Let $E$ and $E'$ be nonisogenous elliptic curves over a field $k$ (characteristic 0) such that $Gal(k(E[p^{\infty}])/k)=Gal(k(E'[p^{\infty}])/k) = SL_2(\mathbb{Z}_p)$ with $p \geq 5$ (where $E[p^{\infty}]$ is the set of $p^n$ torsion points of $E$ for all $n$). Then is it true that $k(E[p^{\infty}])\cap k(E'[p^{\infty}]) = k$, or can someone provide a counterexample?

$\begingroup$ What assumptions are you making on $k$? This question is vacuous in both of the cases that I would consider natural (global fields and local fields). $\endgroup$– Erick KnightSep 8, 2011 at 17:07

$\begingroup$ Sorry  $k$ has characteristic 0. For example $k$ contains the roots of unity or is a countable algebraically closed field. $\endgroup$– Adam HarrisSep 8, 2011 at 17:16

1$\begingroup$ You can't have $k$ algebraically closed because your condition on the Galois groups won't hold. Your statement has a chance to hold for function fields. $\endgroup$– Felipe VolochSep 8, 2011 at 18:48

$\begingroup$ Thanks Felipe  I meant $k$ is finitely generated over an algebraically closed field $\endgroup$– Adam HarrisSep 8, 2011 at 19:05
2 Answers
Since both fields $K(E_{l^\infty})$ and $K(E'_{l^\infty})$ contain the $l$adic cyclotomic extension of $K$, your expectation cannot hold. However, this is almost the only obstruction.
In Propriétés galoisiennes des points d'ordre fini des courbes elliptiques, Invent. Math. 15, 259331 (1972), JP. Serre proved the following Theorem (Theorem 6$''$, p. 325).
Let $K$ be a number field, let $K^{\rm cycl}$ be the (cyclotomic) extension of $K$ generated by all roots of unity. Let $E$ and $E'$ be two elliptic curves such that, over $\bar K$,
(i) $E$ and $E'$ have no complex multiplication;
(ii) $E$ and $E'$ are not isogeneous.
Then, the extensions $K(E_\infty)$ and $K(E^\prime_\infty)$ of $K^{\rm cycl}$ are almost disjoint: $K(E_\infty)\cap K(E'_\infty)$ is finite over $K^{\rm cycl}$.
(By Faltings, hypothesis (ii) is equivalent to the one given by Serre.)

2$\begingroup$ Just a remark: the fact that the Galois group is $\text{SL}_2(\mathbb{Z}_p)$ implies that the $p$adic cyclotomic character on $G_k$ is trivial, and thus must contain the $p$adic cyclotoic extension. $\endgroup$ Sep 8, 2011 at 20:50

$\begingroup$ and thus $k$ must contain (if someone can edit that in, it would be appreciated) $\endgroup$ Sep 8, 2011 at 20:52

$\begingroup$ Apologies  I forgot to put nonCM curves as a hypothesis in my question (but it's probably not good to change this now?) and $p \geq 5$ was just a guess to keep the question more concise, but the theorem of Serre which ACL quoted is exactly what I wanted so thank you all! $\endgroup$ Sep 8, 2011 at 22:17

1$\begingroup$ I don't quite understand the first line, because there was no assumption on $K$, so $K$ could equal $K(\zeta_{l^{\infty}})$ (not if $K$ is a number field, of course). It also seems a little strange to phrase the statement of Serre's theorem in the way you do  by far the hardest part of that statement is Faltings contribution (the Tate conjecture). However, you seemed to have mastered the dark art of divining exactly what the OP wanted to know, rather than what they actually asked! $\endgroup$– MichaelSep 8, 2011 at 23:16

$\begingroup$ To Michael: I changed hypothesis (ii) of Serre to the one I gave here, just because it is more natural, and Serre explicitly mentioned that point. (He had been able to prove it for nonintegral $j$invariants.) To Eric: You're absolutely right! I overlooked that point. $\endgroup$– ACLSep 9, 2011 at 6:04
By the way, I think that under your hypotheses, your question is really about group theory, not about algebraic geometry. Namely: the action of Galois on E[p^infty] x E'[p^infty] gives you a homomorphism
G_K > SL_2(Z_p) x SL_2(Z_p).
Call the image H. By your hypothesis, H projects surjectively onto both copies of SL_2(Z_p). You also know that H is not contained in any conjugate of the diagonal (if it were, E[p^infty] and E'[p^infty] would be isomorphic Galois representations and I'm presuming you're in a situation where Faltings rules that out  you'd better be, if you want an affirmative answer to your question.)
Now what you have to prove is that a subgroup of SL_2(Z_p) x SL_2(Z_p) which projects surjectively onto each direct summand and which is not conjugate to a subgroup of the diagonal must be finiteindex in SL_2(Z_p) x SL_2(Z_p). This is true for SL_2(F_p) by Hall's lemma and I think you can induct from there (but didn't think about it carefully.)

$\begingroup$ Indeed: such arguments are at the heart of the proof of Serre's Theorem quoted above. $\endgroup$– ACLSep 9, 2011 at 6:05