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Cyclotomic units include roots of unity and suitable ratios $(1-{\zeta }^a)/(1-\zeta )$ in cyclotomic fields. Compatible roots $\zeta$ make these ratios norm-compatible along cyclotomic towers. Their closed norm-limit module is a concrete submodule of global units, and its local images generate the power series in the Iwasawa main conjecture.

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.

Write $K=\mathbb{Q}({\mu }_m)$, $L=\mathbb{Q}({\mu }_n)$. For coprime $m,n$, Bézout's identity and powers of a primitive $mn$-th root give $KL=\mathbb{Q}({\mu }_{mn})$. The degrees of these cyclotomic fields are $\phi (m)$, $\phi (n)$ and $\phi (mn)=\phi (m)\phi (n)$. The natural restriction mapis injective and, by these orders, bijective. Under this product description the subgroups fixing $K$ and $L$ are the two complementary factors. Together they generate the whole group, so their common fixed field is $\mathbb{Q}$. By the Fundamental theorem of Galois theory, $K\cap L=\mathbb{Q}$. This also includes $m=1$ or $n=1$.

Let $\alpha =\sqrt[5]{3}$ and let $\zeta$ be a primitive fifth root of unity. Then $L=\mathbb{Q}(\alpha ,\zeta )$. The field degrees $[\mathbb{Q}(\alpha ):\mathbb{Q}]=5$ and $[\mathbb{Q}(\zeta ):\mathbb{Q}]=4$ both divide $[L:\mathbb{Q}]$, while adjoining the two generators gives $[L:\mathbb{Q}]\le 20$. Consequently $[L:\mathbb{Q}]=20$.

The cyclotomic field $\mathbb{Q}(\zeta )/\mathbb{Q}$ is Galois with cyclic group of order $4$. The Fundamental theorem of Galois theory gives a normal subgroupThus $N\cong C_5$ and the chain $\boxed{G▹N▹\{1\}}$ has abelian successive quotients, proving that $G$ is a solvable group. One may refine it by inserting the inverse image of the order-two subgroup of $C_4$, giving cyclic factors $C_2,C_2,C_5$.

A cyclotomic field over $\mathbb{Q}$ is a field $\mathbb{Q}({\zeta }_m)$ generated by an $m$th root of unity of exact order $m$. It contains all roots of $X^m-1$, and is its splitting field. In characteristic zero this polynomial is separable, so the extension is Galois.

Every automorphism sends ${\zeta }_m$ to ${\zeta }_m^a$ with $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$, and is determined by that image. Composition multiplies the exponents modulo $m$, yielding an injective homomorphism into an abelian group. Thus the Galois group is abelian. Using irreducibility of the cyclotomic polynomial, every unit exponent occurs andIf “cyclotomic” is used more broadly for a subfield of such a field, it is still Galois and abelian, because every subgroup of an abelian Galois group is normal.

We use the following standard facts about cyclotomic fields. For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ isAlso $\mathbb{Q}({\zeta }_{{\ell }^r})$ has degree $\phi ({\ell }^r)={\ell }^{r-1}(\ell -1)$, and $\ell$ is totally ramified, with a unique prime and residue field ${\mathbb{F}}_{\ell }$. These are the quadratic subfield of a prime cyclotomic field and total ramification in a prime-power cyclotomic field results.

TakeSince $59\equiv 3(\mathrm{mod}4)$, every $E_r$ contains $K=\mathbb{Q}(\sqrt{-59})$, through its subfield $E_1$. The unique prime above $59$ is totally ramified in $E_r/K$: the absolute ramification index is $58\cdot 59^{r-1}$, the index in $K/\mathbb{Q}$ is $2$, and the relative index is thereforeThe field $K$ has no real places. By part (b), $h_K\mid h_{E_r}$, and part (a) showed $3\mid h_K$. ThusThe ring of integers of a number field is a Dedekind domain, and is a principal ideal domain exactly when its ideal class group is trivial. Hence none of these rings of integers is principal. Finally the degrees $58\cdot 59^{r-1}$ strictly increase, so the fields are pairwise distinct. The explicit infinite family of nonprincipal cyclotomic fields of conductor a prime power is

The multiplicative group of $L={\mathbb{F}}_{q^N}$ is cyclic of order $m=q^N-1$. For a primitive element $\zeta$, $L={\mathbb{F}}_q(\zeta )$, so this is a cyclotomic extension. The cyclotomic character isIts image is the cyclic subgroup $⟨q⟩$, of order $N$.

By Dirichlet theorem on primes in arithmetic progressions, choose a prime $\ell$ withThe cyclotomic field $E=\mathbb{Q}({\zeta }_{\ell })$ is Galois over $\mathbb{Q}$, andwhere the last group is cyclic because the multiplicative group of a finite field is cyclic.

The cyclic group $C_{\ell -1}$ has a unique subgroup $H$ of order $(\ell -1)/r$. By the Galois correspondence, its fixed field $L=E^H$ satisfiesNormality follows because $H$ lies in an abelian group. Thus every $r>1$ occurs as the Galois group of a finite extension of $\mathbb{Q}$. This is the cyclic Galois extension of the rational numbers of every finite degree.

Assume that $\zeta$ is a primitive fifth root of unity. The cyclotomic field $K=\mathbb{Q}(\zeta )$ has degree four and signature $(r_1,r_2)=(0,2)$. The Dirichlet unit theorem therefore gives free rankThe roots of unity in a rational cyclotomic field are the ten elements $\pm {\zeta }^j$, so

For explicit units, $-\zeta$ has order ten, whileHence $1+\zeta$ is a unit. It has infinite order because in the standard complex embeddingwhereas every root of unity has modulus one. Its image in the free $\mathbb{Z}$ factor is therefore nonzero and generates a finite-index subgroup of that factor. Consequentlyhas finite index in ${\mathcal{O}}_K^{\times }$, as described by the unit group of the fifth cyclotomic field.

The primitive root ${\zeta }_n$ is a zero of the cyclotomic polynomial ${\Phi }_n$, of degree $\phi (n)$. Its minimal polynomial over $K$ divides ${\Phi }_n$, sowhere divisibility also follows from the embedding of the Galois group of the cyclotomic field into $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$. Over $\mathbb{Q}$, Gauss's irreducibility theorem for cyclotomic polynomials says that ${\Phi }_n$ is irreducible. Therefore

Fix a primitive $n$th root of unity ${\zeta }_n$. The $n$th cyclotomic polynomial isIts roots are exactly the roots of unity of order $n$.

Partitioning all $n$th roots of unity by exact order gives the cyclotomic factorizationWe prove ${\Phi }_n\in \mathbb{Z}[X]$ by induction on $n$. The assertion is clear for $n=1$. If it holds for every proper divisor of $n$, thenThe denominator is monic and belongs to $\mathbb{Z}[X]$. Exact polynomial division shows that the quotient lies in $\mathbb{Q}[X]$, and Gauss's lemma makes this monic quotient an element of $\mathbb{Z}[X]$.

Now let $K\subseteq \mathbb{C}$. The minimal polynomial of ${\zeta }_n$ over $K$ divides ${\Phi }_n$, so every $K$-conjugate of ${\zeta }_n$ is another primitive root ${\zeta }_n^a$ and lies in $K({\zeta }_n)$. The extension is therefore normal. It is separable because the characteristic is zero, and hence it is Galois.

Every $\sigma \in \mathrm{Gal}(K({\zeta }_n)/K)$ hasfor some $a\in (\mathbb{Z}/n\mathbb{Z}{)}^{\times }$. Since ${\zeta }_n$ generates the extension, this gives an injective homomorphismThe group on the right is abelian, so the Galois group is abelian, as summarized by the cyclotomic field construction.

A number field is totally real when every embedding into the complex numbers has image in the real numbers. The maximal real subfield of a cyclotomic field is a basic example. Ordinary Hilbert class fields split at real places and therefore retain this property.