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.
Best Answer
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,