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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.

Let $\alpha =p^{1/p}$ and let $\zeta$ be a primitive $p$th root of unity. The roots are $\alpha {\zeta }^j$, so $L=\mathbb{Q}(\alpha ,\zeta )$. Eisenstein gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the prime cyclotomic extension has degree $[\mathbb{Q}(\zeta ):\mathbb{Q}]=p-1$. Both degrees divide $[L:\mathbb{Q}]$, and they are coprime. Conversely adjoining $\alpha$ to the cyclotomic field has degree at most $p$. These two bounds giveRestriction to $E=\mathbb{Q}(\zeta )$ gives a normal subgroup $N=\mathrm{Gal}(L/E)$ of order $p$, hence cyclic, with quotient $\mathrm{Gal}(E/\mathbb{Q})\cong C_{p-1}$. Restriction is onto because embeddings of the intermediate field extend in the finite Galois splitting field. Thus $1⊴N⊴G$ has abelian factors, proving $\boxed{G\text{ is solvable}}$. For $p=2$ the cyclotomic field is just $\mathbb{Q}$, and the same argument gives degree two and a cyclic group.

Write $g(x)=x^6+x^3+1$. The second polynomial is $g(-x)$. Negating the roots of a polynomial does not change the field generated by them: each original root is minus a new root, and conversely. Hence this splitting field is again $\mathbb{Q}({\zeta }_9)$ and its field extension degree is six. Equivalently, it is the cyclotomic field $\mathbb{Q}({\zeta }_{18})=\mathbb{Q}({\zeta }_9)$, since ${\zeta }_{18}=-{\zeta }_9^5$.

For a positive integer $m$, the finite ray class group of $\mathbb{Q}$ is $(\mathbb{Z}/m\mathbb{Z}{)}^{\times }/⟨-1⟩$: the unit group permits either sign of a rational generator. Its ray class field is the maximal real subfield of the cyclotomic field $\mathbb{Q}({\zeta }_m)$. Including the real place requires positive principal generators and gives the full group $(\mathbb{Z}/m\mathbb{Z}{)}^{\times }$ and full cyclotomic field. Thus modulus $(5)$ gives $\mathbb{Q}(\sqrt{5})$, while $(5)\infty$ gives $\mathbb{Q}({\zeta }_5)$. The Artin reciprocity map sends a prime $\ell \nmid m$ to the exponent $\ell$, taken modulo $\{\pm 1\}$ in the finite case. Ramification at real places is possible only when those places are included in the modulus of a number field.

A regular heptagon is a regular polygon with seven sides. Its unit-circle vertices have complex coordinates $e^{2\pi ik/7}$ for $k=0,\dots ,6$. It cannot be constructed from a unit segment by a straightedge and compass construction: the first coordinate $\cos (2\pi /7)$ has degree three over $\mathbb{Q}$, since the seventh cyclotomic field has degree six and degree two over its real subfield. Every finite constructible coordinate extension has power-of-two degree, which cannot contain a degree-three subextension by the tower law.

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.