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Let $C$ be the (smooth) curve in $\mathbb{C}^2$ defined by $y^2 = x^4 - 1$, and let $\pi : C \to \mathbb{C}$, $\pi(x,y) = x$. $\pi$ is a ramified cover, ramified over $\pm 1, \pm i$.

$C$ is a non-compact Riemann surface; there is a canonical way to make it a compact RS via $\pi$ by adding "missing" points, here the points over $\pm 1, \pm i$ and $\infty$. If I understand things correctly, this is the same as considering the projective curve $X$ in $\mathbb{P}^2\mathbb{C}$ defined by $y^2z^2 = x^4 - z^4$. This yields a single point at infinity $[0:1:0]$. We can extend $\pi$ as a ramified covering of $\mathbb{P}^1\mathbb{C}$ of degree two, ramifying over $\pm 1, \pm i$ and $\infty$ with ramification index 2 (all of these points have a single antecedent instead of two).

Applying the Riemann-Hurwitz formula, we finally get $2-2g = 4 - (1+1+1+1+1) = -1$, which is obviously wrong. I know the result of compactifying $C$ is supposed to be an elliptic curve, of genus 1, and that the projection should not ramify over infinity, but I can't figure out where I went wrong.

What's strange is, if I started with $C = \{ y^2 = x^3 - 1 \}$ (also an elliptic curve of genus 1), I also get a single point at infinity, and the end result is "right". So where does my method go wrong?

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The point $[0:1:0]$ is a singular point on the projective curve $C^\prime$ defined by $y^2z^2-x^4+z^4=0$. Thus your construction does not give the compactification of the open Riemann surface $C$. In terms of algebraic geometry you have to normalize the projective curve $C^\prime$ to remove the singularity, which will give you two points lying over $[0:1:0]$. In particular there is no ramification at infinity in this case.

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OK, that explains everything. So that means I should for example "blow up" (I'm not very familiar with the concept) the point? – Najib Idrissi Oct 23 '12 at 12:17
Also (sorry for all these questions), if for example I took the curve $y^2z^3 = x^5 - z^5$, I know it is of genus 2. The only point at infinity is also $[0:1:0]$, and it seems to me it's still singular, a triple point actually. And when I apply Riemann-Hurwitz to the projection, I see that there must be a ramification over $\infty$. So does that mean that to normalize, the triple point is split in two, the projection being ramified over one but not the other? Is there a general method? – Najib Idrissi Oct 23 '12 at 12:27
Blowing up a point and normalization are one and the same thing for algebraic curves. As for the latter one takes the ring $R$ of functions holomorphic in some (affine) neighborhood of the singular point. Then one determines the integral closure $S$ of $R$ in the field of meromorphic functions on $C$. $S$ is an extension ring of $R$ and the prime ideals of $S$ lying over the prime ideal of $R$, that corresponds to the singular point, correspond to the "new" points one is looking for. – Hagen Knaf Oct 25 '12 at 10:29

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