The torsion subgroup of principal units $U^{(1)}$

Question

$\newcommand{\U}{U^{(1)}}$ $\newcommand{\O}{\mathcal{O}}$ $\newcommand{\p}{\mathfrak{p}}$ $\DeclareMathOperator{\char}{char}$ $\newcommand{\N}{\mathbb{N}}$

I have a question about the torsion subgroup of principal units $\U$.

Let $K$ be a local field with valuation ring $\O$, maximal ideal $\p$. Let $q = p^f = \#\O/\p$. Let $\char K = 0$. The group of principal units $\U$ is $$ \U := 1 + \p. $$ For $n \in \N$, let $\mu_n$ be a group of $n$-th roots of unity, i.e. $$ \mu_n := \{ x \in K \mid x^n = 1 \}. $$ Then, I want to show that:

The torsion subgroup of $\U$ can be written as $\mu_{p^a}$ for some $a \in \mathbb{N}$.

First, I tried to show that for any $x \in \U$ which has finite order, an order of $x$ can be written as $p^n$ for some $n \in \N$, but I failed.

This question is related to the Proposition (5.7) in Neukirch, "Algebraic Number Theory" at page 140.

Answer 1

Since $U^1$ is a $Z_p$-module of finite type (*), the general theory of modules over principal domains shows that its torsion subgroup is finite, hence cyclic because we are inside the multiplicative group of a field.

(*) Quickest proof: filter $U^1$ by the subgroups $U^n$ = $1$ + $P^n$ . It is easily shown that $U^n$/$U^{n+1}$ is isomorphic to $P^n$/$P^{n+1}$ (non canonically isomorphic to the additive group of the residue field ). Since $U^1$ = projective limit of the quotients $U^1$/$U^n$ , we are done . This even shows that the $Z_p$-rank of $U^1$ is equal to the degree of the field over $Q_p$ .

Answer 2

Suppose $1+x$ is a torsion element of $U^{(1)}$, $x \in \mathfrak p$, killed by $z \in \mathbb Z_p$. That is, $(1+x)^z=1$. If $z=p^au$, $u \in \mathbb Z_p^{\times}$. Let $uv=1$, $v \in \mathbb Z_p^\times$. $$(1+x)^{{p^a}u}=1 \implies (1+x)^{{p^a}uv}=(1+x)^{p^a}=1 \implies 1+x \in \mu_{p^a}.$$