TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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The quotient action is well defined and makes $M / N$ a module

Statement

Let $N \leq M$ be a submodule of a left $R$ -module. The rule

$r (m + N) : = r m + N$

is independent of the representative $m + N$ and, together with the additive quotient group, makes $M / N$ a left $R$ -module.

Facts & Assumptions

Given: A left $R$ -module $M$ and a submodule $N \leq M$ .

[L1]

Additive cosets satisfy $m + N = m^{'} + N$ exactly when $m^{'} - m \in N$ ( $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

[L2]

A submodule is closed under scalar multiplication and is an additive subgroup (Submodule of a module).

[L3]

The additive cosets form a quotient group with the inherited addition (For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

[L4]

The module axioms give distributivity, associativity of scalar action, and $1_{R} m = m$ (Unital left and right modules over a ring; unqualified module means left module).

[L5]

Scalar multiplication preserves additive negatives (In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$ ).

Proof

technique · direct

1.1L1L2L4L5given

If $m + N = m^{'} + N$ , then $m^{'} - m \in N$ by [L1]. Thus $r (m^{'} - m) = r m^{'} - r m \in N$ by [L2, L4, L5], and [L1] gives $r m + N = r m^{'} + N$ ; the proposed scalar action is well defined.

1.2L3L4given

The quotient addition is an abelian group operation: it is a quotient-group operation by [L3], and $(m + N) + (m^{'} + N) = (m + m^{'}) + N = (m^{'} + m) + N = (m^{'} + N) + (m + N)$ .

1.3L4given

For cosets, the module identities in $M$ give $r ((m + N) + (m^{'} + N)) = r (m + m^{'}) + N = (r m + r m^{'}) + N = r (m + N) + r (m^{'} + N)$ , $(r + s) (m + N) = r (m + N) + s (m + N)$ , $(r s) (m + N) = r (s (m + N))$ , and $1_{R} (m + N) = m + N$ .

2.1step 1.1step 1.2step 1.3L4∎

Steps 1.1--1.3 verify a well-defined scalar action on an abelian group satisfying all module axioms; hence $M / N$ is a left $R$ -module.

Depends on

Dependency tree · 2 levels

Sources

McGerty, Algebra II: Rings and Modules, Section 3

The quotient action is well defined and makes M/NM/N a module