Finite fields are perfect - Imported Knowlpedia Content

Let $K$ be a finite field of characteristic $p>0$ . The Frobenius map

\mathrm{Fr}:K\to K,\quad x\mapsto x^p

is a ring endomorphism (see Frobenius endomorphism) and is automatically injective; finiteness forces it to be surjective, hence an automorphism. Therefore every element of $K$ has a unique $p$ -th root in $K$ , and more generally every algebraic extension of $K$ is separable. In other words:

Theorem. Every finite field is perfect.

Equivalent characterizations

Equivalently, every irreducible polynomial over a finite field has distinct roots in an algebraic closure (cf. separable polynomials have distinct roots).

Examples

Prime fields. $\mathbb{F}_p$ is perfect: $\mathrm{Fr}(x)=x^p$ is the identity map on $\mathbb{F}_p$ , hence an automorphism.

Degree- $n$ finite fields. For $K=\mathbb{F}_{p^n}$ , Frobenius is an automorphism of order $n$ , and the extension $K/\mathbb{F}_p$ is Galois with cyclic Galois group generated by Frobenius (see finite-field Galois group is cyclic).

Contrast with an infinite non-perfect field. $\mathbb{F}_p(t)$ is not perfect: $t$ has no $p$ -th root in $\mathbb{F}_p(t)$ , and $x^p-t$ defines an inseparable finite extension (as in perfect implies separable).