LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The vanishing sets define the Zariski topology on the prime spectrum

Statement

Let $R$ be a commutative ring. The subsets $V (I) \subseteq Spec (R)$ , as $I$ ranges over the ideals of $R$ , contain $Spec (R)$ and $\emptyset$ , are closed under arbitrary intersections and finite unions, and therefore define a topology on $Spec (R)$ .

Facts & Assumptions

Given: A commutative ring $R$ .

[L1]

The vanishing sets satisfy $V ((0)) = Spec (R)$ , $V (R) = \emptyset$ , arbitrary intersections, and finite unions (Vanishing-set identities).

[A1]

A family of subsets of a set that contains the whole set and the empty set, is closed under arbitrary intersections, and is closed under finite unions is the family of closed sets of a topology.

Proof

technique · direct

1.1

Fact [L1] gives exactly the four closed-set properties listed in the statement for the family ${V (I)}$ of subsets of $Spec (R)$ .

2.1

By [A1], any family with those four properties is the family of closed sets of a topology on the underlying set. Therefore the subsets $V (I)$ define a topology on $Spec (R)$ .

A1step 1.1

3.1

The vanishing sets are precisely the closed sets of the Zariski topology on $Spec (R)$ .

step 2.1∎

Depends on

Used by

Dependency tree · two levels

6 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

J. S. Milne, A Primer of Commutative Algebra, v4.03, Proposition 14.1 (standard reference, not scraped)
The Stacks Project, Definition 10.17.3 (standard reference, not scraped)