abstract algebra - Lam Multiplication Table

In Lam's A First Course in Noncommutative Rings, he looks at the ring of upper triangular matrices $$A=\begin{pmatrix} R & M \\ 0 & S \end{pmatrix},$$ where $R,S$ are rings and $M$ is an $(R,S)$-bimodule. He then goes on to prove a few facts about ideals in this ring on page 17, Prop 1.17 (which are tedious to verify the 'standard' way). However, he makes these claims using a strange table:

$$\begin{array}{c|ccc}
& R & M & S \\ \hline
R & R & M & 0 \\
M & 0 & 0 & M \\
S & 0 & 0 & S \\
\end{array}$$
He talks about treating the ring $A$ as $R \oplus M \oplus S$ in the 'obvious' way (though to be fair, this to me is actually obvious). However, this claim that the multiplication in $A$ thinking about it as $R \oplus M \oplus S$ is entirely unclear to me. What does this table mean? Could someone break it down? Possibly with an example? I'm not seeing how this table is supposed to work. I tried writing an example down but even then comparing to how I did the computation with how I would for the ring $A'=\begin{pmatrix} R & 0 \\ M & S \end{pmatrix}$, I got the same table, which shouldn't be the case?

Answer

An element of $R$, $M$, or $S$ can be considered as an element of $A$ with only one nonzero entry. The table is then saying that if you take an element of column $X$ and multiply it by an element of row $Y$ in $A$ (where $X,Y\in \{R,M,S\}$), the output is the element of the corresponding cell of the table obtained in the obvious way. That is, it is a "multiplication table" for $A$, except instead of the rows and columns and entries being individual elements of $A$, they are larger subsets of $A$.

For instance, if you take $r\in R$ and $m\in M$ and want to multiply $r\cdot m$ in $A$, you look at the $R$ row and the $M$ column of the table and see that it tells you the output will be in $M$. So $r\cdot m$ will just be the element $rm\in M$ obtained using the $R$-module structure of $M$. In terms of matrices, this is just the fact that $$\begin{pmatrix} r & 0 \\ 0 & 0\end{pmatrix}\cdot\begin{pmatrix} 0 & m \\ 0 & 0\end{pmatrix}=\begin{pmatrix}0 & rm \\ 0 & 0\end{pmatrix}.$$

On the other hand, if you take $r\in R$ and $s\in S$ and want to compute $r\cdot s$ in $A$, the table tells you the output will be $0$. In terms of matrices, this is just the fact that $$\begin{pmatrix} r & 0 \\ 0 & 0\end{pmatrix}\cdot\begin{pmatrix} 0 & 0 \\ 0 & s\end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}.$$

Note that the table would look different if you had $A'=\begin{pmatrix}R & 0 \\ M & S\end{pmatrix}$ instead (for an $(S,R)$-bimodule $M$). For instance, if you took $r\in R$ and $m\in M$, then when you compute $r\cdot m$ in $A'$ you would get $$\begin{pmatrix} r & 0 \\ 0 & 0\end{pmatrix}\cdot\begin{pmatrix} 0 & 0 \\ m & 0 \end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}.$$ So the $(R,M)$ entry of the table would be $0$ instead of $M$, since you always get $0$. On the other hand, the $(M,R)$ entry would be $M$ instead of $0$, since $$\begin{pmatrix} 0 & 0 \\ m & 0\end{pmatrix}\cdot\begin{pmatrix} r & 0 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}0 & 0 \\ mr & 0\end{pmatrix}.$$