ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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.

The polynomial module of a two-by-two operator

Example

For

$T = (\begin{array}{cc} 0 & - 2 \\ 1 & 0 \end{array}) on Q^{2},$

the polynomial module is

$V_{T} ≅ Q [x] / (x^{2} + 2) .$

Its sole invariant factor, minimal polynomial, and characteristic polynomial are all $x^{2} + 2$ , and the displayed matrix is the companion matrix $C (x^{2} + 2)$ .

Facts & Assumptions

Given: The $Q [x]$ -module $V_{T}$ of The $F [x]$ -module $V_{T}$ of an endomorphism, its torsion and annihilator theorem (For finite-dimensional $V$ , $V_{T}$ is finitely generated and torsion, with annihilator generated by the minimal polynomial), and the largest-factor dictionary (On a nonzero finite-dimensional space, the largest invariant factor is the minimal polynomial).

[L1]

In a cyclic power basis the restriction has the companion matrix with ones on the subdiagonal and the negative coefficients in the last column (A vector annihilator gives a power basis and its companion matrix).

Verification

technique · direct

1.1

For $e_{1} = (1, 0)$ , one has $T e_{1} = (0, 1) = e_{2}$ , so $e_{1}, T e_{1}$ is a basis of $Q^{2}$ .

algebra

2.1

Direct multiplication gives $T^{2} e_{1} = - 2 e_{1}$ , and no nonzero polynomial of degree below two kills $e_{1}$ because the vectors in step 1.1 are independent. Thus the vector annihilator is $x^{2} + 2$ , and [L1] gives its displayed companion matrix.

step 1.1L1

3.1

The cyclic map $Q [x] \to V_{T}$ , $p \mapsto p (T) e_{1}$ , has kernel $(x^{2} + 2)$ and is surjective, so it gives the displayed module isomorphism. Consequently the sole invariant factor is $x^{2} + 2$ ; it is the minimal polynomial, and the companion determinant gives the same characteristic polynomial.

step 2.1given∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

A. Apisa, Wisconsin Math 542, cyclic module examples (standard reference, not scraped)
M. Brussel, Finitely Generated Modules over a PID, Section 5 (standard reference, not scraped)