ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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$Z$ and $k [x]$ are Noetherian but not Artinian

Example

Let $k$ be a field. Then $Z$ and $k [x]$ are Noetherian rings, but neither is Artinian.

Facts & Assumptions

Given: A field $k$ .

Verification

technique · direct

1.1

Every ideal of $Z$ is an additive subgroup, so Every subgroup of $(Z, +)$ is $⟨ n ⟩ = n Z$ for exactly one natural number $n$ makes it principal; then A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member shows that $Z$ is Noetherian. Also If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ applied to the field $k$ shows that $k [x]$ is Noetherian.

givenalgebra

2.1

In $Z$ the principal ideals $(2) ⊋ (2^{2}) ⊋ (2^{3}) ⊋ \dots$ form a strict descending chain, because $2^{n} \in (2^{n}) ∖ (2^{n + 1})$ for every $n \geq 1$ . Likewise $(x) ⊋ (x^{2}) ⊋ (x^{3}) ⊋ \dots$ is a strict descending chain of ideals in $k [x]$ . Therefore neither ring is Artinian.

step 1.1givenalgebra

3.1

These two standard examples show that Noetherianity alone does not imply Artinianness.

step 2.1∎

Depends on

Used by

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Dependency tree · two levels

34 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, Sections 3 and 16 (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Example (19.7) (standard reference, not scraped)

ZZ and k[x]k[x] are Noetherian but not Artinian