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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?

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    $\begingroup$ Why should decomposition groups automatically be solvable? They're just stabilizers, I don't see why that should automatically make them solvable. $\endgroup$ Commented May 19, 2017 at 21:12
  • $\begingroup$ This might help : ams.org/journals/tran/1955-080-02/S0002-9947-1955-0075239-5/… $\endgroup$ Commented May 19, 2017 at 21:37
  • $\begingroup$ What is solvable is $D_p/I_p$ ? @AdamHughes $\endgroup$
    – reuns
    Commented May 19, 2017 at 21:41
  • $\begingroup$ @Justin I wonder if there is a modernized exposition of the theorems in this paper... $\endgroup$
    – 54321user
    Commented May 19, 2017 at 21:59
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    $\begingroup$ @MoarCake559, good question. I'm not familiar with anything other than quadratic extensions of local fields. $\endgroup$ Commented May 19, 2017 at 22:06

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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.

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