- Define a perfect field.
- Give an imperfect field and an inseparable polynomial over it.
- Prove that every finite field is perfect.
Solution
A field \(K\) is perfect if every irreducible polynomial over \(K\) is separable. In characteristic \(p>0\), this is equivalent to surjectivity of Frobenius \[ F:K\to K,\qquad x\mapsto x^p. \]
1. Every finite field is perfect.
Let \(K=\mathbb F_{p^m}\). Frobenius is injective because \(x^p=0\) implies \(x=0\). Since \(K\) is finite, every injective self-map is surjective. Hence \(K^p=K\), so \(K\) is perfect.
2. The field \(\mathbb F_p(t)\) is imperfect.
Its \(p\)th powers form \[ \mathbb F_p(t)^p=\mathbb F_p(t^p), \] which does not contain \(t\). Thus \(t\) has no \(p\)th root in \(\mathbb F_p(t)\). The polynomial \[ x^p-t \] is irreducible over \(\mathbb F_p(t)\) (for example by Eisenstein at \(t\) in \(\mathbb F_p[t]\) and Gauss’s lemma), while its derivative is zero. In an algebraic closure it is \[ (x-t^{1/p})^p, \] so it is inseparable. Hence \(\mathbb F_p(t)\) is not perfect.