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Jun 5, 2018 at 16:10 comment added Davide Cesare Veniani Yes, exactly. I wouldn't know how to produce equations with coefficients $\mathbb Q$ for $E \times E'$ (or to prove that it is defined over $\mathbb Q$ in some other way).
Jun 5, 2018 at 15:57 comment added Joe Silverman The abelian surface $E\times E'$ is isomorphic to itself under the action of Gal$(\overline{\mathbb Q}/\mathbb Q)$, since Galois just swaps the factors, so its field of moduli is $\mathbb Q$. I guess maybe one needs to do a bit more to prove that $\mathbb Q$ is a field of definition. Is that what you're worried about?
Jun 5, 2018 at 15:37 comment added Davide Cesare Veniani Take a CM $\mathbb Q$-curve $E$ such that $\mathbb Q(j(E))$ is a quadratic extension of $\mathbb Q$. Let $E'$ be its Galois conjugate. Then $A = E\times E'$ is a singular abelian surface. You mean that $A$ is a good candidate for what I am looking for? How could one prove that $A$ is defined over $\mathbb Q$?
Jun 5, 2018 at 13:49 comment added Ariyan Javanpeykar Yes, you are right. If $A$ is a singular abelian surface over $\mathbb{C}$ which can be defined over $\mathbb{Q}$, then $A$ is isomorphic to $E_1\times E_2$ with $E_1$ and $E_2$ isogenous CM elliptic $\mathbb{Q}$-curves.
Jun 5, 2018 at 13:06 history answered Joe Silverman CC BY-SA 4.0