LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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Localization sections are independent of a distinguished-open presentation

Statement

Assume the Axiom of Choice. The assignment $D (f) \mapsto A_{f}$ , with the localization restrictions, is independent of the representation of a distinguished open and is a sheaf on the distinguished-open basis.

Facts & Assumptions

Given: The Axiom of Choice, a ring $A$ , and an arbitrary cover $D (f) = ⋃_{i \in I} D (g_{i})$ by distinguished opens contained in $D (f)$ .

Proof

technique · direct

1.1

If $D (g) \subseteq D (f)$ , localization universality gives the canonical restriction $A_{f} \to A_{g}$ ; for equal opens the two restrictions are inverse.

given

2.1

Under $D (f) ≅ Spec (A_{f})$ , the cover becomes the distinguished cover by the images of the $g_{i}$ . The spectrum-cover lemma makes those images generate the unit ideal in $A_{f}$ , so a finite subfamily already generates $1$ . The standard localization calculation for a finite unit-ideal cover then glues every compatible family in the $(A_{f})_{g_{i}} ≅ A_{g_{i}}$ uniquely to an element of $A_{f}$ .

step 1.1algebra

2.2

This includes the empty case: if $D (f) = \emptyset$ , then $f$ lies in every prime ideal, so it is nilpotent; hence $A_{f}$ is the zero ring, and the empty compatible family glues uniquely to its sole element.

step 1.1algebra

3.1

Thus gluing and uniqueness hold for every cover of a distinguished open by distinguished opens, not only for finite covers. This is precisely the sheaf axiom for the localization presheaf on the distinguished-open basis, so steps 2.1 and 2.2 prove the claim.

step 2.1step 2.2∎

Depends on

Used by

The localization construction extends to the structure sheaf on Spec A Theorem

Dependency tree · two levels

17 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 26.5.1 (standard reference, not scraped)