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It's a basic fact that a finite Galois extension $L/K$ of a local nonarchimedean field $K$ has solvable (in fact supersolvable [edit: no!]) 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 $K$?

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    $\begingroup$ @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. $\endgroup$ Feb 26, 2020 at 23:22
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    $\begingroup$ @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'). $\endgroup$
    – YCor
    Feb 27, 2020 at 5:21
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    $\begingroup$ Incidentally bumping good questions sometimes leads to some good new answers (e.g., mathoverflow.net/questions/7112 recently) $\endgroup$
    – YCor
    Feb 27, 2020 at 5:24
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    $\begingroup$ I'm glad that this question was bumped as I carelessly state in my course notes that such Galois groups are supersolvable. Mistake fixed. $\endgroup$ Feb 27, 2020 at 9:29
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    $\begingroup$ Link to conversation in chat about this discussion unrelated to the question: chat.stackexchange.com/rooms/10243/conversation/… $\endgroup$
    – YCor
    Feb 27, 2020 at 18:21

5 Answers 5

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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).

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Three comments (which I don't have enough reputation to add as comments):

  • 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.
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    $\begingroup$ Is the Galois group $G$ supersolvable when $p>2$? $\endgroup$ Jan 31, 2023 at 2:13
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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 $p > 2$).

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I would say, it depends on what you mean by: "Is the local inverse Galois problem is solved?"

One way to formulate the local inverse Galois problem is, for any given a finite group $G$ can we decide whether $G$ is the Galois group of a local field. Or more algorithmically:

Does there exist an algorithm that has as input a finite group G and a local field K and outputs whether or not the case that the group G is the Galois group of an extension L/K.

If you formulate it this way the answer is yes. And the reason for this maybe not as nice as one would like.

The reason why the answer is yes is as follows: For a given a group $G$ there are up to isomorphism only finitely many extensions of $L/K$ of degree $\#G$. Indeed, when weighted correctly we even know explicit formula for the number of extensions and there are algorithms that explicitly enumerate these extensions. So the only thing one has to do is compute the Galois group of every extension $L/K$ of degree $|G|$, and see whether it is isomorphic to $G$.

This algorithm can even be made "independent of the characteristic. I.e. for a given group $G$ on can explicitly decide for which characteristics $p$ the group $G$ can be a Galois group of $\mathbb Q_p$. The reason is that when $L/\mathbb Q_p$ is tamely ramified we know very well which groups occur (see Iwasawa's result in Jeremy's answer). So using this it is reduced to doing a finite computation as above for all the primes $p | \#G$ where there can be wild ramification.

So it all depends on how you formulate the question "Is the local inverse Galois problem solved?". Cause it all depends on what way of characterizing the groups that can occur as a local Galois group you are fine with.

Edit I just found out about the existence of the paper The Inverse Galois Problem for p-adic fields by David Roe. His results are only for when the residue characteristic is not $2$ since he uses the explicit description of the absolute Galois group of Jannsen and Wingberg mentioned by Jeremy. In that paper he gives an algorithm to actually count the number of extensions of K with galois group G, which is more efficient than the enumeration I described above. Additionally he gives a set of necessary and a distinct set of sufficient conditions for a group G to be realizable as a Galois group of over K, which might be of independent interest.

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Regarding embedding problems: I have found http://link.springer.com/article/10.1007%2Fs10958-009-9588-7 ("The embedding problem with non-Abelian kernel for local fields", Journal of Mathematical Sciences, September 2009, Volume 161, Issue 4, pp 553-557; Zentralblatt Math: https://zbmath.org/?q=an:05660150), which I unfortunately do not have access to.

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