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Let's consider the ring $R = \begin{bmatrix}\Bbb{Q} & 0\\\Bbb{Q} & \Bbb{Z}\end{bmatrix} = \left\{\begin{bmatrix}q & 0\\p & z\end{bmatrix} {\Big|\,} q,p \in \Bbb{Q}, z \in \Bbb{Z}\right\}$ and the right $R$-module $M = \begin{bmatrix}0 & 0\\\Bbb{Q} & \Bbb{Z}\end{bmatrix} = \left\{\begin{bmatrix}0 & 0\\p & z\end{bmatrix} {\Big|\,} p \in \Bbb{Q}, z \in \Bbb{Z}\right\}$.

I want to prove that $M$ is noetherian, that is, every ascending chain of submodules of $M$ in stationary. But how can I do it? There are infinite submodules of $M$.

mathreadler
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effezeta
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  • Hint: $\mathbb{Z}$ is a noetherian $\mathbb{Z}$-module. – Batominovski Dec 30 '17 at 16:19
  • The $\Bbb{Z}$-module $\Bbb{Z}$ is noetherian because every submodule of $\Bbb{Z}$ is finitely generated, but how can I use this fact to prove that M is noetherian? – effezeta Dec 30 '17 at 17:33
  • You can conclude that every ascending chain $N_1\subseteq N_2\subseteq N_3\subseteq \ldots$ of submodules of $M$, there must exist a fix $d\in\mathbb{Z}_{\geq 0}$ and an index $k$ for which every $N_j$ with $j\geq k$ takes the form $$N_j=\begin{bmatrix}0&0\&d\mathbb{Z}\end{bmatrix},,$$ where $$ is either $0$ or $\mathbb{Q}$. – Batominovski Dec 31 '17 at 02:53

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By this solution you can conclude your ring $R$ is right Noetherian, and $M$, being a right submodule of $R$ obviously has to be right Noetherian as well.

rschwieb
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