LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$

Statement

Let $F_{q}$ be a finite field of order $q$ and let $E / F_{q}$ be an extension of finite fields of degree $n$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension). Then

$∣ E ∣ = q^{n},$

and the relative Frobenius $σ_{q}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields) has order exactly $n$ in $Aut (E / F_{q})$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity). At $n = 1$ this says $σ_{q}$ is the identity, of order one.

Facts & Assumptions

Given: Finite fields $F_{q} \subseteq E$ with $∣ F_{q} ∣ = q$ and $[E : F_{q}] = n$ ; the prime subfield of $E$ is $F_{p}$ with $p$ the characteristic, and $F_{q}$ has the same characteristic, its identity element being that of $E$ .

[L1]

The relative Frobenius is $σ_{q} (x) = x^{q}$ , an $F_{q}$ -automorphism of $E$ , with $σ_{q}^{i} (x) = x^{q^{i}}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields).

[L2]

If $F$ is a field with $q$ elements, then every $a \in F$ satisfies $a^{q} = a$ (A field with $q$ elements is the splitting field of $x^{q} - x$ over its prime subfield).

[L3]

Let $D$ be an integral domain. A nonzero polynomial $f \in D [x]$ of degree $n$ has at most $n$ distinct roots in $D$ (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[L4]

If $F$ is a finite field, then there is a unique prime $p$ and a unique positive integer $m$ with $∣ F ∣ = p^{m}$ ; here $p = char F$ and $m = [F : F_{p}]$ (Every finite field has order $p^{n}$ for a unique prime characteristic $p$ and positive integer $n$ ).

[L5]

For fields $F \subseteq K \subseteq L$ with $K / F$ and $L / K$ finite, $L / F$ is finite and $[L : F] = [L : K] [K : F]$ (Tower law for finite extensions: $[L : F] = [L : K] [K : F]$ ).

Proof

technique · direct

1.1

By [L4] applied to $F_{q}$ , $q = p^{k}$ with $k = [F_{q} : F_{p}]$ ; by [L4] applied to $E$ , $∣ E ∣ = p^{m}$ with $m = [E : F_{p}]$ .

L4given

2.1

The tower $F_{p} \subseteq F_{q} \subseteq E$ and [L5] give $m = [E : F_{q}] [F_{q} : F_{p}] = n k$ , so $∣ E ∣ = p^{n k} = (p^{k})^{n} = q^{n}$ .

step 1.1L5algebra

3.1

Every $x \in E$ satisfies $x^{q^{n}} = x$ , by [L2] applied to $E$ , whose order is $q^{n}$ by step 2.1; by [L1] this says $σ_{q}^{n} = {i d}_{E}$ .

step 2.1L1L2

3.2

For an integer $j$ with $1 \leq j < n$ one has $σ_{q}^{j} \neq {i d}_{E}$ : otherwise every one of the $q^{n}$ elements of $E$ would be a root of the nonzero polynomial $t^{q^{j}} - t \in E [t]$ , whose degree $q^{j}$ is smaller than $q^{n}$ because $q \geq 2$ , contradicting [L3].

step 2.1L1L3algebra

4.1

The least $j \geq 1$ with $σ_{q}^{j} = {i d}_{E}$ is therefore $j = n$ , that is $ord (σ_{q}) = n$ ; for $n = 1$ steps 3.1 and 3.2 say only that $σ_{q} = {i d}_{E}$ , of order one.

step 3.1step 3.2L1∎

Depends on

Used by

Dependency tree · two levels

34 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Finite Fields (expository blurb), Theorem 5.6 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Proposition 4.20 (standard reference, not scraped)

For a degree-nn extension of a field of order qq, the qq-power map has order exactly nn