A doubt about lower nil radical while proving 2-primality of ring.( Baer-McCoy Radical)

Question

I was proving that a reversible ring is 2-Primal for an exercise in T.Y Lam's book, but I got stuck. Here is where I'm stuck: let $a$ be a nilpotent element of $R$ with $a^n=0$. Then using reversibility of ring he concluded $(RaR)^n=0$, but then he says this implies that $RaR\subset \mathrm{Nil}_\ast(R)$ which clearly gives 2-primality as then $\mathrm{Nil}_\ast(R)=$ set of all nilpotent elements.

But why is $RaR\subset \mathrm{Nil}_{*}(R)$?

$\mathrm{Nil}_\ast(R)$ is the set of elements $x$ s.t. every m-system containing $x$ intersects $0$. It is always contained in the set of all nilpotent elements of $R$. "2-primal" means that the nilpotent elements are all contained in $\mathrm{Nil}_\ast(R)$.

please suggest something..

also tell me if we have to calculate $\mathrm{Nil}_{*}(R)$ for some ring what should we aim at. we can't find all m-systems. we know it is a nil ideal so every element is nilpotent and it is contained in $rad(R)$ and intersection of all prime ideals but we also can't find all prime ideals. so what should we do to find $\mathrm{Nil}_{*}(R)$

Answer 1

Consider the $n=1$ case with $aRa=0$. Then $(RaR)^2=RaR\cdot RaR = R\cdot a(R)a \cdot R=0$ since multiplication in the ring is closed.

Answer 2

I am guessing the intent is to prove that $RaR$ is nilpotent to show that it is contained in all prime ideals, and then use the characterization of $Nil_\ast(R)$ as the intersection of all prime ideals to conclude that $RaR\subseteq Nil_\ast(R)$.

If you haven't proven that equivalent definition of $Nil_\ast(R)$, it is highly recommended, and may even be carried out in the text already.

I'm also not totally sure why $(aRa)^n=\{0\}$ would actually help (is that really what's written?) Knowing that $aRaaRaaRaa\ldots =\{0\}$ does not seem to directly imply that $RaRaRa\ldots aR=\{0\}$.

But it is easy enough to show that $aRaRa\ldots aRa=\{0\}$ whenever there are $n$ or more $a$'s. Let $m\geq n$ and $r_i\in R$. Then

$$a^mr_1=0\implies a^{m-1}r_1a=0\\ a^{m-1}r_1ar_2=0\implies a^{m-2}r_1ar_2a=0\\ \ldots\\ a^2r_1ar_2a\ldots r_{m-1}=0\implies ar_1ar_2a\ldots r_{m-1}a=0$$

So $aRaR\ldots aRa=\{0\}$ whenever there are at least $n$ $a$'s, and that means $(RaR)^n=R(aRaRa\ldots aRa)R=\{0\}$. Thus $RaR$ is contained in every prime ideal, and therefore it is contained in $Nil_\ast(R)$.

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Note: an element $a\in R$ is called strongly nilpotent if there exists a nonnegative integer $k$ such that $ar_1ar_2a\ldots ar_{k-1} a=0$ for all choices of $r_i\in R$. Just as the nilradical of a commutative ring is the set of all nilpotent elements, the lower nilradical $Nil_\ast(R)$ is the set of all strongly nilpotent elements.

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Edit Poster has since changed $(aRa)^n$ to $(RaR)^n$. In that case, it looks like you might have overlooked the fact that for an ideal $A$ and prime ideal $P$, $A^n\subseteq P\implies A\subseteq P$. This is all you need to prove that if $RaR$ is nilpotent, it's contained in all prime ideals. It's analogous to the idea that nilpotent elements of commutative rings are contained in all prime ideals.