Completion of the unit group of a local field

Question

Let $K$ be a number field and $\mathfrak{p}$ a finite prime of $K$. Denote the unit group of the ring of integers of the local field $K_\mathfrak{p}$ (i.e. the completion of $K$ via $\mathfrak{p}$) by $U_\mathfrak{p}$. For an archimedian prime we set $U_\mathfrak{p} = K^\times_\mathfrak{p}$.

Now for a rational prime $p$ and an abelian group $A$ denote by $\hat A$ the $p$-completion, i.e.

$$\hat A = \lim_{\begin{matrix} \leftarrow \\ n\end{matrix}} A / p^n A$$

Now the claim is that for $\mathfrak{p} \not\mid p$ the completion $\hat U_\mathfrak{p}$ is finite. Can somebody help me with a hint?

Thank you!

Tom

Answer 1

To put an end to the "quarrel" on notations, let p be a fixed prime number and v a "place" of a number field K (in the usual language of algebraic number theory), $K_v$ the completion of K at v, $k_v$ its residue field, which is a finite field of characterisic q (the prime number underlying v). For any natural integer n , let $U_n$ be the subgroup of $K_v$* defined by : x belongs to $U_n$ iff v(1-x) is at least n. Then $U_0$ is the group of local units, $U_0$/$U_1$ is isomorphic to $k_v$* (multiplicative group) and, for n at least $1$, $U_n$/$U_{n+1}$ is isomorphic to $k_v$ (additive group). This is classical and easily shown by taking residual classes; see e.g. Cassels-Fröhlich's book "Algebraic Number Theory", chapter 1, propos. 4. But $k_v$* is finite and, if q is different from p, by using the above description of $U_n$/$U_{n+1}$ , a straightforward approximation argument shows that the map u -> $u^m$ , m = a power of p , is an automorphism of $U_1$ ; see op. cit. propos. 5. It follows at once that the p-adic completion of $U_0$ is equal to that of $k_v$* , hence is finite .

Answer 2

This was false as originally stated, and a proof is below. The OP has since changed the statement.

Let $K=\mathbb{Q}$, let $p$ be a rational prime, and let $\mathfrak{p}=(p)$. Let $U_p=\mathbb{Q}_p^\times$, and let $q$ be a prime distinct from $p$. Let $n\geq 1$ be arbitrary, consider $\widehat{U}_{p,q,n}=U_p/U_p^{q^n}$, and consider the subgroup $B_n$ of $\widehat{U}_{p,q,n}$ generated by $p$. Since $p$ is not a $q$th power in $\mathbb{Q}_p^\times$, it follows that the subgroup $B_n$ has size $q^n$, and it is isomorphic to $\mathbb{Z}/q^n\mathbb{Z}$. Hence, $$\widehat{U}_{p,q} = \varprojlim U_{p,q,n}$$ has a subgroup $B=\varprojlim B_n$ isomorphic to $\varprojlim \mathbb{Z}/q^n\mathbb{Z}\cong \mathbb{Z}_q$. In particular, $\widehat{U}_{p,q}$ is infinite.