In the proof of Nakayama Lemma, the following proposition is used:

let $R$ be a commutative ring with identity and $M$ be a non-zero finitely generated $R$-module. Then $M$ has a maximal submodule.

Is there any counterexample that $M$ is not finitely generated $R$-module and $M$ has no maximal submodule? Thanks.


1 Answer 1


Consider $\mathbb Z_{p^\infty}$ as $\mathbb Z$-module. It is not hard to see that $\mathbb Z_{p^\infty}$ is not finitely generated $\mathbb Z$-module. Obviously $\mathbb Z_{p^\infty}$ has no maximal subgroup.

  • $\begingroup$ What is $\mathbb{Z}_{p^{\infty}}$? $\endgroup$
    – Shiquan
    Jan 19, 2014 at 8:14
  • $\begingroup$ @RenShiquan $\mathbb{Z}_{p^\infty}=\mathbb{Z}[\frac{1}{p}]/\mathbb{Z}$ $\endgroup$ Jan 19, 2014 at 10:17

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