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 ...

\(\displaystyle mr = 0\) for all \(\displaystyle m \in M\) and \(\displaystyle r \in R\) ... 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 \(\displaystyle A = \begin{pmatrix} R & M \\ 0 & S \end{pmatrix}\)

where I have been thinking that \(\displaystyle M\) and \(\displaystyle R\) are a set of elements (actually a bimodule and left \(\displaystyle R\)-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 \(\displaystyle M\) and \(\displaystyle R\) as matrices ... so you select \(\displaystyle m\) from \(\displaystyle M\) and \(\displaystyle r\) from \(\displaystyle R\) and write:

\(\displaystyle mr = \begin{pmatrix} 0 & m \\ 0 & 0 \end{pmatrix} \begin{pmatrix} r & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\) ...

I do not understand how \(\displaystyle MR\) 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 \(\displaystyle A = \begin{pmatrix} R & M \\ 0 & S \end{pmatrix}\)

where I have been thinking that \(\displaystyle M\) and \(\displaystyle R\) are a set of elements (actually a bimodule and left \(\displaystyle R\)-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 \(\displaystyle M\) and \(\displaystyle R\) as matrices ... so you select \(\displaystyle m\) from \(\displaystyle M\) and \(\displaystyle r\) from \(\displaystyle R\) and write:

\(\displaystyle mr = \begin{pmatrix} 0 & m \\ 0 & 0 \end{pmatrix} \begin{pmatrix} r & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\) ...

I do not understand how \(\displaystyle MR\) 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