example of an Artinian module which is not Noetherian

It is well known, that left (right) Artinian ring is left (right) Noetherian (Akizuki-Hopkins-Levitzki theorem). We will show that this no longer holds for modules.

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

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

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

$$\mathbb{Z}=G_0\subset G_1\subset G_2\subset G_3\subset \mathrm{\cdots },$$

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 $G_n^{\prime }=\pi (G_n)$ is a submodule of $G/\mathbb{Z}$ and

$$0=G_0^{\prime }\subset G_1^{\prime }\subset G_2^{\prime }\subset G_3^{\prime }\subset G_4^{\prime }\subset \mathrm{\cdots }.$$

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

$$G_{n+1}^{\prime }/G_n^{\prime }\simeq \left(G_{n+1}/\mathbb{Z}\right)/\left(G_n/\mathbb{Z}\right)\simeq G_{n+1}/G_n\ne 0,$$

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 $G_n^{\prime }$. Let $N\subseteq G/\mathbb{Z}$ be a proper submodule. Assume that for some $a\in \mathbb{Z}$ and $n\ge 0$ we have

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

We may assume that $\gcd (a,p^n)=1$. 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 $\frac{1}{p^m}+\mathbb{Z}\notin N$. What we showed is that

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

because for every $0\le n\le m-1$ (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

$$N_1\supseteq N_2\supseteq N_3\supseteq \mathrm{\cdots }$$

be a chain of submodules in $G/\mathbb{Z}$. Then there are natural numbers $n_1,n_2,\mathrm{\dots }$ such that $N_i=G_{n_i}^{\prime }$. Note that $G_k^{\prime }\supseteq G_s^{\prime }$ if and only if $k\ge s$. In particular we obtain a sequence of natural numbers

$$n_1\ge n_2\ge n_3\ge \mathrm{\cdots }$$

This chain has to stabilize, which completes the proof. $\mathrm{\square }$

Title	example of an Artinian module which is not Noetherian
Canonical name	ExampleOfAnArtinianModuleWhichIsNotNoetherian
Date of creation	2013-03-22 19:04:18
Last modified on	2013-03-22 19:04:18
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	5
Author	joking (16130)
Entry type	Example
Classification	msc 16D10