PropositionStatement: Literature-sourcedProof: 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.

Localising twice is localising once at the multiplicative set generated by both denominator sets

Statement

Let $S, T \subseteq R$ be multiplicative, let $\overline{T}$ be the image of $T$ in $S^{- 1} R$ , and let $U \subseteq R$ be the multiplicative subset generated by $S \cup T$ . Then there is a unique $R$ -algebra isomorphism ${\overline{T}}^{- 1} (S^{- 1} R) ≅ U^{- 1} R .$ In particular, $(R_{f})_{g} ≅ R_{f g}$ for $f, g \in R$ , where $g$ on the left denotes its image in $R_{f}$ .

Facts & Assumptions

Given: Multiplicative subsets $S, T \subseteq R$ , their generated multiplicative set $U$ , and the image $\overline{T} \subseteq S^{- 1} R$ .

[F1]

A homomorphism out of a localisation is uniquely determined by a map from the original ring that sends the denominator set to units (Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$ ).

[F2]

Two objects with the same localisation universal property are uniquely isomorphic over the original ring (A localisation is unique up to a unique isomorphism compatible with the map from $R$ ).

[F3]

The principal localisation $R_{f}$ inverts the powers of $f$ (Principal localisation $R_{f} = {1, f, f^{2}, \dots}^{- 1} R$ ).

Proof

technique · direct universal-property argument

1.1

A map $h : R \to A$ extends to ${\overline{T}}^{- 1} (S^{- 1} R)$ exactly when it sends $S$ to units and, after the first extension, sends every $t / 1$ with $t \in T$ to a unit.

2.1

The image of $t / 1$ is $h (t)$ , so the condition in step 1.1 is exactly that $h$ send every member of $S \cup T$ to a unit, equivalently every element of the generated set $U$ to a unit.

step 1.1algebra

3.1

Hence ${\overline{T}}^{- 1} (S^{- 1} R)$ and $U^{- 1} R$ have the same universal property over $R$ , so [F2] gives the unique displayed $R$ -algebra isomorphism.

F1F2step 2.1

4.1

For $S = {f^{n} : n \in N}$ and $T = {g^{n} : n \in N}$ , the set $U$ is generated by $f$ and $g$ . A map inverts both precisely when it inverts $f g$ : if $f g$ is a unit, then $f (g (f g)^{- 1}) = 1$ and $g (f (f g)^{- 1}) = 1$ . Thus [F1] identifies this localisation with $R_{f g}$ , including $f = 0$ or $g = 0$ .

F1F3algebra∎

Depends on

Used by

Dependency tree · two levels

8 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, Lemma 10.9.8 (standard reference, not scraped)