# Is $\operatorname{Gal}(\mathbb{Q}_p^{un})\cong \hat{\mathbb{Z}}$?

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 $\mathbb{Z}$...

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