Assume is perfect. ^P2 already covers the case , so assume is a field of prime characteristic with a Frobenius automorphism By ^P1, an inseparable irreducible polynomial must be of the form

for some . But

whence , contradicting irreducibility, so every irreducible polynomial must in fact be separable.

For the converse, assume is imperfect, so is a prime field of characteristic and the Frobenius endomorphism is not surjective. It suffices to construct an inseparable irreducible polynomial.

By non-surjectivity there exists an such that has no roots in and therefore no linear factors. In , on the other hand, does have a root , and in fact, applying the Frobenius endomorphism, we see

which is separable. It follows that an irreducible factor of is separable, as required.