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?
 A: Writing it as $R\oplus M\oplus S$ is a more compact way to write it, and it adequately expresses the addition operation.
But it does nothing to express the multiplication operation: that is what the matrices are for.
Lam says on the page that he is identifying $R,M,S$ with the obvious subgroups of $R\oplus M\oplus S$, i.e. $R\oplus 0\oplus 0$ and so on.
The table records  the result of pairwise matrix multiplications between these subgroups.
A: 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 $$\begin{pmatrix} 0 & 0 \\ m & 0\end{pmatrix}\cdot\begin{pmatrix} r & 0 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}0 & 0 \\ mr & 0\end{pmatrix}.$$
A: More explicitly: if $r_1,r_2 \in R$, and $m_1,m_2 \in M$, and $s_1,s_2 \in S$, we have:
$\begin{bmatrix}r_1&m_1\\0&s_1\end{bmatrix}\begin{bmatrix}r_2&m_2\\0&s_2\end{bmatrix} = \begin{bmatrix}r_1r_2&m_3\\0&s_1s_2\end{bmatrix}$
where $m_3 = r_1m_2 + m_1s_2 \in M$ because $M$ is an $(R,S)$-bimodule.
