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Let $E/\mathbb{Q}$ be an elliptic curve and let $p$ be a prime. Then there is an action of the absolute Galois group of $\mathbb{Q}$ on $E[p]$ that factors through a finite quotient.

Does any finite quotient of the absolute Galois group of $\mathbb{Q}$ arise this way for some $E$ and $p$? What if we consider abelian varieties instead of elliptic curves?

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    $\begingroup$ 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. $\endgroup$ – Wojowu Apr 25 at 16:10
  • $\begingroup$ 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. $\endgroup$ – Chris Wuthrich Apr 25 at 16:44
  • $\begingroup$ 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. $\endgroup$ – Arno Fehm Apr 25 at 18:21
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For elliptic curves, the answer is no, as Chris Wuthrich and Aron Fehm point out in the comments. In fact I think every extension with Galois group a finite simple group not of the form $PSL_2(\mathbb F_p)$ for any $p$ cannot arise this way.

For abelian varieties, the answer is positive. In fact it's sufficient to take just the $2$-torsion, and even just the $2$-torsion of Jacobians of hyperelliptic curves. Given an arbitrary extension, take the minimal polynomial $f$ of a generator $\alpha$ of that extension, form the hyperelliptic curve with equation $y^2 = f(x)$, and take its Jacobian. (If the degree of $f$ is at most $4$, you will need to add some linear factors as well to make it hyper-.) The Galois group of the $2$-torsion of the hyperilliptic curve will be a subgroup of $S_n$, generated by the Galois action on the roots of this polynomial (plus the point at $\infty$, if the degree is odd) and will therefore include the Galois group of your field.

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  • $\begingroup$ What about e.g. abelian threefolds? Do you get all finite quotients? $\endgroup$ – Oniqa Apr 26 at 5:41
  • $\begingroup$ @Oniqa: No bounded dimension will do. $\endgroup$ – Arno Fehm Apr 26 at 6:55

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