ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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The Prüfer $p$ -group is Artinian but not Noetherian

Example

For every prime $p$ , the Prüfer group $Z (p^{\infty}) \leq Q / Z$ is Artinian but not Noetherian as a $Z$ -module. See Noetherian modules: every submodule is finitely generated.

Facts & Assumptions

Given: The hypotheses and objects in the Example.

[L1]

A left $R$ -module $M$ is Noetherian when every submodule of $M$ is finitely generated (def-generated-cyclic-finitely-generated-and-free-modules). This finite-generation definition is the convention; its equivalence with ACC and the maximal condition is proved in thm-equivalent-characterizations-of-noetherian-modules. (Noetherian modules: every submodule is finitely generated).

[L2]

A left $R$ -module $M$ is Artinian when every descending chain $M_{0} \supseteq M_{1} \supseteq \dots$ of submodules stabilizes: there is $N$ such that $M_{n} = M_{N}$ for all $n \geq N$ . This is the descending chain condition. (Artinian modules by the descending chain condition).

[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).

[L4]

The map $j (k) = [(k, 1)]$ is injective and preserves addition, multiplication, and order. Composing with lem-nat-embeds-int embeds $N$ in $Q$ ; we write $k$ for $j (k)$ throughout. (The integers embed in the rationals).

[L5]

For $N \leq M$ , the additive cosets $m + N$ form the quotient module $M / N$ under the well-defined scalar action $r (m + N) : = r m + N .$ (Quotient module $M / N$ with scalar multiplication on additive cosets).

Verification

technique · direct

1.1

Let $C_{n} = {a / p^{n} + Z : a \in Z} \leq Q / Z$ . The class $1 / p^{n} + Z$ generates $C_{n}$ and has order $p^{n}$ , while $C_{n} < C_{n + 1}$ . If an element of $⋃_{k} C_{k}$ has order dividing $p^{n}$ , cancelling its denominator shows that it lies in $C_{n}$ . Hence every cyclic subgroup of order $p^{n}$ is $C_{n}$ , and $Z (p^{\infty}) : = ⋃_{n \geq 0} C_{n}$ is the Prüfer $p$ -group.

L1L2L3L4L5givenalgebra

2.1

The group is not finitely generated, which is the failure of [L1] directly: a finite subset of $⋃_{k} C_{k}$ lies in a single $C_{N}$ , because each of its finitely many members lies in some $C_{k}$ and the $C_{k}$ are nested, so the submodule it generates is contained in $C_{N}$ and is proper. Equivalently, the strict ascending chain $C_{0} < C_{1} < C_{2} < \dots$ of step 1.1 does not stabilize.

step 1.1L1givenalgebra

3.1

If a subgroup $H$ contains elements of unbounded order, then for every $n$ it contains an element whose cyclic subgroup contains the unique $C_{n}$ , so $C_{n} \leq H$ and $H = Z (p^{\infty})$ . Otherwise the element orders in $H$ are bounded by some $p^{n}$ , and step 1.1 gives $H \leq C_{n}$ . The subgroups of the cyclic group $C_{n}$ are the unique $C_{m}$ for $0 \leq m \leq n$ , so every proper subgroup of the Prüfer group is one of these finite cyclic groups.

step 1.1step 2.1givenalgebra

4.1

A descending chain either remains at the whole group or enters some finite $C_{n}$ , after which it stabilizes because $C_{n}$ has only the chain $0 = C_{0} < C_{1} < \dots < C_{n}$ of subgroups. Thus DCC holds. The argument includes $C_{0} = 0$ and is unchanged for $p = 2$ . This proves the stated claim.

step 3.1givenalgebra∎

Depends on

Used by

False statement: every Artinian module is Noetherian False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 40 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Arvind Nair, Algebra I, Lecture 5 (standard reference, not scraped)

The Prüfer pp-group is Artinian but not Noetherian