9.4. MORE ON ALGEBRAIC FIELD EXTENSIONS 235

splitting field of xn−a over F, thus K is a normal extension of F. Then K= F(u) where uis any root of xn−a. The Galois group G(K,F) is Abelian.

Proof: Let u be a root of xn− a and let K equal F(u) . Then let ξ be the nth root ofunity mentioned. Then (

ξku)n

= (ξ n)k un = a

and so each ξku is a root of xn−a and these are distinct. It follows that{

u,ξ u, · · · ,ξ n−1u}

are the roots of xn−a and all are in F(u) . Thus F(u) = K. Let σ ∈ G(K,F) and observethat since σ fixes F,

0 = σ

((ξ

ku)n−a)=(

σ

(ξ

ku))n−a

It follows that σ maps roots of xn−a to roots of xn−a. Therefore, if σ ,α are two elementsof G(K,F) , there exist i, j each no larger than n−1 such that

σ (u) = ξiu, α (u) = ξ

ju

A typical thing in F(u) is p(u) where p(x) ∈ F [x]. Then

σα (p(u)) = p(

ξjξ

iu)= p

(ξ

i+ ju)

ασ (p(u)) = p(

ξiξ

ju)= p

(ξ

i+ ju)

Therefore, G(K,F) is Abelian. ■Thus this one is clearly solvable as noted above. To say a polynomial is solvable by

radicals is expressed precisely in the following definition.

Definition 9.4.48 For F a field, a polynomial p(x) ∈ F [x] is solvable by radicals overF≡ F0 if there are algebraic numbers ai, i = 1,2, ...,k, and a sequence of fields F1 =

F(a1) ,F2 =F1 (a2) , · · · ,Fk =Fk−1 (ak) such that for each i≥ 1,akii ∈Fi−1 and Fk contains

a splitting field K for p(x) over F.

Actually, the only case of interest here is included in the following lemma.

Lemma 9.4.49 In Definition 9.4.48 when the roots of unity are of the form ξk as described

in Lemma 9.4.47, Fk is a splitting field provided you assume F contains all the nth roots of1 for all n≤max{ki}k

i=1.

Proof: by Lemma 9.4.47,

Fk = F(a1,a2, · · · ,ak) = F({

a j1

}k1−1

j=1, ...,

{a j

1

}kk−1

j=1

)and so Fk is the splitting field of ∏

ki=1

(xki −aki

i

). Each ai is a single root of xki−aki

i where

akii ∈ F. ■