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An abelian surface $A$ is called singular if it has maximal Picard number $\rho(A) = 4$.

By work of Shioda-Mitani, any singular abelian surface $A$ is the product $A = E_1 \times E_2$ of two isogenous elliptic curves with complex multiplication. If both $E_1$ and $E_2$ are defined over $\mathbb Q$, then $A$ is of course defined over $\mathbb Q$.

Are there examples of singular abelian surfaces $A$ defined over $\mathbb Q$ which cannot be written as the product of two elliptic curves defined over $\mathbb Q$?

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This is a complement to Joe Silverman's answer and is a bit long for a comment. One seeks to find a $\mathbb{Q}$-curve $E$ so that $\mathbb{Q}(j(E))$ is a quadratic extension of $\mathbb{Q}$. One such $\mathbb{Q}$-curve is $E : y^{2} + \sqrt{2} xy + y = x^{3} + x^{2} + (-2\sqrt{2} - 3) x + \sqrt{2} + 1$, which has CM by $\mathbb{Z}[\sqrt{-6}]$. Moreover, according to the LMFDB database of genus 2 curves, if $C : y^{2} + x^{3} y = x^{3} + 2$, then the Jacobian of $C$ is isogenous to the square of $E$. Is it true that the Jacobian of $C$ is isomorphic to this product? Apparently, Shioda and Mitani proved that if an abelian surface is isogenous to a product of two isogenous elliptic curves with CM, then it is isomorphic to a product of two elliptic curves (which must be isogenous and have CM, necessarily).

Note: I don't think one needs to have a $\mathbb{Q}$-curve whose $j$-invariant is defined over a quadratic extension. Another example is $C : y^{2} = x^{5} - x$ whose Jacobian decomposes as the square of a $\mathbb{Q}$-curve defined over $\mathbb{Q}(\sqrt{2})$ which is not a base change from $\mathbb{Q}$, but whose $j$-invariant is $8000$.

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I think that what you want is contained in the theory of $\mathbb Q$-curves, which are elliptic curves defined over $\overline {\mathbb Q}$ that are isogenous to all of their Galois conjugates. Dick Gross studied CM $\mathbb Q$-curves in his thesis, which was published in [1]. So you'd need to use Gross' work to find a CM $\mathbb Q$-curve $E$ such that $\mathbb Q(j(E))$ is a quadratic extension of $\mathbb Q$.

[1] Gross, Benedict H. Arithmetic on elliptic curves with complex multiplication. With an appendix by B. Mazur. Lecture Notes in Mathematics, 776. Springer, Berlin, 1980. iii+95 pp. MR0563921

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  • $\begingroup$ 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. $\endgroup$
    – Ariyan Javanpeykar
    Commented Jun 5, 2018 at 13:49
  • $\begingroup$ 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 $\mathbb Q$? $\endgroup$
    – Davide Cesare Veniani
    Commented Jun 5, 2018 at 15:37
  • $\begingroup$ 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? $\endgroup$
    – Joe Silverman
    Commented Jun 5, 2018 at 15:57
  • $\begingroup$ 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). $\endgroup$
    – Davide Cesare Veniani
    Commented Jun 5, 2018 at 16:10

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