False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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.

False statement: every subring of a Noetherian ring is Noetherian

Statement

False claim: every subring of a Noetherian ring is Noetherian. See Left and right Noetherian rings.

Facts & Assumptions

Given: The hypotheses and objects in the false claim.

[L1]

A unital ring $R$ is left Noetherian when its left regular module $_{R} R$ is Noetherian, and right Noetherian when the right regular module $R_{R}$ is Noetherian. Unqualified “Noetherian ring” means left Noetherian here; the side is stated whenever both notions occur. (Left and right Noetherian rings).

[L2]

If $D$ is an integral domain, then $D ∖ {0}$ is multiplicative. Its localisation $Frac (D) = (D ∖ {0})^{- 1} D$ is the field of fractions of $D$ . Thus its elements are fractions $a / b$ with $a, b \in D$ and $b \neq 0$ , modulo the localisation equivalence relation. (The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain).

[L3]

For every integral domain $D$ , the localisation $Frac (D)$ is a field. Its canonical map $D ⟶ Frac (D), d ⟼ d / 1,$ is an injective unital ring homomorphism. ( $Frac (D)$ is a field and $d \mapsto d / 1$ embeds the integral domain $D$ ).

Refutation

technique · direct

1.1L1L2L3givenalgebra

Fix a field $F$ and let $R$ consist of polynomials in symbols $x_{0}, x_{1}, \dots$ in which each polynomial contains only finitely many monomials and variables. The usual polynomial operations make $R$ a domain, so it embeds in its fraction field $K$ .

2.1step 1.1givenalgebra

The field $K$ is Noetherian because its only ideals are $0$ and $K$ . In $R$ , the ideals $I_{n} = (x_{0}, \dots, x_{n})$ satisfy $I_{n} < I_{n + 1}$ : setting $x_{0}, \dots, x_{n}$ to zero leaves $x_{n + 1}$ nonzero, so $x_{n + 1} \notin I_{n}$ .

3.1step 2.1givenalgebra∎

Thus the Noetherian ring $K$ contains the non-Noetherian subring $R$ , which refutes the claim. This proves the stated claim.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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

Keith Conrad, Noetherian Modules, Sections 1-2 (standard reference, not scraped)