elliptic curve ${X^3+Y^3=AZ^3}$

Question

consider the elliptic curve $X^3+Y^3=AZ^3$ and $A$ in $K*$ with $O=(1,-1,0)$. show the $j$ invariant of this elliptic curves is $0$. (part d of Silverman exercise page 104 Q3.3 ) I can compute the $j$ invariant of Weierstrass form of elliptic curves but I don't know how to change this elliptic to Weierstrass form and compute $j$ invariant. I search a lot about curves but unfortunately, I can't find!

Answer 1

The Handbook of Elliptic Curves by Ian Connell is a wonderful reference.

http://webs.ucm.es/BUCM/mat/doc8354.pdf

(...after a quick search.)

In section [1.4 Cubic to Weierstrass: Nagell's algorithm] we have the steps of the algorithm that can be applied for any (non-singular) cubic to put it in Weierstraß form. (Characteristic not two, three.)

Then section 1.4.1 gives an application on Selmer curves:

Proposition 1.4.1: Consider the Selmer curve $$au^3+bv^3=c\ ,\qquad abc\ne 0\ ,$$ over some field $K$ of characteristic $\ne 2,3$. Then (after permuting variables) assume $$ \theta=\sqrt[3]{c/b}\in K\ . $$ Then the given Selmer curve is birationally equivalent to the curve $$y^2=x^3-432a^2b^2c^2$$ under the mutually inverse transformations $$ \begin{aligned} u &=-\frac{6b\theta^2x}{y-36abc}\ ,\\ v &=\theta\frac{y+36abc}{y-36abc}\ ,\\[2mm] x &=-\frac{12ab\theta^2u}{v-\theta}\ ,\\ y &=36abc\frac{v+\theta}{v-\theta}\ . \end{aligned} $$

In our case, we can for instance cyclically permute the given equation, rewrite it like $$ -AZ^3+X^3=-Y^3\ , $$ consider $a=-A$, $b=1$, $c=-1$, so that $\theta=\sqrt[3]{c/b}=-1\in\Bbb Q$, then use the transform in the affine space corresponding to $Y=1$: $$ \begin{aligned} Z &=\frac{6x}{y-36A}\ ,\\ X &=-\frac{y+36A}{y-36A}\ ,\ . \end{aligned} $$ and introducing in $-AZ^3+X^3=-1$ we get $$ -A\left(\frac{6x}{y-36A}\right)^3 -\left(\frac{y+36A}{y-36A}\right)^3 =-1\ . $$ Now multiply with the common denominator, and reshape equivalently as: $$ \begin{aligned} -A\cdot 6^3x^3 & -y^3 -3\cdot 6^2Ay^2 -3\cdot 6^4A^2y -6^6A^3 \\ = & -y^3 +3\cdot 6^2Ay^2 -3\cdot 6^4A^2y +6^6A^3\ . \end{aligned} $$ The cancellations can now be followed with bare eyes.

Answer 2

There's already an answer, but here's how I justify the translations: If we set $X = U-V, Y = U+V$ we can eliminate one of the cubes: $$ X^3+Y^3 = 2U^3+ 6UV^2 \\ X^3 + Y^3 - AZ^3 = 2U^3+ 6UV^2 - AZ^3 = 0 $$ And dividing by $U^3$ gives $$ 2 + 6\left(\frac{V}{U}\right)^2 - A\left(\frac{Z}{U}\right)^3 = 0 $$ Setting $\frac{Z}{U} = \frac{n}{6A}, \frac{V}{U} = \frac{m}{6^2A}$ and multiplying across by $6^3A^2$ we get: $$ 432A^2 + m^2 = n^3 $$

which is Weierstrass normal form. I know this isn't the "by-the-book" method, but it's linear rational substitutions only, so you should be able to work it backwards to find the transformation from $(n,m)$ to $(X,Y,Z)$.