# 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
$\left(\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)$
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

$${I}_{n}=\{\left(\begin{array}{cc}\hfill 0\hfill & \hfill \frac{m}{{2}^{n}}\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\mid m\in \mathbb{Z}\}.$$ |

Verify that each ${I}_{n}$ is a left ideal^{} in $R$ and that ${I}_{0}\u228a{I}_{1}\u228a{I}_{2}\u228a\mathrm{\cdots}$.

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

$$\left(\begin{array}{cc}\hfill 0\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill z\hfill \end{array}\right)\text{for some}y,z\in \mathbb{Q}.$$ |

Note that for any $c\in \mathbb{Q}$ and any $\left(\begin{array}{cc}\hfill 0\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill z\hfill \end{array}\right)\in I$, we have $\left(\begin{array}{cc}\hfill 0\hfill & \hfill cy\hfill \\ \hfill 0\hfill & \hfill cz\hfill \end{array}\right)\in I$ since

$$\left(\begin{array}{cc}\hfill 0\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill z\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill cy\hfill \\ \hfill 0\hfill & \hfill cz\hfill \end{array}\right)$$ |

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

Indeed, note that
$V=\{(y,z)\in {\mathbb{Q}}^{2}\mid \left(\begin{array}{cc}\hfill 0\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill z\hfill \end{array}\right)\in I\}$
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 $\left(\begin{array}{cc}\hfill 0\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill z\hfill \end{array}\right)$ 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

$$\left(\begin{array}{cc}\hfill 0\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill z\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill {c}_{1}{y}_{1}+{c}_{2}{y}_{2}\hfill \\ \hfill 0\hfill & \hfill {c}_{1}{z}_{1}+{c}_{2}{z}_{2}\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill {y}_{1}\hfill \\ \hfill 0\hfill & \hfill {z}_{1}\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {c}_{1}\hfill \end{array}\right)+\left(\begin{array}{cc}\hfill 0\hfill & \hfill {y}_{2}\hfill \\ \hfill 0\hfill & \hfill {z}_{2}\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {c}_{2}\hfill \end{array}\right)$$ |

and it follows that $I$ is finitely generated by the set $\{\left(\begin{array}{cc}\hfill 0\hfill & \hfill {y}_{1}\hfill \\ \hfill 0\hfill & \hfill {z}_{1}\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 0\hfill & \hfill {y}_{2}\hfill \\ \hfill 0\hfill & \hfill {z}_{2}\hfill \end{array}\right)\}$ 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

$$\left(\begin{array}{cc}\hfill kn\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill z\hfill \end{array}\right)\text{for some}k\in \mathbb{Z};y,z\in \mathbb{Q}.$$ |

By definition of $n$, there is a matrix of the form $\left(\begin{array}{cc}\hfill n\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)$ in $I$. Since $I$ is a right ideal in $R$ and since $\left(\begin{array}{cc}\hfill n\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill c\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill n\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),$ it follows that $\left(\begin{array}{cc}\hfill n\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ 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

$$\left(\begin{array}{cc}\hfill kn\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\text{for some}k\in \mathbb{Z},y\in \mathbb{Q}.$$ |

Note that $\left(\begin{array}{cc}\hfill kn\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill n\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill k\hfill & \hfill \frac{y}{n}\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$. Hence, $\left(\begin{array}{cc}\hfill n\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ 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

$$\left(\begin{array}{cc}\hfill mn\hfill & \hfill {y}_{1}\hfill \\ \hfill 0\hfill & \hfill {z}_{1}\hfill \end{array}\right)\text{for some}m\in \mathbb{Z};{y}_{1},{z}_{1}\in \mathbb{Q};{z}_{1}\ne 0.$$ |

Since $\left(\begin{array}{cc}\hfill n\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\in I,$ it follows that $\left(\begin{array}{cc}\hfill n\hfill & \hfill {y}_{1}\hfill \\ \hfill 0\hfill & \hfill {z}_{1}\hfill \end{array}\right)\in I.$ Let $\left(\begin{array}{cc}\hfill kn\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill z\hfill \end{array}\right)$ be an arbitrary element of $I$. Since $\left(\begin{array}{cc}\hfill kn\hfill & \hfill y\hfill \\ \hfill 0\hfill & \hfill z\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill n\hfill & \hfill {y}_{1}\hfill \\ \hfill 0\hfill & \hfill {z}_{1}\hfill \end{array}\right)\left(\begin{array}{cc}\hfill k\hfill & \hfill \frac{1}{n}(y-\frac{{y}_{1}z}{{z}_{1}})\hfill \\ \hfill 0\hfill & \hfill \frac{z}{{z}_{1}}\hfill \end{array}\right),$ it follows that $\left(\begin{array}{cc}\hfill n\hfill & \hfill {y}_{1}\hfill \\ \hfill 0\hfill & \hfill {z}_{1}\hfill \end{array}\right)$ 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 |
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Canonical name | ExampleOfARightNoetherianRingThatIsNotLeftNoetherian |

Date of creation | 2013-03-22 14:16:15 |

Last modified on | 2013-03-22 14:16:15 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 18 |

Author | CWoo (3771) |

Entry type | Example |

Classification | msc 16P40 |