1
$\begingroup$

Let $K$ be a quadratic number field, and let $E_1$ and $E_2$ be two isogeneous elliptic curves over $K$. Assume we know that $j(E_1)^\sigma=j(E_2)$ where $\sigma$ is the generator of the Galois group of $K/Q$. Can we then say that some twist of $E_1$ is a $Q$-curve? If so, is there a good way of describing the necessary twist?

$\endgroup$

1 Answer 1

2
$\begingroup$

Yes -- a Q-curve is one whose geometric isogeny class is preserved by Galois, and that's evidently the case here. Of course there is no guarantee that E_1 and its Galois conjugate are isogenous over K. Is that the question you're asking? If so, I think there's a cohomological criterion for this due to Quer -- at least that's what I say in Remark 2.9 of my paper with Chris Skinner about this stuff:

http://www.math.wisc.edu/~ellenber/QcurveF.pdf

$\endgroup$
1
  • $\begingroup$ I was mostly wondering if there is a twist that makes E_1 a Q-curve over K which, as you point out, is not necessarily so, and the criteria for it is given by Quer. $\endgroup$
    – Soroosh
    Sep 2, 2011 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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