I need to calculate the genus of the Fermat Curve, and I'd like to be reviewed on what I have done so far; I'm not secure of my argumentation.

Such curve is defined as the zero locus
$$X=\{[x:y:z]\in\mathbb{P}^2:x^d+y^d+z^d=0\}$$
It is defined by the homogeneous polynomial $F(x,y,z)=x^d+y^d+z^d$ which is obviously non-singular, therefore $X$ is a smooth projective plane curve.
Now, I consider the projection
$$
\begin{array}{cccc}
\pi:&X&\longrightarrow&\mathbb{P}^1\\
&[x:y:z]&\longmapsto&[x:y]
\end{array}
$$
which is a well-defined holomorphic function. I need to calculate $\pi$'s degree, and for this I must calculate the *sum of the multiplicities of the set* $\{p:p\in\pi^{-1}(y)\}$, for any $y$ in $\pi(X)$; since the degree is constant, any such $y$ will do.

We note that, from Miranda's *Riemann Surfaces and Algebraic Curves*' Lemma 4.6 at page 46, that $\pi$ has a branch point at $p\in X$ if and only if $\frac{\partial F}{\partial z}(p)=0$

Having this in mind, to calculate the degree, I chose $[1:0]\in\mathbb{P}^1$; and for this one,
$$\pi^{-1}([1:0])=\{[1:0:z]:1+z^d=0\}$$
Now I can see clearly that $|\pi^{-1}([1:0])|=d$ (the *d-th* roots of $-1$), and each one has multiplicity 1 (because they are not ramification points, as the above mentioned lemma states); therefore the degree of $\pi$ is $d$.

Now from the Lemma again, the ramification points of $\pi$ are those that $z=0$, that is, the points $\{[x:y:0]\in\mathbb{P}^2:x^d+y^d=1\}$ In homogeneous coordinates those points are of the form $[1:y:0]$, where $y$ is a *d-th roots of* $-1$ and from this we conclude that $\pi$ has $d$ points of ramification, and those are
$$
\left[ 1:e^{\dfrac{2\pi i k}{d}}:0\right]
$$

To calculate the multiplicity of each one of these points I used the following argument, **which I'm not sure if it is correct**: for a point $[x:y]\in\mathbb{P}^1$ such that $x^d+y^d=0$, $\pi^{-1}([x:y])=\{[x:y:0]\}$ and therefore $|\pi^{-1}([x:y])|=1$. Since the degree of $\pi$ is $d$, $d$ is also the multiplicity of every point of ramification of $\pi$.

What follows noy is applying Hurwitz's formula (and noting that $g(\mathbb{P}^1)=0$ and a very easy calculation to conclude that $g(x)=\dfrac{(d-1)(d-2)}{2}$.

Now I must apologize for the long, long text, but I will be much grateful for any correction made in my argumentation!