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# example of a right noetherian ring that is not left noetherian

This example, due to Lance Small, is briefly described in *Noncommutative Rings*, by I. N. Herstein, published by the Mathematical Association of America, 1968.

Let $R$ be the ring of all $2\times 2$ matrices $\begin{pmatrix}a&b\\ 0&c\end{pmatrix}$ such that $a$ is an integer and $b,c$ are rational. The claim is that $R$ is right noetherian but not left noetherian.

It is relatively straightforward to show that $R$ is not left noetherian. For each natural number $n$, let

$I_{n}=\{\begin{pmatrix}0&\frac{m}{2^{n}}\\ 0&0\end{pmatrix}\mid m\in\mathbb{Z}\}.$ |

Verify that each $I_{n}$ is a left ideal in $R$ and that $I_{0}\subsetneq I_{1}\subsetneq I_{2}\subsetneq\cdots$.

It is a bit harder to show that $R$ is right noetherian. The approach given here uses the fact that a ring is right noetherian if all of its right ideals are finitely generated.

Let $I$ be a right ideal in $R$. We show that $I$ is finitely generated by checking all possible cases. In the first case, we assume that every matrix in $I$ has a zero in its upper left entry. In the second case, we assume that there is some matrix in $I$ that has a nonzero upper left entry. The second case splits into two subcases: either every matrix in $I$ has a zero in its lower right entry or some matrix in $I$ has a nonzero lower right entry.

CASE 1: Suppose that for all matrices in $I$, the upper left entry is zero. Then every element of $I$ has the form

$\begin{pmatrix}0&y\\ 0&z\end{pmatrix}\text{ for some }y,z\in\mathbb{Q}.$ |

Note that for any $c\in\mathbb{Q}$ and any $\begin{pmatrix}0&y\\ 0&z\end{pmatrix}\in I$, we have $\begin{pmatrix}0&cy\\ 0&cz\end{pmatrix}\in I$ since

$\begin{pmatrix}0&y\\ 0&z\end{pmatrix}\begin{pmatrix}0&0\\ 0&c\end{pmatrix}=\begin{pmatrix}0&cy\\ 0&cz\end{pmatrix}$ |

and $I$ is a right ideal in $R$. So $I$ looks like a rational vector space.

Indeed, note that $V=\{(y,z)\in\mathbb{Q}^{2}\mid\begin{pmatrix}0&y\\ 0&z\end{pmatrix}\in I\}$ is a subspace of the two dimensional vector space $\mathbb{Q}^{2}$. So in $V$ there exist two (not necessarily linearly independent) vectors $(y_{1},z_{1})$ and $(y_{2},z_{2})$ which span $V$.

Now, an arbitrary element $\begin{pmatrix}0&y\\ 0&z\end{pmatrix}$ in $I$ corresponds to the vector $(y,z)$ in $V$ and $(y,z)=(c_{1}y_{1}+c_{2}y_{2},c_{1}z_{1}+c_{2}z_{2})$ for some $c_{1},c_{2}\in\mathbb{Q}$. Thus

$\begin{pmatrix}0&y\\ 0&z\end{pmatrix}=\begin{pmatrix}0&c_{1}y_{1}+c_{2}y_{2}\\ 0&c_{1}z_{1}+c_{2}z_{2}\end{pmatrix}=\begin{pmatrix}0&y_{1}\\ 0&z_{1}\end{pmatrix}\begin{pmatrix}0&0\\ 0&c_{1}\end{pmatrix}+\begin{pmatrix}0&y_{2}\\ 0&z_{2}\end{pmatrix}\begin{pmatrix}0&0\\ 0&c_{2}\end{pmatrix}$ |

and it follows that $I$ is finitely generated by the set $\{\begin{pmatrix}0&y_{1}\\ 0&z_{1}\end{pmatrix},\begin{pmatrix}0&y_{2}\\ 0&z_{2}\end{pmatrix}\}$ as a right ideal in $R$.

CASE 2: Suppose that some matrix in $I$ has a nonzero upper left entry. Then there is a least positive integer $n$ occurring as the upper left entry of a matrix in $I$. It follows that every element of $I$ can be put into the form

$\begin{pmatrix}kn&y\\ 0&z\end{pmatrix}\text{ for some }k\in\mathbb{Z};\ y,z\in\mathbb{Q}.$ |

By definition of $n$, there is a matrix of the form $\begin{pmatrix}n&b\\ 0&c\end{pmatrix}$ in $I$. Since $I$ is a right ideal in $R$ and since $\begin{pmatrix}n&b\\ 0&c\end{pmatrix}\begin{pmatrix}1&0\\ 0&0\end{pmatrix}=\begin{pmatrix}n&0\\ 0&0\end{pmatrix},$ it follows that $\begin{pmatrix}n&0\\ 0&0\end{pmatrix}$ is in $I$. Now break off into two subcases.

*case* 2.1: Suppose that every matrix in $I$ has a zero in its lower right entry. Then
an arbitrary element of $I$ has the form

$\begin{pmatrix}kn&y\\ 0&0\end{pmatrix}\text{ for some }k\in\mathbb{Z},y\in\mathbb{Q}.$ |

Note that $\begin{pmatrix}kn&y\\ 0&0\end{pmatrix}=\begin{pmatrix}n&0\\ 0&0\end{pmatrix}\begin{pmatrix}k&\frac{y}{n}\\ 0&0\end{pmatrix}$. Hence, $\begin{pmatrix}n&0\\ 0&0\end{pmatrix}$ generates $I$ as a right ideal in $R$.

*case* 2.2: Suppose that some matrix in $I$ has a nonzero lower right entry. That is, in $I$
we have a matrix

$\begin{pmatrix}mn&y_{1}\\ 0&z_{1}\end{pmatrix}\text{ for some }m\in\mathbb{Z};\ y_{1},z_{1}\in\mathbb{Q}% ;\ z_{1}\neq 0.$ |

Since $\begin{pmatrix}n&0\\ 0&0\end{pmatrix}\in I,$ it follows that $\begin{pmatrix}n&y_{1}\\ 0&z_{1}\end{pmatrix}\in I.$ Let $\begin{pmatrix}kn&y\\ 0&z\end{pmatrix}$ be an arbitrary element of $I$. Since $\begin{pmatrix}kn&y\\ 0&z\end{pmatrix}=\begin{pmatrix}n&y_{1}\\ 0&z_{1}\end{pmatrix}\begin{pmatrix}k&\frac{1}{n}(y-\frac{y_{1}z}{z_{1}})\\ 0&\frac{z}{z_{1}}\end{pmatrix},$ it follows that $\begin{pmatrix}n&y_{1}\\ 0&z_{1}\end{pmatrix}$ generates $I$ as a right ideal in $R$.

In all cases, $I$ is a finitely generated.

## Mathematics Subject Classification

16P40*no label found*

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