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Kummer theory classifies finite abelian extensions of exponent dividing $n$ by subgroups between $(F^{\times})^{n}$ and $F^{\times}$

Statement

Let $F$ be a field, let $n \geq 1$ , assume $char F ∤ n$ , and assume $μ_{n} \subseteq F$ . Then the assignment

$B ⟼ F (B^{1 / n})$

is an inclusion-preserving bijection between

subgroups $B$ with $(F^{\times})^{n} \subseteq B \subseteq F^{\times}$ and finite quotient $B / (F^{\times})^{n}$ , and
finite abelian extensions $K / F$ whose Galois group has exponent dividing $n$ .

For corresponding objects one has

$Gal (K / F) ≅ Hom (B / (F^{\times})^{n}, μ_{n}) .$

Facts & Assumptions

Given: The field $F$ , the integer $n$ , a subgroup $B$ as above, and a finite abelian extension $K / F$ of exponent dividing $n$ .

[F1]

A Kummer extension is a finite Galois extension of the form $F (B^{1 / n})$ with $μ_{n} \subseteq F$ (Kummer extensions from adjoining $n$ -th roots over a base field containing $μ_{n}$ ).

[L1]

The Kummer pairing for $K = F (B^{1 / n})$ is perfect, in particular bilinear and nondegenerate in both variables (The Kummer pairing $Gal (K / F) \times B / (F^{\times})^{n} \to μ_{n}$ is perfect).

[L2]

Cyclic degree- $d$ extensions over a base containing $μ_{d}$ are exactly the extensions generated by a $d$ -th root whose defining polynomial $x^{d} - a$ is irreducible (If $μ_{n} \subseteq F$ and $char F ∤ n$ , then a degree- $n$ extension is cyclic exactly when it is $F (α)$ with $α^{n} \in F$ and $x^{n} - α^{n}$ irreducible).

[L3]

Artin's fixed-field theorem computes degrees from finite automorphism groups (Artin's fixed-field theorem: $[K : K^{G}] = ∣ G ∣$ and $Aut (K / K^{G}) = G$ ).

[L4]

In a finite Galois extension, normal intermediate fields correspond to quotient Galois groups (Normal subgroups, conjugate fields, and quotient groups in the Galois correspondence).

[L5]

Every finite abelian group is a finite direct product $C_{n_{1}} \times \dots \times C_{n_{r}}$ with $1 < n_{1} ∣ \dots ∣ n_{r}$ (Fundamental theorem of finite abelian groups: invariant-factor form).

Proof

technique · direct

1.1F1choosealgebra

Let $B$ be a subgroup with finite quotient. Because $B / (F^{\times})^{n}$ is finite, choose elements $a_{1}, \dots, a_{r} \in B$ whose classes generate it. Then $F (B^{1 / n}) = F (a_{1}^{1 / n}, \dots, a_{r}^{1 / n}) .$ For each $j$ , the field $F (a_{j}^{1 / n})$ contains all roots of $x^{n} - a_{j}$ because $μ_{n} \subseteq F$ , so it is a finite Galois extension whose automorphism group embeds in $μ_{n}$ by the rule $σ \mapsto σ (a_{j}^{1 / n}) / a_{j}^{1 / n}$ . Thus each one-generator step is cyclic of degree dividing $n$ . The compositum $K : = F (B^{1 / n})$ is therefore finite Galois and abelian, and every automorphism acts on every chosen $n$ -th root by multiplication with an element of $μ_{n}$ . Hence the exponent of $Gal (K / F)$ divides $n$ .

1.2L1

For this extension, the Kummer pairing of [L1] is perfect, so the natural maps $Gal (K / F) ⟶ Hom (B / (F^{\times})^{n}, μ_{n}), B / (F^{\times})^{n} ⟶ Hom (Gal (K / F), μ_{n})$ are isomorphisms because both groups are finite and the pairing is nondegenerate in both variables. In particular, $Gal (K / F) ≅ Hom (B / (F^{\times})^{n}, μ_{n}) .$

1.3algebra

Conversely, let $K / F$ be finite abelian of exponent dividing $n$ and define $B (K) : = {b \in F^{\times} : some n -th root of b lies in K} .$ This contains $(F^{\times})^{n}$ . For $b \in B (K)$ choose $β \in K$ with $β^{n} = b$ and define $χ_{b} (σ) : = \frac{σ (β)}{β} \in μ_{n} (σ \in Gal (K / F)) .$ As in the Kummer-pairing proof, this depends only on the class of $b$ modulo $(F^{\times})^{n}$ , and if $χ_{b} = 1$ then every $σ$ fixes $β$ , so $β \in F$ and $b \in (F^{\times})^{n}$ . Therefore $B (K) / (F^{\times})^{n} ↪ Hom (Gal (K / F), μ_{n}),$ so the quotient is finite.

1.4L2L3L4L5algebrachoose

Let $L : = F (B (K)^{1 / n}) \subseteq K$ and put $G = Gal (K / F)$ . If $σ \in G$ is nontrivial, use [L5] to write $G ≅ C_{n_{1}} \times \dots \times C_{n_{r}}$ . Each $n_{j}$ divides $n$ because the exponent of $G$ divides $n$ . Some coordinate of $σ$ is nonzero, say in $C_{n_{j}}$ . If $ζ \in F$ is a primitive $n$ -th root of unity, then $η = ζ^{n / n_{j}}$ has order $n_{j}$ , so projection to this coordinate followed by a generator-preserving map $C_{n_{j}} \to ⟨ η ⟩$ gives a character $χ : G \to μ_{n}$ with $χ (σ) \neq 1$ . Put $H = \ker χ$ and $d = ∣ im χ ∣$ . Then $H ⊴ G$ , $d ∣ n$ , and [L4] makes $E : = K^{H} / F$ cyclic of degree $d$ . Since $char F ∤ d$ and $μ_{d} \subseteq F$ , [L2] gives $E = F (γ)$ with $γ^{d} \in F$ . Hence $γ^{n} = (γ^{d})^{n / d} \in F$ , so $γ \in L$ . But $χ (σ) \neq 1$ means $σ \notin H = Gal (K / E)$ , so $σ$ acts nontrivially on $E = F (γ)$ and therefore does not fix $L$ . Thus $Gal (K / L) = 1$ , and [L3] gives $[K : L] = 1$ . Therefore $K = F (B (K)^{1 / n})$ .

2.1step 1.4

Starting from a finite abelian extension $K / F$ , step 1.4 shows that the radicals coming from $B (K)$ generate $K$ , so the field-side composite is the identity.

2.2step 1.2algebra

Starting from a subgroup $B$ and putting $K = F (B^{1 / n})$ , one has $B \subseteq B (K)$ by definition. For the reverse inclusion, let $c \in B (K)$ and choose $β_{c} \in K$ with $β_{c}^{n} = c$ . Step 1.2 identifies the perfect pairing with an isomorphism $B / (F^{\times})^{n} ≅ Hom (Gal (K / F), μ_{n}),$ so the character $χ_{c} (σ) = σ (β_{c}) / β_{c}$ equals $χ_{b}$ for some class $\overset{ˉ}{b} \in B / (F^{\times})^{n}$ represented by $β_{b}^{n} = b$ with $b \in B$ . Then for every $σ \in Gal (K / F)$ one has $\frac{σ (β_{c} / β_{b})}{β_{c} / β_{b}} = \frac{χ_{c} (σ)}{χ_{b} (σ)} = 1,$ so $β_{c} / β_{b} \in F$ . Hence $c / b = (β_{c} / β_{b})^{n} \in (F^{\times})^{n} \subseteq B$ , and therefore $c \in B$ . Thus $B (K) = B$ .

3.1step 1.2step 2.1step 2.2∎

Inclusion preservation is immediate from the definitions, and the displayed Galois-group formula is exactly the isomorphism of step 1.2.

Remarks

This page uses the cohomology-free version. The proof is organized through cyclic Kummer subextensions and the concrete pairing, not through $H^{1}$ .

Depends on

Used by

ℚ(ζ₃,³√2,³√3) is a Kummer extension with quotient (ℤ/3)² Example

Dependency tree · two levels

29 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

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

Kummer theory classifies finite abelian extensions of exponent dividing nn by subgroups between (F×)n(F×)n and F×F×