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Let $E$ and $E'$ be elliptic curves over a field $K$ of characteristic zero such that $E$ and $E'$ are non-isogenous over $\bar{K}$. Let $l$ be a large prime and suppose that $K(x(E[l]))=K(x(E'[l]))$ (where these are the fields obtained by adjoining the $x$-coordinates of $l$-torsion points). Then does it follow that $K(E[l])=K(E'[l])$?

Edit: Strengthen the hypothesis so that $K$ contains the roots of unity, $E$ and $E'$ are non-CM, and that the image of Galois on the $l$-adic Tate modules of $E$ and $E'$ is as large as possible i.e $SL_2(\mathbb{Z}_l)$.

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No; if E and E' are quadratic twists, the fields $K(x(E[l]))$ and $K(x(E'[l]))$ are equal, but $K(E[l])$ may not equal $K(E'[l])$.

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Thanks David. I have edited the question so that they are non-isogenous now – Adam Harris Oct 5 '11 at 14:02
To nit-pick -- the answer is still no, for the same reason. Quadratic twists are isomorphic over an extension, but aren't necessarily isogenous over their field of definition. – David Zureick-Brown Oct 5 '11 at 14:22
Adam, quadratic twists are usually not isogenous. Certainly, for any $E$, there exist infinitely many quadratic twists of $E$ that are not isogenous to $E$. – Alex B. Oct 5 '11 at 14:24
Yes - sorry I meant non-isogenous over $\bar{K}$. – Adam Harris Oct 5 '11 at 14:34
Phrased this way, I think this is equivalent to the Frey-Mazur conjecture, which is open. – David Zureick-Brown Oct 5 '11 at 15:42

I think this works: Take two non-isogenous (over $\overline{K}$) curves $E$ and $E'$ with $K(E[\ell])=K(E'[\ell])=K$. Replace $E'$, say, by a quadratic twist. Then $K(E[\ell])=K(x(E[\ell]))=K(x(E'[\ell]))=K$ but $K(E'[\ell])\ne K$.

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Thanks Torsten. Sorry I've had a particular situation in mind and thought it would hold in more generality. Have edited the question and promise that no more hypotheses will be added! – Adam Harris Oct 5 '11 at 17:59

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