Define a perfect field. Prove that every finite field is perfect.
Solution
A field \(F\) is perfect if every irreducible polynomial in \(F[x]\) is separable.
Every finite field is perfect. Let \(F=\FF_q\) have characteristic \(p>0\) and let \[ \operatorname{Fr}:F\to F,\qquad a\mapsto a^p \] be Frobenius. Since \(F\) is finite and Frobenius is injective, it is surjective. Hence every \(a\in F\) has a \(p\)th root in \(F\).
Now let \(f\in F[x]\) be irreducible. If \(f\) were inseparable, then \(f'=0\), so every exponent occurring in \(f\) would be divisible by \(p\) and \[ f(x)=\sum_i a_i x^{pi}. \] Choose \(b_i\in F\) with \(b_i^p=a_i\). In characteristic \(p\), \[ f(x)=\left(\sum_i b_i x^i\right)^p, \] which is reducible unless \(f\) is linear. Thus every nonconstant irreducible polynomial over \(F\) is separable, so \(F\) is perfect.