TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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 elementary tensors of two bases form the product basis of the tensor product

Statement

Let $R$ be a commutative ring. If $M$ is free with basis $(e_{i})_{i \in I}$ and $N$ is free with basis $(f_{j})_{j \in J}$ , then $M \otimes_{R} N$ is free with basis

$(e_{i} \otimes f_{j})_{(i, j) \in I \times J} .$

Equivalently, the canonical map $R^{(I \times J)} \to M \otimes_{R} N$ sending the standard basis vector at $(i, j)$ to $e_{i} \otimes f_{j}$ is an isomorphism. This includes an empty basis in either factor.

Facts & Assumptions

Given: A commutative ring $R$ and free modules $M, N$ with bases indexed by $I, J$ .

[L1]

Tensor products commute with arbitrary direct sums in either variable: $⨁_{i} (M_{i} \otimes_{R} N) ≅ (⨁_{i} M_{i}) \otimes_{R} N$ and $⨁_{i} (N \otimes_{R} M_{i}) ≅ N \otimes_{R} (⨁_{i} M_{i})$ (Tensor products commute with arbitrary direct sums).

[L2]

The regular module is a tensor unit: $R \otimes_{R} R ≅ R$ via $r \otimes s \mapsto r s$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

[L3]

A free module on $X$ is $R^{(X)} = ⨁_{x \in X} R$ , with standard basis and unique finite coordinate expressions, including $X = \emptyset$ (The free module on a set and its standard basis).

[L4]

A map from the basis set of a free module extends uniquely to a module homomorphism (Universal property of the free module on a set).

Proof

technique · direct

1.1givenL3L4

The chosen bases identify $M$ with $⨁_{i \in I} R$ and $N$ with $⨁_{j \in J} R$ , carrying $e_{i}, f_{j}$ to the corresponding standard basis vectors.

2.1step 1.1L1L2L3

Apply [L1] in each variable and then [L2] to obtain canonical isomorphisms $M \otimes_{R} N ≅ ⨁_{i \in I} ⨁_{j \in J} (R \otimes_{R} R) ≅ ⨁_{(i, j) \in I \times J} R = R^{(I \times J)}$ .

3.1step 2.1L3

Tracing a coordinate generator through step 2.1 sends it to $e_{i} \otimes f_{j}$ ; by [L3], those images therefore form a basis and every tensor has a unique finite expansion in them.

3.2step 2.1L3

If $I = \emptyset$ or $J = \emptyset$ , then $I \times J = \emptyset$ , the corresponding factor and the target free module are zero by [L3], and step 2.1 is the unique isomorphism between zero modules.

4.1step 3.1step 3.2∎

This proves the product-basis assertion in all cases.

Depends on

Used by

Dependency tree · two levels

11 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

C. Dennis, Week 1 recap on tensor products (standard reference, not scraped)