Example of Artinian module that is not Noetherian

I've just learned the definitions of Artinian and Noetherian module and I'm now trying to think of examples. Can you tell me if the following example is correct:

An example of a $\mathbb Z$-module $M$ that is not Noetherian: Let $G_{\frac12}$ be the additive subgroup of $\mathbb Q$ generated by $\frac12$. Then $G_{\frac12} \subset G_{\frac14} \subset G_{\frac18} \subset \dots$ is a chain with no upper bound hence $M = G_{\frac12}$ as a $\mathbb Z$-module is not Noetherian.

But $M$ is Artinian: $G_{\frac{1}{2^n}}$ are the only subgroups of $G_{\frac12}$. So every decreasing chain of submodules $G_i$ is bounded from below by $G_{\frac{1}{2^{\min i}}}$.

Edit In Atiyah-MacDonald they give the following example:

Does one have to take the quotient $Q/Z$?

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– Dylan Moreland Jul 20 '12 at 11:56

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You might need to modify the example slightly. $G_{1/4}$ is not a submodule of $G_{1/2}$, so you haven't written down an increasing chain inside of your $M$. – Dylan Moreland Jul 20 '12 at 12:01
@DylanMoreland Thanks for pointing this out. So for rings I wouldn't be able to construct such an example. But is the example in my question about modules right? – Matt N. Jul 20 '12 at 12:02

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– Bruno Stonek Jul 20 '12 at 12:24

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@ClarkKent What you have is an ascending chain in $\mathbb{Q}$. – Joe Johnson 126 Jul 20 '12 at 14:24
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2 Answers

Fix a prime $p$ and let $M_p={\Bbb Z}(\frac1p)/{\Bbb Z}$.

It is not difficult to see that the only submodules of $M_p$ are those generated by $\frac1{p^k}+{\Bbb Z}$ for $k\geq0$. From this it follows that $M_p$ is Artinian but not Noetherian.

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Thank you! So $\mathbb Z (\frac{1}{p}) = \{ \frac{k}{p^n} \mid n \in \mathbb N , k \in \mathbb Z \}$? – Matt N. Jul 20 '12 at 12:09
Right! I think that it can be described as the smallest divisible submodule of ${\Bbb Q}/{\Bbb Z}$ containing $1/p$. – Andrea Mori Jul 20 '12 at 12:16
Do we have to quotient by $\mathbb Z$? Does it not work with subgroups of $\mathbb Q$? – Matt N. Jul 20 '12 at 14:07

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@ClarkKent: $A=\mathbb{Z}(\frac1p)$ is not artinian, as it contains the decreasing sequence of subgroups $A \supset qA \supset q^2A \supset q^3A \supset \cdots$ for any prime $q\neq p$. – Jack Schmidt Jul 20 '12 at 14:59
@JackSchmidt Right, thank you very much. I'm starting to understand it a bit better. – Matt N. Jul 21 '12 at 8:15
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Since links can expire, it is advisable to include a short summary to a link-answer. – Julian Kuelshammer Feb 22 at 14:52

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This link has no new example, it is the famous example "$\mathbb{Z}_{P^{\infty}}$". – AmirHosein SadeghiManesh Feb 22 at 19:04

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It doesn't matter if it's new or not. – Julian Kuelshammer Feb 23 at 8:02
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