CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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.

A localisation is unique up to a unique isomorphism compatible with the map from $R$

Statement

Let $λ : R \to L$ and $λ^{'} : R \to L^{'}$ be unital homomorphisms of commutative rings. Suppose each map sends every $s \in S$ to a unit and has the localisation universal property: every homomorphism from $R$ that inverts $S$ factors uniquely through it. Then there is a unique ring isomorphism $Φ : L \to L^{'}$ such that $Φ λ = λ^{'}$ .

Facts & Assumptions

Given: Two objects $(L, λ)$ and $(L^{'}, λ^{'})$ satisfying the stated universal property for the same pair $(R, S)$ .

[F1]

A homomorphism from $R$ that takes $S$ to units factors uniquely through a localisation map (Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$ ).

Proof

technique · direct universal-property argument

1.1

Apply the universal property of $L$ to $λ^{'}$ and that of $L^{'}$ to $λ$ . This gives unique homomorphisms $Φ : L \to L^{'}$ and $Ψ : L^{'} \to L$ with $Φ λ = λ^{'}$ and $Ψ λ^{'} = λ$ .

2.1

Both $Ψ Φ$ and ${id}_{L}$ compose with $λ$ to give $λ$ . Uniqueness for $L$ gives $Ψ Φ = {id}_{L}$ ; similarly $Φ Ψ = {id}_{L^{'}}$ . Thus $Φ$ is an isomorphism.

F1step 1.1

3.1

Any isomorphism compatible with the maps from $R$ is, in particular, a factorisation of $λ^{'}$ through $λ$ , so it equals $Φ$ by uniqueness.

F1step 1.1∎

Depends on

Universal property of localisation: maps that invert $S$ factor uniquely through $S^{-1}R$

Used by

Dependency tree · two levels

5 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, Proposition 10.9.3 (standard reference, not scraped)

A localisation is unique up to a unique isomorphism compatible with the map from RR