False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

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False statement: every right Noetherian ring is left Noetherian

Statement

False claim: every right Noetherian ring is left 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]

Let $R$ be a commutative ring. Matrices of one shape are added and scaled entrywise. For $A \in M_{m \times n} (R)$ and $B \in M_{n \times p} (R)$ , their product is $A B \in M_{m \times p} (R)$ with $(A B)_{i k} : = \sum_{j < n} a_{i j} b_{j k} .$ The sum is the finite sum in the additive commutative monoid of $R$ , and is $0$ when $n = 0$ . (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[L3]

$(Q, +, \cdot, 0, 1)$ with the operations of def-rat-operations is a field: a commutative ring with $1 \neq 0$ in which every nonzero element has a multiplicative inverse. (The rationals form a field).

Refutation

technique · direct

1.1

Let $T = {(\begin{matrix} a & q \\ 0 & r \end{matrix}) : a \in Z, q, r \in Q}$ . Matrix multiplication is closed and $diag (1, 1)$ is its identity. As a right module, $T = e_{11} T \oplus e_{22} T$ ; the second summand is the simple module $Q_{Q}$ . For a right submodule $J \leq e_{11} T$ , its top-left entries form an ideal $d Z$ , while its intersection with $Q e_{12}$ is either $0$ or all of the simple right $Q$ -module $Q e_{12}$ . Choosing one element of $J$ with top-left entry $d$ shows that $J$ is generated by it and, if needed, $e_{12}$ . Thus $T_{T}$ is Noetherian.

L1L2L3givenalgebra

2.1

For $n \in N$ , the sets $L_{n} = {(\begin{matrix} 0 & q \\ 0 & 0 \end{matrix}) : q \in 2^{- n} Z}$ are left ideals because left multiplication scales $q$ only by the integer top-left entry. The inclusions $L_{0} < L_{1} < L_{2} < \dots$ are strict, witnessed by $2^{- (n + 1)} e_{12}$ . Hence $_{T} T$ is not Noetherian, refuting the claim.

step 1.1givenalgebra∎

Depends on

Used by

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Sources

Abigail C. Bailey and John A. Beachy, On Noncommutative Piecewise Noetherian Rings, Example 1 (standard reference, not scraped)