# 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 $\mathrm{gcd}(a,{p}^{n})=1$. 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 $\frac{1}{{p}^{m}}+\mathbb{Z}\notin N$. 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 |
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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 |