Field theory MOC

Perfect field

Let be a field. is called perfect iff [[Characteristic|]] or is a field of prime characteristic for which the Frobenius endomorphism is an automorphism. field Equivalently,¹

1. every ^irreducible is a separable polynomial;
2. every algebraic extension of is a separable extension.

Proof of equivalence

Assume is perfect. ^P2 already covers the case , so assume is a field of prime characteristic with a Frobenius automorphism By ^P1, an inseparable irreducible polynomial must be of the form

for some . But

whence , contradicting irreducibility, so every irreducible polynomial must in fact be separable.

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

By non-surjectivity there exists an such that has no roots in and therefore no linear factors. In , on the other hand, does have a root , and in fact, applying the Frobenius endomorphism, we see

which is separable. It follows that an irreducible factor of is separable, as required.

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

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Footnotes

1. 2009. Algebra: Chapter 0, §§VII.4.2–4.3 ↩