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Let $p$ be a rational prime number, and let $K$ be a finite extension of $\mathbb Q_p$. Let $A$ be an abelian variety over $K$. For any rational prime number $\ell$, let $K(A[\ell])$ be the field of definition of the $\ell$-torsion points of $A$.

Suppose $A$ and $A'$ are $K$-isogenous abelian varieties over $K$.

Is there an odd prime number $\ell\neq p$ such that $K(A[\ell])=K(A'[\ell])$?

Are there even infinitely many such $\ell$?

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    $\begingroup$ If $f:A \to A'$ is an isogeny, then $f$ induces an isomorphism from $A[l]$ onto $A'[l]$ for any $l$ prime to the degree of $f$. $\endgroup$
    – naf
    Commented Aug 11, 2013 at 10:59
  • $\begingroup$ Thanks for pointing this out. A priori this isomorphism is an isomorphism of abstract groups, is this isomorphism in addition compatible with Galois actions? If yes, this would imply $K(A[\ell])=K(A'(\ell))$ for all $\ell\nmid \deg(f)$? $\endgroup$
    – Oleg Karpin
    Commented Aug 11, 2013 at 11:04
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    $\begingroup$ The isogeny $f$ will be compatible with the Galois actions if it is a $K$-isogeny. $\endgroup$
    – Chandan Singh Dalawat
    Commented Aug 11, 2013 at 11:08
  • $\begingroup$ Yes, it's induced by f, and so compatible with Galois. $\endgroup$
    – abz
    Commented Aug 11, 2013 at 11:08

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Let $f:A\to A′$ be an isogeny over $K$. Then there exists an isogeny $g:A'\to A$ such that the composite of $f$ with $g$ is multiplication by the degree of $f$ (see any book on abelian varieties). It follows that, for every prime $l$ not dividing its degree, $f$ defines an isomorphism from $A[l]$ onto $A'[l]$. This isomorphism automatically commutes with the action of the Galois group.

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  • $\begingroup$ Ok, now everything is clear. Thank you all very much for the answers. $\endgroup$
    – Oleg Karpin
    Commented Aug 16, 2013 at 10:21

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