Let $K$ be a number field. Let $A$ and $B$ be abelian varieties over $K$. Assume that $A$ and $B$ are isogenous over $\bar{K}$, the algebraic closure of $K$. We further assume that the endomorphism ring of $A\_{\bar{K}}$ is $\mathbb{Z}$. The question is:


Does there exist a quadratic extension $L$ over $K$ such that $A$ and $B$ are isogenous over $L$?



Definition field of isogeny between abelian varieties
Yes. The group of isogenies form a locally free module of rank $1$ over the endomorphism ring of $A$, hence are generated by a single isogeny of minimal degree $k$. So every isogeny is that isogeny composed with an endomorphism of $k$, so has degree $n^{2g} k$ for $n \in \mathbb Z$. Hence there are two isogenies with a given degree. The Galois action on the set of isogenies preserves degree, so the stabilizer of any isogeny is an index $2$ subgroup.


Here I'm just thinking of an isogeny as a special kind of subvariety of $A \times B$, from which it is clear that Galois invariance implies definability. 


