A difficulty in solving a question in chapter 8 hungerford. 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  )     
 A: 
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. In fact, the list of right ideals you gave is incomplete. For example, 
$\left\{\begin{bmatrix}0 & a\\ 0 & a\end{bmatrix} \,\middle |\, a\in\mathbb Q\right\}$ 
is a right ideal that was not mentioned. That is not the only one missing. You can take any vector in $\mathbb Q\times\mathbb Q$ and put it in the right hand column, and the $\mathbb Q$ scalar multiples are a right ideal. 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 $0\to \begin{bmatrix}\mathbb Z&0\\0&0\end{bmatrix}\to\begin{bmatrix}\mathbb Z&\mathbb Q\\0&\mathbb Q\end{bmatrix}\to\begin{bmatrix}\mathbb 0&\mathbb Q\\0&\mathbb Q\end{bmatrix}\to 0$ of $R$ modules, and the point is that the right $R$ submodules of the left half are $\mathbb Z$ submodules, and the right $R$ submodules of the right half are $\mathbb Q$ submodules, and in their own rights they are Noetherian. But that means the thing in the middle is Noetherian too.

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