Type and number of points at infinity on Riemann surface Consider a polynomial $P(z) = z^4 \in \mathbb{C}[z]$. Set-theoretically $P(z)$ has one root equal to zero. From algebraic point of view it has four roots: root zero has multiplicity four. Also we can't draw a curve in $\mathbb{C}$ around one of such roots but not around the others.
Now consider a Riemann surface $X$ given by 
$$
   X = \left\{ (z,w) \in \mathbb{C}^2 \colon w^2 - z^4 + 1 = 0 \right\}.
$$
$\hskip1cm$  
Its compactification in $\mathbb{C}P^2$ is a surface $\overline{X}$:
$$
  \overline{X} = \left\{ (\xi : \eta : \zeta) \in \mathbb{C}P^2 \colon \eta^2\zeta^2 - \xi^4 + \zeta^4 = 0 \right\}.
$$
Points at infinity in $\mathbb{C}P^2$ are given by $\zeta = 0$. So if we want to find intersection of $\overline{X}$ with infinity we should solve a system $\eta^2 \zeta^2 - \xi^4 + \zeta^4 = 0$, $\zeta = 0$. It has a solution $\zeta = 0$, $\xi = 0$. Then set-theoretically $\overline{X}$ has one point at infinity $(0:1:0) \in \mathbb{C}P^2$.
Now take in open neighborhood $\eta \neq 0$ of point $(0:1:0)$ as coordinates 
$$
   x = \frac{\xi}{\eta}, \;\;\;y = \frac{\zeta}{\eta}.
$$
Then points at infinity are given by $y = 0$, our surface is given by $y^2 + y^4 - x^4 = 0$ and intersection of our surface with line at infinity is given by the ideal $\langle y, y^2 + y^4 - x^4 \rangle = \langle y, x^4 \rangle \subset \mathbb{C}[x,y]$. Multiplicity of intersection is $\dim_{\mathbb{C}} \frac{\mathbb{C}[x,y]}{\langle y, x^4 \rangle} = 4$. Then our point $(0 : 1 : 0)$ has a multiplicity four and intersection-theoretically there are four points at infinity.
In Dubrovin's book "Riemann surfaces and non-linear equations" author writes that a surface $X$ has two points at infinity like any surface given by equation $w^2 = P_{2m}(z)$ where $P_{2m}(z) \in \mathbb{C}[z]$, $m > 1$ and $P_{2m}(z)$ has mutually distinct roots. I would like to know which theory gives two points at infinity and if there is possible to draw a contour on $\overline{X}$ around one of such points but not around the others. Thank you for answers.
 A: The analytic curve $\bar X$ has only one point $P$ at  infinity (i.e. with $\zeta=0$) but it is not a Riemann surface because $P$ is  a singularity of $\bar X$.
The right compactification of $X$ is the Riemann surface $\bar X^\nu$ obtained by normalizing $\bar X$.
It is obtained by deleting $P$ from $\bar X$,  replacing it by two points $P_1, P_2$, and endowing the resulting set with a suitable structure of Riemann surface.
You then obtain a normalization morphism of analytic  curves $n:\bar X^\nu \to \bar X$ such that $n^{-1}(P)=\lbrace P_1,P_2\rbrace $ and such that the retriction $n_{\text{res}}: \bar X^\nu\setminus \lbrace P_1,P_2\rbrace \to X$ is an isomorphism of Riemann surfaces.  
The Riemann surface $\bar X^\nu$ has genus one and becomes an elliptic curve once you arbitrarily choose a point on it to serve as  the zero point of the elliptic curve.  
To answer your last question, you can certainly choose a contour around each of the two $P_i$'s not enclosing the other one, but this is a general property of Riemann surfaces, having nothing to do with the situation at hand here.
