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.

The one-step submodule criterion; intersections and sums of submodules are submodules

Statement

Let $M$ be a left $R$ -module. A nonempty subset $S \subseteq M$ is a submodule if and only if

$r u + v \in S (r \in R, u, v \in S) .$

Consequently, the intersection of every nonempty family of submodules is a submodule, and, for submodules $A, B \leq M$ ,

$A + B : = {a + b : a \in A, b \in B}$

is a submodule of $M$ .

Facts & Assumptions

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

[L1]

The module axioms include $1_{R} m = m$ and distributivity of scalar multiplication over both additions (Unital left and right modules over a ring; unqualified module means left module).

[L2]

In a module, $(- r) m = - (r m)$ ; taking $r = 1_{R}$ and using [L1] gives $(- 1_{R}) m = - m$ (In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$ ).

[L3]

A nonempty subset of a group is a subgroup exactly when it is closed under $x y^{- 1}$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ).

[L4]

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

Proof

technique · direct

1.1L4given

If $S$ is a submodule, then $u \in S$ implies $r u \in S$ , and then $r u + v \in S$ for $r \in R$ and $u, v \in S$ .

1.2L1L2given

Conversely, suppose the displayed closure condition holds and choose $s \in S$ . With $r = - 1_{R}$ and both elements equal to $s$ , it gives $(- 1_{R}) s + s = 0_{M} \in S$ .

1.3L1L2given

If $u, v \in S$ , the same condition with scalar $- 1_{R}$ , first element $v$ , and second element $u$ gives $(- 1_{R}) v + u = u - v \in S$ .

2.1step 1.2step 1.3L3L4given

The additive subgroup test applies by steps 1.2--1.3; scalar closure follows from the displayed condition with $v = 0_{M}$ . Thus $S$ is a submodule.

3.1step 2.1L4given

For a nonempty family $(N_{i})$ of submodules, $0_{M}$ lies in every $N_{i}$ ; and if $u, v$ lie in their intersection, then $r u + v$ lies in every $N_{i}$ . The criterion proves $⋂_{i} N_{i}$ is a submodule.

4.1step 2.1L1L4given∎

For $x = a + b$ and $y = a^{'} + b^{'}$ in $A + B$ , distributivity gives $r x + y = (r a + a^{'}) + (r b + b^{'}) \in A + B$ ; moreover $0_{M} = 0_{M} + 0_{M} \in A + B$ . The criterion proves $A + B$ is a submodule.

Depends on

Dependency tree · 2 levels

Sources

McGerty, Algebra II: Rings and Modules, Section 3