Finding number fields with certain decomposition group?

Question

I have no reason to expect this to be true, but I thought it wouldn't hurt to ask.

For a Galois number field $K/\mathbb{Q}$, and a rational prime $(p)$ with a prime $\mathfrak{q}_i\subseteq\mathcal{O}_K$ lying above it, the decomposition group is defined as: $$ D = D_{\mathfrak{q}_i\mid p} = \{\sigma\in\text{Gal}(K/\mathbb{Q})\mid \sigma(\mathfrak{q}_i) = \mathfrak{q}_i\} $$ This is just the stabilizer of the action of $\text{Gal}(K/\mathbb{Q})$ on the prime ideals lying above $p$.

Is there anything known about, given a particular group $D$, which pairs of $(K,p)$ have this group as a decomposition group?

I'm interested in the case of $D = \mathbb{Z}/2\mathbb{Z}$ specifically, which may be simple enough that something about it is known.

Answer 1

Let's think about the case when $K=\mathbb{Q}(\sqrt{d})$ is a quadratic field to start with.

To get a non-trivial decomposition group above a rational prime $p$, you need the prime to split or ramify. In other words, applying the Kummer-Dedekind theorem, we want the polynomial $x^2-d$ to be reducible $\bmod{p}$; i.e. $d$ should be a square $\bmod{p}$. Indeed if $\alpha,\beta$ are roots, then your two primes above $p$ are $(\alpha,p)$ and $(\beta,p)$ which are permuted by Galois.

(Note to apply Kummer-Dedekind, I should suppose generally that $p \nmid 4d$ or work properly with the minimal polynomial of a generator for the ring of integers of $K$).

In the general case, note that since the decomposition group is a subgroup, its order must divide the degree $[K:\mathbb{Q}]$. With that done, Chebotarev's density theorem then says that this will happen infinitely often for primes (indeed each element has an equal chance so just count the number of order 2 elements in your Galois group). You are however unlikely to get a full description like the quadratic case.

Answer 2

Your question is a hard one, no complete solution is known yet. General theorems such as Chebotarev's give you only the density of the primes which split completely in your number field, which is far from the kind of "global-local" answer you're looking for. There are however partial known solutions:

1) The case of an abelian extension $K/k$ of number fields (no need for $k$ to be $\mathbf Q$) can be dealt with using CFT, see e.g. Artin-Tate, chap.IX, §2, entitled "Abelian fields with given local behavior". The abelian extensions of $k$ can be characterized as fixed fields of kernels of characters of the idèle class group $C$ of $k$, and the main theorem in A-T reads : Let $S$ be a finite set of places of $k$, let $\chi_v$ be local characters of periods $n_v$ for $v\in S$ and $m$ be the l.c.m. of the $n_v$. There exists a global character $\chi$ on $C$ whose local restrictions are the given $\chi_v$. Its period can be made equal to $2m$ (and even = $m$ outside a so called "special case" due to the particular behvior of the places above $2$).

2) One can go a little beyond the abelian case by considering only pro-$p$-extensions (non abelian). Let $p$ be a fixed prime number, supposed to be odd for simplification, and let $S$ be a finite set of places of $k$ containing the set $S_p$ of $p$-adic places. Let $G_S$ denote the Galois group of the maximal pro-$p$-extension of $k$ which is unramified outside $S$. The maximal abelian pro-$p$-quotient of $G_S$ is a $\mathbf Z_p$-module of finite type, and its $\mathbf Z_p$-torsion $T_S$ can be decomposed as $T_{S_p} \times (\prod W_v)$, where $ v$ runs through $S$ and $W_v$ is the group of $p$-primary roots of unity of the local field $k_v$. The structure of the Galois group $G_v$ of the maximal pro-$p$-extension of $k_v$ is explicitly known by generators and relations, see e.g. H. Koch's book "Galois theory of $p$-extensions". The field $k$ is called $p$-rational if $T_{S_p}$ (obvious notation) is trivial. In this case, $G_S$ is the free pro-$p$-product of the $G_v$ for $v\in S/S_p$. This is an arithmetic analogue of the Riemann existence theorem ./.