ExampleConstruction: AI-generatedVerification: 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.

Decomposition groups in a tower

Example

Let $M = Q (ζ_{5})$ , $L = Q (\sqrt{5})$ , and $K = Q$ . At p=2 there is a unique prime in each field above p. All ramification indices are one, and the residue degrees are four in M/K and two in each step. The decomposition sequence is $1 ⟶ C_{2} ⟶ C_{4} ⟶ C_{2} ⟶ 1,$ and all inertia groups are trivial. The relative Frobenius for M/L is the square of the Frobenius for M/K.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Frobenius in a small cyclotomic field: Let $ζ = ζ_{5}$ be a primitive fifth root of unity and $L = Q (ζ)$ . Then $Gal (L / Q) ≅ (Z / 5 Z)^{\times} = C_{4}$ by $σ_{a} (ζ) = ζ^{a}$ . For every prime $p \neq 5$ , p is unramified and its arithmetic Frobenius is $σ_{p m o d 5}$ . In particular 2 is inert, with $(e, f, g) = (1, 4, 1)$ .

[F2]

Decomposition and inertia in towers: Let M/L/K be a tower of number fields with M/K and L/K finite Galois, and fix nonzero primes $Q ∣ P ∣ p$ . With $H = Gal (M / L)$ , $D (Q / P) = D (Q / p) \cap H, I (Q / P) = I (Q / p) \cap H .$ Restriction gives exact sequences $1 \to D (Q / P) \to D (Q / p) \to D (P / p) \to 1,$ $1 \to I (Q / P) \to I (Q / p) \to I (P / p) \to 1.$ The intersection identities also hold without L/K Galois; the displayed quotient assertions use that hypothesis.

[F3]

Frobenius compatibility in finite towers: Let M/L/K have M/K and L/K finite Galois, and let $Q ∣ P ∣ p$ be nonzero primes with Q unramified over p. Then $Frob (Q / p) ∣_{L} = Frob (P / p), Frob (Q / P) = Frob (Q / p)^{f (P / p)} .$ If $M = L_{1} L_{2}$ with both $L_{i} / K$ finite Galois and Q unramified over p, the Frobenius elements at its two contractions determine $Frob (Q / p)$ uniquely by restriction.

Verification

1.1F1algebra

Put $t = ζ_{5} + ζ_{5}^{- 1}$ . Expanding and using $1 + ζ_{5} + \dots + ζ_{5}^{4} = 0$ gives $t^{2} + t - 1 = 0$ . This polynomial has discriminant 5, not a rational square, so $Q (t) = Q (\sqrt{5})$ has degree two. Exponent 2 sends t to $ζ_{5}^{2} + ζ_{5}^{- 2} = - 1 - t \neq t$ , while exponent 4 fixes t. Thus restriction has kernel the order-two subgroup generated by exponent 4.

2.1F1F2step 1.1

At 2, M/K is unramified and inert with degree four, so there is one prime Q, D(Q/p)=C4 and I(Q/p)=1. Every prime of L above p has a prime of M above it by integral prime factorization, so it is the contraction P of Q. The tower exact sequences now give D(Q/P)=C2 and D(P/p)=C2 with all inertia groups trivial. In the unramified residue isomorphisms these group orders are the residue degrees, giving two in each step.

3.1F3step 1.1step 2.1∎

Compatibility says that the absolute exponent-2 Frobenius restricts to the nonidentity automorphism of L, and the relative Frobenius is its $f (P / p) = 2$ power, namely exponent 4. This realizes the displayed exact sequence.

Depends on

Used by

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Sources

Chapter 8, Propositions 8.13 and 8.15–8.16; Example 8.18 (standard reference, not scraped)