# Elliptic Curve $E/\mathbb{Q}$ with $\Delta_E^{1/3}$ a root of defining cubic

Consider the elliptic curve $$E'\colon y^2 = g(x) = x^3 + \frac{1}{432} .$$ One can check that the discriminant of $E'$ is $-1/432$, that $E'$ has complex multiplication, and that $(-1/432)^{1/3} = (-1/6 \sqrt{2})$ is a root of $g(x)$. Hence there exists a CM elliptic curve over $\mathbb{Q}$ such that the cube root of the discriminant is a root of the defining polynomial.

My question is of the same flavor of this with added assumptions, in particular, we want $$E \colon y^2 = f(x) = x^3 + Ax + B$$ to be a non-CM elliptic curve over $\mathbb{Q}$ in Weierstrass form. Let $\Delta_E$ denote the discriminant of $E$, and for the time being, we want $\Delta_E$ to be a rational square. I want to know whether there exists such an elliptic curve over $\mathbb{Q}$ such that $f(\Delta_E^{1/3}) = 0$. The condition on the discriminant tells us that there are 3 real, distinct roots $f(x)$, which only seems to complicate matters. Any help and/or references would greatly be appreciated.

My guess is that the answer to your question is "no", i.e., there is no elliptic curve without CM with the specified property on the discriminant, and I was very close to prove it... Hopefully you can finish it up somehow. However, there is another CM curve with the same property, namely $E: y^2=x^3-1/16x$!

Let $E: y^2=f(x)=x^3+Ax+B$, for some $A,B\in\mathbb{Q}$ and discriminant $\Delta_E$, be such that $\Delta$ is a perfect square, and $f(\Delta_E^{1/3})=0$. Notice that this means $$0 = \Delta_E +A\Delta_E^{1/3}+B.$$ If $A=0$, then $E$ has CM (with $j$-invariant equal to $0$), so let us assume that $A\neq 0$. Therefore, $\Delta_E^{1/3}=(-B-\Delta_E)/A\in\mathbb{Q}$. It follows that $\Delta_E$ is a cube in addition to a square, so $\Delta_E=\delta^6$ for some $\delta\in\mathbb{Q}$. Thus, $$0 = \delta^6+A\delta^2+B$$ and so $B=-(\delta^6+A\delta^2)$. Moreover, $$\delta^6=\Delta_E=-16(4A^3+27B^2) = -16(4A^3+27(\delta^6+A\delta^2)^2).$$ This condition defines a curve $C:\delta^6+16(4A^3+27(\delta^6+A\delta^2)^2)=0$ which happens to be of genus $1$. There are $4$ points that are easy to find: $$[d,A,z]=[0,0,1],[0,1,0],[1/2,-1/16,1],[-1/2,-1/16,1].$$ The first two correspond to $A=0$ and a solution with $z=0$, so we discard those two. The third and fourth correspond to $E: y^2=x^3-1/16x$, which is a curve with CM (with $j=1728$). Now, $C$ is of genus $1$ and using $P=[0,1,0]$ as the origin point, we can find a (surprisingly simple) Weierstrass equation $$E': y^2=x^3+729$$ and a map $\phi: C\to E'$ given by $$[243d^{10}z + 486d^6Az^4 + 9/16d^4z^7 + 135d^2A^2z^7,-2187d^{11} - 4374d^7Az^3 - 81/16d^5z^6 - 1215d^3A^2z^6, -A^2z^9].$$

It turns out $E'$ has rank $0$ and torsion $\mathbb{Z}/6\mathbb{Z}$, generated by $R=(18,-81)$. The point $[1/2,-1/16,1]$ of $C$ maps to $R$ via $\phi$, and $[-1/2,-1/16,1]$ of $C$ maps to $-R$ via $\phi$. The map $\phi$ is not defined at $[0,1,0]$ or $[0,0,1]$. So the question is... are there points on $C$ that map to $\mathcal{O},2R,3R$, or $4R$? If so, there could be additional solutions to the question.

I leave here the Magma code I used:

S := AffineSpace(Rationals(),2);

f:=-16*(4*A^3+27*(d^6+A*d^2)^2)-d^6;

f;

C := Curve(S,f);

Genus(C);

P:=ProjectiveSpace(Rationals(),2);

C:=Curve(P,-432*d^12 - 864*d^8*A*z^3 - d^6*z^6 - 432*d^4*A^2*z^6 - 64*A^3*z^9);

PointSearch(C,100);

E,phi:=EllipticCurve(C,C![0,1,0]);

E;

Rank(E);

TorsionSubgroup(E);

Generators(E);

P:=phi(C![1/2,-1/16,1]);

2*P;

a,b:=HasPreimage(2*P,phi);

a,b;