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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:

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

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Interesting fact: Artinian rings are Noetherian. [This is not obvious, at least to me.] See here. – 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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In the same direction as Dylan's comment, you might find this question interesting. – 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

up vote 4 down vote accepted

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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Explained Example in PlanetMath

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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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