What are the fibers of a rational morphism to $\mathbb{P}^1$ Given a rational morphsism
$$
f:X \dashrightarrow \mathbb{P}^1
$$
from a projective variety, how can I figure out what the varieties are over a generic point? For example, if I take the rational map
$$
\operatorname{Proj} \left( \frac{\mathbb{C}[x,y,z,w]}{(x^4 + y^4 + z^4 + w^4)} \right) \dashrightarrow \mathbb{P}^1
$$
sending
$$
[x:y:z:w] \mapsto [x:y]
$$
I'm expecting the fibers to be curves, but I'm not sure of what genus.
 A: 1) The rational map $f:X\dashrightarrow \mathbb P^1 $ has poles at the four points $P_j=[0:0:1:\omega_j]$, where the $\omega_j\in \mathbb C$ are the solutions of $z^4+1=0$.
The rational function $f$ is regular on $X_0=X\setminus \{\omega_1,\omega_2,\omega_3,\omega_4\}$ and defines a morphism $$f\vert X_0:X_0\to \mathbb P^1$$  2) The "fibers" of $f$ over $Q=[a:b]$ are the curves $C_Q$ defined by the equations $$C_Q=V(bx-ay, x^4 + y^4 + z^4 + w^4)  \subset X \subset \mathbb P^3                           $$ 
Those curves  all go through the points $P_j$, so that these "fibers" are not disjoint: a bad sign for fibers !
3) The precise meaning of "fiber" is that the actual  fibers of $f\vert X_0$ are the curves $C_Q^0$ obtained from $C_Q$ by deleting the four points $P_j$, and that $C_Q$ is the closure, or projective completion, of $C_Q^0$.
4) An elementary calculation shows that the plane quartic curves $C_Q$ are  smooth  except for the four values  $Q_k=[a_k:b_k]=[1:\omega_k]\in \mathbb P^1$ for which $a^4+b^4=0$ .
These smooth "fibers" have genus 3, like all smooth plane quartics of degree $4$.
5) Each of the four singular "fibers"$$C_{Q_k}=V(y-\omega_k x, x^4 + y^4 + z^4 + w^4)\subset X \subset  \mathbb P^3$$  consists in a  pencil of four lines passing through through the point $A_k=[1:\omega_k:0:0]\in X_0$ and lying in the plane $y=\omega_k x$ .   
A: Suppose pick the point $[1 : a] \in \mathbb P^1$. The fibre above this point is the variety:
$$ (y = ax, \ \  \ x^4 + y^4 + z^4 + w^4 = 0) \subset \mathbb P^3$$
But of course, this is the same variety as this:
$$ ((1 + a^4) x^4 + z^4 + w^4 = 0) \subset \mathbb P^2,$$
where we have dropped the $y$ coordinate, and the $\mathbb P^2$ is parametrised by the $x, z$ and $w$ coordinates.
For generic choices of $a$, this variety is smooth curve in $\mathbb P^2$ of degree $d = 4$. (The smoothness of this variety for generic choices of $a$ is a consequence of Bertini's theorem.) By the genus-degree formula, this plane quartic has genus
$$ g = \frac 1 2 (d - 1)(d - 2) = 3.$$
