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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?

Thanks.

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    Finite dimensional vector spaces would qualify, but these are finite for all intents and purposes, so maybe not.2012-07-20
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    Every PID is noetherian considered as a module over itself. But, say, the field $\mathbb{Q}$ is not a noetherian $\mathbb{Z}$-module!2012-07-20
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    Also, an Abelian group (aka a $\mathbb{Z}$-module) is Noetherian iff it is finitely generated.2012-07-20
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    @ZhenLin But $\mathbb Q$ is a field so it doesn't have any ideals and hence no submodules? No wait. Do submodules of $\mathbb Q$ correspond to subgroups?2012-07-20
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    Confusing. $\mathbb Z$-submodules of $\mathbb Q$ are probably subgroups. But what do subgroups in $\mathbb Q$ look like? Cyclic? Like e.g. the subgroup generated by $\frac12$?2012-07-20
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    @ClarkKent Every fg subgroup of $\mathbb{Q}$ is isomorphic to $\mathbb{Z}$. See this question http://math.stackexchange.com/questions/172699/groups-such-that-every-finitely-generated-subgroup-is-isomorphic-to-the-integer (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.2012-07-20
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    @KReiser Thanks, I think I have it now.2012-07-20
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    Ostrofsky has an example of a module over a finite ring that cannot be given a ring structure compatible with the module structure. The ring is noetherian (being finite), but I don't think the module *itself* is.2012-07-20
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    @ArturoMagidin That sounds like something Barbara *Osofsky* was looking at... but if it really is by a mathematician named Ostrofsky my apologies for interrupting.2012-07-20
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    @rschwieb: No, you're right. I don't know where that extra `tr` came from...2012-07-20
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    @ZhenLin Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-11
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    @JulianKuelshammer My comment is not an answer to the main question. In fact, none of these comments are.2013-06-11

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Take a Noetherian ring $R$. Then $R^n$ is a Noetherian module and all its submodules are Noetherian. For instance, PIDs are Noetherian, because all their ideals are principal, hence finitely generated. You could take $R = \mathbb Z$ or $R = k[x]$ (where $k$ is a field) and consider $M = R^n$. Looking at submodules of $M$, you get more Noetherian modules which are not rings.

Just to completely write out an example, let $R = k[x]$ and consider $\mathfrak m = \{ f(x) \in k[x] \, | \, f(0) = 0 \}$. Then $\mathfrak m \oplus R \trianglelefteq R^2$ is a Noetherian submodule of $R^2$. Its elements are pairs $(f(x), g(x))$ such that $f(0) = 0$.

(I guess this example is a ring without $1$, but for me rings always have a $1$ (because I don't like to think of the ideals of a unital ring as non-unital rings), so I thought it was a good enough example...)

Note that because of the classification of finitely generated modules over a PID (or more generally a Dedekind domain), all finitely generated $R$-modules over such rings will be isomorphic to $$ M \simeq R^{n-1} \oplus \mathfrak a \oplus \mathrm{Tor}(M) $$ where $\mathfrak a$ is a fractional ideal of $R$ (in the PID case, $\mathfrak a = R$) and $$ \mathrm{Tor}(M) \simeq R/\mathfrak p_1^{a_i} \oplus \cdots \oplus R/\mathfrak p_m^{a_m}, $$ so if you want something that is not a commutative ring with $1$, you need to take a ring which is not a Dedekind domain, in particular not a PID. Or if you stick with these rings, you need a non-finitely generated module.

Hope that helps,