2
$\begingroup$

This question could be related to my old and Duality's newer questions.

I am building a $\mathbb{Z}/9\mathbb{Z}$ elliptic curve $E$ over $\mathbb{Q}$:

$$E: y^2+(t^3-3t^2+1)xy + t^3(t-1)^3y=x^2$$

For $t=-16$, Magma computes rank($E$)$=2$ and provides $2$ independent points.

Let $E_K$ represent the elliptic curve $E$ over a cubic field $K$. Note that all numerical coefficients of $E$ and $E_K$ are pairwise the same and rational.

It could be shown using

D. Jeon, C. H. Kim, Y. Lee, Families of elliptic curves over cubic number fields with prescribed torsion subgroups, Mathematics of Computation, V. 80, 273, January 2011, p. 579-591, AMS: S0025-5718-10-02369-0

that over a cubic field $K$ with defining polynomial $f_{18}(x)=(-t + 1)x^3 + (t^2 - 1)x^2 + (-2t^2 + t)x + t^2 - t$, the curve $E_K$ has torsion subgroup $\mathbb{Z}/18\mathbb{Z}$.

For $t=-16$, the root number of $E_K$ is $-1$, which according to the Parity Conjecture means that rank($E_K$)$\ge3$ and is odd.

I am trying to find an additional independent point on $E_K$. Computing over any cubic field $K$ presents a significant challenge for Magma, and independent points of height $h>12$ could hardly be found.

My Magma code is below.

SetClassGroupBounds("GRH");
PR<x> := PolynomialRing(Rationals());
t := -16;
K<z> := NumberField((-t + 1)*x^3 + (t^2 - 1)*x^2 + (-2*t^2 + t)*x + t^2 - t); K;
E := EllipticCurve([t^3-3*t^2+1, 0, t^3*(t-1)^3, 0, 0]); " "; E;
DescentInformation(E : RankOnly); 
EK := BaseChange(E, K); " "; EK;
" "; RootNumber(EK);

Having spent much time working over quadratic fields and with quadratic twists, I wonder about the following.

Question 1: Does there exist some elliptic curve $F$ similar to a (quadratic) $d$-twist, where $d$ is rational over cubic field $K$ (or, better still, $d$ is rational over $\mathbb{Q}$), that would account for the rank difference between $E_K$ and $E$, i.e. rank($E_K$)$=$rank($E$)$+$rank($F$)?

Question 2: If so, how would we determine $d$, write down $F$, and (if an independent point $P_F$ on $F$ is found) map $P_F$ back to a point on $E_K$?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ Three comments: (a) Since the Galois closure $L$ of $K$ is a cubic extension of a quadratic imaginary field, maybe Heegner points will help. (b) magmas RationalPoints is quite efficient in finding points on elliptic curves over number fields, not sure if you have tried it. (c) I doubt this has anything to do with Duality's question that I answered because the curve does not have automorphisms of order 3. Twists don't help here. You could take the quadratic twist with respect to the quadratic subextension in $L$, but that doesn't help usually for finding points over $K$. $\endgroup$
    – Chris Wuthrich
    Commented Feb 3 at 12:43
  • 2
    $\begingroup$ The best you can say is that the rank of $E(K)$ is the sum of the rank of $E(\mathbb{Q})$ and the multiplicity of the two-dimensional irreducible representation $\rho$ of $S_3 = \operatorname{Gal}(L/\mathbb{Q})$ in $E(L)\otimes\mathbb{C}$. But the twist of the Galois representation of $E$ by $\rho$ is not the Galois representation of an elliptic curve. The answer to your question is just "no" I fear. Unless I misunderstood it. $\endgroup$
    – Chris Wuthrich
    Commented Feb 3 at 12:46
  • $\begingroup$ Thank you, Chris! May I please send you an email to ac.uk about the first two comments? $\endgroup$
    – Maksym Voznyy
    Commented Feb 3 at 13:09
  • 1
    $\begingroup$ Sure. My inbox loves interesting email as a rare treat - but I am not the fastest to reply. $\endgroup$
    – Chris Wuthrich
    Commented Feb 3 at 16:54
  • $\begingroup$ @ChrisWuthrich a) Heegner: I always thought that Magma would first need to confirm that the structure (elliptic curve? which one?) has rank exactly $1$. So first I would need help identifying what I am trying to find the Heegner point on (hence I posted this question). And then that rank $1$ confirmation in itself might lead to a deadlock in Magma. Doing similar work over quadratic field on Z2xZ8-->Z8-->Z16 shows that $12$-descent can reach way higher and faster than HeegnerPoint. Too bad I may be out of luck to implement $3,4,12$-descent over cubic fields ($2$-descent tops in Magma over K ). $\endgroup$
    – Maksym Voznyy
    Commented Feb 3 at 17:04

0

Reset to default

You must log in to answer this question.