Alphabeta Math
LemmaStatement: 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.

A prime chain in R extends to a longer chain in R[x]

Statement

Let R be a commutative ring. If

p0pd

is a strict prime chain in R, then

p0R[x]pdR[x]pdR[x]+(x)

is a strict prime chain in R[x]. Consequently dimR[x]dimR+1 whenever dimR is finite.

Facts & Assumptions

Given: A commutative ring R and a strict prime chain p0pd in R.

[L1]

For a prime ideal p, the quotient R/p is an integral domain (R/P is an integral domain if and only if P is a prime ideal).

[L2]

Prime ideals of a quotient correspond to prime ideals containing the quotient ideal (Prime ideals of a quotient ring are exactly the prime ideals containing the ideal).

[L3]

Krull dimension is computed by strict prime chains (Krull dimension of a nonzero ring).

Proof

technique · direct
1.1

For each i, the quotient R[x]/piR[x](R/pi)[x] is a polynomial ring over the domain R/pi, so [L1] and [L2] show that piR[x] is prime. Strictness of the original chain makes the extended chain strict.

L1L2given
2.1

The quotient by pdR[x]+(x) is again R/pd, a domain, so [L1] and [L2] show that pdR[x]+(x) is prime and strictly contains pdR[x].

L1L2step 1.1
3.1

The displayed chain in R[x] therefore has length d+1, and [L3] yields dimR[x]dimR+1 whenever dimR is finite.

L3step 1.1step 2.1

Depends on

Used by

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