If $(yb)^n=0$ with $(yb)^{n-1}\neq 0$, then $y\in l_R(b)=\{r\in R:rb=0\}$.

Question

Let $R$ be a ring with unity and $y,b\in R$. If $(yb)^n=0$ with $(yb)^{n-1}\neq 0$, then $y\in l_R(b)=\{r\in R:rb=0\}$.

This is my attempt:

Suppose that $(yb)^n=0$ with $(yb)^{n-1}\neq 0$. Then $$0=(yb)^n=(yb)(yb)^{n-1}=y(b(yb)^{n-1})$$ so that $$y\in l_R(b(yb)^{n-1})=l_R(b(yb)\cdots by)b)\subseteq l_R(bRb)=l_R(Rb)=l_R(b).$$ That is, $y\in l_R(b)$.

Answer 1

How it is possible? If by contraddiction you have that $y\in l_R(b)$ then

$yb=0$

so

$(yb)^{n-1}=(yb)\cdots (yb)=0$ and it is not possible