Genus of $3x^{3}+4y^{3}+5z^{3}$ I am supposed to calculate the genus of the projective curve $3x^{3}+4y^{3}+5z^{3}$, then use Weil conjecture.
I don't know how to calculate the genus, it seems like plugging in formula just give me $g=2$. In fact, it looks like it is not even a curve, but rather a surface. I am quite lost in this class, so please provide some details.
Someone help please? Thanks.
 A: If you need to use Riemann-Hurwitz you can use the projection mapping $\pi:C\to \Bbb{P}^1, [X:Y:Z]\mapsto [X:Z]$ that simply ignores the $Y$-coordinate. This is well defined everywhere, because nowhere on your curve is $Y$ the only non-zero coordinate.
On the affine chart $Z=1$ using variables $x=X/Z$, $y=Y/Z$, you get the equation
$$
3x^3+4y^3=5.
$$
Here to each $x\in\Bbb{C}$ there are three solutions for $y$ unless $5-3x^3=0$, when there will be three sheets merging. This happens for exactly three distinct values of $x$. Namely at the points $P_k=[x_k:1]$, $x_k=\omega^k\root3\of{5/3}$, 
$\omega=e^{2\pi i/3}, k=0,1,2.$
Above the point at infinity, $[X:Z]=[1:0]$, nothing bad happens. The equation,
written using the dehomogenized variable $u=Y/X$
$$
3+4u^3=0
$$
has three distinct solutions, so no ramification there.
The projective line $\Bbb{P}^1$ aka the Riemann Sphere has genus $g=0$. So Riemann-Hurwitz gives 
$$
2g(C)-2=3(2g(\Bbb{P}^1)-2)+\sum_{k=0}^2(3-1)=3\cdot(-2)+6=0.
$$
Therefore $g(C)=1$ as expected from an elliptic curve.
