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

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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.

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Vanishing sets of finite products

Statement

Let $R$ be a commutative ring, and let $I_{1}, \dots, I_{n} ⊴ R$ with $n \geq 1$ . Then $V (I_{1} \dots I_{n}) = V (I_{1}) \cup \dots \cup V (I_{n})$ .

Facts & Assumptions

Given: A commutative ring $R$ , ideals $I_{1}, \dots, I_{n} ⊴ R$ , and an integer $n \geq 1$ .

[L1]

$V (K)$ is the set of prime ideals containing $K$ (The prime spectrum and vanishing sets).

[L2]

A prime ideal contains a factor whenever it contains a product (Prime ideals and maximal ideals in a commutative ring).

Proof

technique · direct

1.1

If $p \in V (I_{j})$ for some $j$ , then $I_{j} \subseteq p$ . Every element of $I_{1} \dots I_{n}$ lies in $I_{j}$ , so $I_{1} \dots I_{n} \subseteq p$ , and therefore $p \in V (I_{1} \dots I_{n})$ . This proves $V (I_{1}) \cup \dots \cup V (I_{n}) \subseteq V (I_{1} \dots I_{n})$ .

L1givenalgebra

1.2

Conversely, let $p \in V (I_{1} \dots I_{n})$ and suppose that no $I_{j}$ is contained in $p$ . For each $j$ , choose $a_{j} \in I_{j} ∖ p$ . Then $a_{1} \dots a_{n} \in I_{1} \dots I_{n} \subseteq p$ , so repeated use of [L2] forces some $a_{j} \in p$ , a contradiction. Hence $I_{j} \subseteq p$ for some $j$ , and $p \in V (I_{j})$ .

L1L2choosealgebra

2.1

The two inclusions prove $V (I_{1} \dots I_{n}) = V (I_{1}) \cup \dots \cup V (I_{n})$ .

step 1.1step 1.2∎

Depends on

Used by

Vanishing-set identities Lemma

Dependency tree · two levels

7 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)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §13 and §17 (standard reference, not scraped)