ExampleConstruction: Literature-sourcedVerification: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Over Z, the ideal (2) acts surjectively on Z/3Z but does not kill it

Example

Take $R = Z$ , $I = (2)$ , and $M = Z / 3 Z$ . Then $I M = M$ but $M \neq 0$ , and $I ⊈ J (R)$ . So Nakayama's conclusion fails if the Jacobson-radical hypothesis is removed.

Facts & Assumptions

Given: The ring $R = Z$ , the ideal $I = (2)$ , and the $R$ -module $M = Z / 3 Z$ .

[L1]

The Jacobson radical is the intersection of the maximal ideals (The Jacobson radical of a ring).

[L2]

The submodule $I M$ consists of finite sums of products $i m$ (The submodule $I M$ generated by products of elements of an ideal $I$ with elements of a module $M$ ).

[L3]

In $Z / 3 Z$ , multiplication by $2$ is a bijection because $[2]_{3} [2]_{3} = [1]_{3}$ (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

Verification

technique · direct

1.1

By [L3], every class in $Z / 3 Z$ is $2$ times another class, so $I M = M$ .

L2L3

1.2

The module $M$ is nonzero because $[1]_{3} \neq [0]_{3}$ . Moreover $2 \notin J (Z)$ because $2 \notin (3)$ , while $(3)$ is a maximal ideal of $Z$ and [L1] makes $J (Z)$ lie in every maximal ideal.

L1L3algebra

2.1

Thus $I M = M$ and $M \neq 0$ hold with $I ⊈ J (R)$ , exactly exhibiting why the Jacobson-radical hypothesis cannot be dropped.

step 1.1step 1.2∎

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

J. S. Milne, A Primer of Commutative Algebra, v4.03, Lemma 3.9 (standard reference, not scraped)