Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is the absolute Galois group of $\mathbb{Q}_p^{un}$ the profinite completion of $\mathbb{Z}$? I was never quite sure...

In similar cases, it is true. Namely, $\mathbb{C}((t))$ does have absolute Galois group isomorphic to the profinite completion of $\mathbb{Z}$...

share|cite|improve this question
up vote 14 down vote accepted

No. The absolute Galois group of $\mathbb Q_p^{un}$ is the same as the absolute inertia group of $\mathbb Q_p$; I'll denote it by $I_p$. It admits a quotient $I_p^\mathrm{tame}$, corresponding to the extension of $\mathbb Q_p^{un}$ obtained by adjoinng the $n$th roots of $p$ for all $n$ coprime to $p$.

This is analogous to the fact that the algebraic closure of $\mathbb C((t))$ is obtained by adjoining all $n$th roots of $\mathbb Z$. The point in this case is that residue field has char. 0 (it is $\mathbb C$) and so all inertia is tame.

But the map $I_p \to I_p^\mathrm{tame}$ has a non-trivial kernel, which can also be thought of as the pro-$p$-Sylow subgroup of $I_p$. It is non-abelian.

share|cite|improve this answer
Maybe note that $I_p^{tame}\simeq \prod_{\ell\ne p} \mathbb{Z}_\ell$ – Daniel Miller Jul 11 '13 at 22:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.