Unramified cyclic extension of a quadratic field Recently, I was reading a paper on $5-divisibility$ of class number of quadratic fields and there I found the following statement without proof: Every unramified cyclic extension of prime degree $p$ of a quadratic field is normal over $\mathbb{Q}$, and is a $D_p-extension$ of $\mathbb{Q}$, that is, a Galois extension whose Galois group is isomorphic to the Dihedral group $D_p$ of order $2p$. 
I don't understand where I am supposed to use the unramifiedness. The only thing I can think of is to embed it inside the Hilbert Class Field of the given quadratic field but that doesn't seem to help much in proving that it is Galois over $\mathbb{Q}$. Even if it is Galois, how can I show that the Galois group is isomorphic to the Dihedral group of order $2p$?
 A: Your idea about the HCF is exactly what you want to do.  Here are some useful facts:
1)  If $k$ is Galois over $\mathbb{Q}$, then so is its Hilbert Class Field $H$.
2)  In this case, $\mathrm{Gal}(k/\mathbb{Q})$ acts on $\mathrm{Gal}(H/k)$ by "lift-and-conjugate":  given an automorphism $\sigma$ of $k$, you can lift it to an automorphism $\tilde{\sigma}$ of $H$, and then you have $\sigma$ act on elements of $\mathrm{Gal} (H/k)$ by conjugating by $\tilde{\sigma}$.
3) If we choose a field $F$ with $k \subseteq F \subseteq H$, then $\mathrm{Gal}(H/F)$ is a subgroup of $\mathrm{Gal}(H/k)$.  Then $F$ is Galois over $\mathbb{Q}$ if and only if $\mathrm{Gal}(H/F)$ is stable (not necessarily fixed!) under the action from part (2).
4)  If $\mathrm{Cl}_k$ is the class group of $k$, then $\mathrm{Gal} (k/\mathbb{Q})$ also acts on $\mathrm{Cl}_k$ -- just choose an ideal representing a class and act on it.  
5) There is an isomorphism $\phi \colon \mathrm{Cl}_k \rightarrow \mathrm{Gal}(H/k)$, and it respects the action of $\mathrm{Gal}(k/\mathbb{Q})$ on both groups.
Assume now as in your question that $k$ is quadratic, and if $\sigma$ is the element of order 2 in $\mathrm{Gal}(k/\mathbb{Q})$, let us look at how $\sigma$ acts on the class group of $k$.  A given class contains a prime ideal, say $\mathfrak{p}$, diving the rational prime $p$.  If $\mathfrak{p}$ is inert over $\mathbb{Q}$, then $\mathfrak{p}$ is principal and so is its class. Otherwise, $(\sigma \cdot \mathfrak{p}) \mathfrak{p} = (p)$, so $\sigma \cdot \mathfrak{p}$ is in the inverse class of $\mathfrak{p}$.  It follows that $\sigma$ acts on $\mathrm{Cl}_k$ by inversion.
From this, you can show that every subgroup of $\mathrm{Cl}_k$ is stable under the action of $\sigma$, hence by (5), so is every subgroup of $\mathrm{Gal}(H/k)$.  From (3), then, we find that every field $F$ between $k$ and $H$ is Galois over $\mathbb{Q}$.  Moreover, $\mathrm{Gal}(F/k)$ is an abelian, index 2 subgroup of $\mathrm{Gal}(F/\mathbb{Q})$.  And the argument just given shows that the action by conjugation of an element not in this subgroup is by inversion.  It follows that if $\mathrm{Gal}(F/k)$ is actually cyclic, then $\mathrm{Gal}(F/\mathbb{Q})$ is dihedral.
A: The reason why the condition on ramification plays a role here is the following: since $L/K$ is cyclic, $L$ is a class field of $K$ for some conductor ${\mathfrak f}$. If the ideal ${\mathfrak f}$ is not fixed by the Galois group of the quadratic extension $K/{\mathbb Q}$, then the conjugate extension will have a different conductor and therefore cannot be normal over the rationals.
If the conductor is trivial, i.e., if $L$ is contained in the Hilbert class field, then computing the Galois group over the rationals may be done as follows. The generator of the cyclic extension $L/K$ has the form $(\frac{L/K}{c})$ by the surjectivity of the Artin map, where $c$ is a suitable ideal class in Cl$(K)$. Since the Artin symbol is compatible with the Galois action of $K/{\mathbb Q}$, we have
$$ \Big(\frac{L/K}{c}\Big)^\sigma = \Big(\frac{L/K}{c^\sigma}\Big)
    = \Big(\frac{L/K}{c^{-1}}\Big) = \Big(\frac{L/K}{c}\Big)^{-1}, $$
where $\sigma$ is the lift of a generator of $K/{\mathbb Q}$, $\tau^\sigma = \sigma^{-1} \tau \sigma$, and where $c^{1+\sigma} = 1$ since ${\mathbb Q}$ has class number $1$.
