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Let $K/\mathbb{Q}$ be quadratic and let $L/K$ be an (everywhere) unramified Galois extension. If $L/K$ is abelian, then one can show that $L/\mathbb{Q}$ is Galois (eg see here). Is $L/\mathbb{Q}$ necessarily Galois if $L/K$ is not abelian? What if we also assume that $L/K$ has odd degree?

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  • $\begingroup$ Due to generalizations of the Cohen-Lenstra heuristics (arxiv.org/pdf/1702.04644.pdf), for any $G$ and any $\Gamma \le G\wr C_2$ which is generated by involutions outside of the kernel of the projection onto $C_2$, it is conjectured that there are unramified extensions $L/K$ of quadratic number fields such that $Gal(L/K) = G$ and Galois group of the Galois closure over $\mathbb{Q}$ equals $\Gamma$. In many of these cases, $|\Gamma|$ is larger than $2|G|$, e.g. already for the smallest odd order non-abelian group $G=C_7:C_3$, there is a $\Gamma \cong ((C_7^2):C_3):C_2$. $\endgroup$ Nov 5, 2018 at 7:21

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The following result is a special case of a sideresult in my PhD thesis (1995):

Let $k = {\mathbb Q}(\sqrt{d})$ be a quadratic extension with discriminant $d$, let $d = d_1d_2d_3d_4$ be a factorization of $d$ into coprime discriminants, and assume that $$(d_1/p_2) = (d_2/p_1) = (d_3/p_4) = (d_4/p_3) = +1 $$ for all primes $p_j \mid d_j$. Then there exist $\alpha = x_1 + y_1 \sqrt{d_1} \in {\mathbb Z}[\sqrt{d_1}]$ and $\beta = x_3 + y_3 \sqrt{d_3} \in {\mathbb Z}[\sqrt{d_3}]$ satisfying $x_1^2 - d_1y_1^2 = z_1^2d_2$ and $x_3^2 - d_3y_3^2 = z_3^2d_4$ such that $M = {\mathbb Q}(\sqrt{d_1},\sqrt{d_2},\sqrt{d_3},\sqrt{d_4}, \sqrt{\alpha},\sqrt{\beta})$ is an unramified extension of $k$ with Galois group $D_4 \times D_4$.

The subfield $$ L = {\mathbb Q}(\sqrt{d_1d_2},\sqrt{d_1d_3},\sqrt{d_1d_4}, \sqrt{\mu})$$ with $\mu = 2x_1x_3 + 2y_1y_3\sqrt{d_1d_2} + 2z_1z_2\sqrt{d_2d_4}$ has Galois group $D_4$ over $k$, but is not normal over ${\mathbb Q}$.

I have given the concrete example $d = -3 \cdot 13 \cdot 5 \cdot 29$ with $\alpha = (-1 + \sqrt{13}\,)/2$, $\beta = 7 + 2 \sqrt{5}$ and $\mu = -7 + 2\sqrt{65} + 2 \sqrt{-87}$. I just verified with pari that it is indeed unramified:

f=polcompositum(x^2+3*13,x^2+3*5)[1];

f=polcompositum(f,x^2+3*29)[1];

g=x^8 + 28*x^6 + 470*x^4 + 3836*x^2 + 380689;

The compositum $L$ of $f$ and $g$ has degree $16$ and discriminant $3^8 5^8 13^8 29^8$, hence is unramified over $k$.

For odd degree extensions I would try a nontrivial 3-class field tower of a quadratic number field and pick out a suitable subgroup that isn't fixed under the automorphism of order 2. I'm sure some group theorist will be able to come up with an example.

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