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Let $R,S$ be two rings and $M$ be a $(R,S)$-bimodule (i.e. $M$ is a left $R$-module and a right $S$-module such that $(rm)s = r(ms)$ for all $r\in R, s \in S, m \in M$). We can define the ring

$$A= \begin{pmatrix}R \ M\\ 0 \ S\end{pmatrix}$$

with componentwise addition and multiplication of matrices as you would expect it (formal matrix multiplication).

I checked that $A$ becomes an associative ring (with identity if $R$ and $S$ have identity) for these operations.

I have a couple of questions:

(1) Lam says

"First, it is convenient to identify $R,S$ and $M$ as subgroups in $A$ (in the obvious way) and to think of $A$ as $R \oplus M \oplus S$."

Does this mean that we identity for example $S$ with the subgroup $\begin{pmatrix}0 \ 0\\ 0 \ S\end{pmatrix}$ of $A$? Then $A= R \oplus M \oplus S$ indeed becomes true.

(2) Proposition $(1.17)$ states the following:

"The left ideals of $A$ are of the form $I_1 \oplus I_2$ where $I_2$ is a left ideal in $S$ and $I_1$ is a left $R$- submodule of $R \oplus M$ containing $MI_2$."

What does the notation $MI_2$ mean?

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  • $\begingroup$ (1) Yes. (2) I suspect it is the $\mathbb{Z}$-linear span of the set $\left\{ mi \mid m \in M,\ i \in I_2 \right\}$. $\endgroup$ – darij grinberg Sep 27 '19 at 21:58
  • $\begingroup$ With $\mathbb{Z}$-linear span you mean we get sums of elements like that? $\endgroup$ – user661541 Sep 27 '19 at 22:01
  • $\begingroup$ Yes (and $\mathbb{Z}$-multiples, but that's not needed). $\endgroup$ – darij grinberg Sep 27 '19 at 22:04
  • $\begingroup$ And when they say that $I_1$ s a submodule of $R\oplus M$. Is this module structure on $R\oplus M$ usual scalar multiplication on matrices? $\endgroup$ – user661541 Sep 27 '19 at 22:46
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Just as with ideal products, this was not intended to mean simply the set of elementwise products $\{xy\mid x\in M, y\in I_2\}$.

Surely since $I_2\subset S$ and $M$ is a right $S$ module, $MI_2\subseteq M$, but we are talking about left submodules of $M$, so we should expect something that is definitely at least a left $R$ submodule.

Again, using ideal products as our model, $MI_2:=\{\sum m_ia_i\mid m_i\in M, a_i\in I_2\}$, that is, the finite sums of pairwise products, can easily be seen to be a left $R$ submodule of $M$. (This amounts to the same thing mentioned in the comments, the $\mathbb Z$-span of the pairwise products.)

So when I think about this condition "$I_1$ is a left $R$ submodule of $R\oplus M$ containing $MI_2$" I usually think of it as $$\{0\}\oplus MI_2\subseteq I_1\subseteq R\oplus M$$

And when they say that $I_1$ s a submodule of $๐‘…โŠ•๐‘€$. Is this module structure on $๐‘…โŠ•๐‘€$ usual scalar multiplication on matrices?

No: you would have a problem doing that since for some $\alpha, r\in R$ and $m\in M$, $s\in S$, $\alpha\begin{bmatrix}r&m\\0&s\end{bmatrix}=\begin{bmatrix}\alpha r&\alpha m\\0&\alpha s\end{bmatrix}$ would necessitate an action of $R$ on the left of $S$, something which you don't have necessarily.

It is enough to think of $R\oplus M$ as the regular direct sum of left $R$ modules, so that $\alpha(r,m)=(\alpha r, \alpha m)$. It is not safe to think of the left action going all the way across to the third factor $S$, as I mentioned, but you can model it with matrices in the following way.

The action we just described can be reflected in matrix multiplication if $\alpha\begin{bmatrix}r & m \\ 0 & s\end{bmatrix}:=\begin{bmatrix}\alpha&0\\ 0&1\end{bmatrix}\begin{bmatrix}r&m\\0&s\end{bmatrix}$ and likewise on the right hand side, if $\beta\in S$ then $\begin{bmatrix}r & m \\ 0 & s\end{bmatrix}\beta :=\begin{bmatrix}r&m\\0&s\end{bmatrix}\begin{bmatrix}1&0\\ 0&\beta\end{bmatrix}$ for the natural right action of $S$ on $M\oplus S$.

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  • $\begingroup$ Thank you so much for your answer! So to be quite explicit,the theorem says that the left ideals of $A=\begin{pmatrix} R \ M \\ 0 \ S\end{pmatrix}$ are of the form $\begin{pmatrix} I_1\\ 0 \ I_2 \end{pmatrix}$ where $I_1$ is a submodule of $R\oplus M$ containing $MI_2$ and $I_2$ a left ideal of $S$? $\endgroup$ – user661541 Sep 30 '19 at 13:43
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    $\begingroup$ @user661541 Yes: it's awkward to use matrix notation since $I_1$ is "a subset of the top row," but that is the idea. That's why he suggests using the $R\oplus M\oplus S$ notation instead. $\endgroup$ – rschwieb Sep 30 '19 at 13:45
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    $\begingroup$ It feels a little odd at first, but after getting used to it I've always been very impressed with the beauty of the triangular ring construction. Knowing exactly what the left and right ideals are is a very cool thing that you sometimes lose in a construction. $\endgroup$ – rschwieb Sep 30 '19 at 13:46
  • $\begingroup$ Yeah, it seems nice class of rings to get counterexamples! It popped up in my course as a ring that is left Noetherian (Artinian) but not right Noetherian (Artinian). I will check if I understand the proof now I understand the theorem statement. If not, I'll get back to you. If I do understand the proof, I'll gladly accept your answer. Thanks again. $\endgroup$ – user661541 Sep 30 '19 at 13:49
  • $\begingroup$ I'm still confused about the statement $I_1$ is a submodule of $R \oplus M$ containing $MI_2$. How do we see this on matrix level? Does it mean that $I_1$ contains $0 \oplus MI_2$? $\endgroup$ – user661541 Sep 30 '19 at 14:43

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