CorollaryStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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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.

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Every algebra of finite type over a principal ideal domain is a Noetherian ring

Statement

Let $R$ be a principal ideal domain (Principal ideal domain) and let $A$ be a commutative $R$ -algebra of finite type (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras). Then $A$ is a Noetherian ring.

Facts & Assumptions

Given: A principal ideal domain $R$ and a commutative $R$ -algebra $A$ of finite type. A field is a principal ideal domain under the definition in force here: it is an integral domain, and its only ideals are $(0)$ and $(1)$ , both principal. So the field case falls under this statement rather than outside it.

[F1]

An integral domain $R$ is a principal ideal domain (PID) if every ideal $I ⊴ R$ is principal: there is an $a \in R$ with $I = (a)$ (Principal ideal domain).

[L1]

Every principal ideal domain is Noetherian (Every principal ideal domain is Noetherian).

[L2]

Every commutative algebra of finite type over a Noetherian commutative ring is a Noetherian ring (Every algebra of finite type over a Noetherian ring is a Noetherian ring).

Proof

technique · direct

1.1

The base ring $R$ is a principal ideal domain, hence an integral domain and in particular a commutative ring, and hence a Noetherian ring.

F1L1given

2.1

The algebra $A$ is of finite type over the Noetherian commutative ring $R$ , so it is a Noetherian ring.

L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

M. Hochster, Introduction to Commutative Algebra, Math 614, Corollary 5.9 (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §16 (standard reference, not scraped)