TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31

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.

Modules over a commutative ring form a monoidal category

Statement

If $R$ is a commutative ring, then the category $R - M o d$ is monoidal with tensor product $\otimes_{R}$ and unit object $R$ .

Facts & Assumptions

Given: A commutative ring $R$ and $R$ -modules $L, M, N$ .

[L1]

$R$ is a commutative ring in the sense of Commutative ring, and its modules and module homomorphisms form the category $R - M o d$ (Left modules over a fixed ring and module homomorphisms form the large locally small category $R - M o d$ ).

[L2]

Tensor products of compatible bimodules are associative and have the regular module as tensor unit (Associativity of tensor products for compatible bimodules, The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[L3]

Over a commutative ring the tensor product admits the natural symmetry and associativity isomorphisms (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

[L4]

Module homomorphisms induce tensor-product homomorphisms functorially (Module homomorphisms induce tensor-product homomorphisms functorially).

Proof

technique · direct

1.1

By [L1], the modules under discussion already form a category. Because $R$ is commutative, every left $R$ -module is canonically an $(R, R)$ -bimodule, so $M \otimes_{R} N$ is again an $R$ -module.

givenL1

1.2

The associator $((L \otimes_{R} M) \otimes_{R} N) \to L \otimes_{R} (M \otimes_{R} N)$ and the unitors $R \otimes_{R} M \to M$ and $M \otimes_{R} R \to M$ are exactly the isomorphisms supplied by [L2] and [L3].

L2L3

2.1

By [L4], tensoring homomorphisms is functorial in each variable, so $\otimes_{R}$ is a bifunctor on $R - M o d$ . Steps 1.1 and 1.2 therefore provide the data of a monoidal category.

step 1.1step 1.2L1L2L3L4

3.1

Hence modules over a commutative ring form a monoidal category under tensor product.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

The Stacks Project, Section 10.12: Tensor products (standard reference, not scraped)