Answer by Damien Robert for Defining isogenies over smaller fields

2012-10-06T17:07:33Z

There is another obvious obstruction: by definition two twists are isomorphics above an extension, but are not isomorphic above their field of definition (and also not isogenous).

I believe that these are the only two obstructions, meaning that if (1) there exist a K-rational isogeny $f:A \to B$ and (2) $End^0_L(A)=End^0_K(A)$ then every $L$-isogeny $A \to B$ is in fact $K$-rational.

Indeed, since there exist a $K$-isogeny, the $\mathbb{Q}$-rank $r$ of $End^0_K(A)$ is the same as $End^0_K(B)$, and this is also the same as the $\mathbb{Z}$-rank of $Hom_K(A,B)$ (the module of $K$-isogenies), since the endomorphism ring is an order in its endomorphism algebra and $Hom_K(A,B)$ is not empty. Now by hypothesis (2), we have also the the $\mathbb{Z}$-rank of $Hom_L(A,B)$ is of rank $r$. Since $Hom_K(A,B) \subset Hom_L(A,B)$ are of same rank, for every $L$-isogeny $g:A \to B$ there exist $m$ such that $mg$ is $K$-rational. Since $mg=gm$ is $K$-rational and $m$ is $K$-rational, there exist a $K$-rational isogeny $g_2$ such that $gm=g_2 m$ by the universal property of isogenies. But then $m(g-g_2)=0$ which mean that $g=g_2$ and our isogeny $g$ was in fact $K$-rational.

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Answer by Damien Robert for Defining isogenies over smaller fields