Order of poles of y on y^2 = x^4 + a

Question

The complex points on the curve $y^2 = x^4 + a$, together with two additional points $P^+, P^-$, can be viewed as compact Riemann surface $X$. What is the order of poles of the map $(x, y) \mapsto y$ at the $P^+, P^-$? (it seems to equal 2 at both of the points).

Answer 1

The projective closure $\bar X\subset \mathbb P^2(\mathbb C)$ of $X$ is given by the homogeneized equation $y^2z^2 = x^4 + az^4$ and has a unique point at infinity $P=(0:1:0)$, which is unfortunately singular (we suppose $a\neq 0$).
In order to desingularize (= normalize in the case of dimension one ) $\bar X$ we have to replace $P$ by two new points $P^+,P^-$ thus obtaining the required compact Riemann surface $\hat X=X\cup \{P^+,P^-\}$.
A parametrization of $\hat X$ near $P^+$ is given on the disk $|z|\lt \epsilon$ by $ 0 \mapsto P^+$ and $x(z)=\frac {1}{z}, y(z)=\frac {\sqrt {1+az^4}}{z^2}$ for $0\lt |z|\lt \epsilon $ (look here, page 5)
So, as you correctly claim, the regular function $y$ on $X$ extends to a meromorphic function $\hat y$ on $\hat X$ which has a pole of order $2$ at $P^+$.
The same result holds for $P^-$, of course.