ExampleConstruction: AI-generatedVerification: Literature-sourcedprecheck passaudited 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 can strictly lower dimension

Example

Let $R = k [x, y]$ and let $S = R ∖ (x)$ . Then

S^{- 1} R = R_{(x)}

has dimension $1$ , strictly smaller than $\dim R = 2$ .

Facts & Assumptions

Given: A field $k$ , the polynomial ring $R = k [x, y]$ , and the multiplicative set $S = R ∖ (x)$ .

[L1]

Localization does not increase dimension (Localisation does not increase Krull dimension).

[L2]

The polynomial ring $k [x, y]$ has dimension $2$ (A polynomial ring in n variables over a field has dimension n).

[L3]

In the affine domain $k [x, y]$ , the prime $(x)$ satisfies $ht ((x)) + \dim (k [x, y] / (x)) = \dim (k [x, y])$ (Height plus quotient dimension equals ambient dimension in an affine domain).

[L4]

By definition, the height of a prime equals the dimension of the localization at that prime (The height of a prime ideal).

Verification

technique · direct computation

1.1

By [L2], $\dim R = 2$ . Since $R / (x) ≅ k [y]$ has dimension $1$ , [L3] gives $ht ((x)) = 1$ . Therefore [L4] yields $\dim (S^{- 1} R) = \dim (R_{(x)}) = 1$ .

L2L3L4given

2.1

Fact [L1] independently gives $\dim (S^{- 1} R) \leq \dim R = 2$ , so the computed value $1$ is compatible with the general one-sided inequality. Since $1 < 2 = \dim R$ , this localization strictly lowers dimension.

L1step 1.1

3.1

So localization can strictly lower Krull dimension.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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, §§18, 21 (standard reference, not scraped)
Melvin Hochster, Dimension theory and systems of parameters (standard reference, not scraped)