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

Solution 1:

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.