# Division fields of isogenous abelian varieties over a p-adic field

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 $\ell$?

• 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 $f$. – ulrich Aug 11 '13 at 10:59
• 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 $\ell\nmid \deg(f)$? – Oleg Karpin Aug 11 '13 at 11:04
• The isogeny $f$ will be compatible with the Galois actions if it is a $K$-isogeny. – Chandan Singh Dalawat Aug 11 '13 at 11:08
• Yes, it's induced by f, and so compatible with Galois. – abz Aug 11 '13 at 11:08

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.