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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.

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    $\begingroup$ 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
    Commented Jul 10, 2013 at 3:48

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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.

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  • $\begingroup$ 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
    Commented Jul 10, 2013 at 0:04

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