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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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For gcd⁡(n,q)=1 the image of Gal⁡(Fq(μn)/Fq) in (Z/n)× is generated by [q]

Statement

Let Fq be a finite field of order q (Finite fields and their order) and let n≥1 with gcd⁡(n,q)=1 (Coprime integers: gcd⁡(a,b)=1). Then the image of the embedding

Gal⁡(Fq(μn)/Fq)⟶(Z/n)×

of K(μn)/K is Galois and σ↦aσ embeds its Galois group into (Z/n)× is the cyclic subgroup generated by [q]n, and

[Fq(μn):Fq]=ord⁡([q]n),

the multiplicative order of [q]n in (Z/n)× (The order ∣G∣ of a finite group and the order ord⁡(g) of an element, with ord⁡(g)=∞ when no positive power of g is the identity).

Facts & Assumptions

Given: A finite field Fq of order q, of characteristic p with q a power of p (Every finite field has order pn for a unique prime characteristic p and positive integer n), and an integer n≥1 with gcd⁡(n,q)=1; the extension E:=Fq(μn) (The cyclotomic extension K(μn) as a splitting field of tn−1).

[L1]

For a field K and n≥1 with char⁡K∤n, the extension K(μn)/K is finite Galois and σ↦[aσ]n, determined by σ(ζ)=ζaσ on a primitive n-th root of unity, is an injective homomorphism into (Z/n)× with σ(x)=xaσ for every x∈μn (K(μn)/K is Galois and σ↦aσ embeds its Galois group into (Z/n)×).

[L2]

An extension L/Fq of finite fields of degree m is Galois with Gal⁡(L/Fq)=⟨σq⟩ cyclic of order m, where σq(x)=xq (A finite extension of a finite field of order q is Galois with cyclic Galois group generated by x↦xq, The relative Frobenius x↦xq of an extension of finite fields).

[L3]
[L4]

For a finite Galois extension L/F one has ∣Gal⁡(L/F)∣=[L:F] (Equivalent characterizations of a finite Galois extension, The degree [K:F]=dim⁡FK of a finite field extension).

Proof

technique · direct
1.1L1L3given

The characteristic p divides q, and gcd⁡(n,q)=1, so p∤n; hence [L1] applies to K=Fq and E=Fq(μn) is finite Galois over Fq. Also [q]n is a unit of Z/n by [L3].

2.1step 1.1L2

E is a finite field: it is a finite extension of the finite field Fq by step 1.1, so it is a finite-dimensional Fq-vector space over a finite field and therefore has finitely many elements. By [L2], Gal⁡(E/Fq)=⟨σq⟩ with σq(x)=xq.

3.1step 1.1step 2.1L1

The exponent attached to σq by [L1] is [q]n, since σq(ζ)=ζq for a primitive n-th root of unity ζ∈E. Because the embedding is a homomorphism and Gal⁡(E/Fq) is generated by σq, the image is the subgroup of (Z/n)× generated by [q]n.

4.1step 3.1L1L4∎

The embedding is injective, so ∣Gal⁡(E/Fq)∣ equals the order of ⟨[q]n⟩, which is ord⁡([q]n); and [E:Fq]=∣Gal⁡(E/Fq)∣ by [L4]. Hence [E:Fq]=ord⁡([q]n).

Remarks

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