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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.

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  • $\begingroup$ Find a prime ideal $ P$ of norm $N(P)=1+2^{n+l}r$. Let $S=\{a^in O_K,a^{2^lr}\equiv 1\bmod P\}$. Find representatives $J_c$ of the classgroup such that for every product $\prod_i J_{c_i}$ if it is principal then it is generated by an element $\in S$. Let $(\eta_c)=J_cJ_{c^{-1}}$ and $H=\{\frac{aJ_c}{\eta_c},a\in J_{c^{-1}}\cap S\}$. Then $I_K(P)/H$ is generated by $(g^{2^lr})$ where $g$ is a generator of $O_K/P^\times$. Classfield theory says there is an abelian extension $L/K$ with $(\frac{I}{L/K})=\sigma^m$ for any ideal $I \in (g^{2^l r})^m HH^{-1}$ $\endgroup$ – reuns Mar 28 at 21:18
  • $\begingroup$ What does your construction look like when $K$ has class number one, like in the case of $\mathbb Q(\sqrt 5)$ ? I have a hard time following what you are doing. $\endgroup$ – A. Bailleul Mar 29 at 18:23
  • $\begingroup$ For the PID $K=\Bbb{Q}(\sqrt{5})$ let a prime number $p \equiv 1 \bmod 2^n$, $P$ a prime ideal of $O_K$ above $p$ then $O_K/P$ is a field with $1+2^{l+n}r$ elements, then $G = \{ \frac{a}{b}O_K, a\in O_K - P, b \in O_K -P\}$, $H = \{ \frac{a}{b}O_K, a\in O_K, a^{2^l r} \in 1+ P, b \in O_K , b^{2^l r} \in 1+ P\}$ are groups of fractional principal ideals and $G/H$ is cyclic with $2^n$ elements generated by $g^r O_K$ for $g\in O_K$ a generator of $(O_K/P)^\times$. The whole ideal group is $\mathcal{I}_K=P^\Bbb{Z} G$. $\endgroup$ – reuns Mar 30 at 10:12
  • $\begingroup$ Thanks. Did you mean $g\mathcal{O}_K$ for a generator of $G/H$ ? Are you suggesting taking the subgroup $P^{\mathbb Z}H$ then ? Unfortunately I think this subgroup does not contain $P_{K, \mathbb Z}(\mathfrak{m})$, so how do I know the extension I get from class field theory is dihedral over $\mathbb Q$ ? $\endgroup$ – A. Bailleul Mar 30 at 14:46

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