TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Conjugacy of decomposition and inertia groups

Statement

In finite Galois L/K, if $σ P = P^{'}$ above a nonzero p, then $D (P^{'} / p) = σ D (P / p) σ^{- 1}, I (P^{'} / p) = σ I (P / p) σ^{- 1} .$ The residue actions correspond under $κ (P) \to κ (P^{'})$ , $\overset{ˉ}{a} \mapsto \overline{σ a}$ .

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Inertia group of a prime: For finite Galois L/K and a chosen nonzero prime $P ∣ p$ , set $κ (P) = O_{L} / P$ and $κ (p) = O_{K} / p$ . Each $σ \in D (P / p)$ induces a $κ (p)$ -automorphism $\overset{ˉ}{σ}$ of $κ (P)$ : the rule $\overset{ˉ}{a} \mapsto \overline{σ a}$ is independent of the representative because $σ P = P$ . The inertia group is $I (P / p) = \ker ⁣ (D (P / p) ⟶ Gal (κ (P) / κ (p))) .$ Equivalently, $σ \in I (P / p)$ exactly when $σ \in D (P / p)$ and $σ (a) - a \in P$ for every $a \in O_{L}$ . It is a normal subgroup of D.

Proof

1.1F1

For $τ \in G$ , the equality $τ P = P$ is equivalent to $(σ τ σ^{- 1}) P^{'} = P^{'}$ . This proves both subgroup inclusions for D.

2.1F1step 1.1∎

The displayed residue map is well-defined and invertible, with inverse induced by $σ^{- 1}$ . For $τ \in D (P / p)$ and $a \in O_{L}$ , transporting $\overline{τ a}$ gives $\overline{σ τ a}$ , exactly the action of $σ τ σ^{- 1}$ on $\overline{σ a}$ . Therefore the residue action is identity on one side exactly when it is identity on the other. Taking kernels proves the equality for I.

Depends on

Inertia group of a prime

Used by

Decomposition and inertia in towers Theorem
Frobenius elements above a prime are conjugate Theorem

Dependency tree · two levels

3 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

§9.3, Lemma 9.3.2, p.104 (standard reference, not scraped)