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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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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 αK for which

(α, σα, , σn1α)

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] dimFK=[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 GK× 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)={pF[x]:p(T)=0} is a nonzero ideal with a unique monic generator μT, and p(T)=0 if and only if μTp (The annihilator ideal is nonzero and has a unique monic generator; p(T)=0 if and only if μTp, The annihilator set Ann(T)={pF[x]:p(T)=0}; once existence is proved, its unique monic generator μT is the minimal polynomial).

[L4]

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

[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:pF[x]} is all of V (Cyclic subspaces, cyclic vectors, and vector annihilators).

[L7]

With mT,v the unique monic generator of AnnT(v)={p:p(T)v=0} (The vector annihilator is the unique monic generator of AnnT(v) and divides the minimal polynomial) and d=degmT,v, the list (v,Tv,,Td1v) is an ordered basis of Z(v;T) (A vector annihilator gives a power basis and its companion matrix).

Proof

technique · direct
1.1

Tn=idK, because σ has order n in Gal(K/F); so the polynomial xn1 lies in Ann(T) and μTxn1 by [L2]. In particular degμTn.

L2given
1.2

The maps id,σ,σ2,,σn1 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.

L1given
2.1

If p=i<naixi with aiF 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 degμTn.

step 1.2L1L2
3.1

Combining steps 1.1 and 2.1, degμT=n, and since μT is monic and divides the monic xn1 of the same degree, μT=xn1.

step 1.1step 2.1L2algebra
4.1

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

step 3.1L3L4given
5.1

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

step 4.1L5
6.1

Let d=degmT,α. By [L7] the list (α,Tα,,Td1α) is an ordered basis of Z(α;T)=K, which has dimension n, so d=n and (α,σα,,σn1α) is an ordered F-basis of K.

step 5.1L6L7given
7.1

Since Gal(K/F)={id,σ,,σn1}, that list is exactly the family of conjugates of α, so it is a normal basis.

step 1.2step 6.1given

Remarks

  • Why the minimal polynomial is forced to be xn1. The divisibility μTxn1 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 xn1 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

Used by

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Sources