CorollaryStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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.

Localisation does not increase Krull dimension

Let $R$ be a nonzero commutative ring and let $S \subseteq R$ be a multiplicative subset with $0 \notin S$ . Then $\dim (S^{- 1} R) \leq \dim R .$

Given: A nonzero commutative ring $R$ and a multiplicative subset $S \subseteq R$ with $0 \notin S$ .

[L1]

Prime ideals of $S^{- 1} R$ correspond to prime ideals of $R$ disjoint from $S$ , with strict inclusions preserved (Prime ideals of a localization are exactly the primes disjoint from the denominator set).

[L2]

Krull dimension is the supremum of the lengths of strict prime chains (Krull dimension of a nonzero ring).

technique · direct

1.1L1given

By [L1], every strict prime chain in $S^{- 1} R$ contracts to a strict prime chain in $R$ of the same length.

2.1L2step 1.1

Taking suprema and using [L2] gives $\dim (S^{- 1} R) \leq \dim R$ .

3.1step 2.1∎

Hence localization does not increase Krull dimension.

Depends on