Weierstraß normal form

General equation

Karl Theodor Wilhelm Weierstraß (31 Oct 1815 – 19 Feb 1897)

$gz^{3}+hz^{2}w+jzw^{2}+kw^{3}$

${}+mz^{2}+pzw+qw^{2}$

${}+rz+sw$

${}+t\qquad =0.$

Linear transformation

${\begin{bmatrix}z\\w\end{bmatrix}}={\begin{bmatrix}\alpha &\beta \\\gamma &\delta \end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}$

$z=\alpha x+\beta y+\zeta$

$w=\gamma x+\delta y+\eta$

Problem

Substitute $\alpha x+\beta y+\zeta$ and $\gamma x+\delta y+\eta$ for z and w in the general equation, simplify by collecting like terms in respective powers of x and y, and solve for α, β, γ, δ, ζ, η, a and b so that

$y^{2}=x^{3}+ax+b.$

Eliminating the skew term

Sometimes the Weierstraß equation is given in the form

$w^{2}+zw=z^{3}+rz+t$

with an extra skew term $zw$. Let $y=w+{\frac {1}{2}}z$ and complete the square.

$y^{2}-{\frac {1}{4}}z^{2}=z^{3}+rz+t$

$y^{2}=z^{3}+{\frac {1}{4}}z^{2}+rz+t$

Now let $x=z+{\frac {1}{12}}$ and complete the cube, viz.

$x^{3}=z^{3}+{\frac {1}{4}}z^{2}+{\frac {1}{48}}z+{\frac {1}{1728}}$

$y^{2}=z^{3}+{\frac {1}{4}}z^{2}+rz+t$

$y^{2}=x^{3}+\left(r-{\frac {1}{48}}\right)z+t-{\frac {1}{1728}}$

$y^{2}=x^{3}+\left(r-{\frac {1}{48}}\right)\left(x-{\frac {1}{12}}\right)+t-{\frac {1}{1728}}$

$y^{2}=x^{3}+\left[r-{\frac {1}{48}}\right]x+\left[t-{\frac {1}{12}}r+{\frac {1}{864}}\right]$

This is the normal Weierstraß form with $a=r-{\frac {1}{48}}$ and $b=t-{\frac {1}{12}}r+{\frac {1}{864}}$, and obviously rational points are mapped to rational points with the rational linear transformation $y=w+{\frac {1}{2}}z$ and $x=z+{\frac {1}{12}}$.