# example of an Artinian module which is not Noetherian

Let $\mathbb{Z}$ be the ring of integers  and $\mathbb{Q}$ the field of rationals. Let $p\in\mathbb{Z}$ be a prime number  and consider

 $G=\{\frac{a}{p^{n}}\in\mathbb{Q}\ |\ a\in\mathbb{Z};\ n\geq 0\}.$

Of course $G$ is a $\mathbb{Z}$-module via standard multiplication  and addition. For $n\geq 0$ consider

 $G_{n}=\{\frac{a}{p^{n}}\in\mathbb{Q}\ |\ a\in\mathbb{Z}\}.$

Of course each $G_{n}\subseteq G$ is a submodule  and it is easy to see, that

 $\mathbb{Z}=G_{0}\subset G_{1}\subset G_{2}\subset G_{3}\subset\cdots,$

where each inclusion is proper. We will show that $G/\mathbb{Z}$ is Artinian  , but it is not Noetherian.

Let $\pi:G\to G/\mathbb{Z}$ be the canonical projection. Then $G^{\prime}_{n}=\pi(G_{n})$ is a submodule of $G/\mathbb{Z}$ and

 $0=G^{\prime}_{0}\subset G^{\prime}_{1}\subset G^{\prime}_{2}\subset G^{\prime}% _{3}\subset G^{\prime}_{4}\subset\cdots.$

The inclusions are proper, because for any $n>0$ we have

 $G^{\prime}_{n+1}/G^{\prime}_{n}\simeq\big{(}G_{n+1}/\mathbb{Z}\big{)}/\big{(}G% _{n}/\mathbb{Z}\big{)}\simeq G_{n+1}/G_{n}\neq 0,$

due to Third Isomorphism Theorem for modules. This shows, that $G/\mathbb{Z}$ is not Noetherian.

In order to show that $G/\mathbb{Z}$ is Artinian, we will show, that each proper submodule of $G/\mathbb{Z}$ is of the form $G^{\prime}_{n}$. Let $N\subseteq G/\mathbb{Z}$ be a proper submodule. Assume that for some $a\in\mathbb{Z}$ and $n\geq 0$ we have

 $\frac{a}{p^{n}}+\mathbb{Z}\in N.$

We may assume that $\mathrm{gcd}(a,p^{n})=1$. Therefore there are $\alpha,\beta\in\mathbb{Z}$ such that

 $1=\alpha a+\beta p^{n}.$

Now, since $N$ is a $\mathbb{Z}$-module we have

 $\frac{\alpha a}{p^{n}}+\mathbb{Z}\in N$

and since $0+\mathbb{Z}=\beta+\mathbb{Z}=\frac{\beta p^{n}}{p^{n}}+\mathbb{Z}\in N$ we have that

 $\frac{1}{p^{n}}+\mathbb{Z}=\frac{\alpha a+\beta p^{n}}{p^{n}}+\mathbb{Z}\in N.$

Now, let $m>0$ be the smallest number, such that $\frac{1}{p^{m}}+\mathbb{Z}\not\in N$. What we showed is that

 $N=G^{\prime}_{m-1}=\pi(G_{m-1}),$

because for every $0\leq n\leq m-1$ (and only for such $n$) we have $\frac{1}{p^{n}}+\mathbb{Z}\in N$ and thus $N$ is a image of a submodule of $G$, which is generated by $\frac{1}{p^{n}}$ and this is precisely $G_{m-1}$. Now let

 $N_{1}\supseteq N_{2}\supseteq N_{3}\supseteq\cdots$

be a chain of submodules in $G/\mathbb{Z}$. Then there are natural numbers  $n_{1},n_{2},\ldots$ such that $N_{i}=G^{\prime}_{n_{i}}$. Note that $G^{\prime}_{k}\supseteq G^{\prime}_{s}$ if and only if $k\geq s$. In particular we obtain a sequence   of natural numbers

 $n_{1}\geq n_{2}\geq n_{3}\geq\cdots$
Title example of an Artinian module which is not Noetherian ExampleOfAnArtinianModuleWhichIsNotNoetherian 2013-03-22 19:04:18 2013-03-22 19:04:18 joking (16130) joking (16130) 5 joking (16130) Example msc 16D10