# Fields obtained by adjoining x coordinates of torsion points on elliptic curves

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

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 $K(E'[l])$.
• 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
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 $K(E'[\ell])\ne K$.