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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).

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 $L$.

Is this isogeny defined over $K$ if $B$ can be defined over $K$?

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.

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up vote 11 down vote accepted

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 $\mathbb{Q}$.

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 Rational points on varieties.

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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 that 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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