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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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The spectrum of a Noetherian ring is a Noetherian topological space

Statement

Assume the Axiom of Choice.

Let R be a Noetherian commutative ring. Then Spec⁡(R) is a Noetherian topological space.

Facts & Assumptions

Given: A Noetherian commutative ring R and the Axiom of Choice.

[L1]

A topological space is Noetherian exactly when every descending chain of closed subsets stabilizes (Noetherian topological spaces via ACC on opens or DCC on closed subsets).

[L2]

Every Zariski-closed subset has a unique radical defining ideal (Every Zariski-closed subset has a unique radical defining ideal).

Proof

technique · direct
1.1L2givenchoose

Let Z0⊇Z1⊇Z2⊇⋯ be a descending chain of closed subsets of Spec⁡(R). By [L2], each Zn has a unique radical defining ideal an with Zn=V(an).

2.1L2step 1.1algebra

Since Zn+1⊆Zn, every prime ideal in Zn+1 also lies in Zn. Therefore an=⋂p∈Znp⊆⋂p∈Zn+1p=an+1. Thus a0⊆a1⊆a2⊆⋯ is an ascending chain of ideals.

3.1L3step 2.1

By [L3], the ideal chain from step 2.1 stabilizes: there is N such that an=aN for all n≥N. Then Zn=V(an)=V(aN)=ZN for all n≥N. Hence the closed chain stabilizes.

4.1L1step 3.1∎

By [L1], stabilization of every descending closed chain means that Spec⁡(R) is Noetherian.

Depends on

Used by

Dependency tree · two levels

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Sources