Plucker's Formula proof I just read a proof of Plucker's Formula in Rick's Miranda book that says 

A smooth projective plane curve of degree $d$ has genus $\frac{(d-1)(d-2)}{2}$.

To do this he defines map $\pi : X \rightarrow P^1$ such that $\pi[x:y:z]=[x:z]$ and he claims the map has degree $d$, and this is where i have my doubt, i know this is true if $X$ is given by a a function $F$ of the form $x^d+y^d+z^d$ so maybe this is true because we can do some kind of change of variable and assume this ?  Also i have tried to see that its degree $d$ by computing it but got nowhere, since we know $\pi^{-1}[1:0]=[1:y:0]$ such that $F[1:y:0] = 0$ we will have $d$ solutions to this and we know that a point is a ramification point for this map iff $\frac{\partial F}{\partial y}(p)=0$, and we know that $F[x:y:0]=\frac{1}{d}(\frac{\partial F}{\partial x} + y\frac{\partial F}{\partial y})$ but I cant seem to find a reason why $\frac{\partial F}{\partial y} \neq 0$ so any help is welcomed. Or is it because by changing coordinates we can assume that the smooth projective curve will be given by $x^d+y^d+z^d=0$?
Also another thing I dont fully understand, when proving that the degree of a smooth projective curve $X$ will be the degree of an hyperplane divisor we assume that the hyperlane divisor will be given by $G(x,y,z)=x$ and that $[0:0:1]$ is not in $X$, my question is why can we assume both of these things ? I get that by changing the coordinates we could get one but I dont see how we could get the two withouse losing some generality. Thanks in advance!
 A: To find the degree of $\pi$, consider a general point $[\lambda:1] \in \mathbb{P}^1$ i.e. a non-branch point. The preimages of $[\lambda:1]$ would be $[\lambda:y:1]$ such that $F(\lambda,y,1)=0 $. Notice that the point $[0:1:0]$ has been removed from $X$ by a change of coordinates, otherwise the map $\pi$ wouldn't make sense.
Hence, $F(\lambda,y,1)$ is a degree $d$ polynomial in $y$ and therefore has $d$ roots. Let $f(y) = F(\lambda,y,1)$. Note that
$$ \frac{\partial F}{\partial y}(\lambda,y,1) = \frac{\partial f}{\partial y}(y)$$
and, $y_0$ is a repeating root of $f \implies \partial f/\partial y(y_0)=0 \implies \partial F/\partial y(\lambda,y_0,1)=0 \implies [\lambda:y_0:1]$ is a ramification point of $\pi$.
Hence, all the roots of $f$ are distinct and therefore $\pi$ has degree $d$.
For the second part, you can first make a coordinate transformation so that $G(x,y,z)=x$ and then follow up with another transformation of just the $yz$ plane to remove the point $(0,1)$ from the roots of $F(0,y,z)$.
