TITLE: Degree 2 map of elliptic Curve
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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 $C\to\mathbb{P}^1$.