Degree 2 map of elliptic Curve

Question

I'm having a hard time constructing an explicit degree $2$ map $$C \rightarrow \mathbb{P}_k^1$$ where C is given by cutting out the elliptic homogeneous equation $$X^3+Y^3-Z^3=0$$ in $\mathbb{P}_k^2$. Considering the affine open $D(z)$, I see a natural map of function fields $k(t) \hookrightarrow k(t)[s]/(s^3+(t^3-1))$, but this map has degree $3$ I guess, as it gives an algebraic field extension of degree $3$. Sending the generator $t$ to another element might solve that problem, but I don't see how.

Answer 1

Assuming $\operatorname{char}k\neq 2,3$.

If we have a cubic in, for example, Weierstrass form $$ Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XY^2+a_6Z^3 $$ then it is obvious $(X:Y:Z)\mapsto(X:Z)$ is a degree $2$ map to $\mathbb{P}^1$. So let's try to transform our equation to something like that.

First, move a point on our curve, say $(0:1:1)$, to the point $(0:1:0)$ that you would expect for an elliptic curve in Weierstrass form, so \begin{align*} 0&=X^3+Y^3-Z^3\\ &=X^3+(Y-Z)[Y^2+YZ+Z^2]\\ &=X^3+(Y-Z)\left[\frac14(Y-Z)^2+\frac34(Y+Z)^2\right] \end{align*} Hence we have $$ 3(Y-Z)(Y+Z)^2=-4X^3-(Y-Z)^3 $$ So the map $$ (X:Y:Z)\mapsto(X:Y-Z) $$ is a degree 2 map $C\to\mathbb{P}^1$.

Answer 2

I think that the map is

$\phi: C\to \mathbb{P}^1$ such that for each $[X,Y,Z]\in C$ you have that

$\phi([X,Y,Z])=[X,Y]$

It’s well defined because $X,Y\neq 0$ otherwise also $Z=0$ and it is not possibile.