The nilpotent regular element problem (Q2827398)
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scientific article; zbMATH DE number 6641095
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | The nilpotent regular element problem |
scientific article; zbMATH DE number 6641095 |
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19 October 2016
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nilpotent element
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von Neumann regular element
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unit-regular
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Bergman's normal form
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The nilpotent regular element problem (English)
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Given a unital associative ring \(S\), we say that an element \(x\in S\) is von Neumann regular if there exists \(y\in S\) such that \(x=xyx\). The element \(y\in S\) is known as an inner-inverse (or quasi-inverse) of \(x\); if \(y\) is invertible in \(S\), then \(x\) is said to be unit-regular. If \(x=xyx\) and \(y=yxy\), then \(y\) is said to be a generalized inverse of \(x\).NEWLINENEWLINEIn the paper under review, the authors consider the problem of whether a nilpotent von Neumann regular element \(x\) of a ring \(S\) must be unit-regular. This fact plays an important role in the first author's proof of the fact that strongly \(\pi\)-regular rings have stable range one [Proc. Am. Math. Soc. 124, No. 11, 3293--3298 (1996; Zbl 0865.16007)]. Also, it is relevant for the possibility of constructing ring direct limits producing non-separative von Neumann regular rings [the first author et al., Isr. J. Math. 105, 105--137 (1998; Zbl 0908.16002)].NEWLINENEWLINEThe answer to this question is affirmative for nilpotent regular elements of exchange rings [the first author, Proc. Am. Math. Soc. 124, No. 11, 3293--3298 (1996; Zbl 0865.16007), Theorem 2], and for nilpotent elements of an arbitrary ring having all its powers von Neumann regular [\textit{K. I. Beidar} et al., Commun. Algebra 32, No. 9, 3543--3562 (2004; Zbl 1074.16005), Theorem 3.6].NEWLINENEWLINETo settle this question, the authors use a recent description of \textit{G. M. Bergman} [J. Algebra 449, 355--399 (2016; Zbl 1338.16026)] about universally adjoining inner-inverses to arbitrary elements of an algebra over a field. With this tool at hand, they are able to show that the answer is negative, by constructing an example:NEWLINENEWLINELet \(F\) be a field, let \(R=F[x]/\langle x^3\rangle\), and let NEWLINE\[NEWLINES=R\langle q \mid xqx=x, qxq=q\rangleNEWLINE\]NEWLINE the algebra obtained by freely adjoining a generalized inverse \(q\) of \(x\). Then, they use Bergman's normal form to conclude that \(x\) is not unit-regular.NEWLINENEWLINEAs noticed by the authors, \textit{P. P. Nielsen} and \textit{J. Šter} [Trans. Am. Math. Soc. 370, No. 3, 1759--1782 (2018; Zbl 1445.16008)] independently constructed such counterexample. Section 3 in the paper under review is devoted to present this example, and to show that both rings are isomorphic.
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