Existence of an idempotent element $\not = 1$ and $\not = 0$

Question

I have a problem to solve the following problem: If $1=e_1+e_2$ with non-units $e_1,e_2 \in R$ and if $e_1e_2$ is nilpotent, then there is an idempotent element $e\not =0$,$e\not =1$. Perhaps it is a straightforward problem, but I could not solve it so far.

Answer 1

Set $e:=e_1$ so that $e_2=1-e$. Given that $e_1e_2$ is nilpotent there exists $k\in\Bbb{N}$ such that $$0=(e_1e_2)^k=(e(1-e))^k=e^k(1-e)^k,$$ because of course $e$ and $1-e$ commute. Setting $f:=e^k$ and $g:=(1-e)^k$ yields $fg=0$. Moreover $e$ and $1-e$ both divide $f+g-1$ and hence $f+g-1$ is also nilpotent. It follows that $f+g$ is a unit and for $u:=(f+g)^{-1}$ we have $$uf=uf\cdot u(f+g)=(uf)^2+uf\cdot ug=(uf)^2+u^2fg=(uf)^2,$$ where $u$ and $f$ commute because $f$ and $g$ do. This shows that $uf$ is idempotent. Note that $uf\in\{0,1\}$ if and only if $f=0$ or $g=0$, in which case either $e_1$ or $e_2$ is nilpotent, and so the other is a unit, contradicting the assumption that $e_1$ and $e_2$ are non-units.