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Consider a Riemann surface $X$ given by function $F(z,w) = w^2 - z^2 - a^2$, $a \neq 0$: $ X = \left\{ (z,w) \in \mathbb{C}^2 \colon F(z,w) = 0 \right\}. $ To find its points at infinity we first compactify $X$. We can write $ F \left( \frac{\xi}{\zeta}, \frac{\eta}{\zeta} \right) = \frac{\eta^2 - \xi^2 - a^2 \zeta^2}{\zeta ^2} = \frac{ Q(\xi,\eta,\zeta)}{\zeta^2}. $ Then compactification $\overline{X}$ of surface $X$ is given by homogenous polynomial $Q$: $ \overline{X} = \left\{ (\xi : \eta : \zeta ) \in \mathbb{C}P^2 : Q(\xi,\eta,\zeta)=0 \right\} $ Points at infinity are such points of $\overline{X}$ that $\zeta = 0$. So if $\zeta = 0$ we have $\eta^2 = \xi^2$ from what we can derive two points at infinity: $(1:1:0)$ and $(1:-1:0)$.

I try to use the same method to find points at infinity of Riemann surface $ Y = \left\{ (z,w) \in \mathbb{C}^2 \colon w^2 - z^4 + 1 = 0 \right\}. $ Its compactification is a surface $ \overline{Y} = \left\{ (\xi:\eta:\zeta) \in \mathbb{C}P^2 \colon \eta^2\zeta^2 - \xi^4 + \zeta^4 = 0 \right\}. $ In order to find points at infinity we find solutions of system $\eta^2 \zeta^2 - \xi^4 + \zeta^4 = 0$, $\zeta = 0$. It follows that $\xi = 0$ and the unique point at infinity is $(0:1:0)$ but it is a contradiction with theory: surface $Y$ has two points at infinity. Help me please, where is my mistake?

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First the bad news: the theory says that you have $4$ points at infinity, since your curve has degree $4$, so that the situation is twice worse than you think !
But the good news is that there are indeed four points at infinity, if you take the multiplicities into account!

More precisely, in the open neighbourhood $\eta \neq 0$ of $P_\infty=[0:1:0]$ take as coordinates $x=\xi/\eta, y=\zeta/\eta$ so that $P_\infty$ has coordinates $x=y=0$.
The line at infinity is $y=0$ and the curve has equation $y^2-x^4+y^4=0$ in our coordinates.
Their intersection is given by the ideal $\langle y, y^2-x^4+y^4\rangle=\langle y, x^4\rangle \subset \mathbb C [x,y]$ and the multiplicity of the intersection is $\text{dim} _\mathbb C \frac {\mathbb C [x,y]}{\langle y, x^4\rangle}=4$ : all is well that ends well!

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    Th$a$nk you, I asked another question: http://math.stacke$x$ch$a$nge.com/q/201065/162732012-09-23