LemmaStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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Vanishing sets detect radicals

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring and let $I, J ⊴ R$ be ideals. Then

V (I) = V (J) ⟺ \sqrt{I} = \sqrt{J} .

Facts & Assumptions

Given: A commutative ring $R$ , ideals $I, J ⊴ R$ , and the Axiom of Choice.

[L1]

The radical of an ideal is the intersection of the prime ideals containing it (The radical of an ideal is the intersection of the prime ideals containing it).

[L2]

Radical is an idempotent ideal-valued operation (The radical of an ideal is an ideal).

Proof

technique · direct

1.1

If $V (I) = V (J)$ , then the two ideals are contained in exactly the same prime ideals. Applying [L1] to both ideals shows that $\sqrt{I}$ and $\sqrt{J}$ are intersections over the same family of prime ideals, hence $\sqrt{I} = \sqrt{J}$ .

L1given

1.2

Conversely, suppose $\sqrt{I} = \sqrt{J}$ . If $p \in V (I)$ , then $p$ contains $I$ , so [L1] gives $\sqrt{I} \subseteq p$ . Therefore $\sqrt{J} \subseteq p$ , hence $J \subseteq \sqrt{J} \subseteq p$ by [L2], and $p \in V (J)$ . The same argument with $I$ and $J$ reversed shows $V (J) \subseteq V (I)$ .

L1L2given

2.1

Steps 1.1 and 1.2 prove that vanishing sets agree exactly when radicals agree.

step 1.1step 1.2∎

Depends on

Used by

Distinguished-subset covers detect radicals Lemma

Dependency tree · two levels

10 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, §14 The spectrum of a ring (standard reference, not scraped)
The Stacks Project, Section 10.17: The spectrum of a ring (standard reference, not scraped)