TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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

Statement

Let $F$ be a field of characteristic $p > 0$ . The Frobenius map

${Fr}_{F} : F \to F, x \mapsto x^{p},$

is an injective field endomorphism. If $F$ is finite, it is an automorphism. Its $n$ -fold iterate is $x \mapsto x^{p^{n}}$ .

Facts & Assumptions

Given: A field $F$ of positive characteristic $p$ .

[L1]

The binomial theorem holds in every commutative ring (The binomial theorem over an arbitrary commutative ring).

[L2]

For $0 < k < p$ , the prime $p$ divides $(\binom{p}{k})$ (A prime $p$ divides $(\binom{p}{k})$ for $0 < k < p$ ).

[L3]

A positive field characteristic is prime (The characteristic of a field is zero or a prime number).

[L4]

A field homomorphism preserves addition, multiplication, and $1$ (Field homomorphism and embedding).

[L5]

An injection from a finite set to itself is a bijection (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ ).

Proof

technique · direct

1.1givenL1L2L3algebra

By [L1] and [L2], all intermediate terms in $(x + y)^{p}$ have coefficients divisible by $p$ and hence vanish in $F$ , so $(x + y)^{p} = x^{p} + y^{p}$ .

1.2givenL4algebra

Commutativity gives $(x y)^{p} = x^{p} y^{p}$ , and $1^{p} = 1$ , so Frobenius is an endomorphism by [L4].

2.1step 1.1algebra

If $x^{p} = y^{p}$ , then step 1.1 gives $(x - y)^{p} = 0$ . A field has no nonzero nilpotents, so $x - y = 0$ and the map is injective.

3.1step 1.2step 2.1L5

If $F$ is finite, [L5] turns this injection into a bijection, hence an automorphism.

4.1step 1.2algebra∎

Iterating and using $(x^{p^{r}})^{p} = x^{p^{r + 1}}$ gives ${Fr}_{F}^{n} (x) = x^{p^{n}}$ , including $n = 0$ as the identity.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Frobenius x↦xpx↦xp is an injective endomorphism in characteristic pp, and an automorphism for finite fields

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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