Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $SL_2(\mathbb{Z}_l)$.

share|improve this question

2 Answers 2

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 $K(E'[l])$.

share|improve this answer
    
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 $K(E'[\ell])\ne K$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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