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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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Frobenius x↦xp 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→F,x↦xp,

is an injective field endomorphism. If F is finite, it is an automorphism. Its n-fold iterate is x↦xpn.

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 (pk) (A prime p divides (pk) 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).

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=xp+yp.

1.2givenL4algebra

Commutativity gives (xy)p=xpyp, and 1p=1, so Frobenius is an endomorphism by [L4].

2.1step 1.1algebra

If xp=yp, 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 (xpr)p=xpr+1 gives Fr⁡Fn(x)=xpn, including n=0 as the identity.

Depends on

Used by

Dependency tree · two levels

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Sources