# Which groups are Galois over some p-adic field?

Suppose I have some finite $p$-group $G$, or a little extension of it.

How do I know if there exists a prime $l$ and a finite extension $K$ of $\mathbb{Q}_l$ such that $G$ is the Galois group of some finite Galois extension $M/K$?

For instance, I am interested in the case when $G$ is the semidirect product of a cyclic group of order $8$ by its automorphism group, a direct product of two groups of order $2$.

• For the specific part of your question, maybe the methods of ijpam.eu/contents/2015-103-4/13/13.pdf are relevant? – Vladimir Dotsenko Feb 25 '16 at 11:05
• It would sound natural to separate the discussion between $\ell\neq p$ and $\ell=p$. – YCor Feb 25 '16 at 12:30
• See this MO question mathoverflow.net/questions/172569/local-inverse-galois-problem – Laurent Berger Feb 26 '16 at 11:51
• There is a general question and a very specific question (one example). You accepted an answer for the specific answer... it would tend to discourage answers to the general (and more interesting) question (in particular, which $p$-groups are Galois groups over some $p$-adic field?). – YCor Feb 29 '16 at 12:43
• If you are looking for specific examples, then the local fields database here math.la.asu.edu/~jj/localfields is very useful. – Henri Johnston Feb 29 '16 at 13:57

Any finite Galois extension of $\mathbb{Q}_l$ of degree coprime to $l$ is tamely ramified. In particular, its Galois group is an extension of two cyclic groups. This means that your specific $G$ is not realisable as a Galois group when $l \neq 2$.

However, your specific group $G$ is in fact realisable over $\mathbb{Q}_2$. This is the Galois group of the splitting field the polynomial $$x^8 - 6.$$ There are a few ways to see this, and I leave most of the details to you (this can all be easily checked in sage or magma).

First considering this polynomial over $\mathbb{Q}$, it should be hopefully clear that the Galois group of the Galois closure $K$ is $G$ - the cyclic group of order $8$ corresponds to the radical extension $\mathbb{Q}(\sqrt[8]{6})$ and the Klein four group comes from the $8$th roots of unity acting in the way you want. There is a unique prime ideal of $K$ above $2$, hence we see that the decomposition group at $2$ is equal to $G$. Taking the completion of $K$ at this prime ideal then yields a $\mathbb{Q}_2$-extension of the required form.

This construction works for many polynomials of the form $x^8 - n$, but one has to be careful. E.g. one does not obtain the required Galois group when $n=2$ or $n=14$. I'll let you have fun thinking about why this is.

When $p \neq \ell$, if $N/K$ has Galois group $G$ then $N/K$ is tamely ramified. It follows that $N = K(\sqrt[e]{\pi}, \zeta)$ where $e$ is the ramification degree of $N/K$, $\pi$ is some uniformizer of $K$, $\zeta$ is a primitive $(p^f-1)$st root of unity and $f$ is the residue degree. It is not hard to work out $G$ given $e$, $f$ and $\pi$. From this one can work out the list of possible $G$. I'll leave the details to the reader.

The more interesting case is when $p=\ell$, which I now assume. Suppose $K/\mathbb{Q}_p$ is finite, and let $K^p$ be the maximal pro-$p$ extension of $K$ (i.e. the compositum of all finite normal $N/K$ with $(N:K)$ a $p$-power). Then $G_{K,p} := \operatorname{Gal}(K^p/K)$ is known in the literature, and can be written down in terms of generators and relations. Given this, then there is an extension of $K$ with Galois group $G$ (a $p$-group) if and only if there is a surjective homomorphism $G_{K,p} \to G$. For specific $G$, one could use a computer algebra system (like MAGMA) to find all such homomorphisms, and deduce the number of extensions with Galois group $G$.

Note that the theory behind $G_{K,p}$ uses Galois cohomology, and as such is somewhat nonconstructive. In particular, given a homomorphism $\varphi: G_{K,p} \to G$ I don't know of an easy method to produce the corresponding extension $N/K$ (i.e. the fixed field of $\ker \varphi$).

When $K$ does not contain a primitive $p$th root of unity, then $G_{K,p}$ is a free pro-$p$ group with $n+1$ generators where $n=(K:\mathbb{Q}_p)$.

When $K$ does contain a primitive $p$th root of unity, then $G_{K,p}$ is a pro-$p$ group with $n+1$ or $n+2$ generators and one relation. It is known as a Demushkin group and these have been fully classified, although the details are too much for a MO answer. The result of this is that given $K$, one can easily compute $G_{K,p}$. For example, $$D_{\mathbb{Q}_2,2} = \langle a, b, c : a^2 b^4 [b,c] \rangle.$$

The following articles will tell you more:

• H Koch. Galois theory of $p$-extensions. (On the general theory of $p$-extensions, including the full answer in the $p \neq \ell$ case and much of the $p = \ell$ case. See chapter X in particular.)
• J. Labute. Classification of Demushkin groups. PhD thesis, 1965. (Fully describes Demushkin groups in terms of generators and relations.)
• M. Yamagishi. On the number of Galois $p$-extensions of a local field. Proc. Amer. Math. Soc., 1995. (A neat summary of the previous article, and one algorithm for computing the number of extensions with a given Galois $p$-group.)

There is a tower $\mathbb{Q}_\ell\subset\mathbb{Q}_\ell^{nr}\subset\mathbb{Q}_\ell^{tr}\subset\overline{\mathbb{Q}_\ell}$, where the middle two fields are, respectively, the maximal unramified extension and the maximal tamely ramified extension. The Galois groups of the bottom two extensions and how they fit together are well understood, for example $\text{Gal}(\mathbb{Q}_\ell^{nr}/\mathbb{Q}_\ell)\cong\hat{\mathbb{Z}}$. The top group $\text{Gal}(\overline{\mathbb{Q}_\ell}/\mathbb{Q}_\ell^{tr})$ is a pro-$\ell$ group. Anyway, you'll find most (or all) of this in any standard text, such as Serre's Local Fields.

• This is indeed well-known, but I have hard time applying this to specific finite groups. – Pablo Feb 25 '16 at 13:41

See Dalawat & Lee, 2013 for a complete exposition of the theory of tamely ramified extensions of local fields (with finite residue field).