Cyclotomic and Abelian Extensions

The cyclotomic field$\mathbb{Q}({\zeta }_n)$, generated by a primitive $n$th root of unity, is a Galois extension of $\mathbb{Q}$ of degree $\phi (n)$. Its Galois group is abelian, and in fact a group we have already classified: the units modulo $n$. This identification turns the subfields of $\mathbb{Q}({\zeta }_n)$ into a concrete catalogue and, run in reverse, realizes every finite abelian group as a Galois group over $\mathbb{Q}$. The proof uses two facts about Galois extensions: their behavior under composites, and the criterion for an extension to have a single generator.

Composites of Galois extensions

Two facts control how the Galois property and the Galois group transport across a composite.

Translating a Galois extension over a new base field keeps it Galois and can only shrink its group, to the part that acts nontrivially on the piece of $K$ not already in $F^{\prime }$. When both factors are Galois, the composite is Galois and its group is a fiber product.

The converse of the direct-product case is useful: if $K/F$ is Galois with $\mathrm{Gal}(K/F)=G_1\times G_2$, then $K$ is the composite of the two Galois extensions fixed by $G_2$ and $G_1$, meeting in $F$. Finally, every finite separable extension sits inside a smallest Galois extension.

The primitive element theorem

An extension is simple if $K=F(\theta )$ for a single primitive element $\theta$. Whether an extension is simple is decided by counting intermediate fields.

If $K=F(\theta )$, each intermediate field $E$ is generated over $F$ by the coefficients of the minimal polynomial of $\theta$ over $E$, which is a factor of the minimal polynomial over $F$ — and there are finitely many such factors. Conversely, over an infinite field, if $F(\alpha ,\beta )$ had infinitely many candidate generators $\alpha +c\beta$ but only finitely many subfields, two values of $c$ would give the same field, forcing $\beta$ and then $\alpha$ into it.¹ For a Galois extension the theorem is immediate from the Fundamental Theorem: only finitely many subgroups exist, hence only finitely many subfields.

For a Galois extension a primitive element is any combination not fixed by a nontrivial automorphism, since such an element cannot lie in a proper subfield. For example, $\sqrt{2}+\sqrt{3}$ generates $\mathbb{Q}(\sqrt{2},\sqrt{3})$, as no nonidentity element of its Klein four Galois group fixes it.

The Galois group of a cyclotomic field

An automorphism of $\mathbb{Q}({\zeta }_n)$ is determined by where it sends the primitive root ${\zeta }_n$, and it must send ${\zeta }_n$ to another primitive $n$th root of unity — a root of the cyclotomic polynomial${\Phi }_n$, hence ${\zeta }_n^a$ for some $a$ coprime to $n$. Each such $a$ gives an automorphism ${\sigma }_a:{\zeta }_n\mapsto {\zeta }_n^a$, and ${\sigma }_a$ depends only on $a$ modulo $n$.

It is a homomorphism because ${\sigma }_a{\sigma }_b({\zeta }_n)={\sigma }_a({\zeta }_n^b)={\zeta }_n^{ab}={\sigma }_{ab}({\zeta }_n)$, and a bijection because both sides have order $\phi (n)$ and every automorphism is some ${\sigma }_a$.² The $\phi (n)$ primitive roots of unity are the conjugates of ${\zeta }_n$, permuted among themselves by the group.

Subfields from subgroups

Since the group is cyclic when $n=p$ is prime — $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }\cong \mathbb{Z}/(p-1)\mathbb{Z}$ — its subgroups, and therefore the subfields of $\mathbb{Q}({\zeta }_p)$, correspond to the divisors of $p-1$.

More generally, $\mathbb{Q}({\zeta }_p)$ contains $\mathbb{Q}(\sqrt{p^{\ast }})$, where $p^{\ast }=p$ if $p\equiv 1(\mathrm{mod}4)$ and $p^{\ast }=-p$ if $p\equiv 3(\mathrm{mod}4)$.

Worked example: the periods of $\mathbb{Q}({\zeta }_{13})$

For $p=13$ the group $(\mathbb{Z}/13\mathbb{Z}{)}^{\times }\cong \mathbb{Z}/12\mathbb{Z}$ is generated by ${\sigma }_2$ ($2$ is a primitive root mod $13$). Write $\zeta ={\zeta }_{13}$ and $\sigma ={\sigma }_2$. For each subgroup $H\le \mathrm{Gal}(\mathbb{Q}(\zeta )/\mathbb{Q})$, a primitive element of the fixed field is the period

the sum of the $H$-conjugates of $\zeta$. Any $\tau \in H$ permutes the summands, so ${\alpha }_H\in K^H$; and because the primitive $13$th roots of unity $\zeta ,{\zeta }^2,\dots ,{\zeta }^{12}$ form a basis of $\mathbb{Q}(\zeta )$ over $\mathbb{Q}$, no automorphism outside $H$ fixes ${\alpha }_H$, so $\mathbb{Q}({\alpha }_H)$ is the fixed field itself.³ The nontrivial subgroups of $\mathbb{Z}/12\mathbb{Z}$ have orders $2,3,4,6$, generated by ${\sigma }^6,{\sigma }^4,{\sigma }^3,{\sigma }^2$; since ${\sigma }^k(\zeta )={\zeta }^{2^k}$, the periods are explicit sums of powers.

Each row is one intermediate field, and the degrees multiply down the divisor chain: the degree-$2$ field is $\mathbb{Q}(\sqrt{13})$ (as $13\equiv 1(\mathrm{mod}4)$), the degree-$6$ field is the maximal real subfield $\mathbb{Q}(\zeta +{\zeta }^{-1})=\mathbb{Q}\left(2\cos \frac{2\pi }{13}\right)$, and the whole lattice of five subfields mirrors the divisor lattice of $12$ upside down. The period construction gives explicit generators for the subfields of any prime cyclotomic field.

For a composite modulus the group factors by the Chinese Remainder Theorem. If $n=p_1^{a_1}\cdots p_k^{a_k}$, then the cyclotomic fields $\mathbb{Q}({\zeta }_{p_i^{a_i}})$ intersect only in $\mathbb{Q}$ and their composite is $\mathbb{Q}({\zeta }_n)$, so

which reproduces the CRT decomposition of the unit group.⁴

Every abelian group is a Galois group over $\mathbb{Q}$

The subfields of cyclotomic fields realize a large supply of abelian Galois groups — in fact all of them.

The construction uses Dirichlet's theorem that each arithmetic progression $1,1+m,1+2m,\dots$ contains infinitely many primes. Write $G\cong \mathbb{Z}/n_1\times \cdots \times \mathbb{Z}/n_k$ by the fundamental theorem for abelian groups, choose distinct primes $p_i\equiv 1(\mathrm{mod}{n}_i)$, and set $n=p_1\cdots p_k$. Then $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }\cong {\prod }_i\mathbb{Z}/(p_i-1)\mathbb{Z}$ has a quotient isomorphic to $G$ (each factor $\mathbb{Z}/(p_i-1)$ surjects onto $\mathbb{Z}/n_i$), and the fixed field of the corresponding subgroup is Galois over $\mathbb{Q}$ with group $G$.⁵ The converse is a landmark theorem, stated without proof.

Together these say the abelian extensions of $\mathbb{Q}$ are the subfields of cyclotomic fields and nothing else — a complete and explicit description. The analogous problem over a general number field is the subject of class field theory, where roots of unity are replaced by values of more elaborate transcendental functions and the description is far less explicit.

Gauss's constructible polygons

The regular $n$-gon is constructible by straightedge and compass if and only if ${\zeta }_n$ is, and a point is constructible exactly when it lies in a tower of quadratic extensions. Since $[\mathbb{Q}({\zeta }_n):\mathbb{Q}]=\phi (n)$ and its Galois group is abelian, a quadratic tower down to $\mathbb{Q}$ exists precisely when $\phi (n)$ is a power of $2$.

The known Fermat primes are $3,5,17,257,65537$; the constructibility of the $17$-gon was Gauss's discovery at nineteen, and it is the abelian structure of $\mathrm{Gal}(\mathbb{Q}({\zeta }_{17})/\mathbb{Q})\cong \mathbb{Z}/16\mathbb{Z}$ that supplies the chain of quadratic steps.