TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17

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A field is perfect exactly when it has characteristic zero or its Frobenius map is surjective

Statement

A field $F$ is perfect if and only if either $char F = 0$ , or $char F = p > 0$ and the Frobenius map $a \mapsto a^{p}$ is surjective.

Facts & Assumptions

Given: A field $F$ .

[L1]

A field is perfect when all of its nonconstant irreducible polynomials are separable (Perfect fields: every irreducible polynomial is separable).

[L2]

In characteristic $p$ , every irreducible polynomial has a unique form $g (x^{p^{e}})$ with $g$ irreducible and separable (In characteristic $p$ , every irreducible polynomial is uniquely $g (x^{p^{e}})$ with $g$ irreducible and separable).

[L3]

If $a$ is not a $p$ th power, then $x^{p} - a$ is irreducible (If $a$ is not a $p$ th power in a characteristic- $p$ field, then $x^{p^{n}} - a$ is irreducible for every $n \geq 1$ ).

[L4]

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

[L5]

A nonzero polynomial is separable exactly when it is coprime to its derivative (A nonzero polynomial over a field is separable exactly when its gcd with its derivative is $1$ ).

Proof

technique · direct

1.1L1L5algebra

If $char F = 0$ and $f$ is irreducible, then $f^{'} \neq 0$ ; any common nonconstant divisor of $f$ and $f^{'}$ would be associated to $f$ , which is impossible because $\deg f^{'} < \deg f$ . Thus $\gcd (f, f^{'}) = 1$ , so $f$ is separable by [L5].

1.2L2L4

Suppose $char F = p > 0$ and Frobenius is surjective. For irreducible $f = g (x^{p^{e}})$ as in [L2], if $e > 0$ then taking $p^{e}$ th roots of the coefficients through repeated surjectivity and using [L4] would write $f$ as a $p^{e}$ th power of a nonconstant polynomial, contradicting irreducibility. Hence $e = 0$ and every irreducible is separable.

1.3L1L3

Conversely, if Frobenius is not surjective, choose $a \notin F^{p}$ . Then [L3] makes $x^{p} - a$ irreducible, while its derivative is zero, so it is not separable and $F$ is not perfect.

2.1step 1.1step 1.2step 1.3∎

The characteristic-zero argument and the two implications in positive characteristic establish the equivalence.

A field is perfect exactly when it has characteristic zero or its Frobenius map is surjective

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources