Idempotents of triangular matrix ring

Question

Let $A = \begin{bmatrix} R & M \\ 0 & S \\ \end{bmatrix}$ be a formal triangular matrix ring. One easily sees that if $ \begin{bmatrix} e & m \\ 0 & f \\ \end{bmatrix}$ is an idempotent in $A$, then $e=e^2$ and $f=f^2$ and $mf=(1-e)m$. Furthermore, if for all $e=e^2\in R$ and for all $f=f^2\in S$ and for all $m\in M$ we have $em=mf$, we would have $(1-2e)m=0$ and since $(1-2e)^2=1$ we obtain $m=0$. Therefore, with our additional condition, the set of idempotents of $A$ equals that of $\begin{bmatrix} R & 0 \\ 0 & S \\ \end{bmatrix}$.

Is the converse true? Namely, if the set of idempotents of $A$ equals that of $\begin{bmatrix} R & 0 \\ 0 & S \\ \end{bmatrix}$, could one deduce that $em=fm$ for $e=e^2\in R, f=f^2\in S,$ and $m\in M$?

Thanks for any help!

Answer 1

It's true, but for the uninteresting reason that unless $M=0$, neither of the conditions is ever satisfied.

If $0\neq m\in M$ then:

• for $e=0$ and $f=1$, $em\neq mf$, and

• $\begin{bmatrix}0&m\\0&1\end{bmatrix}$ is an idempotent of $A$.

Answer 2

So, also taking $e=1, 0 \neq m \in M$ idempotent of the form $\begin{pmatrix} 1 & m \\ 0 & 0 \end{pmatrix}$