Loosely speaking, are elliptic Kummer extensions big? More concretely:

Let $E$ be an elliptic curve over $\mathbb{Q}$, let $p$ be a prime, and let $F$ be a subfield of $\overline{\mathbb{Q}}$ containing the coordinates of all the $p$-power torsion of $E$. Given $c > 0$, does there exist $N > 0$ (depending only on $E$, $F$, and $p$) such that: If P is a point of $E(\overline{\mathbb{Q}})$ whose image in $(E(\overline{\mathbb{Q}})/E(F))$ has order $p^n$ (in other words, $[p^n]P \in E(F)$, but this does not hold for any smaller power of $p$) with $n \geq N$, then $[F(P):F] \geq c$?


  • I am aware of some Theorems of Basmakov, Ribet, and Bertrand that are roughly in this direction, and in fact more general cases than elliptic curves are considered (e.g. Ribet's paper "Kummer theory on extensions of abelian varieties by tori" -- Duke Math J. 1979). One shortcoming of these excellent papers is that the results are only for almost every prime. I am very interested in the prime 2 (unfortunately!).

  • $\operatorname{Gal}(F(P)/F)$ is naturally a subgroup of $E[p^n]$.

  • The G_m analogue is easy -- the degree of a Kummer extension is as big as it could be.

  • If it's easier, the most important case for me is when $E$ has CM and $F$ is a finite extension of $\mathbb{Q}(E_{tors})$.

  • Does the following make sense: Given a point $Q \in E(F)$, there is a representation $G_F \to (\mathbb{Z}_p)^2$ and more generally $G_\mathbb{Q} \to (\mathbb{Z}_p)^2 \rtimes \operatorname{GL}_2(\mathbb{Z_p})$ coming from the Galois action on points $P$ satisfying $[p^n]P = Q, n \in \mathbb{N}$, and in this context what I'm asking is in the spirit of Serre's Theorem on the large image of the $p$-adic representation, but I'm asking the image to be large uniformly in $Q$?

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    $\begingroup$ What does $[p^{\infty}]$ mean in the formula $(E (\overline{\mathbb Q}/E(F))[p^{\infty}]$? $\endgroup$
    – Will Sawin
    Commented Jan 4, 2014 at 15:55
  • $\begingroup$ Will: I have deleted it. I meant those elements of the quotient which have $p$-power order. Of course this wasn't necessary to include. $\endgroup$ Commented Jan 4, 2014 at 15:59

1 Answer 1


$Gal(F(P)/F)$ is indeed a subgroup of $E[p^{n}]$. We see that it is not contained in $E[p^{n-1}]$, because otherwise $p^{n-1}P\in E(F)$. Thus, since $E[p^{n-1}]$ consists of all the elements of order less than $p^n$, order of $Gal(F(P)/F)$ is at least $p^N$. So we may take any $N \leq \log_p c$

  • $\begingroup$ I agree with your argument, but (both in the question and the answer) I think that the inclusion $\mathrm{Gal}(F(P)/F)\hookrightarrow E[p^n]$ is not natural: it should be the map $\sigma\mapsto P^\sigma-P$, which depends on $P$, unlike $F(P)$. Also, your argument works with this choice and not with other abstract ones, I think - or I do not see how to deduce that the image is not contained in $E[p^{n-1}$. $\endgroup$ Commented Jan 4, 2014 at 18:28
  • $\begingroup$ It is natural because the $p^N$ torsion of $E$ is defined over $F$. Otherwise you would just obtain a class in $ H^1(F,E[p^N])$ $\endgroup$
    – Will Sawin
    Commented Jan 4, 2014 at 19:42
  • $\begingroup$ But what is the map exactly? Isn't it the one I wrote down? $\endgroup$ Commented Jan 4, 2014 at 22:07
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    $\begingroup$ Yes. That map only depends on $P$ up to elements of $E[F]$. Since $E[p^n]$ is contained in $E[F]$, the map is a well-defined homomorphism. Otherwise, it would just be a cocycle, well-defined up to a coboundary. $\endgroup$
    – Will Sawin
    Commented Jan 4, 2014 at 22:38
  • $\begingroup$ I agree with what you write, simply in the question there is only the assumption that $E[p]$ be contained in $F$, not the whole $E[p^n]$. Anyhow I am sorry for the pedantic discussion, it is getting longer than it would deserve... $\endgroup$ Commented Jan 4, 2014 at 23:13

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