# Galois groups of extensions of local fields

We know that Galois groups of extensions of local fields are decomposition groups and hence solvable.

Can we say anything about whether every (finite) solvable group is a Galois group of extensions of local fields?

• Why should decomposition groups automatically be solvable? They're just stabilizers, I don't see why that should automatically make them solvable. Commented May 19, 2017 at 21:12
• This might help : ams.org/journals/tran/1955-080-02/S0002-9947-1955-0075239-5/… Commented May 19, 2017 at 21:37
• What is solvable is $D_p/I_p$ ? @AdamHughes Commented May 19, 2017 at 21:41
• @Justin I wonder if there is a modernized exposition of the theorems in this paper... Commented May 19, 2017 at 21:59
• @MoarCake559, good question. I'm not familiar with anything other than quadratic extensions of local fields. Commented May 19, 2017 at 22:06

## 1 Answer

To simplify, a « local field » $K$ will mean a « finite extension of a $p$-adic field $\mathbf Q_p$. Then, for any finite Galois extension $L/K$, ramification theory tells us that $G=Gal(L/K)$ is solvable, with additional constraints : for instance, the inertia subgroup $G_0$ is the semi-direct product of a cyclic subgroup of order prime to $p$ by a normal $p$-subgroup (property ($R_p$)), and the successive quotients $G_i /G_{i+1}$ are direct products of cyclic groups of order $p$ ; see Serre’s « Local Fields », chapter II, §2, where it is even stated that every group satisfying ($R_p$) can be realized as a $G_0$. But the most convenient approach to your question « whether every (finite) solvable group is a Galois group of an extension of local fields » seems to be through the structure of the absolute Galois group $G_K$ , which is pro-solvable with constraints. It is obtained by the following steps :

Let $V_K$ and $T_K$ resp. be the ramification and inertia subgroups of $G_K$, and let $\mathcal G_K = G_K/V_K = G(K^{tr}/K), \Gamma_K = G_K /T_K = G(K^{nr}/K)$, where $K^{tr}$ (resp. $K^{nr})$ is the maximal tamely ramified (resp. unramified) extension of $K$. The paper of Iwasawa cited by @Chickenmancer was a first attempt to describe $G_K$ as an extension of $\mathcal G_K$ by $V_K$ (beware : these notations are not totally the same as Iwasawa’s). A « modernized exposition » of Iwasawa’s result (as asked by @MoarCake559) can be found in the book by Neukirch, Smith, Wingberg, « Cohomology of Number Fields », chap.VII, §5 : the profinite group $\mathcal G_K$ is topologically generated by 2 generators $\sigma , \tau$ and one relation $\sigma \tau \sigma^{-1} = \tau^q$, where $q$ is the order of the residue field, $\sigma$ is a lift of the Frobenius automorphism which generates $\Gamma_K$ and $\tau$ generates $T_K/V_K$. Moreover $V_k$ is the maximal pro-$p$-subgroup of $G_K$, it is pro-$p$-free of countable infinite rank , and the action of $\mathcal G_K$ on $V_K ^{ab}$ is explicitly known. Finally the extension $1 \to V_K \to G_K \to \mathcal G_K \to$ is split.

The splitting of the above extension suggests to study the maximal pro-$p$-factor $G_K (p)$ of $G_K$ : if N is the degree of $K/\mathbf Q_p$, then $G_K$ is free on (N +1) generators when $K$ does not contain a primitive $p$-th root of unity (Shafarevich), generated by (N + 2) generators and 1 explicit relation otherwise (Demushkin). Using these results, Jannsen and Wingberg (after some work !) have been able to determine completely the structure of $G_K$ : it is generated by (N + 3) generators and 1 explicit relation (op. cit., thm.7.5.14). Thus your problem is solved.