LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Distinguished-subset covers detect radicals

Statement

Assume the Axiom of Choice.

Let $R$ be a commutative ring and let $f, f_{1}, \dots, f_{n} \in R$ with $n \geq 1$ . Then

D (f) \subseteq D (f_{1}) \cup \dots \cup D (f_{n})

if and only if

f \in \sqrt{(f_{1}, \dots, f_{n})} .

Equivalently, some positive power of $f$ lies in the ideal $(f_{1}, \dots, f_{n})$ .

Facts & Assumptions

Given: A commutative ring $R$ , elements $f, f_{1}, \dots, f_{n} \in R$ , an integer $n \geq 1$ , and the Axiom of Choice.

[L1]

Two ideals have the same vanishing set exactly when their radicals agree (Vanishing sets detect radicals).

[L2]

For any $h \in R$ , the principal distinguished subset $D (h)$ is the complement of $V ((h))$ in $Spec (R)$ (Principal distinguished subsets of the prime spectrum), where $(h)$ denotes the principal ideal generated by $h$ (The ideal generated by a subset and principal ideals).

Proof

technique · direct

1.1

Let $J = (f_{1}, \dots, f_{n})$ . If $f \in \sqrt{J}$ and $p \in D (f)$ , then $f \notin p$ . A prime ideal containing $J$ would contain every element of $\sqrt{J}$ and therefore would contain $f$ , so $p$ cannot contain $J$ . Hence at least one $f_{i}$ is omitted by $p$ , which means $p \in D (f_{i})$ for some $i$ . Thus $D (f) \subseteq D (f_{1}) \cup \dots \cup D (f_{n})$ .

L2givenalgebra

1.2

Conversely, assume $D (f) \subseteq D (f_{1}) \cup \dots \cup D (f_{n})$ . If $p$ contains $J$ and omitted $f$ , then $p$ would lie in the left-hand side and hence in some $D (f_{i})$ , contradicting $f_{i} \in J \subseteq p$ . Therefore every prime ideal containing $J$ also contains $f$ , so $V (J) = V (J + (f))$ . By [L1], the radicals of $J$ and $J + (f)$ agree; since $f \in J + (f)$ , this forces $f \in \sqrt{J}$ .

L1L2given

2.1

Steps 1.1 and 1.2 prove the equivalence. The final sentence is just the definition of membership in a radical ideal.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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, Section 10.17: The spectrum of a ring (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §14 The spectrum of a ring (standard reference, not scraped)