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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. –  Miha Habič Jul 20 '12 at 10:12
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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! –  Zhen Lin Jul 20 '12 at 10:13
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Also, an Abelian group (aka a $\mathbb{Z}$-module) is Noetherian iff it is finitely generated. –  Miha Habič Jul 20 '12 at 10:18
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@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
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@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

1 Answer 1

up vote 3 down vote accepted

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,

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