Let R be a ring with an identity element.R∈IBN means that R<sup>m</sup>■R<sup>n</sup> implies m=n,R ∈IBN<sub>1</sub> means that R<sup>m</sup> ■R<sup>n</sup&...Let R be a ring with an identity element.R∈IBN means that R<sup>m</sup>■R<sup>n</sup> implies m=n,R ∈IBN<sub>1</sub> means that R<sup>m</sup> ■R<sup>n</sup>⊕K implies m≥n,and R ∈IBN<sub>2</sub> means that R<sup>m</sup>■R<sup>m</sup>⊕K implies K=0.In this paper we give some characteristic properties of IBN<sub>1</sub> and IBN<sub>2</sub>,with orderings o the Grothendieck groups.In addition,we obtain the following results:(1)If R ∈IBM<sub>1</sub> and all finitely generated projective left R-modules are stably free,then the Grothendieck group K<sub>o</sub>(R)is a totally ordered abelian group.(2)If the pre-ordering of the Grothendieck group K<sub>o</sub>(R)of a ring R is a partial ordering,then R ∈IBM<sub>1</sub> or K<sub>o</sub>(R)=0.展开更多
Let T=RM NS (θ,φ) and θ=0. We use a complete different technique to obtain the generalize results for K 0(T), i.e., K 0(T/I)K 0(R)K 0(S/NM) and K 0(T/J(T)) K 0(R/J(R))K(S/J(S)).
A ring R is called orthogonal if for any two idempotents e and f in R, the condition that e and f are orthogonal in R implies the condition that [eR] and [fR] are orthogonal in K0(R)+, i.e., [eR]∧[fR] = 0. In this pa...A ring R is called orthogonal if for any two idempotents e and f in R, the condition that e and f are orthogonal in R implies the condition that [eR] and [fR] are orthogonal in K0(R)+, i.e., [eR]∧[fR] = 0. In this paper, we shall prove that the K0-group of every orthogonal, IBN2 exchange ring is always torsion-free, which generalizes the main result in [3].展开更多
基金Supported by National Nature Science Foundation of China.
文摘Let R be a ring with an identity element.R∈IBN means that R<sup>m</sup>■R<sup>n</sup> implies m=n,R ∈IBN<sub>1</sub> means that R<sup>m</sup> ■R<sup>n</sup>⊕K implies m≥n,and R ∈IBN<sub>2</sub> means that R<sup>m</sup>■R<sup>m</sup>⊕K implies K=0.In this paper we give some characteristic properties of IBN<sub>1</sub> and IBN<sub>2</sub>,with orderings o the Grothendieck groups.In addition,we obtain the following results:(1)If R ∈IBM<sub>1</sub> and all finitely generated projective left R-modules are stably free,then the Grothendieck group K<sub>o</sub>(R)is a totally ordered abelian group.(2)If the pre-ordering of the Grothendieck group K<sub>o</sub>(R)of a ring R is a partial ordering,then R ∈IBM<sub>1</sub> or K<sub>o</sub>(R)=0.
文摘Let T=RM NS (θ,φ) and θ=0. We use a complete different technique to obtain the generalize results for K 0(T), i.e., K 0(T/I)K 0(R)K 0(S/NM) and K 0(T/J(T)) K 0(R/J(R))K(S/J(S)).
基金the National Natural Science Foundation of China (No. 10571080) the Natural Science Foundation of Jiangxi Province (No. 0611042) the Science and Technology Projiet Foundation of Jiangxi Province (No. G[20061194) and the Doctor Foundation of Jiangxi University of Science and Technology.
文摘A ring R is called orthogonal if for any two idempotents e and f in R, the condition that e and f are orthogonal in R implies the condition that [eR] and [fR] are orthogonal in K0(R)+, i.e., [eR]∧[fR] = 0. In this paper, we shall prove that the K0-group of every orthogonal, IBN2 exchange ring is always torsion-free, which generalizes the main result in [3].