Can one describe explicitly the Galois group $G=\operatorname{Gal}(\overline{\mathbb Q_p}/\mathbb Q_p)$?

I only know the most basic stuff: unramified extensions of $\mathbb Q_p$ are equivalent to extensions of $\mathbb F_p$, so the Galois group of the unramified part is $\hat{\mathbb Z}$, generated by the Frobenius automorphism.

Then I have a tame knowledge of tamely ramified extensions, given by adjoining $p^{1/n}$, $(p,n)=1$. Elements of $G$ send $p^{1/n}$ to $\alpha p^{1/n}$, $\alpha^n=1$. Alltogether, I saw the tamely ramified Galois group described neatly as the profinite group with generators $\sigma,\tau$ ($\sigma$ a lift of Frobenius) and relation $\sigma\tau\sigma^{-1}=\tau^p$.

But I know wildly nothing about wildly ramified extensions, i.e. about the kernel of the projection $G\to\langle \sigma,\tau|\sigma\tau\sigma^{-1}=\tau^p\rangle$ (besides perhaps that it's a pro-$p$ group).


As you say, the wild ramification group

$$\mathfrak{w}_p = \operatorname{Aut}( \overline{\mathbb{Q}_p} / \mathbb{Q}_p^{\operatorname{tame}})$$

is a pro-$p$-group. It is also topologically finitely generated, i.e., a quotient of the pro-$p$-completion of a free group of finite rank. Thus in theory we can describe this group -- i.e., $\operatorname{Aut}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$ -- completely by giving relations.

Added: The above assertion that $\mathfrak{w}_p$ is topologically finitely generated is false, as has been pointed out to me by Chandan Singh Dalawat and Daniel Litt. The absolute Galois group of $\mathbb{Q}_p$ is finitely generated. I believe what I had in mind is that quotients of topologically finitely generated topological groups are topologically finitely generated. Which is true -- but the absolute Galois group of the maximal totally tamely ramified extension of $\mathbb{Q}_p$ is a subgroup, and subgroups of noncommutative topological groups do not automatically inherit finite generation. In fact it is a theorem of Iwasawa (Transactions of the AMS, 1955) that $\mathfrak{w}_p$ is a free pro-$p$ group of countably infinite rank. Mea culpa.

For $p > 2$, the determination of the relations for

$$\mathfrak{g}_p = \operatorname{Aut}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$$

has been done! The set of relations is finite for each $p$ and is (for $p > 2$) explicitly known, due to work of Jannsen and Wingberg. This wikipedia article gives references.

However, to the best of my knowledge (and my knowledge here is not the best), this kind of generators and relations description has not been especially useful. (To be sure, it is a nice result. It happens to be the case that not all nice results are especially useful...) For instance, the Local Langlands Conjecture is a deep result about finite-dimensional representations of the absolute Galois group of $\mathbb{Q}_p$. So far as I know (insert further provisos...), knowing a presentation for the Galois group is utterly unhelpful in establishing results like this.

Probably M. Emerton will come along shortly and tell you more. I hope so...

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  • 3
    $\begingroup$ I've seen people use explicit presentations of p-adic Galois groups in deformation theory, for instance in seminar talks by Gebhard Boeckle. $\endgroup$ – David Loeffler Sep 17 '13 at 6:45
  • $\begingroup$ Thanks a lot! As your prophecy about an even greater mind coming after you to answer this question didn't come true, I'm gladly accepting your answer. I looked up the generators and (pretty complicated) relations in the reference and find it quite amazing (an absolute layman as I am) that such a thing is known. $\endgroup$ – user8268 Sep 17 '13 at 17:46
  • $\begingroup$ @user8268: You're welcome. (And I still think he will appear; give him time...) $\endgroup$ – Pete L. Clark Sep 17 '13 at 22:24
  • $\begingroup$ @David: That sounds neat. Please do post an answer if you know or can link to any further information. $\endgroup$ – Pete L. Clark Sep 17 '13 at 22:26
  • $\begingroup$ I'm a little confused by this answer; do you mean to say that $G_K$ is topologically finitely generated? Looking at the paper of Iwasawa mentioned in user82218's answer, it seems to me that the wild ramification group is free pro-$p$ on countably many generators... $\endgroup$ – Daniel Litt Jun 22 '16 at 18:03

Your question is about the group $P=\mathrm{Gal}(\bar{\mathbf Q}_p|V)$, where $V$ is the maximal tamely ramified extension of $\mathbf{Q}_p$. In his On Galois Groups of Local Fields, Transactions of the American Mathematical Society Vol. 80, No. 2 (Nov., 1955), pp. 448-469, Iwasawa proved that $P$ is the pro-$p$ completion of the free group on countably many free generators (Theorem 2(ii)). He also proved that the short exact sequence $$ 1\to P\to \mathrm{Gal}(\bar{\mathbf Q}_p|\mathbf{Q}_p) \to \Gamma\to1 $$ splits (Theorem 2(iii)), where $\Gamma=\mathrm{Gal}(V|\mathbf{Q}_p)$. If we choose a splitting, we get the conjugation action of $\Gamma$ on $P$. Iwasawa did not describe this action but he describes the resulting action on the maximal abelian quotient of $P$ using local class field theory ($\S$3.3).

His results are valid not just for $\mathbf{Q}_p$ but for any finite extension of $\mathbf{Q}_p$.

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