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

Statement

Let $R$ be a commutative ring and let $f, g \in R$ . Then

D (0) = \emptyset, D (1) = Spec (R), D (f g) = D (f) \cap D (g) .

Moreover, for every integer $n \geq 1$ one has

D (f^{n}) = D (f) .

Facts & Assumptions

Given: A commutative ring $R$ , elements $f, g \in R$ , and an integer $n \geq 1$ .

[L1]

$D (h)$ is the set of prime ideals that do not contain $h$ (Principal distinguished subsets of the prime spectrum).

Proof

technique · direct

1.1

Every prime ideal contains $0$ , so no prime lies in $D (0)$ . Every prime ideal is proper, so it does not contain $1$ ; hence every prime lies in $D (1)$ . Therefore $D (0) = \emptyset$ and $D (1) = Spec (R)$ .

L1given

1.2

Let $p \in Spec (R)$ . Then $p \in D (f g)$ exactly when $f g \notin p$ . Because $p$ is prime, this is equivalent to saying that neither $f$ nor $g$ lies in $p$ , that is, $p \in D (f) \cap D (g)$ . The same prime-ideal property shows that $f^{n} \in p$ exactly when $f \in p$ , so $D (f^{n}) = D (f)$ .

L1givenalgebra

2.1

Steps 1.1 and 1.2 prove the stated identities for principal distinguished subsets.

step 1.1step 1.2∎

Depends on

Used by

The zero ring has empty prime spectrum Example

Dependency tree · two levels

5 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 The Spectrum of a Ring (standard reference, not scraped)