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Apr 25, 2021 at 21:50 answer added Will Sawin timeline score: 4
Apr 25, 2021 at 18:21 comment added Arno Fehm Even something stronger holds: The field obtained by adjoing all $E[p]$ to $\mathbb{Q}$, for all ellliptic curves $E$ and all $p$, is still very far away from the algebraic closure $\overline{\mathbb{Q}}$ (in fact it is a Hilbertian field). I know of no similar result when running over all abelian varieties though.
Apr 25, 2021 at 16:44 comment added Chris Wuthrich If your finite group has a 2-dimensional representation, then en.wikipedia.org/wiki/Serre%27s_modularity_conjecture implies that it comes from a modular form if it is odd etc. But only few of those come from elliptic curves.
Apr 25, 2021 at 16:10 comment added Wojowu Every such action is odd, meaning the complex conjugation is mapped to multiplication by $-1$ (this is certainly true for elliptic curves and I believe also for abelian varieties). So any quotient which identifies complex conjugation and identity won't arise this way.
Apr 25, 2021 at 16:03 review First posts
Apr 25, 2021 at 17:27
Apr 25, 2021 at 15:56 history asked Leray J CC BY-SA 4.0