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

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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.
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The $Z$ -module $Z / p^{k}$ has length $k$

Example

For a prime $p$ and $k \in N$ , the $Z$ -module $Z / p^{k} Z$ has length $k$ . In particular, $k = 0$ gives the zero module. See Composition series and length of a module.

Facts & Assumptions

Given: The hypotheses and objects in the Example.

[L1]

A composition series of a left $R$ -module $M$ is a finite chain $0 = M_{0} < M_{1} < \dots < M_{n} = M$ whose factors $M_{i} / M_{i - 1}$ are simple. If such a series exists, the length $ℓ_{R} (M)$ is its number $n$ of factors; thm-jordan-holder-theorem-for-modules proves independence of the chosen series. The zero module has the empty series and length $0$ . (Composition series and length of a module).

[L2]

Any two composition series of a module have the same length, and their simple factors agree up to permutation and isomorphism. (Jordan–Hölder theorem for modules).

[L3]

For a short exact sequence $0 \to N \to M \to Q \to 0$ , the module $M$ has finite length if and only if $N$ and $Q$ do, and then $ℓ_{R} (M) = ℓ_{R} (N) + ℓ_{R} (Q) .$ . (Module length is additive in short exact sequences).

[L4]

Let $G$ be a finite group such that the positive integer $∣ G ∣$ is prime. Then every $g \neq e$ has order $∣ G ∣$ , satisfies $⟨ g ⟩ = G$ , and hence generates $G$ . In particular, $G$ is cyclic. (A finite group of prime order is cyclic and every nonidentity element generates it).

Verification

technique · direct

1.1

For $k \geq 1$ , the chain $0 < p^{k - 1} Z / p^{k} Z < \dots < p Z / p^{k} Z < Z / p^{k} Z$ is strict, and every successive quotient has order $p$ and is therefore the simple $Z$ -module $Z / p$ .

L1L2L3L4givenalgebra

2.1

This composition series has $k$ factors, so Jordan–Hölder gives length $k$ . For $k = 0$ , $Z / p^{0} Z = 0$ has the empty series; for $k = 1$ , the displayed module itself is simple.

step 1.1givenalgebra

3.1

The calculation concerns the finite-length quotient module and does not require the ambient ring $Z$ to be Artinian. This proves the stated claim.

step 2.1givenalgebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 64 results over 13 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 ZZ-module Z/pkZ/pk has length kk