<P> If the Frobenius endomorphism x → x p (\ displaystyle x \ to x ^ (p)) of F is not surjective, there is an element a ∈ F (\ displaystyle a \ in F) which is not a pth power of an element of F. In this case, the polynomial X p − a (\ displaystyle X ^ (p) - a) is irreducible and inseparable . Conversely, if there exists an inseparable irreducible (non-zero) polynomial f (X) = ∑ a i X i p (\ displaystyle \ textstyle f (X) = \ sum a_ (i) X ^ (ip)) in F (X), then the Frobenius endomorphism of F cannot be an automorphism, since, otherwise, we would have a i = b i p (\ displaystyle a_ (i) = b_ (i) ^ (p)) for some b i (\ displaystyle b_ (i)), and the polynomial f would factor as ∑ a i X i p = (∑ b i X i) p . (\ displaystyle \ textstyle \ sum a_ (i) X ^ (ip) = \ left (\ sum b_ (i) X ^ (i) \ right) ^ (p).) </P> <P> If K is a finite field of prime characteristic p, and if X is an indeterminate, then the field of rational functions over K, K (X), is necessarily imperfect, and the polynomial f (Y) = Y − X is inseparable (its formal derivative in Y is 0). More generally, if F is any field of (non-zero) prime characteristic for which the Frobenius endomorphism is not an automorphism, F possesses an inseparable algebraic extension . </P> <P> A field F is perfect if and only if all irreducible polynomials are separable . It follows that F is perfect if and only if either F has characteristic zero, or F has (non-zero) prime characteristic p and the Frobenius endomorphism of F is an automorphism . This includes every finite field . </P> <P> Let E ⊇ F (\ displaystyle E \ supseteq F) be a field extension . An element α ∈ E (\ displaystyle \ alpha \ in E) is separable over F if it is algebraic over F, and its minimal polynomial is separable (the minimal polynomial of an element is necessarily irreducible). </P>

Prove that every finite separable extension of a field is necessary a simple extension