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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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Every finite cyclic extension has a normal basis

Statement

Let K/F be a finite Galois extension whose Galois group is cyclic, say Gal⁡(K/F)=⟨σ⟩ of order n=[K:F]. Then K/F has a normal basis (Normal bases of a finite Galois extension): there is α∈K for which

(α, σα, …, σn−1α)

is an ordered F-basis of K (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

Facts & Assumptions

Given: A finite Galois extension K/F with Gal⁡(K/F)=⟨σ⟩ cyclic of order n; by [L6] dim⁡FK=[K:F]=n, and σ is an F-linear endomorphism of the F-vector space K, written T when regarded as such. Evaluation of a polynomial at T is that of Polynomial evaluation at an endomorphism: p(T)=∑kakTk.

[L1]

Let G be a group and K a field. Every finite family of distinct group homomorphisms G→K× is linearly independent over K as a family of functions (Dedekind's linear independence theorem for distinct characters).

[L2]

For every endomorphism T of a finite-dimensional F-vector space, Ann⁡(T)={p∈F[x]:p(T)=0} is a nonzero ideal with a unique monic generator μT, and p(T)=0 if and only if μT∣p (The annihilator ideal is nonzero and has a unique monic generator; p(T)=0 if and only if μT∣p, The annihilator set Ann⁡(T)={p∈F[x]:p(T)=0}; once existence is proved, its unique monic generator μT is the minimal polynomial).

[L4]

μT∣χT for every endomorphism T of a finite-dimensional vector space (The minimal polynomial divides the characteristic polynomial, μT∣χT).

[L5]

An endomorphism T of a finite-dimensional vector space has a cyclic vector if and only if μT=χT (A cyclic vector exists exactly when the minimal and characteristic polynomials agree); v is a cyclic vector when Z(v;T)={p(T)v:p∈F[x]} is all of V (Cyclic subspaces, cyclic vectors, and vector annihilators).

[L7]

With mT,v the unique monic generator of Ann⁡T(v)={p:p(T)v=0} (The vector annihilator is the unique monic generator of Ann⁡T(v) and divides the minimal polynomial) and d=deg⁡mT,v, the list (v,Tv,…,Td−1v) is an ordered basis of Z(v;T) (A vector annihilator gives a power basis and its companion matrix).

Proof

technique · direct
1.1L2given

Tn=idK, because σ has order n in Gal⁡(K/F); so the polynomial xn−1 lies in Ann⁡(T) and μT∣xn−1 by [L2]. In particular deg⁡μT≤n.

1.2L1given

The maps id,σ,σ2,…,σn−1 are pairwise distinct elements of Gal⁡(K/F), and their restrictions to K× are pairwise distinct group homomorphisms K×→K×, since two field automorphisms of K agreeing on K× agree on K.

2.1step 1.2L1L2

If p=∑i<naixi with ai∈F satisfies p(T)=0, then ∑i<naiσi is the zero function on K, hence on K×, so [L1] applied to the family of step 1.2 forces every ai to be 0; thus no nonzero polynomial of degree less than n annihilates T, and deg⁡μT≥n.

3.1step 1.1step 2.1L2algebra

Combining steps 1.1 and 2.1, deg⁡μT=n, and since μT is monic and divides the monic xn−1 of the same degree, μT=xn−1.

4.1step 3.1L3L4given

By [L3] the polynomial χT is monic of degree dim⁡FK=n, and μT∣χT by [L4]; two monic polynomials of the same degree, one dividing the other, are equal, so μT=χT.

5.1step 4.1L5

By [L5] there is a cyclic vector α∈K for T, that is Z(α;T)=K.

6.1step 5.1L6L7given

Let d=deg⁡mT,α. By [L7] the list (α,Tα,…,Td−1α) is an ordered basis of Z(α;T)=K, which has dimension n, so d=n and (α,σα,…,σn−1α) is an ordered F-basis of K.

7.1step 1.2step 6.1given∎

Since Gal⁡(K/F)={id,σ,…,σn−1}, that list is exactly the family of conjugates of α, so it is a normal basis.

Remarks

  • Why the minimal polynomial is forced to be xn−1. The divisibility μT∣xn−1 is cheap; the content is the lower bound on its degree, and that is exactly Dedekind's independence of characters. Without it the minimal polynomial could be a proper divisor of xn−1 and no cyclic vector would be available.

  • The hypothesis is on the group, not on the base field. No finiteness or infiniteness of F is used, so this proof also covers cyclic extensions of infinite fields, for which Every finite Galois extension of an infinite field has a normal basis gives a second and quite different argument.

Depends on

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Sources