---
title: Cyclotomic and Abelian Extensions
module: Galois Theory
moduleNumber: 11
lessonNumber: 3
order: 1103
summary: >
  The Galois group of the $n$th cyclotomic field over $\mathbb{Q}$ is the unit
  group $(\mathbb{Z}/n\mathbb{Z})^\times$, which makes cyclotomic fields the
  worked catalogue of abelian extensions of $\mathbb{Q}$. The isomorphism
  identifies subfields with subgroups, realizes every finite abelian group as a
  Galois group over $\mathbb{Q}$, and leads to Kronecker–Weber. Composites of
  Galois extensions and the primitive element theorem supply the machinery.
topics: [Galois Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 14 — Galois Theory; §14.4 Composite and Simple Extensions"
  - book: Dummit & Foote
    ref: "§14.5 Cyclotomic Extensions and Abelian Extensions over Q"
draft: false
---

The [cyclotomic field](/abstract-algebra/field-theory/separable-and-cyclotomic-extensions)
$\mathbb{Q}(\zeta_n)$, generated by a primitive $n$th root of unity, is a Galois extension
of $\mathbb{Q}$ of degree $\varphi(n)$. 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.

> **Proposition (Translation).** If $K/F$ is Galois and $F'/F$ is any extension, then the
> composite $KF'/F'$ is Galois, and restriction to $K$ gives an isomorphism
> $\operatorname{Gal}(KF'/F') \cong \operatorname{Gal}(K/K \cap F')$, a subgroup of
> $\operatorname{Gal}(K/F)$.

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 $F'$.
When both factors are Galois, the composite is Galois and its group is a **fiber product**.

> **Proposition (Composite of Galois extensions).** If $K_1/F$ and $K_2/F$ are Galois,
> then $K_1 \cap K_2$ and $K_1 K_2$ are Galois over $F$, and
> $$
> \operatorname{Gal}(K_1 K_2 / F) \cong \{(\sigma, \tau) : \sigma|_{K_1 \cap K_2} =
> \tau|_{K_1 \cap K_2}\} \leq \operatorname{Gal}(K_1/F) \times \operatorname{Gal}(K_2/F),
> $$
> the pairs of automorphisms that agree on the intersection. When $K_1 \cap K_2 = F$, this
> is the full direct product $\operatorname{Gal}(K_1/F) \times \operatorname{Gal}(K_2/F)$.

$$
% caption: The composite of two Galois extensions sits atop the diamond; the center node
% $M = K_1 \cap K_2$ is the intersection. The Galois group is the fiber product of the two
% groups over their action on $M$, and the full direct product when $M = F$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node (top) at (0,3) {$K_1 K_2$};
\node (k1) at (-2,1.5) {$K_1$};
\node (k2) at (2,1.5) {$K_2$};
\node (cap) at (0,1.5) {$M$};
\node (F) at (0,0) {$F$};
\draw[black] (k1) -- (top);
\draw[black] (k2) -- (top);
\draw[black] (F) -- (k1);
\draw[black] (F) -- (k2);
\draw[acc, dashed] (F) -- (cap);
\draw[acc, dashed] (cap) -- (top);
\end{tikzpicture}
$$

The converse of the direct-product case is useful: if $K/F$ is Galois with
$\operatorname{Gal}(K/F) = G_1 \times G_2$, 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.

> **Definition (Galois closure).** For a finite separable extension $E/F$, the **Galois
> closure** is the smallest Galois extension of $F$ containing $E$, obtained as the
> composite of the splitting fields of the minimal polynomials of a basis of $E$.

## 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.

> **Proposition (Simple $\Leftrightarrow$ finitely many subfields).** A finite extension
> $K/F$ is simple if and only if there are only finitely many fields between $F$ and $K$.

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.[^df-simple] For a Galois extension the theorem
is immediate from the Fundamental Theorem: only finitely many subgroups exist, hence only
finitely many subfields.

> **Theorem (Primitive Element Theorem).** Every finite separable extension is simple. In
> particular, every finite extension of a field of characteristic $0$ is simple.

$$
% caption: The primitive element theorem collapses a two-generator separable extension
% $F(a,b)$ into a simple extension $F(t)$ generated by a single well-chosen combination
% $t = a + cb$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% left: two generators
\node (Fab) at (0,2.6) {$F(a)(b)$};
\node (Fa) at (-1.3,1.3) {$F(a)$};
\node (Fb) at (1.3,1.3) {$F(b)$};
\node (F1) at (0,0) {$F$};
\draw[black] (F1) -- (Fa) -- (Fab);
\draw[black] (F1) -- (Fb) -- (Fab);
% right: one generator
\node (Ft) at (6,2.6) {$F(t)$};
\node (F2) at (6,0) {$F$};
\draw[black, draw=acc] (F2) -- (Ft);
% equality link
\draw[<->, acc, thick] (1.1,2.6) -- (4.9,2.6);
\node[font=\footnotesize, anchor=south] at (3,2.65) {$t = a + cb$};
\end{tikzpicture}
$$

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](/abstract-algebra/field-theory/separable-and-cyclotomic-extensions)
$\Phi_n$, 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 $n$.

> **Theorem (Cyclotomic Galois group).** The map
> $$
> (\mathbb{Z}/n\mathbb{Z})^\times \longrightarrow \operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}),
> \qquad a \pmod n \longmapsto \sigma_a,
> $$
> is an isomorphism. In particular $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois of degree
> $\varphi(n)$ with abelian group $(\mathbb{Z}/n\mathbb{Z})^\times$.

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
$\varphi(n)$ and every automorphism is some $\sigma_a$.[^df-cyclo] The $\varphi(n)$
primitive roots of unity are the conjugates of $\zeta_n$, permuted among themselves by the
group.

$$
% caption: The Galois group of $\mathbb{Q}(\zeta_5)$ is $(\mathbb{Z}/5\mathbb{Z})^\times$
% acting on the primitive 5th roots of unity $z, z^2, z^3, z^4$; the automorphism
% $\sigma_a$ sends $z \mapsto z^a$, permuting them cyclically.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) circle (2);
\node[draw, circle, inner sep=1.3pt, fill=black!8] (r0) at (90:2)  {$1$};
\node[draw, circle, inner sep=1.3pt, fill=acc!12] (r1) at (18:2)  {$z$};
\node[draw, circle, inner sep=1.3pt, fill=acc!12] (r2) at (306:2) {$z^{2}$};
\node[draw, circle, inner sep=1.3pt, fill=acc!12] (r3) at (234:2) {$z^{3}$};
\node[draw, circle, inner sep=1.3pt, fill=acc!12] (r4) at (162:2) {$z^{4}$};
\draw[->, acc, thick] (r1) to[bend right=15] (r2);
\draw[->, acc, thick] (r2) to[bend right=15] (r4);
\node[font=\footnotesize, acc] at (0,0) {$C_4$ acts};
\end{tikzpicture}
$$

### 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 $p - 1$.

> **Worked example.** For $p = 5$, $(\mathbb{Z}/5\mathbb{Z})^\times \cong
> \mathbb{Z}/4\mathbb{Z}$, generated by $\sigma_2$. Its unique proper subgroup
> $\{1, \sigma_4\} = \{1, \sigma_{-1}\}$ fixes
> $$
> \alpha = \zeta_5 + \zeta_5^{-1} = 2\cos\tfrac{2\pi}{5}.
> $$
> From $1 + \zeta_5 + \zeta_5^2 + \zeta_5^3 + \zeta_5^4 = 0$, the two sums
> $\alpha = \zeta_5 + \zeta_5^4$ and $\beta = \zeta_5^2 + \zeta_5^3$ satisfy $\alpha +
> \beta = -1$ and $\alpha\beta = \zeta_5 + \zeta_5^2 + \zeta_5^3 + \zeta_5^4 = -1$, so both
> solve $t^2 + t - 1 = 0$. Hence $\alpha = \tfrac{1}{2}(-1 + \sqrt{5})$, and the quadratic
> subfield is $\mathbb{Q}(\sqrt{5})$.

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

$$
% caption: The cyclic Galois group $C_4$ of $\mathbb{Q}(\zeta_5)$ has a single proper
% subgroup $C_2$, matching the single intermediate field, the quadratic $\mathbb{Q}(w)$
% with $w = \sqrt{5}$. Here $z = \zeta_5$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% fields
\node (K) at (0,3) {$\mathbb{Q}(z)$};
\node (M) at (0,1.5) {$\mathbb{Q}(w)$};
\node (Q) at (0,0) {$\mathbb{Q}$};
\draw[black] (Q) -- node[right, font=\footnotesize]{$2$} (M);
\draw[black] (M) -- node[right, font=\footnotesize]{$2$} (K);
% groups
\node (one) at (5,3) {$1$};
\node (H) at (5,1.5) {$C_2$};
\node (G) at (5,0) {$C_4$};
\draw[black] (G) -- (H) -- (one);
% links
\draw[<->, dashed, black] (K) -- (one);
\draw[<->, dashed, black] (M) -- (H);
\draw[<->, dashed, black] (Q) -- (G);
\end{tikzpicture}
$$

### 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 $H \leq \operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$,
a primitive element of the fixed field is the **period**
$$
\alpha_H = \sum_{\tau \in H} \tau(\zeta),
$$
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.[^df-periods] 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.

| Subgroup | Order | Period generating the fixed field | Degree over $\mathbb{Q}$ |
| --- | --- | --- | --- |
| $\langle \sigma^6 \rangle$ | $2$ | $\zeta + \zeta^{-1}$ | $6$ |
| $\langle \sigma^4 \rangle$ | $3$ | $\zeta + \zeta^3 + \zeta^9$ | $4$ |
| $\langle \sigma^3 \rangle$ | $4$ | $\zeta + \zeta^8 + \zeta^{12} + \zeta^5$ | $3$ |
| $\langle \sigma^2 \rangle$ | $6$ | $\zeta + \zeta^4 + \zeta^3 + \zeta^{12} + \zeta^9 + \zeta^{10}$ | $2$ |

Each row is one intermediate field, and the degrees multiply down the divisor chain: the
degree-$2$ field is $\mathbb{Q}(\sqrt{13})$ (as $13 \equiv 1 \pmod 4$), the degree-$6$
field is the maximal real subfield $\mathbb{Q}(\zeta + \zeta^{-1}) = \mathbb{Q}\bigl(2\cos\tfrac{2\pi}{13}\bigr)$,
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](/abstract-algebra/ring-theory/fractions-and-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
$$
\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong \prod_{i} (\mathbb{Z}/p_i^{a_i}\mathbb{Z})^\times \cong (\mathbb{Z}/n\mathbb{Z})^\times,
$$
which reproduces the CRT decomposition of the unit group.[^df-crt]

## 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.

> **Corollary.** For every finite abelian group $G$ there is a subfield $K$ of a
> cyclotomic field with $\operatorname{Gal}(K/\mathbb{Q}) \cong G$.

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](/abstract-algebra/products-and-group-structure/direct-products-and-finite-abelian-groups),
choose distinct primes $p_i \equiv 1 \pmod{n_i}$, 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 $G$.[^df-abelian] The converse is a landmark theorem, stated
without proof.

> **Theorem (Kronecker–Weber).** Every finite abelian extension of $\mathbb{Q}$ is
> contained in a cyclotomic field $\mathbb{Q}(\zeta_n)$ for some $n$.

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](/abstract-algebra/field-theory/straightedge-and-compass-constructions)
if and only if $\zeta_n$ is, and a point is constructible exactly when it lies in a tower
of quadratic extensions. Since $[\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n)$ and its
Galois group is abelian, a quadratic tower down to $\mathbb{Q}$ exists precisely when
$\varphi(n)$ is a power of $2$.

> **Theorem (Gauss).** The regular $n$-gon is constructible with straightedge and compass
> if and only if $\varphi(n)$ is a power of $2$, equivalently $n = 2^k p_1 \cdots p_r$ for
> distinct **Fermat primes** $p_i = 2^{2^s} + 1$.

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
$\operatorname{Gal}(\mathbb{Q}(\zeta_{17})/\mathbb{Q}) \cong \mathbb{Z}/16\mathbb{Z}$ that
supplies the chain of quadratic steps.

[^df-simple]: **Dummit & Foote**, _Abstract Algebra_, §14.4, Proposition 24 and Theorem 25 (Primitive Element Theorem) — a finite extension is simple iff it has finitely many intermediate fields; finite separable extensions are simple.
[^df-cyclo]: **Dummit & Foote**, _Abstract Algebra_, §14.5, Theorem 26 — $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\times$ via $a \mapsto \sigma_a$, with $\sigma_a(\zeta_n) = \zeta_n^a$.
[^df-periods]: **Dummit & Foote**, _Abstract Algebra_, §14.5, Example 2 — the periods $\sum_{\tau \in H} \tau(\zeta_p)$ generate the fixed fields of subgroups $H$; worked for $\mathbb{Q}(\zeta_{13})$ with generator $\sigma_2$.
[^df-crt]: **Dummit & Foote**, _Abstract Algebra_, §14.5, Corollary 27 — the prime-power cyclotomic fields intersect in $\mathbb{Q}$ and compose to $\mathbb{Q}(\zeta_n)$, giving the CRT factorization of the Galois group.
[^df-abelian]: **Dummit & Foote**, _Abstract Algebra_, §14.5, Corollary 28 and the Kronecker–Weber theorem — every finite abelian group is the Galois group of a subfield of a cyclotomic field, and conversely every abelian extension of $\mathbb{Q}$ lies in a cyclotomic field.