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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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The degree [F(an):F] is the order of a(F×)n in F×/(F×)n

Statement

Let F be a field, let n1, assume charFn, assume μnF, and let aF×. If the class of a in F×/(F×)n has order d, then

[F(an):F]=d.

Facts & Assumptions

Given: The field F, the integer n, the element aF×, and the order d of the class aˉ=a(F×)n in F×/(F×)n.

[L1]

The Kummer pairing on the one-generator extension is bilinear and nondegenerate in both variables (The Kummer pairing Gal(K/F)×B/(F×)nμn is perfect).

Proof

technique · direct
1.1

Let K=F(an). The subgroup of F×/(F×)n generated by aˉ is cyclic of order d, and [L1] gives a homomorphism ϕ ⁣:Gal(K/F)μn,ϕ(σ)=σ,aˉ. If ϕ(σ)=1, then σ lies in the left kernel of the pairing, so σ=1 by [L1]. Thus ϕ is injective.

L1
2.1

Let H=ϕ(Gal(K/F))μn. If H=m<d, then every hH satisfies hm=1, so bilinearity makes σ,aˉm=σ,aˉm=1 for every σGal(K/F). Then aˉm lies in the right kernel, contradicting [L1] because aˉ has order d. Therefore H=d. Since ϕ is injective, this gives Gal(K/F)=d.

step 1.1L1algebra
3.1

The extension K/F is finite Galois in this one-generator Kummer situation, so its degree equals the order of its Galois group. Hence [K:F]=d.

step 2.1L1algebra

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources