Finite fields are perfect

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

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 .

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

Examples

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

2. 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 ).

3. 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 ).