关于理想上环和模的性质(A)

Q4 Mathematics
S. Bouchiba, Y. Arssi
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引用次数: 0

摘要

本文引入并研究了环$R$或$R$-模$M$沿$R$的理想$I$的性质($mathcalA$)的概念。例如,满足属性($mathcal A$)的$R$以上的任何模块$M$,都满足$R$中的任何理想$I$的属性($mathcal A$)。我们也对理想$I$感兴趣,它们本身就是$mathcalA$-模块。特别地,我们证明了如果$I$包含在$R$的幂零根中,那么任何$R$-模都是沿着$I$的$mathcal A$-模,因此,$I$本身就是$mathccal A$-模块。此外,我们给出了一个环$R$具有理想$I$的例子,该理想$I$I本身是$mathcal a$-模,而$I$不是$mathccal a$-模块。此外,在$IsubseteqZ(R)$和$InsobsteqZ(R)$的两种情况下,我们完全刻画了环$R$满足性质($mathcalA$)和理想$I$。最后,我们研究了性质($mathcalA$)沿着理想相对于直积的行为。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
On Property (A) of rings and modules over an ideal
This paper introduces and studies the notion of Property ($mathcal A$) of a ring $R$ or an $R$-module $M$ along an ideal $I$ of $R$. For instance, any module $M$ over $R$ satisfying the Property ($mathcal A$) do satisfy the Property ($mathcal A$) along any ideal $I$ of $R$. We are also interested in ideals $I$ which are $mathcal A$-module along themselves. In particular, we prove that if $I$ is contained in the nilradical of $R$, then any $R$-module is an $mathcal A$-module along $I$ and, thus, $I$ is an $mathcal A$-module along itself. Also, we present an example of a ring $R$ possessing an ideal $I$ which is an $mathcal A$-module along itself while $I$ is not an $mathcal A$-module. Moreover, we totally characterize rings $R$ satisfying the Property ($mathcal A$) along an ideal $I$ in both cases where $Isubseteq Z(R)$ and where $Insubseteq Z(R)$. Finally, we investigate the behavior of the Property ($mathcal A$) along an ideal with respect to direct products.
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来源期刊
Journal of Algebra and Related Topics
Journal of Algebra and Related Topics Mathematics-Discrete Mathematics and Combinatorics
CiteScore
0.60
自引率
0.00%
发文量
0
审稿时长
16 weeks
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