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 $r\in R$.

Note, you don't want that $sx$ nilpotent implies $sxr$ is nipotent, but that every element of $Rx$ is nilpotent implies $sxr$ is nilpotent for any $s,r\in R$. — Thomas Andrews, Jul 19 '14 at 19:04

score 5 · Answer 1 · answered Jul 19 '14 at 18:42

5

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.

answered Jul 19 '14 at 18:42

Manny Reyes

2,992

also can you clear my this doubt that let x be any element of a nil left ideal say I. then (x)=RxR is nil. is it true? – Bhaskar Vashishth Jul 19 '14 at 18:47
A positive answer to your question is equivalent to Koethe's conjecture: http://en.m.wikipedia.org/wiki/Radical_of_a_ring#The_upper_nil_radical_or_K.C3.B6the_radical – Manny Reyes Jul 19 '14 at 19:05

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

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