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 $\Gamma \cong ((C_7^2):C_3):C_2$. $\endgroup$– Joachim KönigCommented 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.
answered Dec 22, 2015 at 16:39