S-unit theorem

Define $\phi :G\to {\mathbb{Z}}^S$ by $\phi (u)=(v_{\mathfrak{p}}(u){)}_{\mathfrak{p}\in S}$; it is a group homomorphism because $v_{\mathfrak{p}}(xy)=v_{\mathfrak{p}}(x)+v_{\mathfrak{p}}(y)$ for all nonzero fractional ideals.

The kernel of $\phi$ is ${\mathcal{O}}_K^{\times }$: a unit $u\in G$ has $\phi (u)=0$ exactly when $v_{\mathfrak{p}}(u)=0$ for every $\mathfrak{p}\in S$, and by definition $v_{\mathfrak{p}}(u)=0$ for every $\mathfrak{p}\mathrm{\notin }S$, so $\phi (u)=0$ exactly when $\mathrm{div}(u)=0$, that is $u\in \ker \mathrm{div}={\mathcal{O}}_K^{\times }$.

Put $h:=\mathrm{\mid }\mathrm{Cl}({\mathcal{O}}_K)\mathrm{\mid }\ge 1$, a finite number; for $\mathfrak{p}\in S$ let $[\mathfrak{p}]$ be its class, so $[\mathfrak{p}{]}^h=1$ by [F8], and hence the divisor $h[\mathfrak{p}]$ lies in the kernel of $\mathrm{cl}$, i.e. is principal: there is ${\pi }_{\mathfrak{p}}\in K^{\times }$ with $({\pi }_{\mathfrak{p}})={\mathfrak{p}}^h$; since ${\mathfrak{p}}^h$ is an integral ideal, ${\pi }_{\mathfrak{p}}\in {\mathcal{O}}_K\setminus \{0\}$.

For $\mathfrak{p},\mathfrak{q}\in S$ one has $v_{\mathfrak{q}}({\pi }_{\mathfrak{p}})=v_{\mathfrak{q}}({\mathfrak{p}}^h)=h{v}_{\mathfrak{q}}(\mathfrak{p})=h{\delta }_{\mathfrak{p}\mathfrak{q}}$, using additivity of valuations and $v_{\mathfrak{q}}(\mathfrak{p})=1$ for $\mathfrak{q}=\mathfrak{p}$ and $0$ otherwise.

Consequently $h{\mathbb{Z}}^S\subseteq \phi (G)$, because $\phi ({\pi }_{\mathfrak{p}})=h{e}_{\mathfrak{p}}$ for each $\mathfrak{p}\in S$ and the elements $h{e}_{\mathfrak{p}}$ generate $h{\mathbb{Z}}^S$.

The image $M:=\phi (G)$ is a subgroup of ${\mathbb{Z}}^S$, hence free of rank $s\le \mathrm{\mid }S\mathrm{\mid }$; since $h{\mathbb{Z}}^S$ is a subgroup of $M$ isomorphic to ${\mathbb{Z}}^{\mathrm{\mid }S\mathrm{\mid }}$, the same rank bound gives $\mathrm{\mid }S\mathrm{\mid }\le s$, so $s=\mathrm{\mid }S\mathrm{\mid }$ and $M\cong {\mathbb{Z}}^{\mathrm{\mid }S\mathrm{\mid }}$; also $M$ has finite index in ${\mathbb{Z}}^S$, because it contains $h{\mathbb{Z}}^S$.

Since $M$ is free abelian, the surjection $\phi :G\twoheadrightarrow M$ splits: choose a $\mathbb{Z}$-basis $m_1,\dots ,m_{\mathrm{\mid }S\mathrm{\mid }}$ of $M$ and preimages $u_1,\dots ,u_{\mathrm{\mid }S\mathrm{\mid }}\in G$ with $\phi (u_i)=m_i$, and define the homomorphism $\sigma :M\to G$ by $\sigma (m_i)=u_i$; then $\phi \circ \sigma ={\mathrm{i}\mathrm{d}}_M$, and every $g\in G$ is written as $g=\sigma (\phi (g))\cdot (g\sigma (\phi (g){)}^{-1})$ with $g\sigma (\phi (g){)}^{-1}\in \ker \phi$ and $\sigma (\phi (g))\in \mathrm{im}\sigma$, while $\ker \phi \cap \mathrm{im}\sigma =\{1\}$ because $\phi (\sigma (m))=m$; hence $G\cong \ker \phi \times M$.

By [F2] and step 5.1, $G\cong (\mu (K)\times {\mathbb{Z}}^{r_1+r_2-1})\times {\mathbb{Z}}^{\mathrm{\mid }S\mathrm{\mid }}\cong \mu (K)\times {\mathbb{Z}}^{r_1+r_2-1+\mathrm{\mid }S\mathrm{\mid }}$, so $G$ is finitely generated, its free rank is $r_1+r_2-1+\mathrm{\mid }S\mathrm{\mid }$, and the torsion subgroup of $G$ is the torsion subgroup of $\mu (K)\times {\mathbb{Z}}^{r_1+r_2-1+\mathrm{\mid }S\mathrm{\mid }}$, namely $\mu (K)$.

Directly: if $g\in G\subseteq K^{\times }$ has finite order then $g$ is a root of unity and so lies in $\mu (K)$; conversely every root of unity lies in $\mu (K)\subseteq {\mathcal{O}}_K^{\times }\subseteq G$; hence the torsion subgroup of $G$ is $\mu (K)$, finite, in agreement with step 6.1.

Choice accounting: the unit theorem [F2], class-group finiteness input [F3], principal-divisor exact sequence [F4], and ideal-factorisation and valuation inputs [F5] assume AC. The selections performed here are finitely many (the elements ${\pi }_{\mathfrak{p}}$ for $\mathfrak{p}\in S$ and the lifts $u_i$ of a basis of $M$), so these selections need only finite choice; the splitting is noncanonical because it depends on those lifts.