TITLE: Do Abelian varieties isomorphic to all their conjugates descend?
QUESTION [5 upvotes]: Suppose $A$ is an abelian variety over $\overline{\mathbb{Q}}$ of dimension $g$, such that $A$ is isomorphic to all of its Galois conjugates. Note that I'm not including any polarization data.
Can I conclude that $A$ descends to $\mathbb{Q}$, or at least to a field $K$ of bounded degree over $\mathbb{Q}$?
Thanks!

REPLY [3 votes]: 1.Theorem (Shimura 1971, "On the Field of Rationality for an Abelian Variety"):  If $g$ = dim $A$ is odd and $Aut(A)=\{\pm1\}$ then $A$ has a model over its field of moduli.  Proof:  $\{1\} = H^1(\overline{\mathbb{Q}}^*) \rightarrow  H^2(\pm1) \rightarrow H^2(\overline{\mathbb{Q}}^*)$ and $\{\pm1\}$ acts faithfully on $H^0(A_{\overline{\mathbb{Q}}}, \Omega^g) = \overline{\mathbb{Q}}$ since $g$ is odd, so the image of the obstruction in $H^2(Aut(H^0(A_{\overline{\mathbb{Q}}}, \Omega^g))) = H^2(\overline{\mathbb{Q}}^*)$ can be identified as the obstruction to the descent of a one-dimenstional $\overline{\mathbb{Q}}$ vector space to $\mathbb{Q}$, which is therefore trivial.
2.Of interest:
Maus 1973, "On the Field of Moduli of an Abelian Variety with Complex Multiplication"
Shimura 1982, "Models of an Abelian Variety with Complex Multiplication Multiplication over Small Fields"
Shioda 1977, "Some Remarks on Abelian Varieties"