Let $K$ be a fixed quadratic number field, say $K=\mathbb Q(\sqrt 5)$. For any integer $n \geq 3,$ I would like to build a number field $D_n$ such that $D_n/\mathbb Q$ is Galois, with Galois group isomorphic to $D_{2^{n-1}}$ (dihedral group of order $2^n$), $K \subset D_n$ and with small discriminant.

I know class field theory could help me here, since I'm basically trying to construct a cyclic extension of $K$ with conditions on the ramification but I am not too familiar with its use.

Proposition 1 of this article of Louboutin, Park and Lefeuvre gives me exactly what I need, provided I can find a subgroup $H$ of $\mathcal I_K(\mathfrak{m})$ (= the group of fractional ideals prime to $\mathfrak{m}$) containing $P_{K, \mathbb Z}(\mathfrak{m})$ (= the subgroup of $\mathcal I_K(\mathfrak{m})$ of principal ideals of the form $(\alpha)$, with $\alpha \equiv a$ mod $\mathfrak{m}$, for some $a \in \mathbb Z$ prime with $\mathfrak{m}$) such that $\mathcal I_K(\mathfrak{m})/H$ is cyclic of order $2^n$, for some modulus $\mathfrak{m}$ which is invariant under conjugation of $K/\mathbb Q$, and having small enough prime factors.

In the case $K = \mathbb Q(\sqrt 5)$, the class number is one, so we can see non-zero fractional ideals of $K$ as elements of $K^{\times}$ up to units of $\mathcal O_K$. I have been told to look for modulus given by primes $p \equiv 1$ mod $2^n$ which split in $K$, i.e. $\left(\frac{5}{p}\right)=1$, but I still have no idea how to show that such a subgroup $H$ exists.