How can I prove that the ring of all $2*2$ matrices $$S=\begin{equation*} \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} \qquad \end{equation*}$$

Such that $a$ is an integer and $b,c$ are rationals is right Noetherian but not left Noetherian. I have seen a lot of discussions here about it but I still did not understand it, could anyone give me a clear and simple answer to it please?

Edit My professor answer was as follows:

These are the right ideals of $S$ (and I do not know why), $\begin{equation*} \begin{bmatrix} n\mathbb{Z} & Q \\ 0 & 0 \end{bmatrix} \qquad \end{equation*}$, $\begin{equation*} \begin{bmatrix} n\mathbb{Z} & Q \\ 0 & Q \end{bmatrix} \qquad \end{equation*}$, $\begin{equation*} \begin{bmatrix} 0 & Q \\ 0 & 0 \end{bmatrix} \qquad \end{equation*}$, $\begin{equation*} \begin{bmatrix} 0 & Q \\ 0 & Q \end{bmatrix} \qquad \end{equation*}$, $\begin{equation*} \begin{bmatrix} 0 & 0 \\ 0 & Q \end{bmatrix} \qquad \end{equation*}$, $\begin{equation*} \begin{bmatrix} n\mathbb{Z} & 0 \\ 0 & 0 \end{bmatrix} \qquad \end{equation*}$, and she said and because $\mathbb{Z}$ is Noetherian $S$ is right Noetherian.

Then she calculated the following (and I do not know why):

$\begin{equation*} \begin{bmatrix} \mathbb{Z} & Q \\ 0 & Q \end{bmatrix} \qquad \end{equation*}$$\begin{equation*} \begin{bmatrix} n & q \\ 0 & p \end{bmatrix} \qquad \end{equation*} = $\begin{equation*} \begin{bmatrix} n\mathbb{Z} & q\mathbb{Z} + pQ \\ 0 & pQ \end{bmatrix} \qquad \end{equation*}

And she said so if $q \in Q,$ $\begin{equation*} \begin{bmatrix} 0 & q\mathbb{Z} \\ 0 & 0 \end{bmatrix} \qquad \end{equation*}$ is a left ideal of $S$.

Then she added if $n >1,$

$$\frac{\mathbb{Z}}{n} \subsetneq \frac{\mathbb{Z}}{n^2} \subsetneq ...... $$

Really I did not understand how she is thinking and how she is organizing her answer, could anyone explain her answer to me (she is inpatient professor this is why I did not ask her )

  • $\begingroup$ See also the solution here. $\endgroup$ Jan 8, 2018 at 10:55
  • 1
    $\begingroup$ You asked for a hint at the duplicate, and got links to two questions which give you everything you need to solve the problem. If you have a question about the links you should ask in a comments of those questions, not just post the same question again asking for a complete answer instead of a hint. $\endgroup$
    – rschwieb
    Jan 8, 2018 at 11:47
  • $\begingroup$ @rschwieb my professor answered it in a so different way this is why I am shocked ..... I think the better thing that I should do is to post my professor answer so that I could discuss it with others because I am not fully convinced with it. $\endgroup$
    – Emptymind
    Jan 8, 2018 at 16:51
  • $\begingroup$ @Intuition Sure, if you thought the proof had a gap you could probably formulate a post around the problem. $\endgroup$
    – rschwieb
    Jan 8, 2018 at 17:27
  • 2
    $\begingroup$ @Intuition Thank you for improving your question. $\endgroup$
    – rschwieb
    Jan 9, 2018 at 14:43

1 Answer 1


These are the right ideals (I do not know why)

As I mentioned when you asked this question previously, this solution explains what all the right ideals are. Just follow the description of right ideals at the link, and you can see.

It is a good exercise to prove that characterization of right ideals too. It simply amounts to reasoning what a right ideal must contain if you are multiplying with things like $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&0\\0&1\end{bmatrix}$ and $\begin{bmatrix}0&1\\0&1\end{bmatrix}$ on the right.

Then finally, after studying that solution, see why $R$ is right Noetherian: you have an exact sequence $0\to \begin{bmatrix}0&0\\0&\mathbb Q\end{bmatrix}\to\begin{bmatrix}\mathbb Z&\mathbb Q\\0&\mathbb Q\end{bmatrix}\to\begin{bmatrix}\mathbb Z&\mathbb Q\\ 0&0\end{bmatrix}\to 0$ of $R$ modules, and the point is that both halves are Noetherian. But that means the thing in the middle is Noetherian too. (Hint: any nonzero submodule of the right hand module is generated by $\begin{bmatrix}n&0\\ 0 & 0\end{bmatrix}$ and $\begin{bmatrix}0&1\\ 0 & 0\end{bmatrix}$ for some $n$. The left hand module is a $\mathbb Q$ module via the action of $R$ and is 1-dimensional.)

Then she calculated the following (and I do not know why)

That computation is shorthand for what it looks like to multiply something of the form $\begin{bmatrix}n&q\\0&p\end{bmatrix}$ on the left by an arbitrary element of the ring. If you consider the subset of elements with $n=p=0$, then this proves that $ \begin{bmatrix} 0 & q\mathbb{Z} \\ 0 & 0 \end{bmatrix} $ is a left ideal of $S$, since it is an additive subgroup which absorbs multiplication on the left. That much is reasonable.

Then she added if $n >1,$ $\frac{\mathbb{Z}}{n} \subsetneq \frac{\mathbb{Z}}{n^2} \subsetneq ...... $

This, without more explanation, is a little unclear, but the goal should be apparent: we must find a sequence of strictly increasing left ideals of $R$. The chain written here is not a sequence of ideals in anything we are looking at... it looks like a sequence of quotients of $\mathbb Z$. Based on the last observation, all we need to do is find a strictly increasing sequence of left ideals of the form $ \begin{bmatrix} 0 & q\mathbb{Z} \\ 0 & 0 \end{bmatrix} $ and this is easily obtained by letting $q\in\{\frac1n,\frac1{n^2},\frac1{n^3},\frac1{n^4}\ldots \}$

  • $\begingroup$ And thank you for your patience in answering me ...... I am from a developing country where the people(including me) are scientifically poor ..... forgive me if sometimes my questions lack of context ...... in our countries we learn thinking accidentally and not in a rigor way. $\endgroup$
    – Emptymind
    Jan 9, 2018 at 15:59
  • $\begingroup$ If $R$ in your answer is the same ring $S$ in the question, I do not see how $\begin{bmatrix}\mathbb Z&0\\0&0\end{bmatrix}$ is a right $R$-module. $\endgroup$
    – khashayar
    Aug 28, 2023 at 21:38
  • $\begingroup$ @khashayar You are right! That is a mistake. I think i have repaired it. $\endgroup$
    – rschwieb
    Aug 30, 2023 at 2:10

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