CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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.

Each projection in the primary decomposition is a polynomial in the endomorphism

Statement

In the primary decomposition $V = ⨁_{i < r} V_{i}$ associated with $μ_{T} = \prod_{i < r} f_{i}$ , the projection $E_{i} : V \to V_{i}$ along the other primary components is a polynomial in $T$ . More precisely, if $g_{i} = μ_{T} / f_{i}$ and $a_{i} f_{i} + b_{i} g_{i} = 1$ , then

$E_{i} = b_{i} (T) g_{i} (T) .$

Facts & Assumptions

Given: The primary decomposition and the polynomials $f_{i}, g_{i}$ in the Statement.

[L1]

The irreducible-power factors of $μ_{T}$ give the direct sum $V = ⨁_{i} \ker f_{i} (T)$ (Primary decomposition: the irreducible-power factors of $μ_{T}$ split $V$ into their invariant kernels).

[L2]

Coprime polynomials satisfy a Bézout identity (Bézout identity and the Euclidean algorithm for polynomials over a field).

[L3]

Polynomial evaluation sends $p (x) = \sum a_{k} x^{k}$ to the endomorphism $p (T) = \sum a_{k} T^{k}$ (Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ ).

Proof

technique · direct

1.1L2L3chooseconstruct

The polynomials $f_{i}$ and $g_{i}$ are coprime, so choose $a_{i}, b_{i}$ with $a_{i} f_{i} + b_{i} g_{i} = 1$ by [L2], and set $E_{i} = b_{i} (T) g_{i} (T)$ using [L3].

2.1step 1.1L1algebra

On $V_{i} = \ker f_{i} (T)$ the evaluated Bézout identity gives $E_{i} = I$ . On $V_{j}$ for $j \neq i$ , the polynomial $g_{i}$ is divisible by $f_{j}$ , so $E_{i} = 0$ .

3.1step 2.1L1∎

By the unique decomposition in [L1], the operator described in step 2.1 is exactly projection onto $V_{i}$ along the sum of the other components.

Depends on

Used by

Dependency tree · two levels

11 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

Anthony W. Knapp, Basic Algebra, 2nd ed., Ch. V, §5, Theorem 5.19 (standard reference, not scraped)