Defining isogenies over smaller fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:31:54Z http://mathoverflow.net/feeds/question/108945 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108945/defining-isogenies-over-smaller-fields Defining isogenies over smaller fields Harry 2012-10-05T18:53:47Z 2012-10-06T17:20:43Z <p>I'm having some issues with abelian varieties and fields of definition. This already became clear in my previous question on Jacobians. Here's another question. If somebody can explain some nice facts on fields on definition this would help me a lot (because these aren't the only questions I have concerning fields of definition). </p> <p>Let $A$ be an abelian variety over a number field $K$. Let $L/K$ be a finite field extension. Suppose that there exists an isogeny $A_L\to B$ defined over $L$. </p> <p>Is this isogeny defined over $K$ if $B$ can be defined over $K$?</p> <p>I think the answer is negative, but I have to admit that my set of examples is very scarce and, therefore, I can't find an easy counterexample. </p> http://mathoverflow.net/questions/108945/defining-isogenies-over-smaller-fields/108946#108946 Answer by Daniel Loughran for Defining isogenies over smaller fields Daniel Loughran 2012-10-05T19:09:45Z 2012-10-05T19:09:45Z <p>No: Consider the elliptic curve $E: y^2 = x^3 + x$ defined over $\mathbb{Q}$. Then the isogeny $y \mapsto iy$ and $x \mapsto -x$ is defined over $\mathbb{Q}(i)$ but obviously not over $\mathbb{Q}$.</p> <p>In general if $K \subset L$ is Galois, then a morphism defined over $L$ comes from one over $K$ if and only if it is invariant under the action of $Gal(L/K)$. This follows from the so-called "theory of descent". See for example Bjorn Poonen's notes <a href="http://www-math.mit.edu/~poonen/papers/Qpoints.pdf" rel="nofollow">Rational points on varieties.</a></p> 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>