MHB Example on Triangular Rings - Lam, Example 1.14

[Math Amateur]

I am reading T. Y. Lam's book, "A First Course in Noncommutative Rings" (Second Edition) and am currently focussed on Section 1:Basic Terminology and Examples ...

I need help with an aspect of Example 1.14 ... ...

Example 1.14 reads as follows: https://www.physicsforums.com/attachments/5984
https://www.physicsforums.com/attachments/5985I cannot follow why the results in Table 1.16 follow ...

For example, according to Table 1.16 ...

$$mr = 0$$ for all $$m \in M$$ and $$r \in R$$ ... but why

Similarly I don't follow the other entries in the Table ...

Can someone please help ...

Peter

 

[Deveno]

We have:

$\begin{pmatrix}0&m\\0&0\end{pmatrix}\begin{pmatrix}r&0\\0&0\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}$

 

[Math Amateur]

We have:

$\begin{pmatrix}0&m\\0&0\end{pmatrix}\begin{pmatrix}r&0\\0&0\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}$

Hi Deveno,

Thanks for the help ... but I do not follow you ...

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

where I have been thinking that $$M$$ and $$R$$ are a set of elements (actually a bimodule and left $$R$$-module) that we select elements from ... and then multiply ... that is, M and R are sets not actually matrices themselves ...

... ... BUT ... you seem to have interpreted $$M$$ and $$R$$ as matrices ... so you select $$m$$ from $$M$$ and $$r$$ from $$R$$ and write:

$$mr = \begin{pmatrix} 0 & m \\ 0 & 0 \end{pmatrix} \begin{pmatrix} r & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$ ...

I do not understand how $$MR$$ in the table becomes a matrix multiplication ...Can you please clarify ...?

Peter

 

[Deveno]

Hi Deveno,

Thanks for the help ... but I do not follow you ...

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

where I have been thinking that $$M$$ and $$R$$ are a set of elements (actually a bimodule and left $$R$$-module) that we select elements from ... and then multiply ... that is, M and R are sets not actually matrices themselves ...

... ... BUT ... you seem to have interpreted $$M$$ and $$R$$ as matrices ... so you select $$m$$ from $$M$$ and $$r$$ from $$R$$ and write:

$$mr = \begin{pmatrix} 0 & m \\ 0 & 0 \end{pmatrix} \begin{pmatrix} r & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$ ...

I do not understand how $$MR$$ in the table becomes a matrix multiplication ...Can you please clarify ...?

Peter

$R$ is a ring, $S$ is a ring, and $M$ is an $(R,S)$-bimodule. A typical element of $A$ is:

$\begin{pmatrix}r&m\\0&s\end{pmatrix}$ with $r\in R, m\in M$, and $s\in S$.

We have an isomorphism (of abelian groups): $A \to R \oplus M \oplus S$ given by:

$\begin{pmatrix}r&m\\0&s\end{pmatrix} \mapsto (r,m,s)$.

We can thus regard our matrices as $\Bbb Z$-linear combinations of $(r,0,0),(0,m,0)$ and $(0,0,s)$ or equivalently as $\Bbb Z$-linear combinations of the matrices:

$\begin{pmatrix}r&0\\0&0\end{pmatrix}, \begin{pmatrix}0&m\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\0&s\end{pmatrix}$

Formally, then, using the distributive law of matrices, we have:

$(r,m,s)(r',m',s') = [(r,0,0) + (0,m,0) + (0,0,s)][(r',0,0) + (0,m'0) + (0,0,s')]$

$=(r,0,0)(r',0,0) + (r,0,0)(0,m',0) + (r,0,0)(0,0,s') + (0,m,0)(r',0,0) + (0,m,0)(0,m',0) + (0,m,0)(0,0,s') + (0,0,s)(r',0,0) + (0,0,s)(0,m',0) + (0,0,s)(0,0,s')$

so in order to completely determine the multiplication in $A$, we need to know what these 9 terms are. The 3x3 table is a mnemonic SCHEMATIC, to remember which abelian subgroup ($R,M$ or $S$) each term is.

Note that the $1,2$ entry in the matrix is: $rm' + ms'$, which is in $M$ since $M$ is an $(R,S)$-bimodule.

That is, $RM \subseteq M$, and $MS \subseteq M$, as (left or right) scalar product sets; for example:

$RM = \{rm\mid r\in R, M \in M\} \subseteq M$, since $M$ is a (left) $R$-module.

 

[Math Amateur]

$R$ is a ring, $S$ is a ring, and $M$ is an $(R,S)$-bimodule. A typical element of $A$ is:

$\begin{pmatrix}r&m\\0&s\end{pmatrix}$ with $r\in R, m\in M$, and $s\in S$.

We have an isomorphism (of abelian groups): $A \to R \oplus M \oplus S$ given by:

$\begin{pmatrix}r&m\\0&s\end{pmatrix} \mapsto (r,m,s)$.

We can thus regard our matrices as $\Bbb Z$-linear combinations of $(r,0,0),(0,m,0)$ and $(0,0,s)$ or equivalently as $\Bbb Z$-linear combinations of the matrices:

$\begin{pmatrix}r&0\\0&0\end{pmatrix}, \begin{pmatrix}0&m\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\0&s\end{pmatrix}$

Formally, then, using the distributive law of matrices, we have:

$(r,m,s)(r',m',s') = [(r,0,0) + (0,m,0) + (0,0,s)][(r',0,0) + (0,m'0) + (0,0,s')]$

$=(r,0,0)(r',0,0) + (r,0,0)(0,m',0) + (r,0,0)(0,0,s') + (0,m,0)(r',0,0) + (0,m,0)(0,m',0) + (0,m,0)(0,0,s') + (0,0,s)(r',0,0) + (0,0,s)(0,m',0) + (0,0,s)(0,0,s')$

so in order to completely determine the multiplication in $A$, we need to know what these 9 terms are. The 3x3 table is a mnemonic SCHEMATIC, to remember which abelian subgroup ($R,M$ or $S$) each term is.

Note that the $1,2$ entry in the matrix is: $rm' + ms'$, which is in $M$ since $M$ is an $(R,S)$-bimodule.

That is, $RM \subseteq M$, and $MS \subseteq M$, as (left or right) scalar product sets; for example:

$RM = \{rm\mid r\in R, M \in M\} \subseteq M$, since $M$ is a (left) $R$-module.

Thanks for for the help, Deveno ...

Just working through your post and reflecting on what you have said ...

Thanks again ...

Peter