Showing that the range of a degree 2 morphism from a plane quartic is an elliptic curve

Question

I apologize in advance, because I believe I am making a horrible mistake that I am entirely unaware of, but here we go:

Suppose we have a degree $2$ map $\phi: C \to E$ defined by $[x:y:z] \to [x:y:z^2]$ where $C$ is a smooth plane quartic of the form $z^4 + a_2(x, y)z^2 + a_4(x, y)$ where $a_i(x,y)$ is a homogeneous degree $d$ polynomial in $x$ and $y$.

I have been told that $E$ will be an elliptic curve defined by the equation $w^2 + a_2(x, y)w + a_4(x, y)$ ($w = z^2$). However, I am having a hard time understanding why this equation actually gives us an elliptic curve. I must be doing something horribly wrong, but I can’t even see how the genus of $E$ would be $1$. Based on the genus-degree formula, we would have $g = 3$.

Any hints/resources would be appreciated. Thanks in advance.

Answer 1

Note that $w^2+a_2(x,y)w+a_4(x,y)$ is not homogeneous under the standard grading with $\deg x = \deg y = \deg w$, as $\deg w^2=2 \neq 4 = \deg a_4(x,y)$, so the usual genus-degree formula doesn't apply here. Instead, the correct location for this curve to live is the weighted projective space $\Bbb P(1,1,2)$ with $\deg x = \deg y = 1$ and $\deg w = 2$.

We'll verify that $E$ is smooth and use Riemann-Hurwitz to see what its genus must be. The homogeneous Jacobian of $C$ is $$\left(\frac{da_2(x,y)}{dx}z^2+\frac{da_4(x,y)}{dx}, \frac{da_2(x,y)}{dy}z^2+\frac{da_4(x,y)}{dy}, 4z^3+2za_2(x,y)\right),$$ while the homogeneous Jacobian of $E$ is $$\left(\frac{da_2(x,y)}{dx}w+\frac{da_4(x,y)}{dx}, \frac{da_2(x,y)}{dy}w+\frac{da_4(x,y)}{dy}, 2w+a_2(x,y)\right).$$ The first two entries are the same up to $w=z^2$, while the third entry of the Jacobian of $C$ can be written as $4z^3+2za_2(x,y) = 2z(z^2+2a_2(x,y))$, so nonvanishing of the Jacobian of $C$ implies nonvanishing of the Jacobian of $E$, hence smoothness of $C$ implies smoothness of $E$.

To compute the ramification points of $\phi$, we look for points on $C$ where $z=0$. These are exactly solutions of $a_4(x,y)$, which must be distinct: if $a_4(x,y)$ has a repeated root, then the Jacobian of $C$ must vanish there, which would make $C$ nonsmooth. Therefore there are 4 ramification points of $\phi$, and the ramification index is 2 at each one, so Riemann-Hurwitz gives $$ 4 = 2g(C)-2 = 2(2g(E)-2) + \deg R = 4(g(E)-1) + 4,$$ or that $g(E)=1$ and $E$ is an elliptic curve.