# 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 $G_{1/2}$ 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 $\mathbb{Q}/\mathbb{Z}$?

• Interesting fact: Artinian rings are Noetherian. [This is not obvious, at least to me.] See here. – Dylan Moreland Jul 20 '12 at 11:56
• 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? – Rudy the Reindeer Jul 20 '12 at 12:02
• In the same direction as Dylan's comment, you might find this question interesting. – Bruno Stonek Jul 20 '12 at 12:24
• @ClarkKent What you have is an ascending chain in $\mathbb{Q}$. – Joe Johnson 126 Jul 20 '12 at 14:24

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.
• Thank you! So $\mathbb Z (\frac{1}{p}) = \{ \frac{k}{p^n} \mid n \in \mathbb N , k \in \mathbb Z \}$? – Rudy the Reindeer 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 $\mathbb Q$? – Rudy the Reindeer Jul 20 '12 at 14:07
• @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