Number of branch points for a projection to $\Bbb{CP}^1$

Question

Based on a helpful response to my previous post which offered advice on how to split the image of a map into affine and non-affine components, I've come up with a solution to the following problem. However the solutions provided by the TA indicate that there are in fact 9 branching points (my solution gives exactly 4).

I would be very grateful if someone could point out where I've made my mistake?

The problem

Given a projective curve C defined by $$Z^2Y^2 = X^4 + Y^4 + Z^4$$ Consider the holomorphic map$$ f: [X:Y:Z] \mapsto [X:Y] = \mathbb{CP}^1 $$

What are the branching points of f?

My solution

For $[X:Y] \in \mathbb{CP}^1$, let $[U:V]:=[X:Y] $. First we consider the preimage of the affine component $\{[U:V]\ | \ U = 1 \}$. We have that:

$$ f^{-1} (\ [1:V]\ ) = C \ \cap [1:Y:Z] $$

Which gives us the corresponding affine curve:

$$ z^2y^2 - 1 - y^4 - z^4 = 0 $$

Rewriting, we have:

$$ (-1)(z^2)^2 + (y^2)(z^2) + (- y^4 - 1) = 0 $$

Which gives us coefficients $ a = -1, \ b=y^2,\ c = -y^4 - 1$ for the quadratic formula:

$${z^2=\frac{-y^2\pm\sqrt{-3y^4 -4}}{-2}}$$

Writing $\sqrt[4]{-1} = \omega_i $ for a 4th root of negative unity, we then have four branching points

$$y \in \{ \omega_1, \omega_2, \omega_3, \omega_4 \}$$ each with brancing index 4.

Finally, consider the preimage of the non-affine component:

$$f^{-1} (\ [0:V]\ ) = C \ \cap [0:Y:Z] $$

which gives the corresponding curve

$$z^2y^2 - y^4 - z^4 = 0.$$

Now, since $[0:0:0] \notin \mathbb{CP}^2$, it follows that $y=z=0$ is not a solution, and we can therefore scale the co-ordinates $[0:Y:Z]$ to $[0:1:Z]$ by multiplication by some nonzero factor. This then gives us:

$$ z^4 -1 -z^2 = 0$$

Which clearly has four distinct roots $z_1, z_2, z_3, z_4$ , with:

$$ f(z_i) = [0:V] = \infty $$

It follows that the preimage of $\infty$ contains four points, each with ramification index 1 and therefore it is not a branching point. Since there are no more possible candidates for branching points, we are done.

Answer 1

Claim.
The map $f\colon C \rightarrow \mathbb{CP}^1; [X:Y:Z]\mapsto [X:Y]$ has $12$ ramification points, each of multiplicity $2$. (You seem to call ramification points "branching points". In the terminology I prefer, branching points (aka branch points) lie in the codomain, while ramification points lie in the domain.)

The ramification points of $f$ are the $12$ points in the set $$\big\{[\lambda:\pm \sqrt{2}:1] \in \mathbb{CP}^2 \ \vert \ \lambda^4=-3\big\}\cup\big\{[1:\omega:0] \in \mathbb{CP}^2 \ \vert \ \omega^4=-1\big\}.$$

Consequently, the set of branch points of $f$ is the $12$-element set $$\big\{[\lambda:\pm \sqrt{2}] \in \mathbb{CP}^1 \ \vert \ \lambda^4=-3\big\}\cup\big\{[1:\omega] \in \mathbb{CP}^1 \ \vert \ \omega^4=-1\big\}.$$

Proof.
Consider the homogeneous polynomial $F(X,Y,Z)\colon=X^4+Y^4+Z^4-Z^2Y^2.$ A point $p\in C$ is a ramification point of $f$ if and only if $\frac{\partial F}{\partial Z}(p)=0.$ For a proof see here.

Thus, a point $[X:Y:Z]\in \mathbb{CP}^2$ is a ramification point of $f$ if and only if $4Z^3-2ZY^2=0$ and $X^4+Y^4+Z^4-Z^2Y^2=0.$

Firstly, let $p=[X:Y:Z]\in \mathbb{CP}^2$ with $Z\neq 0$. Consider the representative of $p$ with $Z=1$. We see that $p$ is a ramification point of $f$ if and only if $Y^2=X^4+Y^4+1$ and $4=2Y^2$. Solving for $X$ and $Y$ shows that the ramification points $[X:Y:Z]$ of $f$ with $Z\neq 0$ are precisely the eight points in $\big\{[\lambda:\pm \sqrt{2}:1] \in \mathbb{CP}^2 \ \vert \ \lambda^4=-3\big\}$.

Secondly, let $q=[X:Y:Z]\in \mathbb{CP}^2$ with $Z=0$. We see that $q$ is a ramification point of $f$ if and only if $X^4+Y^4=0$. This implies that for any ramification point $q$ with $Z=0$ we have $X\neq 0$. For otherwise we would have $Y=0$, contradicting the definition of $\mathbb{CP}^2$. Thus, we can restrict our attention to $q=[X:Y:Z]\in \mathbb{CP}^2$ with $X=1$. We then immediately see that the ramification points $[X:Y:Z]$ of $f$ with $Z=0$ are precisely the four points in $\big\{[1:\omega:0] \in \mathbb{CP}^2 \ \vert \ \omega^4=-1\big\}$.

To determine the multiplicities of the ramification points of $f$, denote by $d$ the degree of the smooth projective plane curve $C$. Recall that the degree of the ramification divisor of $f$ is equal to $d(d-1)$: $$\sum_{p\in C}\big(\operatorname{mult}_p(f)-1\big)=d(d-1).$$ For a proof sketch of this result see my answer here.

In this case $d=4$, so that $\sum_{p\in C}\big(\operatorname{mult}_p(f)-1\big)=12$. Therefore, since $f$ has twelve ramification points, each ramification point has multiplicity $2$.