemann surface, if $(a,b) \in f$, then based on which of the $\frac{\partial f}{\partial x}\vert_{(a,b)}$ or $\frac{\partial f}{\partial y}\vert_{(a,b)}$ are zero,one of the projection maps $x$ or $y$ is the coordinate map on a neighborhood of $(a,b)$. and the inverse of these maps are of the form $(x,\phi(x))$ or $(\psi(y) ,y)$, depending $\phi$ and $\psi$ are holomorphic functions on a an open set in $\mathbb{C}$.
Note that also in this context, for the homogenization of $f$,$F$, the set $\hat{C}$ which is the set of the solutions of $F=0$ in $\mathbb{P}^{2}$ is also a Riemann surface. for $\hat{C}$ the coordinate maps are obtained when you intersect it with the planes $X=1$ or $Z=1$ or $Y=1$ which are all biholomorphic to $\mathbb{C}$ and then use the projection maps. Now for $\hat{C}$ the map $ x:C \mapsto \mathbb{C}$ defines a holomorphic map, $X$, from $\hat{C}$ to $\mathbb{P}^{1}$. The degree of this map is equal to the degree $f$ which we call it $d$. So the field of meromorphic functions on $\hat{C}$ would be an algebraic extension of degree $d$ of $\mathbb{C}(z)$, Let's call it $K$. Then $K = \mathbb{C}(z)[w]/P$ for some $w \in K$ and $P$ an irreducible polynomial with coefficeints in $\mathbb{C}(z)$. As $y$ as a function extends to $\frac{y}{z}$ on $\hat{C}$, and as $K \subset \mathbb{C}(\frac{x}{z},\frac{y}{z})$ and also $f(\frac{x}{z},\frac{y}{z})=0$ so:
$$ K = \mathbb{C}(\frac{x}{z},\frac{y}{z})/<f>.$$
Where $<f>$ is the ideal generated by $f$ in $\mathbb{C}(\frac{x}{z})$. or it could be seen as the fraction field of $\frac{C[\frac{x}{z},\frac{y}{z}]}{<f>}$. So in this context when you are defining the $\text{ord}_p$ as a valuation on the $K$, then in this context local the local of point a $[a : b : 1]$ equals to $\mathbb{C}\{x_{[a:b:1]}\}$. Where $x_{[a:b:1]}$ is any coordinate map around $[a:b:1]$ and $\mathbb{C}\{x_{[a:b:1]}\}$ is the ring of convergent power series with coefficients in $\mathbb{C}$ of $x_{[a:b:1]}$, which on the other hand ,I think, it should be isomorphic the local ring that you mentioned. So whenever you find a uniformizer, you are finding a coordinate map for that point. Thus a uniformizer should be a coordinate map for that point and a function of order $1$ at that point.
