CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 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.

Change of rings: $N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$

Statement

Let $R \to S$ be a homomorphism of commutative rings, let $N$ be a right $S$ -module, and let $M$ be a left $R$ -module. There is a natural group isomorphism

$N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$

given by

$n \otimes m ⟼ n \otimes (1_{S} \otimes m),$

with inverse $n \otimes (s \otimes m) \mapsto n s \otimes m$ .

Facts & Assumptions

Given: A ring map $R \to S$ , a right $S$ -module $N$ , and a left $R$ -module $M$ .

[L1]

Restriction makes $N$ a right $R$ -module and makes $S$ an $(S, R)$ -bimodule used in $S \otimes_{R} M$ (Restriction of scalars and extension of scalars $S \otimes_{R} M$ along a ring homomorphism $R \to S$ ).

[L2]

Compatible bimodules have the associativity isomorphism $(N \otimes_{S} S) \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$ (Associativity of tensor products for compatible bimodules).

[L3]

The tensor-unit isomorphism identifies $N \otimes_{S} S$ with $N$ by $n \otimes s \mapsto n s$ (The regular module is a tensor unit: $R \otimes_{R} N ≅ N$ and $M \otimes_{R} R ≅ M$ ).

Proof

technique · direct

1.1

Apply [L2] to $N$ , the $(S, R)$ -bimodule $S$ , and $M$ , then use [L3] on the left factor to obtain $(N \otimes_{S} S) \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$ and hence $N \otimes_{R} M ≅ N \otimes_{S} (S \otimes_{R} M)$ .

givenL1L2L3

2.1

Tracing elementary tensors through step 1.1 gives $n \otimes m \mapsto n \otimes (1 \otimes m)$ ; tracing the inverse gives $n \otimes (s \otimes m) \mapsto n s \otimes m$ .

step 1.1L2L3

3.1

The two formulas are mutually inverse on elementary tensors. The inverse after the forward map sends $n \otimes m$ to $n 1_{S} \otimes m = n \otimes m$ . In the other direction, balance over $S$ gives $n s \otimes (1_{S} \otimes m) = n \otimes s (1_{S} \otimes m) = n \otimes (s \otimes m)$ . The universal properties therefore make the composites identities.

step 2.1L1L2L3

3.2

Each construction commutes with homomorphisms in $N$ and $M$ because its elementary-tensor formula does, so the isomorphism is natural.

step 2.1algebra

4.1

This proves the change-of-rings isomorphism and its stated inverse.

step 3.1step 3.2∎

Depends on

Used by

Extension of scalars carries flat modules to flat modules Proposition
Flatness is transitive under a flat change of rings Proposition

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 10 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 4 on tensor products and flatness (standard reference, not scraped)

Change of rings: N⊗RM≅N⊗S(S⊗RM)N⊗R​M≅N⊗S​(S⊗R​M)