4
$\begingroup$

Let $l$ be a prime $\geq 5$. Does there exist a pair $E,E'$ of elliptic curves, both defined over the same number field $K$, which are not $l$-isogenous over $K$, but are $l$-isogenous over a quadratic extension?

I feel that the answer is yes, though I cannot come up with an example.

I would also like to ask what happens if in addition one assumes that $E$ has CM. I am hoping that the answer in this case is no. In this case both curves will have the same CM field.

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ If you are looking for a concrete non-CM example, take $l=11$, take $E$ to be "121a1", and $E'$ to be "3025c2". Then $E$ and $E'$ are not $11$-isogenous, but they become isogenous over $\mathbb{Q}(\sqrt{5})$. If you want a CM example, take $l=11$, $E=$"121b1", and $E'=$"3025a2". $\endgroup$
    – Álvaro Lozano-Robledo
    Jul 10, 2013 at 3:48

1 Answer 1

Reset to default
5
$\begingroup$

Take an elliptic curve $E$ (say, over the rationals) with CM and $\ell$ a prime that splits in the CM field. Now take $E'$ to be a twist of $E$, so $E, E'$ are not isomorphic over the rationals and not $\ell$ isogenous either, as such an isogeny, composed with a self-isogeny will force them to be isomorphic. Now, over the field where they become isomorphic, they are also $\ell$ isogenous.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Or if you want to think in terms of $L$-series equality, here $a_p(E)=\chi_d(p) a_p(E')$ for all primes $p$ with the (nonsquare) twisting factor $d$, and $a_p=0$ for primes that are inert in $K$, and over $K$ one has $a_{\frak p}(E)=\chi_d(p)a_{\overline{\frak p}}(E)$ and similarly for $E'$, so $\{a_{\frak p}(E),a_{\overline{\frak p}}(E)\}=\{a_{\frak p}(E'),a_{\overline{\frak p}}(E')\}$. $\endgroup$
    – v08ltu
    Jul 10, 2013 at 0:04

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.