CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The degree $[F (\sqrt[n]{a}) : F]$ is the order of $a (F^{\times})^{n}$ in $F^{\times} / (F^{\times})^{n}$

Statement

Let $F$ be a field, let $n \geq 1$ , assume $char F ∤ n$ , assume $μ_{n} \subseteq F$ , and let $a \in F^{\times}$ . If the class of $a$ in $F^{\times} / (F^{\times})^{n}$ has order $d$ , then

$[F (\sqrt[n]{a}) : F] = d .$

Facts & Assumptions

Given: The field $F$ , the integer $n$ , the element $a \in F^{\times}$ , and the order $d$ of the class $\overset{ˉ}{a} = a (F^{\times})^{n}$ in $F^{\times} / (F^{\times})^{n}$ .

[L1]

The Kummer pairing on the one-generator extension is bilinear and nondegenerate in both variables (The Kummer pairing $Gal (K / F) \times B / (F^{\times})^{n} \to μ_{n}$ is perfect).

Proof

technique · direct

1.1L1

Let $K = F (\sqrt[n]{a})$ . The subgroup of $F^{\times} / (F^{\times})^{n}$ generated by $\overset{ˉ}{a}$ is cyclic of order $d$ , and [L1] gives a homomorphism $ϕ ⁣ : Gal (K / F) \to μ_{n}, ϕ (σ) = ⟨ σ, \overset{ˉ}{a} ⟩ .$ If $ϕ (σ) = 1$ , then $σ$ lies in the left kernel of the pairing, so $σ = 1$ by [L1]. Thus $ϕ$ is injective.

2.1step 1.1L1algebra

Let $H = ϕ (Gal (K / F)) \leq μ_{n}$ . If $∣ H ∣ = m < d$ , then every $h \in H$ satisfies $h^{m} = 1$ , so bilinearity makes $⟨ σ, {\overset{ˉ}{a}}^{m} ⟩ = ⟨ σ, \overset{ˉ}{a} ⟩^{m} = 1$ for every $σ \in Gal (K / F)$ . Then ${\overset{ˉ}{a}}^{m}$ lies in the right kernel, contradicting [L1] because $\overset{ˉ}{a}$ has order $d$ . Therefore $∣ H ∣ = d$ . Since $ϕ$ is injective, this gives $∣ Gal (K / F) ∣ = d .$

3.1step 2.1L1algebra∎

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

Depends on

The Kummer pairing $\operatorname{Gal}(K/F)\times B/(F^\times)^n\to\mu_n$ is perfect

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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

B. Conrad, Kummer Theory, Theorem 5.5 and Theorem 5.12 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Theorem 5.30 (standard reference, not scraped)

The degree [F(an):F][F(na​):F] is the order of a(F×)na(F×)n in F×/(F×)nF×/(F×)n