Transforming $X^3 + Y^3 + Z^3 + 3XYZ = 0$ into an elliptic curve

Question

I am trying to transform $X^3 + Y^3 + Z^3 + 3XYZ = 0$ into an equation of the form $y^2 = x^3 + Ax + B$. I tried dividing by $Z^3$ and then letting $X/Z = u + v$ and $Y/Z = u - v$. That gave me the equation $2u^3 + 6uv^2 + 3(u^2 - v^2) + 1 = 0$. I then tried dividing by $v^2$ and letting $y/36 = v/u$ and $-x/6 = 1/u$. This gave me $y^2 = x^3 + (6^3/2)x - (3/6^2)xy^2 - 2(6^3)$. Unfortunately, I cannot figure out a way to get rid of the $xy$ factor. I would appreciate it if someone could give me a hint on how to proceed from here, or to try a different approach. Thanks.

Answer 1

I will follow Nagell's algorithm, as described for instance in the excellent Handbook of Ian Connell, §1.4, Cubic to Weierstraß, page 115 and following, a link found quickly is

Ian Connell, Elliptic Curve Handbook, 1999

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$(0)$ We start with the given curve, and let us set $Z=1$ (and pass to an affine piece of the curve) for an easy typing, so $$ X^3 + Y^3 + 3XY+1=0\ . $$ We need it in the form $(\unicode{xB6})$, so that $(0,0)$ is a point on it. For this we use some point we can figure out, and translate it. I will use $(X,Y)=(0,-1)$, thus use $X$ further and the substitution $Y_1=Y+1$, $Y=Y_1-1$. We obtain: $$ X^3 + Y_1^3 - 3Y_1^2 + 3Y_1 + 3XY_1 - 3X = 0\ . $$ The boxed coefficient is our coefficient $s_9$ from loc. cit., and $(X,Y_1)$ the variables used here instead of $(u,v)$ used in $(\unicode{xB6})$.

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$(1)$ Yes, $s_9=3\ne 0$. The curve is not singular in $(X,Y_1)=(0,0)$.

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$(2)$ The homogenized form, taking $X=U/W$, $Y_1=V/W$ is $$ 0 = F := \underbrace{U^3 + V^3}_{F_3} + \underbrace{3V(U-V)}_{F_2}W + \underbrace{3(V-U)}_{F_1}W^2\ . $$ The tangent line in the point $P=[U:V:W]=[0:0:1]$ has the equation $F_1=0$, i.e. $V=U$, and it intersects the cubic in $2U^3=0$, with solution $U=0$ with multiplicity $3$, which points to the point $P$, so the "other" two intersection points are $P$, $P$, counted with multiplicity. The flex point case. And it is already the origin (in the $W=1$ affine component). We do not need any change, and go back to the affine coordinates, which are back $(X,Y_1)$ again: $$ X^3 + Y_1^3 +3Y_1(X-Y_1) + 3(Y_1 - X) = 0\ . $$

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$(3)$ Write $t$ for $Y_1/X$, and we use now $(X,t)$ as variables. So $Y_1$ is replaced by $tX$. We obtain $$ X^3 (1+ t^3) + 3X^2t(1-t) + 3X(t-1)=0\ . $$ We work in the affine world $X\ne 0$, so write instead: $$ X^2 (1+ t^3) + 3Xt(1-t) + 3(t-1)=0\ . $$ We solve for $X$, and obtain: $$ \tag{$*$} X = \frac 1{2(1+t^3)}\left(3t(t-1)\pm \sqrt{\underbrace{9t^2(t-1)^2-12(t-1)(t^3+1)}_{:=\delta}^{\phantom\delta }}\right) \ . $$ One obvious solution for the vanishing of the radical is $t_0=1$. So we substitute $t = t_0+\frac 1\tau=1+\frac 1\tau$, and obtain $$ \delta = -3 \, \left(t^3 + 3 \, t^2 + 4\right)(t - 1) \\ = \underbrace{ -3(8\tau^3 + 9\tau^2 +6\tau + 1) }_{=:\rho} \frac 1{\tau^4}\ , $$ and $\rho$ is a cubic in $\tau$.

Imagine now that in the above $(*)$ relation we isolate the radical on one side, and all the rest on the other side.

Take squares on both sides.

Clear the denominator $\tau^4$ on the $\delta$-side. On the other side we have a square, depending on $X,\tau$, let us use the letter $\nu$ for it. We obtained thus: $$ \nu^2 = \underbrace{ -3(8\tau^3 + 9\tau^2 +6\tau + 1) }_{=:\rho} \ . $$ We are done. But let us complete the cosmetic steps remained, and have a final check with the computer.

That coefficient $-3\cdot 2^3$ of $\tau^3$ is maybe not the one we want, we multiply both sides with $9$, and use $y=3\nu$, $x=-6\tau$ to obtain the equation: $$ y^2 = x^3 -\frac{27}4x^2 +27x-27\ . $$

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Computer check. Let us bring both the cubic we started with, and the result above in the short Weierstraß form, using a minimal model.

R.<X,Y,Z> = PolynomialRing(QQ)

cubic = X^3 + Y^3 + Z^3 + 3*X*Y*Z
f, g = WeierstrassForm(cubic)
E = EllipticCurve(QQ, [f, g])
M = E.minimal_model()

print(E)
print(M)

EE = EllipticCurve(QQ, [0, -27/4, 0, 27, -27])

This gives the elliptic curve $E$, and its minimal model $M$

Elliptic Curve defined by y^2 = x^3 + 189/16*x + 351/32 over Rational Field
Elliptic Curve defined by y^2 + x*y = x^3 - x^2 + 12*x + 8 over Rational Field

And we can ask for the short Weierstraß form of the last curve:

Elliptic Curve defined by y^2 = x^3 + 15309*x + 511758 over Rational Field

Is our result the same result? Yes:

sage: E.is_isomorphic(EE)
True

This can be checked by replacing $x$ by $x'+9/4$ in $y^2 = x^3 -\frac{27}4x^2 +27x-27$, just a translation, and from our computed curve we obtain the equation with $f=189/16$, $g= 351/32$.