LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck 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

1.1L1given

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 $D (1) = Spec (R)$ .

1.2L1givenalgebra

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

2.1step 1.1step 1.2∎

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

Distinguished-subset identities

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources