Poles and Zeros on $C=\{[X,Y,Z] \in \mathbb P^2\mathbb C \mid X^4 + XY^3 + Z^4=0\}$ of $f=\frac X Z$. 
Problem: Let $X,Y,Z$ be homogeneous coordinates in $\mathbb P^2\mathbb C$ and
$$C=\{[X,Y,Z] \in \mathbb P^2\mathbb C \mid X^4 + XY^3 + Z^4=0\}.$$
Let $f$ be the meromorphic function $f=\frac X Z$.

*

*Calculate poles and zeros for $f$ with their order.

*Calculate ramification points with their index, and calculate the genus $g$ of $C$ using Hurwitz Formula.

*Find 3 linearly independent holomorfic differentials on $C$.


Thoughts: I checked the derivatives and the curve is smooth. Thus $g=(d-1)(d-2)/2=3$, but we are asked to use Hurwitz Formula to find $g$. We try to find zeros and poles for $f$: In the chart $Z \not = 0$ our curve is $x^4 + xy^3 + 1=0$ and $f=X/Z=x$. We need $x=0$ but no point of this kind is on the curve in this chart. Since $f$ gives us the first coordinate $x$, we see that $y^3= \frac{-1-x^4}{x}$ gives us a ramification point iff $RHS=0$ ie we find $4$ ramification points of index 3. We need to check the remaining points with $Z=0$. Namely $P=[0,1,0]$ and $Q_i=[1,a_i,0]$ for $i=1,2,3$ and $a_i$ the third roots of $-1$. We look for $P$ in the chart $Y\not = 0$. We get that the curve is $x^4 + x +z^4=0$ and $f=\frac X Z = \frac X Y \cdot \frac Y Z = \frac x z = \frac {-z^3}{x^3+1}$ using the curve, thus $P$ is a zero of order 3 and $Q_i$ are 3 poles of order 1. Hurwirz Formula states: $$2n + 2g - 2= \sum_{p \in C}(e_p(f)-1).$$ We set $n=3$ because we found $3$ poles, and $RHS=3(1-1)+(3-1)+4\cdot (3-1)$ for the $3$ simple poles, a zero of order 3 and 4 points of ramification. Thus $g=3$.
I don't know if this is correct in any way, and how to solve question 3!
Thanks!
 A: This is a pretty sketchy method (at least I think so), but it works.  I've never completely understood why it works so if someone can illuminate that I would appreciate it.
For question 3, let's look at a chart, say $z=1$.  There, the module of differentials is generated freely by $dx$ and $dy$ with relation $(4x^3+y^3)dx=-3xy^2dy.$  Now, smoothness implies that $4x^3+y^3$ and $3xy^2$ do not simultaneously vanish on $C$ in this chart, so $-\frac{dx}{3xy^2}=\frac{dy}{4x^3+y^3}$ is a holomorphic differential without poles on this chart.
If you look at its image in the chart $y=1$ I believe you get $\frac{-z^2dx-xzdz}{3x}=
\frac{-zdz}{4x^3+1}$.  This has no poles on this chart.  Since these two charts cover the curve, this gives a global holomorphic differential.
To get two others that are linearly independent, do a similar trick but start by looking at the relation between $dx$ and $dz$ in the chart $y=1$, and then the relation between $dy$ and $dz$ in the chart $x=1$.  I found that you get three linearly independent forms this way.
