Past exam of the algebraic curves course of the University of Oxford / 2023 / 2 / c / Solution

First make the invertible linear change of homogeneous coordinatesThe original equation becomesThis takes the flex $[0:0:1]$ to the point at infinity $[0:1:0]$ of a Weierstrass model.

Put $h=\omega -1$ and choose $s\in \mathbb{C}$ with $s^2=h^3$. Both are nonzero. DefineSince $U=hX+Z$, the three factors on the right are $hX$, $h(X-Z)$ and $h(X-\lambda Z)$, whereThe resulting equation is preciselythe Legendre form of an elliptic curve. In particular, $\lambda \ne 0,1$.

Combining the two changes gives an explicit projective transformation from the original coordinates:Its determinant is $-\sqrt{3}/(hs)\ne 0$, so this is a valid transformation of the whole projective plane, including the points where an affine coordinate would be undefined.