transform into weierstrass-form

Question

How can I transform the elliptic curve $E/\mathbb{C}$ of the form $$y^2=4(x-e_1)(x-e_2)(x-e_3)$$ with $e_1>e_2>e_3\in\mathbb{R}$ roots of $E$ into a Weierstrass-Form $y^2=x^3+ax+b$?

Answer 1

First change $y=2y'$ and $x=x'$. That will leave the model in the form $$y'^2 = (x'-e_1)(x'-e_2)(x'-e_3),$$ after cancelling $4$'s on both sides. Now expand the polynomial, in $x'$, so that you have $$y'^2 = x'^3+Ax'^2+Bx'+C.$$ Finally, a change of the form $x' = X-A/3$ and $y'=Y$ gets rid of the $X^2$ term, and leaves $$Y^2 = X^3+DX+E.$$