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.

Title example of a right noetherian ring that is not left noetherian ExampleOfARightNoetherianRingThatIsNotLeftNoetherian 2013-03-22 14:16:15 2013-03-22 14:16:15 CWoo (3771) CWoo (3771) 18 CWoo (3771) Example msc 16P40