If $Rx$ is nil then $Rxr$ is nil for any $r \in R$

Question

i am studying kothe's conjecture, ad got stuck here. if $R$ is any non commutative ring, then how is it true that if the ideal $Rx$ is nil then $Rxr$ is nil for any $r \in R$.

let $sx\in Rx$, then $(sx)^n=0$ for some $n$, but how is $sxr$ nilpotent for any $r\in R$.

Answer 1

Notice that $$(sxr)^n = (sxr)(sxr)\cdots(sxr) = sx(rsx)\cdots(rsx)r = sx(rsx)^{n-1}r.$$ Because $rsx \in Rx$ and this left ideal is nil, for sufficiently large $n$ the expression above is zero.