# Noetherian and Artinian modules

I don't understand clearly what is meant by Noetherian and Artinian modules. I tried explaining it to myself using the definitions but it is still not clear. Can someone please help explain it to me using examples etc. Thank you!

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Did you read Wikipedia for instance? –  lhf May 8 '12 at 12:00
Perhaps you could tell us which definition you've read? Some definite a Noetherian module as a module in which every submodule is finitely generated. –  Eugene Jun 11 '12 at 2:19

$\mathbb Z$:

Note that $\mathbb Z$ is a $\mathbb Z$-module. We claim that it is Noetherian. To see this, note that submodules of $\mathbb Z$ correspond to subgroups of it so that all submodules are of the form $(n)$, the ideal generated by $n$ since $\mathbb Z$ is a PID. Now if you have $(n) \subset (m)$ then $m$ must divide $n$ so that $(n_1) \subset (n_2) \subset (n_3) \subset \dots$ stabilises for $k$ large enough since $n_1$ only has a finite number of divisors. Hence every ascending chain in $\mathbb Z$ must be stationary which means that $\mathbb Z$ is Noetherian.

On the other hand, observe that for any $n > 1$ you have $(n) \supset (n^2) \supset (n^3) \supset \dots$ which is a decreasing chain of submodules and hence $\mathbb Z$ is not Artinian.

A finite abelian group $G$:

Since $G$ only has finitely many elements, every increasing and every decreasing sequence will eventually be stationary and hence a finite abelian group is always both, Artinian and Noetherian.

$\mathbb Q$ (as a $\mathbb Z$-module):

Note that $\mathbb Q$ contains $\mathbb Z$ as a submodule hence $\mathbb Q$ cannot be Artinian.

Also note that if $(\frac1p)$ denotes the submodule of $\mathbb Q$ generated by $\frac1p$ then the following is an increasing sequence of submodules (and hence $\mathbb Q$ is not Noetherian): $(\frac1p) \subset (\frac{1}{p^2}) \subset (\frac{1}{p^3}) \subset \dots$

$\mathbb Q / \mathbb Z$ (as $\mathbb Z$-module):

This is not Noetherian since for example $(\frac1p) \subset (\frac{1}{p^2}) \subset \dots$ is an increasing chain of submodules that is not stationary.

On the other hand, subgroups of $\mathbb Q / \mathbb Z$ look like $(\frac{1}{n})$, the subgroup generated by $\frac1n$. If we have $(\frac1n) \supset (\frac1m)$ we know that $m$ divides $n$ so if $(\frac{1}{n_1}) \supset (\frac{1}{n_2}) \supset \dots$ is a decreasing chain it eventually becomes stationary because there are only a finite number of divisors of $n_1$. Hence $\mathbb Q / \mathbb Z$ is Artinian.

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You can find a decent explanation on this Wiki page and also this Wiki page.

I have several examples to keep in mind as you develop a feel for these conditions:

• The ring of integers $\mathbb{Z}$ considered as a right $\mathbb{Z}$-module.
• Any finite dimensional vector space over a field $\mathbb{F}$ (an $\mathbb{F}$ module, of course.)
• Any infinite dimensional vector space over a field.
• For a fixed prime $p$, the set $\{x\in \mathbb{Q}/\mathbb{Z} \mid \exists n\in\mathbb{N}, x{p^n}=0 \}$ considered as a $\mathbb{Z}$ module.

These are examples of modules which are/aren't Noetherian/Artinian in all mixtures, and if you can work out which is which then you will have a better idea of what's going on.

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