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Does there exist an unramified $PSL_2(\mathbb{F}_p)$ extension of a quadratic field $K$ for $p\geq 5$?

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    $\begingroup$ By Minkowsky bound, every proper extension of Q is ramified, so what do you mean by that? $\endgroup$ – Lior Bary-Soroker Oct 29 '18 at 8:37
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    $\begingroup$ For $p=5$ (hence $\operatorname{PSL}_2(\mathbb{F}_p)\cong \mathfrak{A}_5 $), yes for infinitely many quadratic fields: this is a result of K. Uchida, Tôhoku Math. J. (2) 22 (1970), 220-224. I doubt that much more is known. $\endgroup$ – abx Oct 29 '18 at 14:27
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    $\begingroup$ A suggestion: Let $E$ be an elliptic curve, and let $M = \mathbb{Q}(E[p])$. It is quite common that $\mathrm{Gal}(M/\mathbb{Q}) = GL_2(F_p)$. Let $Z = \{ \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} \}$ and $\Gamma \subset GL_2(F_p)$ be those $g$ where $\det g$ is a QR mod $p$. The quotient $\Gamma/Z$ is isomorphic to $PSL_2(F_p)$, and the fixed field of $\Gamma$ is the quadratic subfield of $\mathbb{Q}(\zeta_p)$. Surely, experts know enough about the ramification of this example to compute whether the fixed field of $Z$ is unramified over that of $\Gamma$? $\endgroup$ – David E Speyer Oct 29 '18 at 19:29
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    $\begingroup$ For $p=7$ (e.g. worldscientific.com/doi/abs/10.1142/S179304211850118X) and $p=11$ (similar arguments) also yes for infinitely many, and a few more small $p$ would probably be doable in a similar way. For arbitrary $p$, this is probably wide open. $\endgroup$ – Joachim König Nov 5 '18 at 6:11

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