Currently working through Silverman and Tate's Rational Points on Elliptic Curves.

Right now I'm at Weierstrass form, and I'm interested in seeing some examples of curves being transformed into the Weierstrass form, showing most substitutions, calculations, and maybe some information about what each substitution "does" to the curve, that is, how does it change the curve but maintain its algebraic structure?

  • $\begingroup$ What do you mean by "calculations"? Do you want to see an example of how the transformation is found? $\endgroup$ Apr 28, 2017 at 15:08

1 Answer 1


The following is an excerpt from this book (Example 2.2.3):

Let $d\in\mathbb{Z}$, $d\neq 0$ and let $E$ be the elliptic curve given by the cubic equation $$X^3+Y^3 = dZ^3$$ with $\mathcal{O} = [1,-1,0]$. The reader should verify that $E$ is a smooth curve. We wish to find a Weierstrass equation for $E$. Note that if we change $X=U+V$, $Y=-V$, $Z=W$, then we obtain a new equation \begin{align} U^3+3U^2V+3UV^2=dW^3. \end{align} Since this equation is quadratic in $V$, and cubic in $W$, with no other cubic monomials that involve $W$, the variable $W$ will end up playing the role of $x$, and the variable $V$ will play the role of $y$ in our Weierstrass model. Next, we change variables to obtain a coefficient of $1$ in front of $V^2$ and $W^3$. If we multiply the previous equation through by $d^2$, we obtain \begin{align}\label{eq-ex223b} d^2U^3+3d^2U^2V+3d^2UV^2=d^3W^3,\end{align} and now we change variables $x=3dW$, $y=9dV$, and $z=U$. Then, we obtain \begin{align}\label{eq-ex223c} d^2z+\frac{dyz}{3}+\frac{y^2z}{27}=\frac{x^3}{27},\end{align} or, equivalently, $y^2z+9dyz=x^3-27d^2z$, which is a Weierstrass equation. Thus, $[x,y,z] = [3dW,9dV,U]=[3dZ,-9dY,X+Y]$ and we have found a change of variables $\psi:E\to \widehat{E}$ given by $$\psi([X,Y,Z]) = [3dZ,-9dY,X+Y]$$ such that the image lands on the curve in Weierstrass equation $\widehat{E}: y^2z+9dyz=x^3-27d^2z$. The map $\psi$ is invertible; the inverse map $\psi^{-1}: \widehat{E} \to E$ is $$\psi^{-1}([x,y,z]) = \left[\frac{9dz+y}{9d},\ -\frac{y}{9d},\ \frac{x}{3d}\right].$$ In affine coordinates, the change of variables is going from $X^3+Y^3=d$ to the curve $y^2+9dy=x^3-27d^2$ via the maps: \begin{eqnarray*} \psi(X,Y) &=& \left(\frac{3d}{X+Y},-\frac{9dY}{X+Y}\right),\\ \psi^{-1}(x,y) &=& \left(\frac{9d+y}{3x},-\frac{y}{3x}\right). \end{eqnarray*} We leave it as an exercise for the reader to verify that the model can be further simplified to the form $y^2=x^3-432d^2$.


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