CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-13

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.

Every module is a quotient of a free module

Statement

For every left $R$ -module $M$ , the free module $R^{(M)}$ on its underlying set admits a canonical surjection $ε_{M} : R^{(M)} \to M$ , determined by $ε_{M} (e_{m}) = m$ . Consequently $M ≅ R^{(M)} / \ker ε_{M}$ .

Facts & Assumptions

Given: A left $R$ -module $M$ .

[L1]

Every set map from a basis set to a module extends uniquely to a homomorphism from the free module (Universal property of the free module on a set).

[F1]

The quotient module $F / K$ consists of additive cosets and carries the induced module operations (Quotient module $M / N$ with scalar multiplication on additive cosets).

Proof

technique · constructive

1.1L1construct

Apply [L1] to the identity set map on the underlying set of $M$ ; this defines $ε_{M} : R^{(M)} \to M$ with $ε_{M} (e_{m}) = m$ .

2.1step 1.1

Every $m \in M$ equals $ε_{M} (e_{m})$ , so $ε_{M}$ is surjective, including when $M = 0$ .

2.2step 1.1F1construct

Define $ϕ : R^{(M)} / \ker ε_{M} \to M$ by $ϕ (x + \ker ε_{M}) = ε_{M} (x)$ . Equality of cosets makes this well defined, and [F1] makes it a homomorphism.

3.1step 2.1step 2.2

The map $ϕ$ is surjective by step 2.1 and injective because $ϕ (x + \ker ε_{M}) = 0$ exactly when $x \in \ker ε_{M}$ . Hence it is an isomorphism.

4.1step 3.1discharge-construct∎

Thus $M$ is canonically a quotient of a free module.

Every module is a quotient of a free module

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources