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8 hours ago comment added Laurent Moret-Bailly @ChrisWuthrich Agreed, thanks!
16 hours ago history became hot network question
19 hours ago vote accept curious math guy
20 hours ago comment added Chris Wuthrich @LaurentMoret-Bailly If $j(\mathcal{A})$ is in $\mathbb{Q}$ there is an elliptic curve $E/\mathbb{Q}$ such that $E \times \mathbb{Q}_p$ is a twist of $\mathcal{A}$. If $j\not \in \{0,1728\}$, there is a $D\in\mathbb{Q}_p^{\times}$ such that $\mathcal{A}$ is the quadratic twist $E^D$. Now approximate $D$ by a rational $D'$ such that $\mathbb{Q}(\sqrt{D}) = \mathbb{Q}(\sqrt{D'})$. Then $A=E^{D'}$ will do. Similar with sextic, cubic and quartic twists.
23 hours ago answer added Daniel Loughran timeline score: 9
23 hours ago comment added Vik78 You might be able to say something about Brauer-Manin obstructions for various moduli spaces of abelian varieties, especially modular curves. See math.mit.edu/~poonen/papers/heuristic.pdf
yesterday comment added Laurent Moret-Bailly @ChrisWuthrich: This condition is necessary, but if $j(\mathcal{A})\in\mathbb{Q}$ I can only conclude, a priori, that there is $A$ over $\mathbb{Q}$ that becomes isomorphic to $\mathcal{A}$ over $\overline{\mathbb{Q}_p}$.
yesterday comment added Chris Wuthrich For an elliptic it is simply the question if $j(\mathcal{A})\in \mathbb{Q}$.
yesterday history asked curious math guy CC BY-SA 4.0