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 $2\times2$ matrices 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 , we have 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 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 in $I$ corresponds to the vector $(y,z)$ in $V$ and $ (y,z)=(c_1y_1+c_2y_2,c_1z_1+c_2z_2)$ for some $c_1, c_2\in\mathbb{Q}$ Thus $$ \begin{pmatrix} 0 & y\\ 0 & z \end{pmatrix} = \begin{pmatrix} 0 & c_1y_1+c_2y_2\\ 0 & c_1z_1+c_2z_2 \end{pmatrix} = \begin{pmatrix} 0 & y_1\\ 0 & z_1 \end{pmatrix} \begin{pmatrix} 0 & 0\\ 0 & c_1 \end{pmatrix} + \begin{pmatrix} 0 & y_2\\ 0 & z_2 \end{pmatrix} \begin{pmatrix} 0 & 0\\ 0 & c_2 \end{pmatrix} $$ and it follows that $I$ is finitely generated by the set 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 in $I$ Since $I$ is a right ideal in $R$ and since it follows that 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 . Hence, 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 it follows that Let be an arbitrary element of $I$ Since it follows that generates $I$ as a right ideal in $R$

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