LemmaStatement: AI-generatedProof: AI-generated

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.

Common projective complements and cancellation by free modules

Statement

Let $A$ be a commutative ring with unit. Call a finite projective $A$ -module $P$ cancellative if, for every finite projective $Q$ and every integer $N \geq 0$ , $P \oplus A^{N} ≅ Q \oplus A^{N}$ implies $P ≅ Q$ . This is equivalent to cancellation of an arbitrary common finite projective summand: $P \oplus T ≅ Q \oplus T$ implies $P ≅ Q$ .

Facts & Assumptions

Given: The data and hypotheses in the statement.

Goal: Establish the stated conclusion.

[F1]

The choice-free module clause states that every finite projective $T$ has a finite projective complement $T'$ with $T \oplus T' ≅ A^{N}$ . Finite projective complements

Proof

1.1F1given

If $P$ is cancellative in the free-summand sense and $P \oplus T ≅ Q \oplus T$ , add the complement $T'$ to both sides. Associativity converts this into $P \oplus A^{N} ≅ Q \oplus A^{N}$ , so $P ≅ Q$ .

2.1givenstep 1.1∎

Conversely $A^{N}$ is finite projective and can itself be used as the common summand $T$ . Thus cancellation of arbitrary common summands implies the free-summand definition. This also proves the definition has the claimed precise meaning.

Depends on

Finite projective complements

Used by · 1

Local application interface for a critical-rank cancellation theorem

Dependency tree · 3 levels

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