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

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!

• In pari/gp run ellfromeqn(x^3+y^3-a) Commented Jul 3, 2019 at 19:34

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.

• thanks for your compeletly respond! Commented Jul 31, 2019 at 13:53

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 $$(X,Y,Z)$$.