TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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.

Localisation of modules is extension of scalars

Statement

Let $R$ be a commutative ring, let $S \subseteq R$ be multiplicative, and let $M$ be a left $R$ -module. The map $Φ : (S^{- 1} R) \otimes_{R} M ⟶ S^{- 1} M, Φ ((a / s) \otimes m) = a m / s,$ is an isomorphism of $S^{- 1} R$ -modules. Its inverse is $Ψ : S^{- 1} M ⟶ (S^{- 1} R) \otimes_{R} M, Ψ (m / s) = (1 / s) \otimes m .$

Facts & Assumptions

Given: A commutative ring $R$ , a multiplicative subset $S \subseteq R$ , and a left $R$ -module $M$ .

[L1]

A balanced pairing on $(S^{- 1} R) \times M$ induces a unique homomorphism from $(S^{- 1} R) \otimes_{R} M$ (Universal property of the tensor product for balanced maps into abelian groups, A formula on elementary tensors defines a homomorphism exactly when its underlying pairing is balanced).

[L2]

Over a commutative ring, a tensor product carries the scalar action $r (x \otimes m) = (r x) \otimes m$ (Over a commutative ring, $M \otimes_{R} N$ is an $R$ -module with $r (m \otimes n) = (r m) \otimes n = m \otimes (r n)$ ).

[L3]

The localisation map $M \to S^{- 1} M$ is universal for maps into $S^{- 1} R$ -modules (Universal property of localisation for modules).

[L4]

In $S^{- 1} R$ , fraction arithmetic is well defined and every $s / 1$ with $s \in S$ is a unit with inverse $1 / s$ (The localisation relation is an equivalence relation and fraction arithmetic is well defined).

Proof

technique · direct

1.1L1L4algebra

The pairing $q ((a / s), m) = a m / s$ is balanced because it is additive in each variable and $q (((a / s) r), m) = a r m / s = q (a / s, r m)$ for every $r \in R$ .

1.2L2L4algebra

The map $i : M \to (S^{- 1} R) \otimes_{R} M$ , $i (m) = (1 / 1) \otimes m$ , is $R$ -linear, and every $s \in S$ acts invertibly on the target because $(s / 1)^{- 1} = 1 / s$ in $S^{- 1} R$ .

2.1step 1.1L1construct

By [L1], step 1.1 induces a unique homomorphism $Φ : (S^{- 1} R) \otimes_{R} M \to S^{- 1} M$ with $Φ ((a / s) \otimes m) = a m / s$ .

2.2step 1.2L3construct

By [L3], step 1.2 induces a unique $S^{- 1} R$ -linear map $Ψ : S^{- 1} M \to (S^{- 1} R) \otimes_{R} M$ with $Ψ (m / s) = (1 / s) \otimes m$ .

3.1step 2.1step 2.2

For every $m / s \in S^{- 1} M$ , $(Φ Ψ) (m / s) = Φ ((1 / s) \otimes m) = m / s$ .

3.2step 2.1step 2.2L2algebra

For every elementary tensor $(a / s) \otimes m$ , $(Ψ Φ) ((a / s) \otimes m) = Ψ (a m / s) = (1 / s) \otimes a m = (a / s) \otimes m$ by the tensor scalar action of [L2].

4.1step 3.1step 3.2∎

Steps 3.1 and 3.2 show that $Φ$ and $Ψ$ are inverse $S^{- 1} R$ -linear isomorphisms.

Localisation of modules is extension of scalars

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources