Remark 2

Let $e$ and $f$ be idempotents in A. By Proposition 13, b), we have $Ae= Af$ if and and only if $ef=e$ and $f e=f$. Consequently, if the ring A is commutative, then the relation $Ae= Af$ is equivalent to $e=f$. This does not hold in general, as shown by the example $A =\mathbf{M}_2(\mathbf{Z}), e = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, and $f = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$.

We say that idempotents $e$ and $e'$ in the ring A are orthogonal if $ee'=$ $e'e= 0$. Let $(e_i)_{i\in I}$ be a finite family of pairwise orthogonal idempotents in A. Since we have

the element $\sum_{i\in I}e_i$ of A is idempotent.

A partition of an idempotent $e$ in A is a finite family $(e_i)_{i\in I}$ of pairwise orthogonal idempotents in A such that $e=\sum_{i\in I}e_i$. We say that an idempotent $e$ in A is decomposable if there exists a partition of $e$ consisting of pairwise orthogonal idempotents distinct from 0 and $e$; in the opposite case, we say that it is indecomposable. Observe that 0 is a decomposable idempotent.