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

1.1

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.

F1algebra

2.1

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.

F1F2step 1.1

3.1

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.

F3step 1.1step 2.1∎

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)