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Oct 29, 2021 at 5:34 vote accept Alice
Oct 29, 2021 at 5:02 answer added KConrad timeline score: 6
Oct 27, 2021 at 13:32 comment added F Zaldivar I think the argument in Neukirch works: The ${\mathbb Z}_p$-submodule $V\subseteq U^{(1)}$ is free of finite rank $d$ (and therefore closed); the isomorphism $V\simeq {\mathbb Z}_p^d$ sends (topological) generators to (topological) generators.
Oct 27, 2021 at 12:44 comment added YCor A compact $d$-dimensional abelian $p$-adic Lie group is always isomorphic to $\mathbf{Z}_p^d\times F$ for some finite abelian group $F$. Once this is granted, things boil down to understand the torsion part. I assume $U^{(1)}$, which can be written $\mathbf{Z}_p^d\times F'$ with $d=[K:\mathbf{Q}_p]$, is by definition the kernel of $O_K^\times\to F_q^*$. An element in the kernel has the form $1+m$ with $m$ in the maximal ideal. One sees easily using the binomial formula that $(1+m)^p$ tends to zero. So it already follows that $F'$ is a $p$-group.
Oct 27, 2021 at 12:37 history edited YCor
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S Oct 27, 2021 at 12:21 review First questions
Oct 27, 2021 at 15:11
S Oct 27, 2021 at 12:21 history asked Alice CC BY-SA 4.0