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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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Frobenius xxp 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

FrF:FF,xxp,

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

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
1.1

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.

givenL1L2L3algebra
1.2

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

givenL4algebra
2.1

If xp=yp, then step 1.1 gives (xy)p=0. A field has no nonzero nilpotents, so xy=0 and the map is injective.

step 1.1algebra
3.1

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

step 1.2step 2.1L5
4.1

Iterating and using (xpr)p=xpr+1 gives FrFn(x)=xpn, including n=0 as the identity.

step 1.2algebra

Depends on

Used by

Dependency tree · next 3 levels

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Sources