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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 $G_{1/2}$ be the additive subgroup of $\mathbb Q$ generated by $\frac12$. Then $G_{1/2} \subset G_{1/4} \subset G_{1/8} \subset \dotsb$ is a chain with no upper bound hence $M = G_{1/2}$ as a $\mathbb Z$-module is not Noetherian.

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

Edit In Atiyah-MacDonald they give the following example:

Let $G$ be the subgroups of $\mathbb{Q}/\mathbb{Z}$ consisting of all elements whose order is a power of $p$, where $p$ is a fixed prime. Then $G$ has exactly one subgroup $G_n$ of order $p^n$ for each $n \geq 0$, and $G_0 \subset G_1 \subset \dotsb \subset G_n \subset \dotsb$ (strict inclusions) so that $G$ does not satisfy the a.c.c. On the other hand the only proper subgroups of $G$ are the $G_n$, so that $G$ does satisfy d.c.c.

(Original images here and here.)

Does one have to take the quotient $\mathbb{Q}/\mathbb{Z}$?

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    $\begingroup$ Interesting fact: Artinian rings are Noetherian. [This is not obvious, at least to me.] See here. $\endgroup$
    – Dylan Moreland
    Commented Jul 20, 2012 at 11:56
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    $\begingroup$ 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 $M$. $\endgroup$
    – Dylan Moreland
    Commented Jul 20, 2012 at 12:01
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    $\begingroup$ @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? $\endgroup$
    – Rudy the Reindeer
    Commented Jul 20, 2012 at 12:02
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    $\begingroup$ In the same direction as Dylan's comment, you might find this question interesting. $\endgroup$
    – Bruno Stonek
    Commented Jul 20, 2012 at 12:24
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    $\begingroup$ @ClarkKent What you have is an ascending chain in $\mathbb{Q}$. $\endgroup$
    – J126
    Commented Jul 20, 2012 at 14:24

1 Answer 1

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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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    $\begingroup$ Thank you! So $\mathbb Z (\frac{1}{p}) = \{ \frac{k}{p^n} \mid n \in \mathbb N , k \in \mathbb Z \}$? $\endgroup$
    – Rudy the Reindeer
    Commented Jul 20, 2012 at 12:09
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    $\begingroup$ Right! I think that it can be described as the smallest divisible submodule of ${\Bbb Q}/{\Bbb Z}$ containing $1/p$. $\endgroup$
    – Andrea Mori
    Commented Jul 20, 2012 at 12:16
  • $\begingroup$ Do we have to quotient by $\mathbb Z$? Does it not work with subgroups of $\mathbb Q$? $\endgroup$
    – Rudy the Reindeer
    Commented Jul 20, 2012 at 14:07
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    $\begingroup$ @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 $q\neq p$. $\endgroup$
    – Jack Schmidt
    Commented Jul 20, 2012 at 14:59
  • $\begingroup$ @JackSchmidt Right, thank you very much. I'm starting to understand it a bit better. $\endgroup$
    – Rudy the Reindeer
    Commented Jul 21, 2012 at 8:15

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