Transform $X^3+Y^3+Z^3-3\mu XYZ = 0$ into standard form of elliptic curve

Question

How to transform the projective equation $$X^3+Y^3+Z^3-3\mu XYZ = 0$$ into the standard form of elliptic curve $$ y^2=x^3+px+q? $$

Answer 1

Here's a really terrible answer: ask a computer. I asked a computer ($\textsf{Magma}$, to be specific), and here's what it told me:

> QQu<u> := RationalFunctionField(Rationals());
> PP<X,Y,Z> := ProjectiveSpace(QQu,2);
> R<T0,T1,T2> := PolynomialRing(QQu,3);
> f := T0^3 + T1^3 + T2^3 - 3*u*T0*T1*T2;
> C := Curve(PP,f);
> E, phi := EllipticCurve(C);
> E;
Elliptic Curve defined by y^2 + 3*u/(u^3 - 1)*x*y - 9/(u^6 - 2*u^3 + 1)*y = x^3 - 9*u^2/(u^6 - 2*u^3 + 1)*x^2 + 27*u/(u^9 - 3*u^6 + 3*u^3 - 1)*x - 27/(u^12 - 
    4*u^9 + 6*u^6 - 4*u^3 + 1) over Univariate rational function field over Rational Field
> phi;
Mapping from: CrvPln: C to CrvEll: E
Composition of Mapping from: CrvPln: C to Curve over Univariate rational function field over Rational Field defined by
X^3 - 3*u^2/(u^3 - 1)*X^2*Z + 3*u/(u^3 - 1)*X*Y*Z - 3/(u^3 - 1)*Y^2*Z + 3*u/(u^3 - 1)*X*Z^2 - 3/(u^3 - 1)*Y*Z^2 - 1/(u^3 - 1)*Z^3
with equations : 
Y
-X
X + u*Y + Z and
Mapping from: Curve over Univariate rational function field over Rational Field defined by
X^3 - 3*u^2/(u^3 - 1)*X^2*Z + 3*u/(u^3 - 1)*X*Y*Z - 3/(u^3 - 1)*Y^2*Z + 3*u/(u^3 - 1)*X*Z^2 - 3/(u^3 - 1)*Y*Z^2 - 1/(u^3 - 1)*Z^3 to CrvEll: E
with equations : 
3/(u^3 - 1)*X
-9/(u^6 - 2*u^3 + 1)*Y
Z
and inverse
-3/(u^3 - 1)*$.1
$.2
-9/(u^6 - 2*u^3 + 1)*$.3
> Ew, psi, psi_inv := WeierstrassModel(E);
> Ew;
Elliptic Curve defined by y^2 = x^3 + (-2187*u^4 - 17496*u)/(u^12 - 4*u^9 + 6*u^6 - 4*u^3 + 1)*x + (39366*u^6 - 787320*u^3 - 314928)/(u^18 - 6*u^15 + 15*u^12 
    - 20*u^9 + 15*u^6 - 6*u^3 + 1) over Univariate rational function field over Rational Field
> psi;
Elliptic curve isomorphism from: CrvEll: E to CrvEll: Ew
Taking (x : y : 1) to (36*x - 81*u^2/(u^6 - 2*u^3 + 1) : 216*y + 324*u/(u^3 - 1)*x - 972/(u^6 - 2*u^3 + 1) : 1)
> psi_inv;
Elliptic curve isomorphism from: CrvEll: Ew to CrvEll: E
Taking (x : y : 1) to (1/36*x + 9/4*u^2/(u^6 - 2*u^3 + 1) : 1/216*y - 1/24*u/(u^3 - 1)*x + (9/8*u^3 - 9/2)/(u^9 - 3*u^6 + 3*u^3 - 1) : 1)

(Here $u$ is $\mu$.) Let me try to translate this into English. $\textsf{Magma}$ first gave me a curve $E$ in long Weierstrass form: \begin{align*} y^2 + \frac{3u}{u^3 - 1} xy &- \frac{9}{u^6 - 2u^3 + 1} y =\\ &x^3 - 9 \frac{u^2}{u^6 - 2u^3 + 1} x^2 + \frac{27 u}{u^9 - 3u^6 + 3u^3 - 1} x - \frac{27}{u^{12} - 4u^9 + 6u^6 - 4u^3 + 1} \end{align*} It also gave me an isomorphism $\phi: C \overset{\sim}{\to} E$ as a composition of maps.

Next, I asked for a curve in short Weierstrass form, and $\textsf{Magma}$ gave me the curve $E_w$: \begin{align*} y^2 &= x^3 + \frac{-2187 u^4 - 17496 u}{u^{12} - 4 u^9 + 6 u^6 - 4 u^3 + 1}x + \frac{39366 u^6 - 787320 u^3 - 314928}{u^{18} - 6 u^{15} + 15 u^{12} - 20 u^9 + 15 u^6 - 6 u^3 + 1} \end{align*} as well as an isomorphism $\psi: E \overset{\sim}{\to} E_w$ and its inverse. This looks pretty horrible, and there may very well be a much nicer answer. I also tried asking for a minimal model, but it would take me a while to get it to fit it on the screen. One last word of warning: $\textsf{Magma}$ required that the base ring be a field, so $\mu$ is not allowed to be $0$ (despite the fact that $X^3 + Y^3 + Z^3 = 0$ defines a perfectly good elliptic curve), although the equations might still work out.