236 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

At this point, it is a good idea to recall the big fundamental theorem mentioned abovewhich gives the correspondence between normal subgroups and normal field extensionssince it is about to be used again.

F≡ F0 ⊆ F1 ⊆ F2 · · · ⊆ Fk−1 ⊆ Fk ≡KG(F,F) = {ι} ⊆ G(F1,F) ⊆ G(F2,F) · · · ⊆ G(Fk−1,F) ⊆ G(Fk,F)

Theorem 9.4.50 Let K be a splitting field for a separable polynomial p(x) ∈ F [x]. Let{Fi}k

i=0 be the increasing sequence of intermediate fields between F and K. Then each ofthese is a normal extension of F and the Galois group G

(F j−1,F

)is a normal subgroup of

G(F j,F). In addition to this,

G(F j,F)≃ G(K,F)/G(K,F j)

where the symbol ≃ indicates the two spaces are isomorphic.

Theorem 9.4.51 Let f (x) be a separable polynomial in F [x] where F contains all nth rootsof unity for each n∈N or for all n≤m and the roots of unity are of the form ξ

k as describedin Lemma 9.4.47. Let K be a splitting field of f (x) . If f (x) is solvable by radicals over F,or solvable by radicals over F with the ki ≤ m in Definition 9.4.48, then the Galois groupG(K,F) is a solvable group.

Proof: Using the definition given above for f (x) to be solvable by radicals, there is asequence of fields

F0 = F⊆ F1 ⊆ ·· · ⊆ Fk, K⊆ Fk,

where Fi = Fi−1 (ai), akii ∈ Fi−1, and each field extension is a normal extension of the pre-

ceding one. By Lemma 9.4.49, Fk is the splitting field of a polynomial having coefficientsin F j−1. This follows from the Lemma 9.4.49 above. Then it follows from Theorem 9.4.50,letting F j−1 play the role of F, that

G(F j,F j−1

)≃ G

(Fk,F j−1

)/G(Fk,F j)

By Lemma 9.4.47, the Galois group G(F j,F j−1

)is Abelian and so this and the above iso-

morphism requires that G(Fk,F) is a solvable group since the quotient groups are Abelian.By Theorem 9.4.43, it follows that, since G(K,F) is a subgroup of G(Fk,F) , it must

also be solvable. ■Now consider the equation

p(x) = xn−a1xn−1 +a2xn−2 + · · ·±an, p(x) ∈ F [x] , n≥ 5

and suppose that p(x) has distinct roots, none of them in F. Let K be a splitting field forp(x) over F so that p(x) = ∏

nk=1 (x− ri) . Then it follows that ai = si (r1, · · · ,rn) where the

si are the elementary symmetric functions defined in Definition 9.1.3. For σ ∈G(K,F) youcan define σ̄ ∈ Sn by the rule σ̄ (k) ≡ j where σ (rk) = r j. Recall that the automorphismsof G(K,F) take roots of p(x) to roots of p(x). This mapping σ → σ̄ is onto, a homomor-phism, and one to one and onto because the symmetric functions si are unchanged whenthe roots are permuted. Thus a rational function in s1,s2, · · · ,sn is unaffected when theroots rk are permuted. It follows that G(K,F) cannot be solvable if n≥ 5 because Sn is notsolvable.