Let $R$ be a ring with unity. An element $a\in R$ is said to be a quasi nilpotent element of $R$ if $1-ax$ is unit for all $x\in comm(a)$ where $comm(a)=\lbrace x \in R | ax=xa\rbrace $. It is obvious that all nilpotent elements are quasi nilpotent elements, but converse need not be true. I am having hard time finding a good example.
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$\begingroup$ The elements you describe are exactly the elements of the Jacobson radical, which is the intersection of all maximal ideals. The nilpotent elements are the elements of the nilradical, the intersection of all prime ideals. So any ring in which the Jacobson radical is larger than the nilradical gives a counterexample. $\endgroup$– Sergey GuminovCommented Feb 7, 2023 at 19:05
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1$\begingroup$ @IDC Maybe that is not quite the case. the poster's condition is $1-ax$ is a unit $\textbf{for all }\mathbf {x\in comm(a)}$ etc, etc. Of course, that means everything in the Jacobson radical, but in principle maybe there is an example could contain something more than the Jacobson radical. $\endgroup$– rschwiebCommented Feb 7, 2023 at 19:43
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1$\begingroup$ @rschwieb, you are right, I forgot to mention that my comment was about the commutative case. $\endgroup$– Sergey GuminovCommented Feb 7, 2023 at 20:06
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1 Answer
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Let $R=\mathbb{R}[[x]]$ be the ring of formal power series with coefficients in $\mathbb{R}$. This is a commutative ring, and $a_0+\sum_{n=1}^{\infty}a_nx^n$ is a unit if and only if $a_0\neq 0$.
In this ring, $x$ is not nilpotent; but for any element $s\in \mathbb{R}[[x]]$, we have that the constant term of $1-sx$ is $1$, so $1-sx$ is a unit. Thus, $x$ is quasi-nilpotent but not nilpotent.
answered Feb 7, 2023 at 18:53