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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The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$

Statement

Let $F_{q}$ be a finite field of order $q$ , let $n \geq 1$ , and let $F_{q^{n}}$ be a finite field having $F_{q}$ as a subfield with $[F_{q^{n}} : F_{q}] = n$ (The degree $[K : F] = \dim_{F} K$ of a finite field extension). For a positive divisor $d$ of $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ) put

$F_{q^{d}} : = {x \in F_{q^{n}} : x^{q^{d}} = x} .$

Then the fields $F_{q^{d}}$ , as $d$ runs over the positive divisors of $n$ , are exactly the intermediate fields of $F_{q^{n}} / F_{q}$ , with distinct divisors giving distinct fields;

$[F_{q^{d}} : F_{q}] = d, ∣ F_{q^{d}} ∣ = q^{d};$

and for positive divisors $d, e$ of $n$ ,

$F_{q^{d}} \subseteq F_{q^{e}} ⟺ d ∣ e .$

The two ends of the lattice are instances: $d = 1$ gives the base field $F_{q}$ and $d = n$ gives $F_{q^{n}}$ .

Facts & Assumptions

Given: Finite fields $F_{q} \subseteq E : = F_{q^{n}}$ with $∣ F_{q} ∣ = q$ and $[E : F_{q}] = n \geq 1$ , and the relative Frobenius $σ_{q} (x) = x^{q}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields), whose $i$ -th iterate is $x \mapsto x^{q^{i}}$ .

[L1]

$E / F_{q}$ is Galois and $Gal (E / F_{q}) = ⟨ σ_{q} ⟩$ is cyclic of order $n$ (A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ ).

[L2]

In a cyclic group $⟨ g ⟩$ of finite order $n$ , for each positive divisor $c$ of $n$ the subgroup $⟨ g^{n / c} ⟩$ has order $c$ and is the unique subgroup of that order, every subgroup has this form for exactly one such $c$ , and $⟨ g^{n / c} ⟩ \subseteq ⟨ g^{n / c^{'}} ⟩$ if and only if $c ∣ c^{'}$ (A finite cyclic group has exactly one subgroup of each order dividing its own).

[L3]

For $K / F$ finite Galois with $G = Gal (K / F)$ , the assignments $H \mapsto K^{H}$ and $F^{'} \mapsto Gal (K / F^{'})$ are mutually inverse inclusion-reversing bijections between subgroups $H \leq G$ and intermediate fields $F \subseteq F^{'} \subseteq K$ , and $[K : K^{H}] = ∣ H ∣$ , $[K^{H} : F] = [G : H]$ (The fundamental theorem of finite Galois theory).

[L4]

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

[L5]

For an extension of finite fields $L / F_{q}$ of degree $m$ one has $∣ L ∣ = q^{m}$ (For a degree- $n$ extension of a field of order $q$ , the $q$ -power map has order exactly $n$ ).

[L6]

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

[L7]

For a finite group $G$ and $H \leq G$ one has $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

Write $G : = Gal (E / F_{q}) = ⟨ σ_{q} ⟩$ , cyclic of order $n$ by [L1], and for a positive divisor $d$ of $n$ put $H_{d} : = ⟨ σ_{q}^{d} ⟩$ .

2.1

By [L2] applied to $G$ with generator $σ_{q}$ and $c = n / d$ , the subgroup $H_{d} = ⟨ σ_{q}^{n / (n / d)} ⟩$ has order $n / d$ ; every subgroup of $G$ is $H_{d}$ for exactly one positive divisor $d$ of $n$ ; and $H_{e} \subseteq H_{d}$ if and only if $d ∣ e$ , since $H_{e} \subseteq H_{d}$ reads $⟨ σ_{q}^{n / (n / e)} ⟩ \subseteq ⟨ σ_{q}^{n / (n / d)} ⟩$ , which by [L2] says $(n / e) ∣ (n / d)$ , that is $d ∣ e$ .

step 1.1L2algebra

2.2

The fixed field of $H_{d}$ is $E^{H_{d}} = {x \in E : σ_{q}^{d} (x) = x} = {x \in E : x^{q^{d}} = x} = F_{q^{d}}$ , because an element fixed by $σ_{q}^{d}$ is fixed by all its powers and conversely.

step 1.1L6given

3.1

By [L3] the map $H \mapsto E^{H}$ is a bijection from the subgroups of $G$ onto the intermediate fields of $E / F_{q}$ ; composing with the bijection of step 2.1 between positive divisors of $n$ and subgroups, the fields $F_{q^{d}} = E^{H_{d}}$ are exactly the intermediate fields, distinct divisors giving distinct fields.

step 2.1step 2.2L3

3.2

Degrees: [L3] gives $[F_{q^{d}} : F_{q}] = [E^{H_{d}} : F_{q}] = [G : H_{d}]$ , and [L7] with step 2.1 turns this into $∣ G ∣ / ∣ H_{d} ∣ = n / (n / d) = d$ ; then [L5] gives $∣ F_{q^{d}} ∣ = q^{d}$ .

step 2.1step 2.2L3L5L7algebra

4.1

Inclusions: [L3] makes the correspondence inclusion-reversing, so $F_{q^{d}} = E^{H_{d}} \subseteq E^{H_{e}} = F_{q^{e}}$ exactly when $H_{e} \subseteq H_{d}$ , which by step 2.1 holds exactly when $d ∣ e$ . At $d = 1$ one has $H_{1} = G$ and $F_{q^{1}} = E^{G} = F_{q}$ by [L4], and at $d = n$ one has $H_{n} = {i d}$ and $F_{q^{n}} = E$ ; with steps 3.1 and 3.2 this proves every clause.

step 2.1step 2.2step 3.1step 3.2L3L4∎

Remarks

Why the lattice is exactly the divisor lattice. Uniqueness of the subgroup of each order in a cyclic group is what leaves no choice: had $Gal (E / F_{q})$ been the Klein four-group, three distinct subgroups of order two would have produced three intermediate fields of the same degree, and no indexing by divisors could exist.

Depends on

Used by

Dependency tree · two levels

39 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 5.2 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Proposition 4.23 and Corollary 4.21 (standard reference, not scraped)

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