2.9 Solvability by Radicals

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For the sake of simplicity, we shall assume throughout this section that all

fields have characteristic 0. Let F be a field and let E be an extension of

F. If E = F(_) for some _ 2 E satisfying _n 2 F, for some integer n, then

E is called a simple radical extension of F. Let f(x) 2 F[x], and let E be a

splitting field for f(x) over F. Assume also that there is a sequence

F = F0 _ F1 _ · · · _ Fr _ E,

where each Fk is a simple radical extension of Fk−1. (We call the tower

F = F0 _ F1 _ · · · _ Fr a root tower .) Then we say that the polynomial

f(x) is solvable by radicals .

Lemma 2.9.1 Assume that the polynomial f(x) 2 F[x] is solvable by radicals,

and let E be a splitting field for f(x) over F. Then there exists a root

tower

F = F0 _ F1 _ · · · _ Fr _ E

where Fr is Galois over F.

Lemma 2.9.2

(a) Let E = F(a) be a simple radical extension, where an 2 F. Assume

that the polynomial xn − 1 splits completely in F[x]. Then Gal(E/F)

is cyclic.

(b) Let E _ F be a Galois extension of prime degree q, and assume that

xq − 1 splits completely in F[x]. Then E is a simple radical extension

of F.

We are now in a position to state E. Galois’ famous result.

Theorem 2.9.3 Let F be a field of characteristic 0, and let f(x) 2 F[x],

with Galois group G. Then f(x) is solvable by radicals if and only if G is a

solvable group.

70 CHAPTER 2. FIELD AND GALOIS THEORY

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