S-units of Q

The localisation $R$ embeds in $\mathbb{Q}$ by $\phi (a\mathrm{/}b)=a\mathrm{/}b$: the map is well defined because if $a\mathrm{/}b=a^{\mathrm{\prime }}\mathrm{/}{b}^{\mathrm{\prime }}$ in $R$, then $u(a{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}b)=0$ for some $u\in M$ by [F1] and hence $a{b}^{\mathrm{\prime }}=a^{\mathrm{\prime }}b$ by [F2], so the two rationals agree; it respects the fraction arithmetic of [F1], which is the arithmetic of the field $\mathbb{Q}$; and it is injective, since $\phi (a\mathrm{/}b)=0$ forces $a=0$, and then $1\cdot (a\cdot 1-0\cdot b)=0$ gives $a\mathrm{/}b=0\mathrm{/}1$ by [F1]. Hence $R$ is identified with the subring $\{a\mathrm{/}b\in \mathbb{Q}:a\in \mathbb{Z},b\in M\}$ of $\mathbb{Q}$, in which $1\mathrm{/}s$ is the inverse of $s$ for every $s\in M$.

Fix $x\in {\mathbb{Q}}^{\times }$ and write its prime factorisation as $x=\pm {\prod }_{q\mathrm{p}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}}{q}^{e_q(x)},e_q(x)\in \mathbb{Z},$ with all but finitely many exponents zero; existence and uniqueness of the triple (sign, exponent vector) is [F3], applied to the numerator and denominator of $x$ in lowest terms. Writing $x=a\mathrm{/}b$ in lowest terms with $b>0$ and fixing a prime $q$, one has $e_q(x)=v_q(a)-v_q(b)$; if $q\nmid b$ this is the nonnegative integer $v_q(a)$, while if $q\mid b$ then $v_q(b)>0$ and $v_q(a)=0$ by coprimality, so $e_q(x)<0$. Thus $e_q(x)\ge 0$ if and only if $q\nmid b$.

For a prime $q$, the localisation ${\mathbb{Z}}_{(q)}$ consists of the fractions $m\mathrm{/}n$ with $q\nmid n$ by [F5], so for $x=a\mathrm{/}b\in {\mathbb{Q}}^{\times }$ in lowest terms with $b>0$ one has $x\in {\mathbb{Z}}_{(q)}$ if and only if $q\nmid b$: if $q\nmid b$ then $x=a\mathrm{/}b$ has the required form, while if $x=a\mathrm{/}b=m\mathrm{/}n$ with $q\nmid n$, then $an=bm$, and $q\mid b$ would give $q\mid an$ and hence $q\mid a$, contradicting coprimality.

For a nonzero prime ideal $\mathfrak{p}$ of $\mathbb{Z}$ and $x\in {\mathbb{Q}}^{\times }$, [F5] writes $v_{\mathfrak{p}}(x)=v_{\mathfrak{p}}((x))$ as the unique integer $n$ with $x{\mathbb{Z}}_{\mathfrak{p}}=(x{)}_{\mathfrak{p}}={\mathfrak{p}}^n{\mathbb{Z}}_{\mathfrak{p}}$. The powers ${\mathfrak{p}}^n{\mathbb{Z}}_{\mathfrak{p}}$ with $n\ge 0$ are exactly the ideals of the discrete valuation ring ${\mathbb{Z}}_{\mathfrak{p}}$, and for $n<0$ the fractional ideal ${\mathfrak{p}}^n{\mathbb{Z}}_{\mathfrak{p}}$ properly contains ${\mathbb{Z}}_{\mathfrak{p}}$; hence $v_{\mathfrak{p}}(x)\ge 0$ if and only if $x{\mathbb{Z}}_{\mathfrak{p}}\subseteq {\mathbb{Z}}_{\mathfrak{p}}$, that is, if and only if $x\in {\mathbb{Z}}_{\mathfrak{p}}$.

An element $x\in {\mathbb{Q}}^{\times }$ lies in $R$ if and only if $e_q(x)\ge 0$ for every prime $q\mathrm{\notin }S$. If $x=a\mathrm{/}b$ with $a\in \mathbb{Z}$ and $b\in M$, then $b$ has no prime divisor outside $S$, so for $q\mathrm{\notin }S$ one has $e_q(x)=v_q(a)-v_q(b)=v_q(a)-0\ge 0$, where the exponents are those of the factorisations of $a$ and $b$; conversely, if $e_q(x)\ge 0$ for all $q\mathrm{\notin }S$ and $x=a\mathrm{/}b$ is in lowest terms with $b>0$, then a prime $q\mid b$ satisfies $v_q(a)=0$ and $v_q(x)=-v_q(b)<0$, so $q\in S$; hence every prime divisor of $b$ lies in $S$, that is $b\in M$, and $x=a\mathrm{/}b\in R$.

For the comparison with the $S$-unit theorem, evaluate its torsion factor and its signature input. $\mathbb{Q}$ is a number field with signature $(1,0)$ and ${\mathcal{O}}_{\mathbb{Q}}=\mathbb{Z}$ by [F8]; the roots of unity in $\mathbb{Q}$ are $\pm 1$, since $x=a\mathrm{/}b\in {\mathbb{Q}}^{\times }$ in lowest terms with $x^n=1$ gives $a^n=b^n$, whence $\mathrm{\mid }a{\mathrm{\mid }}^n=\mathrm{\mid }b{\mathrm{\mid }}^n$ and, by uniqueness of prime factorisation, $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }b\mathrm{\mid }$, while coprimality forces $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }b\mathrm{\mid }=1$; conversely $\pm 1$ are roots of unity. Thus $\mu (\mathbb{Q})=\{\pm 1\}$ by [F9].

The same group is obtained from the $S$-unit theorem. By [F5] the ideals $(p_1),\dots ,(p_m)$ are nonzero prime ideals of ${\mathcal{O}}_{\mathbb{Q}}=\mathbb{Z}$ by [F8], so $S_{\mathbb{Q}}:=\{(p_1),\dots ,(p_m)\}$ is a finite set of nonzero prime ideals. The two rings agree on ${\mathbb{Q}}^{\times }$, and both contain $0$. Indeed, let $0\mathrm{\ne }x=a\mathrm{/}b\in R$ with $a\in \mathbb{Z}$ and $b\in M$, and let $\mathfrak{p}\mathrm{\notin }{S}_{\mathbb{Q}}$ be a nonzero prime ideal: then $b\mathrm{\notin }\mathfrak{p}$, because otherwise some prime $p_i$ with $p_i\mid b$ lies in $\mathfrak{p}$, so $(p_i)\subseteq \mathfrak{p}$, and maximality of $(p_i)$ with $\mathfrak{p}$ proper forces $\mathfrak{p}=(p_i)\in S_{\mathbb{Q}}$, a contradiction. Hence $b\mathrm{/}1$ is a unit of ${\mathbb{Z}}_{\mathfrak{p}}$ and $x=(a\mathrm{/}1)(b\mathrm{/}1{)}^{-1}\in {\mathbb{Z}}_{\mathfrak{p}}$, so $v_{\mathfrak{p}}(x)\ge 0$ by step 1.4 and $x\in {\mathcal{O}}_{\mathbb{Q},S_{\mathbb{Q}}}$. Conversely, let $0\mathrm{\ne }x=a\mathrm{/}b\in {\mathcal{O}}_{\mathbb{Q},S_{\mathbb{Q}}}$ be in lowest terms with $b>0$, and suppose a prime $q\mathrm{\notin }S$ divides $b$; then $(q)$ is a nonzero prime ideal of $\mathbb{Z}$ and $(q)\mathrm{\notin }{S}_{\mathbb{Q}}$ by the distinctness in [F5], so $v_{(q)}(x)\ge 0$ and step 1.4 gives $x\in {\mathbb{Z}}_{(q)}$, contradicting step 1.3. Therefore every prime divisor of $b$ lies in $S$, that is $b\in M$ and $x=a\mathrm{/}b\in R$. Thus ${\mathcal{O}}_{\mathbb{Q},S_{\mathbb{Q}}}=R$ as subrings of $\mathbb{Q}$, and their unit groups inside $\mathbb{Q}$ coincide: ${\mathcal{O}}_{\mathbb{Q},S_{\mathbb{Q}}}^{\times }=R^{\times }$.

Consequently an element $x\in {\mathbb{Q}}^{\times }$ is a unit of $R$ if and only if $e_q(x)=0$ for every prime $q\mathrm{\notin }S$: a unit of $R$ lies in $R$ together with its inverse, so step 2.1 gives $e_q(x)\ge 0$ and $e_q(x^{-1})=-e_q(x)\ge 0$ for $q\mathrm{\notin }S$, and conversely the two inequalities $e_q(x)\ge 0$, $-e_q(x)\ge 0$ for $q\mathrm{\notin }S$ place both $x$ and $x^{-1}$ in $R$.

Therefore $R^{\times }=\{x\in {\mathbb{Q}}^{\times }:e_q(x)=0$ for all $q\mathrm{\notin }S\}=\{\pm p_1^{n_1}\cdots p_m^{n_m}:n_i\in \mathbb{Z}\}$: for the first description, an element with vanishing exponents outside $S$ has prime factorisation involving only primes of $S$ and a sign, and conversely a monomial $\pm p_1^{n_1}\cdots p_m^{n_m}$ has all exponents outside $S$ equal to zero; the sign and the exponents in such an expression are unique by the uniqueness clause of [F3].

The assignment $(\epsilon ,n_1,\dots ,n_m)\mapsto \epsilon p_1^{n_1}\cdots p_m^{n_m}$ is an isomorphism $\{\pm 1\}\times {\mathbb{Z}}^m\to R^{\times }$: it is a group homomorphism because the exponents add, it is surjective by step 4.1, and it is injective because $\epsilon p_1^{n_1}\cdots p_m^{n_m}=1$ forces $\epsilon =1$ and $n_1=\cdots =n_m=0$ by the uniqueness of the factorisation of the positive integer obtained after moving negative exponents to the other side.

By the $S$-unit theorem [F7] applied to $K=\mathbb{Q}$ and $S_{\mathbb{Q}}$, ${\mathcal{O}}_{\mathbb{Q},S_{\mathbb{Q}}}^{\times }\cong \mu (\mathbb{Q})\times {\mathbb{Z}}^{r_1+r_2-1+\mathrm{\mid }{S}_{\mathbb{Q}}\mathrm{\mid }}=\{\pm 1\}\times {\mathbb{Z}}^{1+0-1+m}=\{\pm 1\}\times {\mathbb{Z}}^m,$ of rank $m$. Combined with step 2.3 this agrees with the explicit computation of steps 4.1 and 5.1, and the $m$ generators are the classes of $p_1,\dots ,p_m$.

Scope and boundary cases: for $m=0$ the set $M=\{1\}$ and $R=\mathbb{Z}$, and the computation returns ${\mathbb{Z}}^{\times }=\{\pm 1\}\cong \{\pm 1\}\times {\mathbb{Z}}^0$ by [F4], while $S_{\mathbb{Q}}=\mathrm{\varnothing }$ has rank $0$; for $m\ge 1$ the exponents $n_i$ range over $\mathbb{Z}$ and negative exponents are allowed, the generators being units because $p_i\mathrm{/}1\in M$ has inverse $1\mathrm{/}{p}_i$ in $R$. Choice enters through the $S$-unit theorem [F7] and through the AC-qualified Dedekind interface of [F5]; the identification of $R$ with a subring of $\mathbb{Q}$, the exponent bookkeeping and the enumeration of signs use no choice.