Does $a=ea$ and $ae=eae$ for some $a\in R$, imply that $a=e$ for any idempotent $e$ in $R$?

Question

Let $R$ be a ring with unity and $e^2=e\in R$. If $a=ea$ and $ae=eae$ for some $a\in R$, then prove that $a=e$.

Solution: Suppose that $a=ea$ and $ae=eae$, then $a^2=eaea=e(eae)a=eeaea=eaea=eaa=ea^2$. That is, $a^2=ea^2$. If $e\neq 1,0$, then $a=e$ applies. If $a=1,0$, here it is not making sense.

Question: Is the argument above valid?

Answer 1

Note that if $a=ea$, then automatically $ae=eae$, so the second hypothesis adds nothing.

As it stands it’s definitely false. Let $R$ be the ring of $2\times 2$ real matrices, and take $e=\pmatrix{1&0\\0&0}$ and $a=\pmatrix{1&1\\0&0}$. Then $e^2=e$ and $a=ea$, but $e\ne a$.

Answer 2

No. The proposition is false in any ring $\mathcal R$ with unity $1_{\mathcal R}$ and additive identity $0_{\mathcal R}$. Consider $e=1_{\mathcal R}$. Clearly, $1_{\mathcal R}\cdot r=r$ for all $r\in\mathcal R$ but not every ring element is $1_{\mathcal R}$ because $1_{\mathcal R}\neq 0_{\mathcal R}$.