论文标题

干净,弱清洁和微弱的清洁交换组戒指

On clean, weakly clean, and feebly clean commutative group rings

论文作者

Li, Yuanlin, Zhong, Qinghai

论文摘要

如果可以将$ r $的每个元素写为单位和一个愿意的总和,则据说是$ r $。如果$ r $的每个元素都是单位和一个单位的差异,则$ r $的干净,如果每个元素$ r $都可以写成$ r = u+r = u+e_1-e_2 $,则$ r $是微弱的,其中$ us $ us $ us是一个单位和$ e_2 $,e_2 $ is e_2 $ is or thogogonal iDempotents。显然清洁的戒指是弱清洁的环,两者都很清洁。在最近的一篇文章(J. AlgebraAppl。17(2018),1850111(5页))中,McGoven的特征是当组戒指$ \ Mathbb z _ {(p)} [C_Q] $变得脆弱且轻巧清洁,其中$ p,q $是独特的。在本文中,我们考虑了一个更一般的环境。令$ k $为代数数字字段,$ \ mathcal o_k $它的整数戒指,$ \ Mathfrak p \ subset \ subset \ mathcal o $ a nonzer prime理想和$ \ MATHCAL O _ {\ MATHFRAK P} $ $ \\ MATHCAL O $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ \ MATHFRAK P $。我们调查何时零环$ \ MATHCAL O _ {\ MATHFRAK P} [G] $弱干净且微弱地干净,其中$ G $是一个有限的Abelian组,并在$ k = \ nathbb q(q embb q(emmate)$ q的情况下,为这种组弱且微弱地清洁此类组的表征明确的表征q(\ sqrt {d})$是二次字段。

A ring $R$ is said to be clean if each element of $R$ can be written as the sum of a unit and an idempotent. $R$ is said to be weakly clean if each element of $R$ is either a sum or a difference of a unit and an idempotent, and $R$ is said to be feebly clean if every element $r$ can be written as $r=u+e_1-e_2$, where $u$ is a unit and $e_1,e_2$ are orthogonal idempotents. Clearly clean rings are weakly clean rings and both of them are feebly clean. In a recent article (J. Algebra Appl. 17 (2018), 1850111(5 pages)), McGoven characterized when the group ring $\mathbb Z_{(p)}[C_q]$ is weakly clean and feebly clean, where $p, q$ are distinct primes. In this paper, we consider a more general setting. Let $K$ be an algebraic number field, $\mathcal O_K$ its ring of integers, $\mathfrak p\subset \mathcal O$ a nonzero prime ideal, and $\mathcal O_{\mathfrak p}$ the localization of $\mathcal O$ at $\mathfrak p$. We investigate when the group ring $\mathcal O_{\mathfrak p}[G]$ is weakly clean and feebly clean, where $G$ is a finite abelian group, and establish an explicit characterization for such a group ring to be weakly clean and feebly clean for the case when $K=\mathbb Q(ζ_n)$ is a cyclotomic field or $K=\mathbb Q(\sqrt{d})$ is a quadratic field.

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