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\begin{document}

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Let $\ok{K}^\times$ denote the multiplicative group of units in $K$.
\medskip

If $S$ is a finite set of places containing $S_\infty$, the archimedean places of K,
then define the S-units of K to be
$$\ok{K,S}^\times = \{ x \in K : |x|_\nu = 1, \forall \nu \in M_K - S \}.$$
\medskip

Note that $\ok{K,S_\infty}^\times = \ok{K}^\times$.
\bigskip

\begin{lem}
\label{L:II.4.1}
Let $0 < c \leq C < \infty$, \\
$V = \{ x \in \ok{K,S}^\times : c \leq |x_\nu|_\nu \leq C, \forall \nu \in S \}$.
Then $V$ is finite.
\end{lem}

\begin{proof}
Let $W = \{ \alpha \in J_K : |\alpha_\nu|_\nu = 1, \forall \nu \in M_K - S$ and
$c \leq |\alpha_\nu|_\nu \leq C, \forall \nu \in S \}$.
As it is a product of compact sets, $W$ is compact.
\medskip

$V = W \cap K^\times$, and $K^\times \subset J_K$ is discrete, so $V$ is compact and discrete.
Hence, $V$ is finite.
\end{proof}
\bigskip

\begin{lem}
\label{L:II.4.2}
\{roots of unity in $K\} = \{ x \in K : |x|_\nu = 1, \forall \nu \}$
\end{lem}

\begin{proof}
Let $\zeta \in K$ such that $\zeta^n = 1$.
$1 = |\zeta^n|_\nu = |\zeta|_\nu^n$, so $|\zeta|_\nu = 1$.
\medskip

$\{ x \in K : |x|_\nu = 1, \forall \nu \}$ is a subgroup of $K^\times$, and by
Lemma \ref{L:II.4.1} with $c = C = 1$, a finite subgroup.
Hence, $\exists n$ such that $\forall \zeta$,
$\zeta^n = 1$.
\end{proof}
\bigskip

\begin{thm} (Dirichlet) \\
\label{T:II.4.2}
$\ok{K,S}^\times \cong $(finite cyclic group)$\times \Z^{s-1}$, where $s = \#S$.
\end{thm}

\begin{proof}
Let $J_{K,S} = \{ \alpha \in J_K : |\alpha_\nu|_\nu = 1, \forall \nu \in M_K - S \}$,
an open subgroup of $J_K$.
Note that $\ok{K,S}^\times = J_{K,S} \cap K^\times = J_{K,S}^1 \cap K^\times$, by the
product formula.
Let $J_{K,S}^1 = J_{K,S} \cap J_K^1$, an open subgroup of $J_K^1$.
An open subgroup of a topological group is closed, so $J_{K,S}^1$ is closed in  $J_K^1$.
\medskip

Since natural maps are open, $J_{K,S}^1/\ok{K,S}^\times =
J_{K,S}^1/(J_{K,S}^1 \cap K^\times)$ is a closed subset of
$J_K^1/K^\times$, which is compact by Theorem \ref{T:II.3.1}.  Hence,
$J_{K,S}^1/\ok{K,S}^\times$ is compact.
\medskip

Define $\lambda: J_{K,S} \longrightarrow \R^s$ by
$\lambda(\alpha) = (\log|\alpha_1|_1, \ldots, \log|\alpha_s|_s)$,
where \\ $\alpha_1, \ldots, \alpha_r$ are $\alpha$'s archimedean components.
Note that \\ $\lambda(J_{K,S}) = \R^r \times \prod_{i = r+1}^s \Z \log N_{p_i}$.
\medskip

Let $T = \lambda(J_{K,S}^1)$.
If $\alpha \in J_{K,S}^1$, then $\prod_{S} |\alpha_\nu|_\nu = 1$, i.e.
$\sum_{S} \log |\alpha_\nu|_\nu = 0$.
\medskip

Let $\mu: (x_1, \ldots, x_s) \longmapsto (x_2, \ldots, x_s)$.  Then
$\mu$ is surjective on $T$:  if $\alpha \in J_{K,S}$ such that
$\log |\alpha_1| = x_2, \ldots, \log |\alpha_s| = x_s$, and $\alpha_1 =
1/\prod_{i=2}^s |\alpha_i|_i$, then $\alpha \in J_{K,S}^1$.
Hence, $T \cong \R^{r-1} \times \Z^{s-r}$.
\medskip

Also, we have a surjective continuous homomorphism \\
$J_{K,S}^1/\ok{K,S}^\times \longrightarrow T/\lambda(\ok{K,S}^\times)$.
Hence, $T/\lambda(\ok{K,S}^\times)$ is compact.
\medskip

$\lambda(\ok{K,S}^\times)$ is discrete, since
$\{x \in \ok{K,S}^\times : \frac{1}{2} \leq |x|_2 \leq 2\}$ is finite, by
Lemma \ref{L:II.4.1}; and any discrete subgroup $\Lambda$ of
$T \cong \R^{r-1} \times \Z^{s-r}$ such that $T/\Lambda$ is compact
is a full lattice in $T$.
Hence, $\Lambda = \lambda(\ok{K,S}^\times) \cong \Z^{(s-r) + (r-1)} = \Z^{s-1}$.
\medskip

Since, by Lemma \ref{L:II.4.2}, $\{x \in \ok{K,S}^\times : \lambda(x) = 0\} =
\{ x \in K : |x|_\nu = 1, \forall \nu \} = \{$roots of unity in $K\}$
is a finite cyclic group, we have a split exact sequence
$$0 \longrightarrow \hbox{ (finite cyclic group) } \longrightarrow \ok{K,S}^\times
\longrightarrow \Lambda \cong \Z^{s-1} \longrightarrow 0.$$
\medskip

Hence, $\ok{K,S}^\times \cong $(finite cyclic group)$\times \Z^{s-1}$.
\end{proof}

\end{document}