TITLE: Is $\operatorname{Gal}(\mathbb{Q}_p^{un})\cong \hat{\mathbb{Z}}$?
QUESTION [10 upvotes]: 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 $\mathbb{Z}$...

REPLY [14 votes]: 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.