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

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.

In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$

Statement

For every left $R$ -module $M$ , scalar $r \in R$ , and element $m \in M$ ,

$0_{R} m = 0_{M}, r 0_{M} = 0_{M}, (- r) m = - (r m), r (- m) = - (r m) .$

Facts & Assumptions

Given: A left $R$ -module $M$ , $r \in R$ , and $m \in M$ .

[L1]

The module action distributes over both addition operations, and $(M, +, 0_{M})$ is an abelian group (Unital left and right modules over a ring; unqualified module means left module).

[L2]

The additive structure $(R, +, 0_{R})$ is an abelian group, so $r + (- r) = 0_{R}$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L3]

Cancellation holds in every group, hence in the additive groups of $R$ and $M$ (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution).

Proof

technique · direct

1.1

Since $0_{R} = 0_{R} + 0_{R}$ , distributivity gives $0_{R} m = 0_{R} m + 0_{R} m$ ; cancellation yields $0_{R} m = 0_{M}$ .

L1L3given

1.2

Since $0_{M} = 0_{M} + 0_{M}$ , distributivity gives $r 0_{M} = r 0_{M} + r 0_{M}$ ; cancellation yields $r 0_{M} = 0_{M}$ .

L1L3given

2.1

From $(r + (- r)) m = 0_{R} m = 0_{M}$ , distributivity and step 1.1 give $r m + (- r) m = 0_{M}$ , so $(- r) m = - (r m)$ .

step 1.1L1L2L3given

3.1

From $r (m + (- m)) = r 0_{M} = 0_{M}$ , distributivity and step 1.2 give $r m + r (- m) = 0_{M}$ , so $r (- m) = - (r m)$ .

step 1.2L1L3given∎

Depends on

Used by

Dependency tree · two levels

12 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

McGerty, Algebra II: Rings and Modules, Section 3 (standard reference, not scraped)

In a module, 0Rm=0M0R​m=0M​, r0M=0Mr0M​=0M​, (−r)m=−(rm)(−r)m=−(rm) and r(−m)=−(rm)r(−m)=−(rm)