User damien robert - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:48:57Z http://mathoverflow.net/feeds/user/26737 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108945/defining-isogenies-over-smaller-fields/109014#109014 Answer by Damien Robert for Defining isogenies over smaller fields Damien Robert 2012-10-06T17:07:33Z 2012-10-06T17:20:43Z <p>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).</p> <p>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.</p> <p>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. </p>