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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Orders of decomposition and inertia groups

Statement

For finite Galois L/K and nonzero $P ∣ p$ , writing e and f for its ramification index and residue degree, $∣ D (P / p) ∣ = e f, ∣ I (P / p) ∣ = e, ∣ D (P / p) / I (P / p) ∣ = f .$ The prime P is unramified over p if and only if its inertia group is trivial.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Decomposition inertia exact sequence: For finite Galois L/K and fixed nonzero $P ∣ p$ , reduction gives the exact sequence $1 ⟶ I (P / p) ⟶ D (P / p) ⟶ Gal (κ (P) / κ (p)) ⟶ 1.$ In particular $D (P / p) / I (P / p)$ is canonically the residue Galois group.

[F2]

Decomposition group and completion: Let L/K be finite Galois and $P ∣ p$ nonzero primes. Then $L_{P} / K_{p}$ is finite Galois of degree $e (P / p) f (P / p)$ . Continuous extension gives a canonical isomorphism $D (P / p) ≅ Gal (L_{P} / K_{p}),$ whose inverse restricts an automorphism to the embedded copy of L.

[F3]

A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x \mapsto x^{q}$ : Let $F_{q}$ be a finite field of order $q$ and let $E$ be a finite field having $F_{q}$ as a subfield, with $[E : F_{q}] = n$ (def-extension-degree-and-finite-extension). Then $E / F_{q}$ is a finite Galois extension (def-finite-galois-extension-and-galois-group) and $Gal (E / F_{q}) = ⟨ σ_{q} ⟩$ is cyclic of order $n$ , generated by the relative Frobenius $σ_{q} ⁣ : x \mapsto x^{q}$ (def-relative-frobenius-of-a-finite-field-extension).

Proof

1.1

The local correspondence gives $∣ D ∣ = e f$ . The residue extension has cyclic Galois group of order f, so the exact sequence gives $∣ D / I ∣ = f$ and $∣ I ∣ = ∣ D ∣ / f = e$ .

F1F2F3

2.1

Finite residue extensions are separable. Thus here unramified means e=1, which implies $∣ I ∣ = 1$ and I trivial. Conversely trivial I has order one, so e=1 and the prime is unramified.

F3step 1.1∎

Depends on

Used by

Decomposition inertia in a quadratic field Example
Good polynomial reduction kills inertia Lemma
Decomposition and inertia fixed fields Theorem
Decomposition and inertia in towers Theorem
Unramified frobenius element exists uniquely Theorem

Dependency tree · two levels

18 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

§9.3.2, Corollary 9.3.7, p.106 (standard reference, not scraped)