S-integers and S-units of a number field

Let $K$ be a number field (Number field), and let $S$ be a finite set of nonzero prime ideals of ${\mathcal{O}}_K$ (Ring of integers); only finite primes belong to $S$. Write $v_{\mathfrak{p}}$ for the prime-ideal valuation of a nonzero fractional ideal (Prime-ideal valuations on fractional ideals) and, for $x\in K^{\times }$, write $v_{\mathfrak{p}}(x):=v_{\mathfrak{p}}((x))$ for the principal fractional ideal $(x)$ (Fractional ideals).

The ring of $S$-integers of $K$ is

${\mathcal{O}}_{K,S}=\{0\}\cup \{x\in K^{\times }:v_{\mathfrak{p}}(x)\ge 0\text{ for every nonzero prime }\mathfrak{p}\mathrm{\notin }S\},$

and the group of $S$-units is

${\mathcal{O}}_{K,S}^{\times }=\{x\in K^{\times }:v_{\mathfrak{p}}(x)=0\text{ for every nonzero prime }\mathfrak{p}\mathrm{\notin }S\}$

with the group structure inherited from $K^{\times }$ (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times }$ is a group under multiplication; $0\in R^{\times }$ only in the zero ring).

No infinite place belongs to $S$. Only nonzero prime ideals, that is finite places, are admitted into $S$; a real or complex place is never a member. The archimedean places are already carried by the logarithmic embedding, so admitting them into $S$ would count them twice. Consequently the rank formula of the $S$-unit theorem is $r_1+r_2-1+\mathrm{\mid }S\mathrm{\mid }$, with $\mathrm{\mid }S\mathrm{\mid }$ the number of finite primes chosen.

The description via the fractional ideal is the one used below. ${\mathcal{O}}_{K,S}$ consists of $0$ and the nonzero elements of $K$ whose principal fractional ideal involves no prime outside $S$ in a denominator, and an element $x$ of ${\mathcal{O}}_{K,S}^{\times }$ is precisely an element of $K^{\times }$ whose principal fractional ideal involves no prime outside $S$ at all, in numerator or denominator. This intrinsic description, and not a localisation, is the one the S-unit theorem consumes; it also makes visible why $S$ is a set of finite primes only.

Put $R={\mathcal{O}}_K$. By Clearing denominators for an algebraic number, every $x\in K$ has $mx\in R$ for a positive integer $m$, so $K=\mathrm{Frac}(R)$. For $x\mathrm{\ne }0$, $(x)=xR$ is consequently a nonzero fractional ideal: it is an $R$-submodule of $K$ and $m(x)\subseteq R$.

The Dedekind property can also be established without Choice. By The ring of integers has rank the degree, $R\cong {\mathbb{Z}}^n$ additively, where $n=[K:\mathbb{Q}]\ge 1$. Every subgroup of ${\mathbb{Z}}^n$ has a finite integer basis (the finite induction in that theorem, steps 1.2, 2.3 and 3.2). Thus every ideal of $R$ is finitely generated over $R$, so $R$ is Noetherian. It is an integrally closed domain by The integral closure of a domain in a field extension is integrally closed. For every nonzero prime $\mathfrak{p}$, $R\mathrm{/}\mathfrak{p}$ is a finite domain by A nonzero number-field ideal has finite quotient, hence a field: multiplication by any nonzero element is injective on this finite set and therefore surjective. Thus every nonzero prime is maximal. Moreover $R\mathrm{/}2R$ has $2^n>1$ elements; among its proper ideals one of largest cardinality is maximal, and its inverse image is a nonzero prime of $R$. Together with the prime $(0)$ this proves $\dim R=1$, so $R$ is Dedekind (Dedekind domains).

To justify the local valuation without the Choice-qualified general invertibility theorem, use the local calculation in Integral ideal factorisation in a number field, in ZF, steps 2.1--6.1, with the nonzero integral ideal $\mathfrak{a}=\mathfrak{p}$. It proves that $R_{\mathfrak{p}}$ has maximal ideal $(\pi )$ and each nonzero element is uniquely $u{\pi }^k$, with $u$ a unit and $k\ge 0$. Its fraction field is $K$, so each $x\in K^{\times }$ is uniquely $u{\pi }^j$ with $j\in \mathbb{Z}$. Hence $(x{)}_{\mathfrak{p}}={\pi }^j{R}_{\mathfrak{p}}$, exactly the valuation used in the Definition. Changing $\pi$ by a unit does not change $j$. Multiplication adds these exponents, inversion negates them, and $v_{\mathfrak{p}}(x)\ge 0$ is equivalent to $x\in R_{\mathfrak{p}}$. Consequently ${\mathcal{O}}_{K,S}$ is the intersection of these local subrings over $\mathfrak{p}\mathrm{\notin }S$, and $x$ is a unit of that intersection precisely when both $x$ and $x^{-1}$ belong to it, equivalently all those exponents vanish. This verifies the ring and group assertions without selecting uniformizers simultaneously; the construction uses no Choice.