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 )
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.)
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.
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 $q\in\{\frac1n,\frac1{n^2},\frac1{n^3},\frac1{n^4}\ldots \}$