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I've been trying to think of an $R$-module that is Noetherian, not finite and is not a ring.

Examples that I know are:

1 A finite Abelian group is a Noetherian $\mathbb Z$-module (of course it satisfies a.c.c. because it's finite as a set)

2 $\mathbb Z$ is a Noetherian $\mathbb Z$-module. It's submodules correspond to ideals $I$ and $\mathbb Z $ is a PID. So every chain of ideals will eventually end in prime ideals. (the chain might branch into several maximal ideals)

3 Similar to 2, $k[x]$ where $k$ is a field is also a PID and by the same argument as in 2 also Noetherian.

But two of these three are rings and one is finite.

What are more interesting examples of Noetherian $R$-modules?

And is every PID (=principal ideal domain) a Noetherian module?


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Finite dimensional vector spaces would qualify, but these are finite for all intents and purposes, so maybe not. –  Miha Habič Jul 20 '12 at 10:12
Every PID is noetherian considered as a module over itself. But, say, the field $\mathbb{Q}$ is not a noetherian $\mathbb{Z}$-module! –  Zhen Lin Jul 20 '12 at 10:13
Also, an Abelian group (aka a $\mathbb{Z}$-module) is Noetherian iff it is finitely generated. –  Miha Habič Jul 20 '12 at 10:18
@ClarkKent Every fg subgroup of $\mathbb{Q}$ is isomorphic to $\mathbb{Z}$. See this question math.stackexchange.com/questions/172699/… (specifically the top answer) for classification of all subgroups of $\mathbb{Q}$. As for your confusion, it is important to remember where we are: In the example discussed, we're working in $\mathbb{Z}$-mod, so submodules are subgroups of abelian groups. –  KReiser Jul 20 '12 at 10:26
@JulianKuelshammer My comment is not an answer to the main question. In fact, none of these comments are. –  Zhen Lin Jun 11 '13 at 21:49

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