LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Adjoining roots of unity to a finite Galois extension adds an abelian kernel and preserves solvability

Statement

Let $K / F$ be a finite Galois extension, let $n \geq 1$ , assume $char F ∤ n$ , and put $L = K (μ_{n})$ . Then $L / F$ is finite Galois, restriction gives a surjective homomorphism

$Gal (L / F) ↠ Gal (K / F),$

and its kernel is $Gal (L / K)$ , which is abelian. Consequently $Gal (L / F)$ is solvable if and only if $Gal (K / F)$ is solvable.

Facts & Assumptions

Given: A finite Galois extension $K / F$ , the cyclotomic extension $F (μ_{n}) / F$ , and the compositum $L = K (μ_{n})$ .

[L1]

The cyclotomic extension $F (μ_{n}) / F$ is finite Galois with abelian Galois group when $char F ∤ n$ (The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ , The Galois group of a cyclotomic extension is abelian, $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity).

[L2]

A finite extension is Galois exactly when it is the splitting field of a separable polynomial (Equivalent characterizations of a finite Galois extension).

[L3]

In a finite Galois tower, normal intermediate fields correspond to normal subgroups and quotient Galois groups (Normal subgroups, conjugate fields, and quotient groups in the Galois correspondence).

[L4]

Subgroups and quotients of solvable groups are solvable, and an extension of solvable groups is solvable (Subgroups and quotients of solvable groups are solvable, Extensions and finite direct products of solvable groups are solvable).

Proof

technique · direct

1.1

Because $K / F$ is finite Galois, it is the splitting field of a separable polynomial $f \in F [x]$ by [L2]. By [L1], $F (μ_{n})$ is the splitting field of the separable polynomial $x^{n} - 1$ over $F$ . Therefore $L = K (μ_{n})$ is the splitting field over $F$ of the separable polynomial $f (x) (x^{n} - 1)$ , so $L / F$ is finite Galois by [L2].

L1L2

1.2

Since $K / F$ is Galois and $K$ is an intermediate field of the finite Galois extension $L / F$ , [L3] gives a surjective restriction map $ρ ⁣ : Gal (L / F) ↠ Gal (K / F)$ with kernel $Gal (L / K)$ .

2.1

The kernel extension $L / K$ is the cyclotomic extension $K (μ_{n}) / K$ , so [L1] makes $Gal (L / K)$ abelian. If $Gal (L / F)$ is solvable, then its quotient $Gal (K / F)$ is solvable by [L4]. Conversely, if $Gal (K / F)$ is solvable, then [L4] applied to the exact sequence with abelian kernel from step 1.2 makes $Gal (L / F)$ solvable.

step 1.2L1L4

3.1

This proves every part of the statement.

step 1.1step 1.2step 2.1∎

Depends on

Used by

In characteristic 0, a polynomial solvable by radicals has a solvable Galois group Theorem

Dependency tree · two levels

45 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

J. Ash, Basic Abstract Algebra, Section 6.8 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Lemma 5.33 (standard reference, not scraped)