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A CM-field is a totally imaginary number field that is a quadratic extension of a totally real number field. Its nontrivial automorphism over that real subfield acts as complex conjugation in every complex embedding. Imaginary quadratic number fields are examples. An arbitrary totally imaginary number field need not have such a real subfield of index two.

The hypotheses make $K$ a CM-field, with nontrivial automorphism $c$ over the totally real number field $k$. In every complex embedding of $K$, this automorphism acts as complex conjugation. Write $E_K={\mathcal{O}}_K^{\times }$, $E_k={\mathcal{O}}_k^{\times }$ and ${\mu }_K$ for the finite cyclic group of roots of unity in $K$.

For any $u\in E_K$, the element $u/c(u)$ is an algebraic integer, indeed a unit, and every conjugate has absolute value one. The Kronecker theorem on algebraic integers in the unit disk makes it a root of unity. For clarity, the essential proof is finite: the conjugates of all positive powers have modulus one, so the integer coefficients of their monic minimal polynomials are bounded by fixed binomial coefficients. There are only finitely many such polynomials and roots; two powers coincide, forcing a root of unity.

Thusis a homomorphism. Its kernel is exactly $E_k$, since a conjugation-fixed element lies in $k$ and both it and its inverse are integral there. For $\zeta \in {\mu }_K$, $c(\zeta )={\zeta }^{-1}$ and $\varphi (\zeta )={\zeta }^2$. Consequently $\varphi$ induces an injectionTo check injectivity explicitly, if $\varphi (u)={\zeta }^2$, then $\varphi (u/\zeta )=1$, so $u/\zeta \in E_k$. The cyclic group ${\mu }_K$ has even order because it contains $-1$, and its quotient by squares has order two. Therefore the unit index of a CM-field satisfies

Put $C={\bigcup }_{r\ge 1}\mathbb{Q}({\mu }_{p^r})$, where ${\mu }_{p^r}$ is the group of $p^r$th roots of unity. The compatible Galois groups of the cyclotomic fields giveFor odd $p$, the Teichmüller character givesThe p-adic logarithm identifies $1+p{\mathbb{Z}}_p$ with $p{\mathbb{Z}}_p$, which is topologically isomorphic to ${\mathbb{Z}}_p$. More explicitly, $a\mapsto (1+p{)}^a$ is an isomorphism from the additive p-adic integers onto $1+p{\mathbb{Z}}_p$: the logarithm of $1+p$ has valuation one and generates $p{\mathbb{Z}}_p$. Take the fixed field of ${\mu }_{p-1}$ in $C$.

For $p=2$, use insteadThe p-adic logarithm identifies the second factor with $4{\mathbb{Z}}_2$, and $a\mapsto 5^a$ identifies it with ${\mathbb{Z}}_2$. The fixed field of $\{\pm 1\}$ in $C$ is the union of the maximal real subfields of the cyclotomic fields of $2$-power conductor. Thus in both cases the cyclotomic Zp-extension exists:where ${\Delta }_p={\mu }_{p-1}$ for odd $p$ and ${\Delta }_2=\{\pm 1\}$.

The closed subgroup $p^n{\mathbb{Z}}_p$ is the unique subgroup of index $p^n$, so the Galois correspondence supplies the unique degree-$p^n$ field $\mathbb{Q}[p^n]$. In the odd case it is the ${\Delta }_p$-fixed field in $\mathbb{Q}({\mu }_{p^{n+1}})$; in the even case it is the real subfield of $\mathbb{Q}({\mu }_{2^{n+2}})$. These finite layers are all totally real number fields.

Now set $A=\mathbb{Q}[p^n]$, $B=\mathbb{Q}[p^m]$, and let $H_A$ be the ordinary Hilbert class field of $A$. Thus $H_A/A$ is abelian, unramified at finite primes, and split at every real place, with degree $h(A)$. If $m=n$ there is nothing to prove. Otherwise the unique prime over $p$ is totally ramified in $B/A$, by the permitted total-ramification fact and multiplicativity of the ramification index. Any nontrivial intermediate field of $B/A$ is ramified at that prime. HenceIt follows that the two extensions are linearly disjoint field extensions, andUnramified extensions remain unramified after base change. The real places also remain split: $H_A$ and $B$ are totally real, so their compositum is totally real. Therefore $B{H}_A/B$ lies inside the ordinary Hilbert class field $H_B/B$. Its degree divides $[H_B:B]=h(B)$, provingThis is class number divisibility with split real places; using a class-field statement restricted to imaginary fields would unnecessarily omit the present real layers.

There is no infinite component in the modulus $(5)$. The ordinary ideal class group of $\mathbb{Q}$ is trivial, and its unit group is $\{\pm 1\}$. The ray class exact sequence therefore givesConcretely, a fractional ideal prime to $5$ has a rational generator $r=a/b$ with $5\nmid ab$. Reducing $r$ modulo $5$ is well-defined up to multiplication by the units $\pm 1$. Its ray class is trivial precisely when one of the two generators $r,-r$ is congruent to one. Thus the quotient has the two classes $\{1,4\}$ and $\{2,3\}$. No positivity restriction is imposed on a generator.

To identify the corresponding field, put $E=\mathbb{Q}({\zeta }_5+{\zeta }_5^{-1})=\mathbb{Q}(\sqrt{5})$. The maximal real subfield of the fifth cyclotomic field is fixed by complex conjugation, the element $-1$ of $(\mathbb{Z}/5\mathbb{Z}{)}^{\times }$. For a prime $\ell \ne 5$, the Frobenius automorphism of $\mathbb{Q}({\zeta }_5)/\mathbb{Q}$ is ${\zeta }_5\mapsto {\zeta }_5^{\ell }$. Its restriction to $E$ thus corresponds exactly to the residue class of $\ell$ modulo $\{\pm 1\}$. Multiplicativity extends this to the Artin reciprocity map on all fractional ideals prime to $5$, whose kernel is exactly $P_{\mathbb{Q},1}((5))$. Its field discriminant $5$ also confirms that $E$ is unramified outside $5$; moreover $E$ is a totally real number field. The uniqueness in the Main theorem of class field theory now gives

Including the real place changes the ray class group: a principal generator must then be positive as well as congruent to one. The residue and signature group of a modulus givesHere the isomorphism sends $(u,\epsilon )$ to $\epsilon u$. Consequently ${\mathbb{Q}}_{(5)\infty }=\mathbb{Q}({\zeta }_5)$, of degree four. The finite modulus gives the real quadratic field; the modulus with the real place gives the full cyclotomic field. The printed finite-modulus assertion needs no correction.

A quadratic number field is a number field of degree of a field extension two, necessarily of the form $\mathbb{Q}(\sqrt{d})$ for a square-free integer $d\ne 0,1$. Its two field embeddings send $a+b\sqrt{d}$ to $a\pm b\sqrt{d}$. If $d>0$, both images lie in the real numbers, giving a totally real number field; if $d<0$, they form a pair related by complex conjugation, giving an imaginary quadratic field. The ring of integers of a quadratic field depends on $d$ modulo four.^[ref]