# How to prove that a Klein quartic curve has genus 3?

I want to prove that the Klein quatic curve $X\subset \mathbb{CP}^2$ determined by $xy^3+yz^3+zx^3=0$ has genus 3. I know I should use the Hurwitz formula $$2g(X)-2=\deg(F)(2g(Y)-2)+\sum_{p}(\text{mult}_p(f)-1)$$

But I have trouble determining what $Y$ and $f~$ I should use here:

A standard approach for such problems is by setting $Y=\mathbb{CP}^1$ and $f([x:y:z])=[x:z]$, but this map is not well-defined here(when $x=z=0$). So how do we find the genus of $X$?

As Qiaochu mentioned, the genus-degree formula tells us immediately that the genus of the Klein quartic is three. Nonetheless, it might be helpful to think about how your proposed method via Hurwitz's theorem would work.

I know you said that your $f : X \to \mathbb {CP}^1$ sending $[x:y:z]\mapsto [x:z]$ is ill-defined at $[0:1:0]$, but this is not true. Any rational map from a smooth curve to projective space can be extended to a regular map (see here). In your example, you can extend $f$ by defining$$f([x:y:z]) = [x/z: 1] = [-z^2/y^2 - x^3 /y^3 : 1 ],$$which is manifestly regular in an open neighbourhood of the point $[0:1:0]$.

Let's analyse the covering map $f : X \to \mathbb {CP}^1$. For the time being, we will work in the open affine chart $\{ [1: u] \mid u \in \mathbb C \}\subset \mathbb {CP}^1$. For a fixed value of $u$, the preimage $f^{-1}([1 : u])$ is the set of points, $$f^{-1}([1 : u]) = \{ [1:v:u] \in \mathbb {CP}^2 \ | \ v^3 + u^3v + u = 0 \} .$$ For generic choices of $u$, the cubic $v^3 + u^3v + u$ has three distinct roots. So the degree of $f$ is $3$.

Where are the ramification points for this covering map? These occur at values of $u$ where the cubic equation for $v$ has repeated roots. To find them, let's look at the discriminant of the cubic, $$\Delta = - 4(u^3)^3 - 27(u)^2 = u^2(-4u^7 - 27).$$ When $u = 0$, all three roots of the cubic in $v$ are equal. When $u$ is any of the seven roots of $-4u^7 - 27$, the cubic in $v$ has two equal roots and a distinct root. For any other $u$, the cubic has three distinct roots. (To distinguish "three equal roots" from "two equal roots", use the fact that the sum of the roots is $-u^3$ and the product of the roots is $-u$.)

Thus $f^{-1}([1:0])$ is a single point. If $u$ is any of the seventh roots of $- 27 / 4$, then $f^{-1}([1:u])$ consists of two points. For any other $u$, $f^{-1}([1:u])$ consists of three distinct points.

We should not forget to consider $[0:1] \in \mathbb{CP}^1$, which is not covered by our affine chart. $f^{-1}([0:1])$ consists of two points: $[0:0:1]$ and $[0:1:0]$. To find the latter, we really must write $f$ as $f([x:y:z]) = [-z^2/y^2 - x^3 /y^3 : 1 ]$.

To summarise, $f$ is a degree $3$ map with one triple ramification point and eight double ramification points on $X$. Hurwitz's theorem gives $$2g(X) - 2 = 3\times (2\times 0 - 2) + 1 \times (3-1) + 8 \times (2-1),$$ which implies that $g(X) = 3$.