CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Fields of characteristic zero, finite fields, and algebraically closed fields are perfect

Statement

Every field of characteristic zero is perfect. Every finite field is perfect, and every algebraically closed field is perfect.

Facts & Assumptions

Given: A field $F$ in one of the classes named in the Statement.

[L1]

Perfectness is equivalent to characteristic zero or, in characteristic $p > 0$ , surjectivity of Frobenius (A field is perfect exactly when it has characteristic zero or its Frobenius map is surjective).

[L2]

Frobenius is an automorphism of every finite field (Frobenius $x \mapsto x^{p}$ is an injective endomorphism in characteristic $p$ , and an automorphism for finite fields).

[L3]

In an algebraically closed field, every nonconstant polynomial has a root (An algebraically closed field: every nonconstant polynomial has a root in the field).

Proof

technique · direct

1.1

The characteristic-zero case is immediate from [L1].

1.2

If $F$ is finite of characteristic $p$ , [L2] makes Frobenius surjective, so [L1] makes $F$ perfect.

L1L2

2.1

If $F$ is algebraically closed of characteristic $p$ , then for every $a \in F$ the polynomial $x^{p} - a$ has a root by [L3]; hence every $a$ is a $p$ th power and [L1] makes $F$ perfect.

L1L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 49 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

P. L. Clark, Field Theory, Chapters 3 to 5 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Chapters 2, 3, and 5 (standard reference, not scraped)