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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

1.1

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

L1construct

2.1

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

step 1.1

2.2

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.

step 1.1F1construct

3.1

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.

step 2.1step 2.2

4.1

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

step 3.1discharge-construct∎

Depends on

Used by

Dependency tree · two levels

7 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. Kleshchev, Lectures on Abstract Algebra for Graduate Students, sections 3.6, 3.14, and 3.15 (standard reference, not scraped)
The Stacks Project, Algebra (standard reference, not scraped)
P. Hekmati, Homological Algebra, section 3.1 (standard reference, not scraped)