Unit ranks by signature

Given: The Axiom of Choice and a number field $K$ of signature $(r_1,r_2)$ (Number field, Archimedean embeddings and signature).

The signature satisfies $r_1+2r_2=[K:\mathbb{Q}]$, with $r_1$ the number of real embeddings and $r_2$ the number of complex conjugate pairs (Archimedean embeddings and signature).

${\mathcal{O}}_K^{\times }\cong \mu (K)\times {\mathbb{Z}}^{r_1+r_2-1}$; the rank of ${\mathcal{O}}_K^{\times }$ is $r_1+r_2-1$, and when $r_1+r_2-1=0$ one has ${\mathcal{O}}_K^{\times }=\mu (K)$ (Dirichlet unit theorem).

For a finite field extension $K\mathrm{/}\mathbb{Q}$ one has $[K:\mathbb{Q}]=1$ if and only if $K=\mathbb{Q}$ (A finite extension has degree one if and only if the two fields are equal). Moreover ${\mathcal{O}}_{\mathbb{Q}}$, the integral closure of $\mathbb{Z}$ in $\mathbb{Q}$ (Ring of integers), equals $\mathbb{Z}$: a rational number integral over $\mathbb{Z}$ is an integer (The rational algebraic integers are exactly the integers), and every integer $n$ is a root of the monic polynomial $X-n$, so ${\mathcal{O}}_{\mathbb{Q}}=\mathbb{Z}$ with unit group ${\mathcal{O}}_{\mathbb{Q}}^{\times }={\mathbb{Z}}^{\times }=\{\pm 1\}$ ($(\mathbb{Z},\cdot ,1)$ is a commutative monoid whose group of units is $\{1,-1\}$; equivalently $u\mid 1$ holds exactly for $u=1$ and $u=-1$).

If $x\in \mathbb{R}$ and $x^n=1$ for some integer $n\ge 1$, then $x=\pm 1$: the inequalities $x>1$, $0<x<1$ are preserved by taking $n$-th powers, and $x<0$ with $x^n=1$ forces $n$ even and $(-x{)}^n=1$, so $-x=1$ (Monotonicity of $x\mapsto x^n$ and of $n\mapsto a^n$, Sign rules for products and monotonicity of multiplication, The group ${\mu }_n(K)$ of $n$-th roots of unity in a field, and primitive $n$-th roots of unity).

For a nonzero squarefree integer $d\mathrm{\ne }1$, $d$ is not a rational square by unique prime factorisation (The fundamental theorem of arithmetic: every integer $n\ge 1$ is a product of primes, and the factorisation is unique up to order — if ${\prod }_{i<r}{p}_i={\prod }_{j<s}{q}_j$ with every $p_i$ and $q_j$ prime, then $r=s$ and $q_i=p_{\pi (i)}$ for some $\pi \in \mathrm{Sym}(r)$), so $X^2-d$ is irreducible over $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{d})$ has degree $2$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element, An element is algebraic over $F$ if and only if its simple extension $F(a)\mathrm{/}F$ is finite). Its embeddings correspond exactly to the two roots $\pm \sqrt{d}$ ($F$-embeddings of $F(\alpha )$ into an algebraically closed field correspond to the distinct roots of $m_{\alpha }$): if $d>0$ both embeddings are real, and if $d<0$ neither is real and they are complex conjugates. The signatures are therefore $(2,0)$ and $(0,1)$ respectively (Archimedean embeddings and signature).

In a finite tower, the intermediate degree divides the total degree (The degree of an intermediate field divides the degree of a finite extension). For an algebraic element $\alpha$, $[\mathbb{Q}(\alpha ):\mathbb{Q}]$ is the degree of its monic irreducible minimal polynomial (An element is algebraic over $F$ if and only if its simple extension $F(a)\mathrm{/}F$ is finite, The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

Every positive integer has a unique prime factorisation (The fundamental theorem of arithmetic: every integer $n\ge 1$ is a product of primes, and the factorisation is unique up to order — if ${\prod }_{i<r}{p}_i={\prod }_{j<s}{q}_j$ with every $p_i$ and $q_j$ prime, then $r=s$ and $q_i=p_{\pi (i)}$ for some $\pi \in \mathrm{Sym}(r)$); consequently every nonzero rational $\mathrm{\Delta }$ can be written as $q^{2}d$ with $q\in {\mathbb{Q}}^{\times }$ and $d$ a nonzero squarefree integer, by writing each prime exponent of $\mathrm{\Delta }$ as $2k+e$ with $e\in \{0,1\}$ and retaining the sign in $d$.

The Axiom of Choice is assumed; it is used only through the AC-qualified unit theorem [F2] (The Axiom of Choice).