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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 )

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  • $\begingroup$ See also the solution here. $\endgroup$ Jan 8, 2018 at 10:55
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    $\begingroup$ 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
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    $\begingroup$ @Intuition Thank you for improving your question. $\endgroup$
    – rschwieb
    Jan 9, 2018 at 14:43

1 Answer 1

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

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  • $\begingroup$ 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

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