Hostname: page-component-cd9895bd7-mkpzs Total loading time: 0 Render date: 2024-12-25T06:38:08.682Z Has data issue: false hasContentIssue false

MORPHIC RINGS AS TRIVIAL EXTENSIONS

Published online by Cambridge University Press:  31 January 2005

JIANLONG CHEN
Affiliation:
Department of Mathematics, Southeast University, Nanjing, 210096, P. R. China e-mail: jlchen@seu.edu.cn
YIQIANG ZHOU
Affiliation:
Department of Mathematics and Statistics, Memorial University of Newfoundland St. John's, NL A1C 5S7, Canada e-mail: zhou@math.mun.ca
Rights & Permissions [Opens in a new window]

Abstract

Core share and HTML view are not available for this content. However, as you have access to this content, a full PDF is available via the ‘Save PDF’ action button.

A ring $R$ is called left morphic if, for every $a\in R$, $R/Ra\cong {\bf l}(a)$ where ${\bf l}(a)$ denotes the left annihilator of $a$ in $R$. Right morphic rings are defined analogously. In this paper, we investigate when the trivial extension $R\propto M$ of a ring $R$ and a bimodule $M$ over $R$ is (left) morphic. Several new families of (left) morphic rings are identified through the construction of trivial extensions. For example, it is shown here that if $R$ is strongly regular or semisimple, then $R\propto R$ is morphic; for an integer $n>1$, ${\mathbb Z}_n\propto {\mathbb Z}_n$ is morphic if and only if $n$ is a product of distinct prime numbers; if $R$ is a principal ideal domain with classical quotient ring $Q$, then the trivial extension $R\propto {Q}/{R}$ is morphic; for a bimodule $M$ over $\mathbb Z$, ${\mathbb Z}\propto M$ is morphic if and only if $M\cong{\mathbb Q}/{\mathbb Z}$. Thus, ${\mathbb Z}\propto {\mathbb Q}/{\mathbb Z}$ is a morphic ring which is not clean. This example settled two questions both in the negative raised by Nicholson and Sánchez Campos, and by Nicholson, respectively.

Type
Research Article
Copyright
2005 Glasgow Mathematical Journal Trust