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Frobenius is an injective endomorphism in characteristic , and an automorphism for finite fields
Statement
Let be a field of characteristic . The Frobenius map
is an injective field endomorphism. If is finite, it is an automorphism. Its -fold iterate is .
Facts & Assumptions
Given: A field of positive characteristic .
The binomial theorem holds in every commutative ring (The binomial theorem over an arbitrary commutative ring).
For , the prime divides (A prime divides for ).
A positive field characteristic is prime (The characteristic of a field is zero or a prime number).
A field homomorphism preserves addition, multiplication, and (Field homomorphism and embedding).
An injection from a finite set to itself is a bijection (A subset of a finite set is finite, with , and equality holds if and only if ).
Proof
By [L1] and [L2], all intermediate terms in have coefficients divisible by and hence vanish in , so .
Commutativity gives , and , so Frobenius is an endomorphism by [L4].
If , then step 1.1 gives . A field has no nonzero nilpotents, so and the map is injective.
If is finite, [L5] turns this injection into a bijection, hence an automorphism.
Iterating and using gives , including as the identity.
Depends on
- The binomial theorem over an arbitrary commutative ring
- A prime $p$ divides $\binom pk$ for $0<k<p$
- The characteristic of a field is zero or a prime number
- Field homomorphism and embedding
- A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$, and equality holds if and only if $B = A$
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 89 results over 22 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
- K. Conrad, Finite Fields, Sections 1-2 (standard reference, not scraped)
- J. S. Milne, Fields and Galois Theory, Propositions 4.19-4.24 (standard reference, not scraped)