Perfect field

Let $𝐾$ be a field. $𝐾$ is called perfect iff $c h a r 𝐾 = 0$ or $𝐾$ is a field of prime characteristic for which the Frobenius endomorphism is an automorphism. #m/def/field Equivalently,¹

every irreducible polynomial $𝑓 (𝑥) \in 𝐾 [𝑥]$ is a separable polynomial;
every algebraic extension of $𝐾$ is a separable extension.

Proof of equivalence

Assume $𝐾$ is perfect. ^P2 already covers the case $c h a r 𝐾 = 0$ , so assume $𝐾$ is a field of prime characteristic $𝑝$ with a Frobenius automorphism $𝜎 : 𝐾 \to 𝐾$ By ^P1, an inseparable irreducible polynomial $𝑓 (𝑥) \in 𝐾 [𝑥]$ must be of the form

𝑓 (𝑥) = 𝑛 \sum 𝑖 = 0 𝑎 𝑖 (𝑥 𝑝) 𝑖

for some ${𝑎 𝑖} 𝑛 𝑖 = 0 \subset 𝐾$ . But

𝜎 - 1 (𝑓 (𝑥)) = 𝑛 \sum 𝑖 = 0 𝜎 - 1 (𝑎 𝑖) 𝑥 𝑖 : = 𝑔 (𝑥)

whence $𝑔 (𝑥) 𝑝 = 𝑓 (𝑥)$ , contradicting irreducibility, so every irreducible polynomial must in fact be separable.

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

By non-surjectivity there exists an $𝛼 \in 𝐾$ such that $𝑓 (𝑥) : = 𝑥 𝑝 - 𝛼 \in 𝐾 [𝑥]$ has no roots in $𝐾$ and therefore no linear factors. In $―― 𝐾$ , on the other hand, $𝑓 (𝑥)$ does have a root $𝛽$ , and in fact, applying the Frobenius endomorphism, we see

𝑓 (𝑥) = (𝑥 - 𝛽) 𝑝

which is separable. It follows that an irreducible factor of $𝑓 (𝑥)$ is separable, as required.

Note that by the elementary Pigeonhole principle, every Galois field is perfect.

#state/tidy | #lang/en | #SemBr

2009. Algebra: Chapter 0, §§VII.4.2–4.3 ↩

Perfect field

Footnotes