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Let $M$ be a (left) module over an associative division ring $R$. Then it has the following properties.

1) For every submodule $N$ of $M$, there exists a submodule $L$ such that $M = N + L$ and $M \cap L = 0$.

2) Every finitely generated submodule has a composition series.

Now let $M \neq 0$ be a (left) faithful module over an associative ring $R$ with unity 1. Suppose $M$ satisfies the above conditions. Is $R$ necessarily a division ring?

EDIT Related question:Module over a ring which satisfies Whitehead's axioms of projective geometry

Community
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  • Is the condition you actually want that every module $M$ has the above properties? – Qiaochu Yuan Sep 20 '14 at 04:07
  • If so, I think this is true iff $R$ is semisimple. In particular $R$ can be a matrix ring over a division ring. – Qiaochu Yuan Sep 20 '14 at 04:14
  • @QiaochuYuan No. I'm asking if an existence of one such $M$ implies that $R$ is a division ring. However, as you wrote, this is not the case. I'm looking for sufficient conditions on $M$ for $R$ to be a division ring. –  Sep 20 '14 at 05:45
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    There are no such conditions which can be stated purely internally to the category of modules. The reason is that any such conditions will be Morita invariant, but being a division ring is not a Morita invariant property. More concretely, categorical statements about modules can't distinguish division rings from matrix algebras over division rings, which have the same categories of modules. – Qiaochu Yuan Sep 22 '14 at 06:47
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    @QiaochuYuan: May I humbly suggest that you promote that to an answer? – Jyrki Lahtonen Sep 22 '14 at 06:49
  • Also notice that condition $1$ just says "$M$ is semisimple," and then all submodules are semisimple, including the finitely generated ones. Finitely generated semisimple modules of course have composition series. So what I'm getting at is that 1$\implies$2. – rschwieb Sep 22 '14 at 13:16

2 Answers2

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No. If $R$ is the ring of $2\times2$ matrices over a field $K$ and $M=K^2$ is the left $R$-module of column vectors, then

  • $M$ has no non-trivial submodules, so conditions 1)+2) are vacuously true, and
  • $R$ is the ring of endomorphisms of $M$ as a $K$-module, so it is a faithful module, but
  • $R$ is not a division ring.
Jyrki Lahtonen
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Let $R=\mathbb{Z}$ so that $R$-modules are abelian groups and submodules are subgroups. Let $M$ be a cyclic group of prime order. Then $M$ is simple, so 1) is satisfied, and every finite group has a composition series, so 2) is satisfied. However, $R$ is not a division ring.

Jared
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  • If the version of the question you're answering is "suppose every $M$ satisfies..." then I don't see what you're doing. Take $M = \mathbb{Z}$. It's finitely generated but it doesn't have a composition series because it doesn't have a maximal submodule. – Qiaochu Yuan Sep 20 '14 at 04:10
  • So 2) fails. And 1) also fails because e.g. in $\mathbb{Z}/p^2\mathbb{Z}$ the submodule $\mathbb{Z}/p\mathbb{Z}$ fails to have a complement. – Qiaochu Yuan Sep 20 '14 at 04:12
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    @QiaochuYuan: I'm answering the question as written, but you're probably right to suggest that the question is meant to be "suppose every $R$-module $M$ satisfies...," in which case this answer is incorrect. – Jared Sep 20 '14 at 04:13
  • Sorry. I forgot to impose the obvious condition that $M$ is a faithful module over $R$. –  Sep 22 '14 at 06:41