# Dihedral extension of $\mathbb Q$ with small discriminant

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.

• 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 $I \in (g^{2^l r})^m HH^{-1}$ – reuns Mar 28 at 21:18
• 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. – A. Bailleul Mar 29 at 18:23
• 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 $\mathcal{I}_K=P^\Bbb{Z} G$. – reuns Mar 30 at 10:12
• 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 $\mathbb Q$ ? – A. Bailleul Mar 30 at 14:46