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The elements of a finite extension fixed by the $q$ -power map are exactly the base field

Statement

Let $F_{q}$ be a finite field of order $q$ and let $E$ be a finite field having $F_{q}$ as a subfield. Then

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

Equivalently, the fixed field of the cyclic group generated by the relative Frobenius $σ_{q}$ (The relative Frobenius $x \mapsto x^{q}$ of an extension of finite fields) is the base field:

$E^{⟨ σ_{q} ⟩} = F_{q} .$

Facts & Assumptions

Given: Finite fields $F_{q} \subseteq E$ with $∣ F_{q} ∣ = q$ (Finite fields and their order), and the set $S : = {x \in E : x^{q} = x}$ .

[L1]

The relative Frobenius is $σ_{q} (x) = x^{q}$ , an $F_{q}$ -automorphism of $E$ (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]

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

Proof

technique · direct

1.1L2given

Applying [L2] to the field $F_{q}$ , which has exactly $q$ elements, every $a \in F_{q}$ satisfies $a^{q} = a$ ; hence $F_{q} \subseteq S$ .

1.2L3given

$S$ is the set of roots in $E$ of the polynomial $t^{q} - t \in E [t]$ , which is nonzero of degree $q$ ; a field is an integral domain, so [L3] gives $∣ S ∣ \leq q$ .

2.1step 1.1step 1.2given

Since $F_{q} \subseteq S$ , $∣ F_{q} ∣ = q$ and $∣ S ∣ \leq q$ , the finite sets $F_{q}$ and $S$ coincide: ${x \in E : x^{q} = x} = F_{q}$ .

3.1step 2.1L1L4∎

An element of $E$ is fixed by every power of $σ_{q}$ precisely when it is fixed by $σ_{q}$ itself, so $E^{⟨ σ_{q} ⟩} = {x \in E : x^{q} = x}$ by [L1] and [L4], and step 2.1 identifies this with $F_{q}$ .

Remarks

What forces equality is a count, not an inclusion. The inclusion $F_{q} \subseteq S$ is immediate; the content is that $t^{q} - t$ cannot have more than $q$ roots, so the $q$ elements of $F_{q}$ already use them all up. The same count identifies the intermediate fields of an extension of finite fields (The intermediate fields of $F_{q^{n}} / F_{q}$ are the $F_{q^{d}}$ , one for each positive divisor $d$ of $n$ ).

Depends on

Used by

Dependency tree · two levels

16 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, Roots and Irreducibles (expository blurb), Lemma 4.2 (standard reference, not scraped)
K. Conrad, Finite Fields (expository blurb), Section 5 (standard reference, not scraped)

The elements of a finite extension fixed by the qq-power map are exactly the base field