# Local inverse Galois problem

It's a basic fact that a finite Galois extension $$L/K$$ of a local nonarchimedean field $$K$$ has solvable (in fact supersolvable) Galois group $$G$$. One sees this by using the ramification filtration $$G_i=\{g\in G: g\beta\equiv \beta \pmod {\varpi_K^i}\}$$.

More nontrivially, local class field theory tells us if $$L/K$$ is abelian then $$G$$ must be a quotient of $$K^\times$$, whose structure as an abelian group is known explicitly.

In any case, this means the inverse Galois problem over $$K$$ is quite circumscribed. So is it known precisely which finite groups are Galois groups over $$K$$?

• Has anyone tried to solve embedding problems? – user19475 Jun 25 '14 at 16:29
• @YCor What's the point of editing all these old questions? I also saw that you did this with one of my questions. Please don't do it again. – Martin Brandenburg Feb 26 at 23:22
• @MartinBrandenburg the point is (mainly) to improve tagging. I understand bumping is inconvenient, so I refrain to to too much (without refraining it would be easily 30 a day!). The reason why it came to your questions is that these were good questions with many upvotes, and I treated in priority such questions (in that case, removing the deprecated tag 'abstract-algebra'). – YCor Feb 27 at 5:21
• I'm glad that this question was bumped as I carelessly state in my course notes that such Galois groups are supersolvable. Mistake fixed. – Laurent Berger Feb 27 at 9:29

The short answer is (as far as I am aware) no, but there is a lot that is known. Jannsen and Wingberg have given an explicit presentation for $Gal(\overline{K}/K)$ in the case that the residue characteristic is not $2$ (published in Inventiones Math in 1982/1983), and Volker (1984, Crelle) handles the case when $K$ has residue characteristic $2$ and $\sqrt{-1} \in K$. This does not, however, make it trivial to determine which finite groups are quotients of $Gal(\overline{K}/K)$. Some more information can be obtained from Section VII.5 of "Cohomology of Number Fields" by Neukirch, Schmidt and Wingberg. Here's a paraphrase.

If $K$ is a local nonarchimedean field with residue field of characteristic $p$ (and order $q$), let $G = Gal(\overline{K}/K)$, $T$ be the inertia group, and $V$ be the ramification group. Then $G/T \cong \hat{\mathbb{Z}}$, $T/V \cong \prod_{\ell \ne p} \mathbb{Z}_{\ell}$, and $V$ is a free pro-$p$ group of countably infinite rank. Iwasawa showed that $G/V$ is a profinite group with two generators $\sigma$ and $\tau$ so that $\sigma \tau \sigma^{-1} = \tau^{q}$. Also, the maximal pro-$\ell$ quotient of $G$ is known for all $\ell$. For example, if $\mu_{\ell} \not\subseteq K$ and $\ell \ne p$, the maximal pro-$\ell$ quotient is $\mathbb{Z}_{\ell}$ (i.e. for each positive integer $k$, there is a unique Galois extension $L/K$ of degree $\ell^{k}$, namely the unramified one).

• The parenthetical claim in the statement of the question is false: Galois groups of local fields need not be supersolvable. For example, $A_4$ is not supersolvable but it is the Galois group of http://www.lmfdb.org/LocalNumberField/2.4.6.7. Another example is $\mathrm{GL}_2(3)$, which is the Galois group of http://www.lmfdb.org/LocalNumberField/2.8.10.2.
• Even if we restrict to extensions of $\mathbb{Q}_p$, giving a precise answer to this question is likely to be difficult. For example, $C_2^4$ is not the Galois group of any extension of $\mathbb{Q}_p$ (it has too many index-2 subgroups), nor is $\mathrm{SL}_2(3)$ (as proved by Weil in his "Dyadic Exercises" paper), even though $\mathrm{GL}_2(3)$ is.
• In general one knows that $G=\mathrm{Gal}(L/K)$ must have a cyclic series of the form $$W\unlhd I\unlhd G$$ in which $W$ is a $p$-group that is also normal in $G$ and $I$ has order prime to $p$ (one can add further constraints imposed by the filtration of $W$ and the action of Frobenius). But as the examples above show, these necessary conditions are not sufficient.
One restriction is, e.g. that the groups occuring as a Galois extension of $K/\mathbf{Q}_p$ have to be generated by $\leq n(K)$ elements for some $n(K) \in \mathbf{N}$ depending on $K$. More precisely, one has $n(K) \leq N + 3$ with $N = [K:\mathbf{Q}_p]$ by [Neukirch-Schmidt-Wingberg], Theorem (7.5.14) (for $p > 2$).