TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$

Statement

Let $R$ be a commutative ring and let $M, N$ be $R$ -modules. The abelian group $M \otimes_{R} N$ has a unique $R$ -module structure for which

$r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$

for every $r \in R$ , $m \in M$ , and $n \in N$ .

Facts & Assumptions

Given: A commutative ring $R$ and $R$ -modules $M, N$ , each regarded on either side by the common scalar action.

[L1]

Balanced pairings induce unique homomorphisms from tensor products (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

In a commutative ring, $r s = s r$ for all $r, s \in R$ (Commutative ring).

[L3]

An elementary-tensor formula descends exactly when its underlying pairing is balanced (A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

Proof

technique · direct

1.1

For fixed $r \in R$ , the pairing $(m, n) \mapsto (r m) \otimes n$ is additive in both variables and balanced: $(r (s m)) \otimes n = ((r s) m) \otimes n = ((s r) m) \otimes n = (r m) \otimes (s n)$ by [L2].

givenL2algebra

2.1

By [L1] and [L3], step 1.1 induces an additive endomorphism $x \mapsto r x$ of $M \otimes_{R} N$ satisfying $r (m \otimes n) = (r m) \otimes n$ . The tensor balance relation also gives $(r m) \otimes n = m \otimes (r n)$ .

step 1.1L1L3

3.1

On elementary tensors, $(r + s) x = r x + s x$ , $r (x + y) = r x + r y$ , $(r s) x = r (s x)$ , and $1_{R} x = x$ follow from the module axioms of $M$ . In each identity the two additive maps induce the same balanced pairing, so uniqueness in [L1] makes the identity hold for every tensor.

step 2.1L1algebra

3.2

Any $R$ -module structure with the displayed formula has, for each $r$ , a scalar-multiplication endomorphism inducing the same balanced pairing as step 2.1. Uniqueness in [L1] therefore makes the structure unique.

step 2.1L1

4.1

Steps 2.1, 3.1 and 3.2 give the asserted unique $R$ -module structure and both elementary-tensor formulas.

step 2.1step 3.1step 3.2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 40 results over 17 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

Over a commutative ring, M⊗RNM⊗R​N is an RR-module with r(m⊗n)=(rm)⊗n=m⊗(rn)r(m⊗n)=(rm)⊗n=m⊗(rn)