For a ring with idempotent $e$, is $r \mapsto ere$ a homomorphism?

Question

Let $R$ be a ring with idempotent $e$, then $eRe$ (the corner ring) is also a ring. Then is the map $\varphi(r) := ere$ a homomorphism?

Surely if $e$ is central, then we have $e(rs)e = e^2(rs)e = erese = (ere)(ese)$, but what if $e$ is not neccessarily central?

Answer 1

For a concrete counterexample, consider $R=M_2(K)$ and the idempotent $e=\begin{pmatrix} 1 & 0 \cr 0 & 0 \end{pmatrix}$. Let $r=\begin{pmatrix} a & b \cr c & d \end{pmatrix}$ and $s=\begin{pmatrix} a' & b' \cr c' & d' \end{pmatrix}$, then we have $$ \phi(r)\phi(s)=\begin{pmatrix} aa' & 0 \cr 0 & 0 \end{pmatrix}\neq \begin{pmatrix} aa'+bc' & 0 \cr 0 & 0 \end{pmatrix}=\phi(rs). $$

Answer 2

Consider the ring $R$ of $n \times n$ matrices over a field, with $n > 1$. You know that this is a simple ring.

Choose $e = e_{11}$ to be the matrix that has zero everywhere, except for a $1$ in the $(1, 1)$ position.

Then $e R e$ consists of the matrices which are zero everywhere, except possibly in the $(1, 1)$ position.

There is no way $e R e$, a commutative, non-trivial ring, can be isomorphic to a proper quotient of the simple, non-commutative ring $R$.