Linear Algebra, Theory And Applications, 2012a

474 FIELDS AND FIELD EXTENSIONS
Definition F.8.6 A field F is called perfect if every polynomial p (x) having coefficients in
F is separable.
The above shows that fields of characteristic 0 are perfect. The above theory about
Galois groups and fixed fields all works for perfect fields. What about fields of characteristic
p where p is a prime? The following interesting lemma has to do with a nonzero a ∈ F
having a p th root in F.
Lemma F.8.7 Let F be a field of characteristic p. Let a ≠0where a ∈ F. Then either
x p − a is irreducible or there exists b ∈ F such that x p − a =(x − b) p .
Proof: Suppose that x p − a is not irreducible. Then x p − a = g (x) f (x) wherethe
degree of g (x) ,k is less than p and at least as large as 1. Then let b be a root of g (x). Then
b p − a = 0. Therefore,
x p − a = x p − b p =(x − b) p .
That is right. x p − b p =(x − b) p just like many beginning calculus students believe. It
happens because of the binomial theorem and the fact that the other terms have a factor of
p. Hence
x p − a =(x − b) p = g (x) f (x)
and so g (x) divides (x − b) p which requires that g (x) =(x − b) k since g (x) has degree k.
It follows, since g (x) is given to have coefficients in F, thatb k ∈ F. Alsob p ∈ F. Sincek, p
are relatively prime, due to the fact that k