Unramified extensions of quadratic fields 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? 
 A: 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.
