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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.

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    $\begingroup$ Have you considered applying Andre-Oort? Does it say anything about this? $\endgroup$
    – Watson Ladd
    Commented Feb 14, 2018 at 18:49
  • $\begingroup$ No, I have not, and unfortunately I am not very familiar with it $\endgroup$
    – Vlad
    Commented Feb 14, 2018 at 19:02
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    $\begingroup$ 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
    Commented Feb 14, 2018 at 22:06

2 Answers 2

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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.

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  • $\begingroup$ 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
    Commented Feb 14, 2018 at 19:03
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    $\begingroup$ My comment above is essentially an algebraic version of this that doesn't require the random points. $\endgroup$
    – Felipe Voloch
    Commented Feb 14, 2018 at 22:09
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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).

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