PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Every intermediate field of $Q (μ_{n}) / Q$ is Galois over $Q$ with abelian Galois group

Statement

Let $n \geq 1$ and let $F$ be an intermediate field of $Q (μ_{n}) / Q$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ). Then $F / Q$ is a finite Galois extension (Finite Galois extensions and $Gal (K / F)$ ) and $Gal (F / Q)$ is abelian.

Facts & Assumptions

Given: An integer $n \geq 1$ and an intermediate field $Q \subseteq F \subseteq Q (μ_{n})$ ; $Q$ is an ordered field (The rationals form a totally ordered field), so $char Q = 0$ (The characteristic of a ring: the least $n \geq 1$ with $n \cdot 1_{R} = 0$ when one exists, and $0$ otherwise) and divides no positive integer (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ). Write $G : = Gal (Q (μ_{n}) / Q)$ .

[L1]

$Q (μ_{n}) / Q$ is finite Galois ( $K (μ_{n}) / K$ is Galois and $σ \mapsto a_{σ}$ embeds its Galois group into $(Z / n)^{\times}$ ) with $G ≅ (Z / n)^{\times}$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ), and $G$ is abelian (The Galois group of a cyclotomic extension is abelian).

[L2]

For $K / F_{0}$ finite Galois with group $G$ , the maps $H \mapsto K^{H}$ and $E \mapsto Gal (K / E)$ are mutually inverse bijections between subgroups of $G$ and intermediate fields (The fundamental theorem of finite Galois theory).

[L3]

With $K / F_{0}$ finite Galois, $G = Gal (K / F_{0})$ , $H \leq G$ and $E = K^{H}$ : the extension $E / F_{0}$ is Galois exactly when $H$ is normal in $G$ (Normal subgroup: invariance under conjugation), and then restriction gives $Gal (E / F_{0}) ≅ G / H$ (Normal subgroups, conjugate fields, and quotient groups in the Galois correspondence, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

Proof

technique · direct

1.1

By [L1] the extension $Q (μ_{n}) / Q$ is finite Galois and $G$ is abelian.

2.1

By [L2] there is a subgroup $H \leq G$ with $F = Q (μ_{n})^{H}$ .

step 1.1L2

3.1

Since $G$ is abelian, $g H g^{- 1} = H$ for every $g \in G$ , so $H$ is normal in $G$ ; hence $F / Q$ is Galois and $Gal (F / Q) ≅ G / H$ by [L3].

step 1.1step 2.1L3

4.1

A quotient of an abelian group is abelian, since the images of two commuting elements commute and every element of $G / H$ is such an image; so $Gal (F / Q)$ is abelian.

step 1.1step 3.1L3∎

Remarks

This is the proved half of the Kronecker–Weber picture on this page. Every subfield of a rational cyclotomic field is abelian over $Q$ ; the converse is recorded separately as Recorded, not proved: every finite abelian extension of $Q$ lies in a cyclotomic field ‡.

Depends on

Used by

Recorded, not proved: every finite abelian extension of ℚ lies in a cyclotomic field Remark

Dependency tree · two levels

62 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Cyclotomic Extensions (expository blurb), Section 4 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Chapter 5, cyclotomic extensions (standard reference, not scraped)

Every intermediate field of Q(μn)/QQ(μn​)/Q is Galois over QQ with abelian Galois group