5.7. Proof of Galois’s solvability theorem.

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Recall that a polynomial f(X) ∈ F[X] is said to be solvable if there is a tower of fields

F = F0 ⊂ F1 ⊂ F2 ⊂ ・ ・ ・ ⊂ Fm

such that

(a) Fi = Fi−1[αi], where αmi

i

∈ Fi−1 for some mi;

(b) Fm splits f(X).

Theorem 5.23. Let F be a field of characteristic 0. A polynomial f ∈ F[X] is solvable

if and only if its Galois group Gf is solvable.

Before proving the sufficiency, we need a lemma.

Lemma 5.24. Let f ∈ F[X] be separable, and let F_ be an extension field of F. Then the

Galois group of f as an element of F_[X] is a subgroup of that of f as an element of F[X].

Proof. Let E_ be a splitting field for f over F_, and let α1, . . . , αm be the roots of

f(X) in E_. Then E = F[α1, ..., αm] is a splitting field of f over F. Any element of

Gal(E_/F _) permutes the αi and so maps E into itself. The map σ → σ|E is an injection

Gal(E_/F _) → Gal(E/F).

Proof. (Gf solvable =⇒ f solvable). Let f ∈ F[X] have solvable Galois group. Let

F_ = F[ζ] where ζ is a primitive nth root of 1 for some large n—for example, n = (degf)!

will do. The lemma shows that the Galois group G of f as an element of F_[X] is a subgroup

of Gf , and hence is solvable. This means that there is a sequence of subgroups

G = Gm ⊃ Gm−1 ⊃ ・ ・ ・ ⊃ G1 ⊃ G0 = {1}

such that each Gi is normal in Gi+1 and Gi+1/Gi is cyclic (even of prime order, but we don’t

need this). Let E be a splitting field of f(X) over F_, and let Fi = EGi . We have a sequence

of fields

F ⊂ F[ζ] = F_ ⊂ F1 ⊂ F2 ⊂ ・ ・ ・ ⊂ Fm = E

with Fi Galois over Fi−1 with cyclic Galois group. According to (5.20b), Fi = Fi−1[αi] with

α[Fi :Fi−1]

i

∈ Fi−1. This shows that f is solvable.

Before proving the necessity, we need to make some observations. Let Ω be a Galois

extension of F, and let E be an extension of F contained in Ω. The Galois closure _ E of E

in Ω is the smallest subfield of Ω containing E that is Galois over F. Let G = Gal(Ω/F )

and H = Gal(Ω/E). Then _ E will be the subfield of Ω corresponding to the largest normal

subgroup of G contained in H (Galois correspondence 3.17), but this is

_

σ∈G σHσ−1 (see

Groups 4.10), and σHσ−1 corresponds to σE. Hence (see 3.18) _ E is the composite of the

fields σE, σ ∈ G. In particular, we see that if E = F[α1, . . . , αm], then _ E is generated over

F by the elements σαi, σ ∈ G.

Proof. (f solvable =⇒ Gf solvable). It suffices to show that Gf is a quotient of a

solvable group. Hence it suffices to find a Galois extension _ E of F with Gal( _ E/F) solvable

and such that f(X) splits in _ E[X].

We are given that f splits in an extension Fm of F with the following property: Fm =

F[α1, . . . , αm] and, for all i, there exists an mi such that αmi

i

∈ F[α1, . . . , αi−1]. By (5.1)

we know Fm = F[γ] for some γ. Let g(X) be the minimum polynomial of γ over F, and let

46 J.S. MILNE

Ω be a splitting field of g(X)(Xn −1) for some suitably large n. We can identify Fm with a

subfield of Ω. Let G = {σ1 = 1, σ2, . . . } be the Galois group of Ω/F and let ζ be a primitive

nth root of 1 in Ω. Choose _E to be the Galois closure of Fm[ζ] in Ω. According to the above

remarks, _ E is generated over F by the elements

ζ,α1, α2, . . . , αm, σ2α1, . . . , σ2αm, σ3α1, . . . .

When we adjoin these elements one by one, we get a sequence of fields

F ⊂ F[ζ] ⊂ F[ζ,α1] ⊂ ・ ・ ・ ⊂ F_ ⊂ F__ ⊂ ・ ・ ・ ⊂ _ E

such that each field F__ is obtained from its predecessor F_ by adjoining an rth root of an

element of F_. According to (5.20a) and (5.7), each of these extensions is Galois with cyclic

Galois group, and so G has a normal series with cyclic quotients. It is therefore solvable.

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