Question

Is true statement over arbitrary ring: "If R is a commutative ring or a left noetherian ring, then any finitely generated artinian left R-module M has finite length."

Answer 1

We construct here a cyclic artinian left module $M$ of infinite length over some (noncommutative, non left-noetherian) ring $R$. In the triangular ring $R = \begin{pmatrix} \mathbb{Q} & 0 \\ \mathbb{Q} & \mathbb{Z} \end{pmatrix}$, the idempotent $e = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ generates the left ideal $Re = \begin{pmatrix} \mathbb{Q} & 0 \\ \mathbb{Q} & 0 \end{pmatrix}$ (which is in fact an ideal). We express this module in the simpler form $\begin{pmatrix} \mathbb{Q} \\ \mathbb{Q} \end{pmatrix}$, and consider its submodule $\begin{pmatrix} 0 \\ \mathbb{Z}_{(p)} \end{pmatrix}$, where $Z_{(p)}$ denotes the localization of $Z$ at a prime ideal $(p)$.

Since $\left( \begin{array}{cc}a & 0\\ b & c \end{array} \right)\left( \begin{array}{cc}0 & 0\\ x & 0 \end{array} \right) = \left( \begin{array}{cc}0 & 0\\ cx & 0 \end{array} \right)$ ($a, b \in \mathbb{Q}; c \in \mathbb{Z}$), the ideal $Re$ acts trivially on $\begin{pmatrix} 0 \\ \mathbb{Q} \end{pmatrix}$ so the $R$-submodules of $\begin{pmatrix} 0 \\ \mathbb{Q} \end{pmatrix}$ are just $\begin{pmatrix} 0 \\ G \end{pmatrix}$ where $G$ is any subgroup of $\mathbb{Q}$. Now $\mathbb{Q} / \mathbb{Z}_{(p)}$ is isomorphic to the Prüfer $p$-group (the group of $p^n$-th roots of unity for $n \in \mathbb{N}$), which is of infinite length as a $\mathbb{Z}$-module. Therefore, the cyclic $R$-module $M := \begin{pmatrix} \mathbb{Q} \\ \mathbb{Q} \end{pmatrix} / \begin{pmatrix} 0 \\ \mathbb{Z}_{(p)} \end{pmatrix} \supseteq M' = \begin{pmatrix} 0 \\ \mathbb{Q} \end{pmatrix} / \begin{pmatrix} 0 \\ \mathbb{Z}_{(p)} \end{pmatrix}$ is also of infinite length. Now $M / M' \sim \begin{pmatrix} \mathbb{Q} \\ \mathbb{Q} \end{pmatrix} / \begin{pmatrix} 0 \\ \mathbb{Q} \end{pmatrix}$ is a simple $R$-module, and $M'$ is an artinian $\mathbb{Z}$-module (and hence an artinian $R$-module). It follows that $M$ is also an artinian $R$-module (of infinite length), as desired.