5
$\begingroup$

Let $E$ be an elliptic curve defined over $\mathbb{Q}$ by the equation $$E: y^2=x^3-Ax+B=:f(x).$$

Consider the abelian variety $E^3:=E \times E \times E \subset \mathbb{P}^2 \times \mathbb{P}^2 \times \mathbb{P}^2$. Let $(x_1,y_1,x_2,y_2,x_3,y_3)$ denote the affine coordinates for a point in $E^3$. Inside $E^3$ we consider the curve $\mathcal{C}$ defined by the equations

$$\mathcal{C}:\begin{cases} y_1^2=f(x_1);\\ y_2^2=f(x_2);\\ y_3^2=f(x_3);\\ y_1=x_2^m;\\ y_2=x_3^n, \end{cases}$$ where $m$ and $n$ are positive integers, which can be taken as large as desired, if necessary.

I would like to know how one can show that $\mathcal{C}$ is not contained in any abelian subvariety of $E^3$ of dimension two. I would hope that once $m$ and $n$ are sufficiently large, one can prove this, but I cannot make any progress with it at this stage.

$\endgroup$
  • 1
    $\begingroup$ Have you considered applying Andre-Oort? Does it say anything about this? $\endgroup$ – Watson Ladd Feb 14 '18 at 18:49
  • $\begingroup$ No, I have not, and unfortunately I am not very familiar with it $\endgroup$ – Vlad Feb 14 '18 at 19:02
  • 3
    $\begingroup$ Let $\phi_i, i=1,2,3$ be the three maps $C \to E$ and $\omega_i = \phi_i^* (dx/y)$, try to prove that the $\omega_i$ are linearly independent over $\mathbb{C}$. If true, this should be doable by local calculations. $\endgroup$ – Felipe Voloch Feb 14 '18 at 22:06
2
$\begingroup$

I just want to make a comment, but the system does not allow that due to my low reputation.

Consider the complex points $E\times E\times E$ as $\mathbb{C}^3/\Lambda$. A curve $C$ in the product is not contained in any (translated) two-dimensional abelian subvariety if and only if there exist three smooth points on $C$, such that the tangent vectors of $C$ at the three points are linearly independent. If you can find three random points on each $C_{m, n}$, then you may expect the tangent vectors (after translating to the origin) are linearly independent.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I agree, this was the only thing I have actually tried, but I do not know how to "produce" the three suitable points. Maybe I am missing something easy $\endgroup$ – Vlad Feb 14 '18 at 19:03
  • 3
    $\begingroup$ My comment above is essentially an algebraic version of this that doesn't require the random points. $\endgroup$ – Felipe Voloch Feb 14 '18 at 22:09
1
$\begingroup$

Are you trying to prove this for a general f(x), or for a specific f(x)? If it is the latter, you can pick a couple of values for m and n, compute the zeta function of the curve over a small finite field F_p and show that the numerator of the zeta function is not divisible by that of E.

This is of course experimental, and for this to work you need a bit of luck. But sounds like you believe that this is true (for large m, n), in which case a Chebotarev type argument would suggest that this would work for a random p (unless you pick a really bad --- interesting?!! --- f(x), e.g. a CM curve).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.