TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$

Statement

Let $F_{q}$ be a finite field of order $q$ and let $E$ be a finite field having $F_{q}$ as a subfield, with $[E : F_{q}] = n$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension). Then $E / F_{q}$ is a finite Galois extension (Finite Galois extensions and $Gal (K / F)$ ) and

$Gal (E / F_{q}) = ⟨ σ_{q} ⟩$

is cyclic of order $n$ , generated by the relative Frobenius $σ_{q} ⁣ : x \mapsto x^{q}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields).

Facts & Assumptions

Given: Finite fields $F_{q} \subseteq E$ with $∣ F_{q} ∣ = q$ and $[E : F_{q}] = n$ , and the subgroup $G : = ⟨ σ_{q} ⟩$ of $Aut (E / F_{q})$ .

[L1]

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

[L2]

${x \in E : x^{q} = x} = F_{q}$ , that is $E^{⟨ σ_{q} ⟩} = F_{q}$ (The elements of a finite extension fixed by the $q$ -power map are exactly the base field).

[L3]

$∣ E ∣ = q^{n}$ and $σ_{q}$ has order exactly $n$ in $Aut (E / F_{q})$ (For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ ).

[L4]

If $G$ is a finite group of automorphisms of $K$ , then $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ (Artin's fixed-field theorem: $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ ).

[L5]

For a finite extension $K / F$ with $G = Aut (K / F)$ , the conditions " $K / F$ is Galois", " $K$ is the splitting field over $F$ of a separable polynomial", " $∣ G ∣ = [K : F]$ " and " $K^{G} = F$ " are equivalent (Equivalent characterizations of a finite Galois extension).

[L6]

$K^{G} = {x \in K : σ (x) = x for every σ \in G}$ (The fixed field $K^{G}$ of a group of field automorphisms).

Proof

technique · direct

1.1

$G = ⟨ σ_{q} ⟩$ is a cyclic group of automorphisms of $E$ , finite of order $n$ by [L1] and [L3].

L1L3

1.2

Its fixed field is $E^{G} = F_{q}$ by [L2] and [L6].

L2L6

2.1

Applying [L4] to the finite automorphism group $G$ of $E$ and using step 1.2, $[E : F_{q}] = [E : E^{G}] = ∣ G ∣ = n$ and $Aut (E / F_{q}) = Aut (E / E^{G}) = G$ .

step 1.1step 1.2L4

3.1

Hence $∣ Aut (E / F_{q}) ∣ = n = [E : F_{q}]$ , so $E / F_{q}$ is Galois by [L5], and $Gal (E / F_{q}) = Aut (E / F_{q}) = ⟨ σ_{q} ⟩$ is cyclic of order $n$ by step 2.1 and step 1.1. At $n = 1$ the group is trivial and $E = F_{q}$ .

step 1.1step 2.1L5∎

Remarks

Separability and normality are never argued separately. The usual route checks that $E$ is a splitting field of $t^{q^{n}} - t$ and that this polynomial has no repeated root. Routing through Artin's fixed-field theorem: $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ instead replaces both checks by one count: an automorphism group of order $n$ whose fixed field is $F_{q}$ already forces $∣ Aut ∣ = [E : F_{q}]$ , which is one of the equivalent Galois conditions.

Depends on

Used by

Dependency tree · two levels

29 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)

A finite extension of a finite field of order qq is Galois with cyclic Galois group generated by x↦xqx↦xq