Use characteristic-p powers as automorphisms

Try

In characteristic two, expand (a+b)². The middle coefficient is two and therefore vanishes, leaving a²+b². More generally, in characteristic p every intermediate binomial coefficient in (a+b)ᵖ is divisible by p. Raising to the pth power preserves addition as well as multiplication.

This map, called Frobenius, is special to positive characteristic. Over the rationals, squaring does not preserve addition. Record the characteristic before applying the identity.

Predict

If aᵖ=bᵖ in a field, then (a−b)ᵖ=0. Predict why this forces a=b. Does injectivity alone imply surjectivity? It does on a finite field, but an infinite field requires a separate argument and can provide a counterexample.

Experiment

In the four-element field with α²=α+1, compute the Frobenius map and its second iterate.

Element	Square	Fourth power
0	0	0
1	1	1
α	α+1	α
α+1	α	α+1

Frobenius fixes the prime field and exchanges the two other elements. Its square is identity, while it is not itself identity. Compare with the reducible quotient from the preceding lesson, where a nonzero nilpotent squares to zero and injectivity fails.

Understand

In any characteristic-p field, Frobenius preserves addition, multiplication and one, and is injective by the zero-power argument. On a finite field it is therefore an automorphism. On F_p(t), it is not onto: the element t cannot be a pth power of a rational function because the exponent of t in a pth power’s numerator-minus-denominator factorization is divisible by p, whereas t has exponent one.

For a field with pⁿ elements, every nonzero element satisfies a^(pⁿ−1)=1 by Lagrange in the multiplicative group. Including zero, every element satisfies a^(pⁿ)=a, so the nth Frobenius iterate is identity. No earlier positive iterate k<n is identity: otherwise all pⁿ elements would be roots of x^(pᵏ)−x, contradicting its smaller degree. Thus Frobenius has order n.

The multiplicative group of a finite field is cyclic. By the finite abelian structure theorem, its exponent d is attained as the order of an element: choose generators in each primary cyclic factor with maximal exponent. Every group element satisfies xᵈ=1, so the field root bound gives pⁿ−1≤d. Lagrange gives d≤pⁿ−1. Equality proves that chosen element generates the entire multiplicative group. This proof uses both group structure and the field root bound.

A field F_(pⁿ) has a unique subfield of size pᵈ precisely when d divides n. Necessity follows from the tower law over F_p. For sufficiency, use the root-field construction proved in the next splitting-field lesson: the pᵈ roots of x^(pᵈ)−x form a field. If d divides n, every such root is also fixed by the nth Frobenius iterate, so lies in F_(pⁿ) inside a common algebraic closure. Any subfield of that size consists of exactly those roots, proving uniqueness.

More generally, the fixed field of the kth Frobenius power has p^gcd(n,k) elements. Since Frobenius has order n, the subgroup generated by its kth power equals that generated by its gcd(n,k)th power, by Bézout on exponents. Their fixed sets agree, and the preceding root-field description counts them.

Apply

Find the fixed field size for the fourth Frobenius iterate in F_(2⁶). Compute gcd(6,4)=2 before applying the formula. Explain why this counts four fixed elements, not four automorphisms or an extension degree of four.

Keep the logical dependencies visible: the automorphism-order proof requires only finite field arithmetic and root bounds; the existence and uniqueness of every subfield uses the splitting-field construction next. A count alone does not construct a missing field.

Continue

Splitting fields provide all roots at once and prove that finite fields of every prime-power size exist and are unique up to isomorphism. This also clarifies the common algebraic closure used to compare their subfields.