By S.K. Jain

ISBN-10: 019966451X

ISBN-13: 9780199664511

This exact and entire quantity offers an up to date account of the literature as regards to making a choice on the constitution of jewelry over which cyclic modules or right cyclic modules have a finiteness or a homological estate. The finiteness stipulations and homological houses are heavily interrelated within the experience that both speculation induces the opposite in a few shape. this can be the 1st e-book to convey all of this significant fabric at the topic together.

Over the final 25 years or extra quite a few mathematicians have investigated earrings whose issue jewelry or issue modules have a finiteness or a homological estate. They made very important contributions resulting in new instructions and questions, that are indexed on the finish of every bankruptcy for the good thing about destiny researchers. there's a wealth of fabric at the subject that is mixed during this publication, it comprises greater than 2 hundred references and isn't claimed to be exhaustive.

This booklet will attract graduate scholars, researchers, and pros in algebra with an information of simple noncommutative ring idea, in addition to module idea and homological algebra, reminiscent of a one-year graduate direction within the thought of jewelry and modules.

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**Extra info for Cyclic Modules and the Structure of Rings**

**Sample text**

Suppose that the module R/B contains a nonzero injective submodule C/B. Therefore R/B = C/B ⊕ D/B, say. Since R/D ∼ = C/B, the module R/D is injective. Since B is countably generated and D/B is cyclic, D is countably generated. 2, D is ﬁnitely generated. Therefore D is a cyclic direct summand of RR . Since B is essential in RR , we have D = R. Therefore C/B = 0, a contradiction to our assumption. This shows that the module R/(A ⊕ (1 − e)R) does not contain a nonzero injective submodule. 4 (Osofsky’s Theorem) For a ring R, the following conditions are equivalent.

Now assume R is not a domain. Then there exists a nonzero element a ∈ R such that annr (a) = 0. Now R/annr (a) cannot be isomorphic to R because R is uniform. Thus R/annr (a) is proper cyclic and hence artinian. As R/annr (a) ∼ = aR, we have that aR is artinian. Note that by assumption, R/aR is artinian. Hence R is artinian, a contradiction. Thus R must be a domain. 3 The following are equivalent statements for a ring R: 1. Every proper cyclic right R-module is artinian. 2. Every cyclic right R-module R/I with I = 0 is artinian.

Thus R/S has ﬁnite right Goldie dimension, and hence R has ﬁnite right Goldie dimension. So R contains independent uniform right ideals U1 , . . , Uk such that U = U1 ⊕ · · · ⊕ Uk ⊂e RR . Thus R/U is an artinian right R-module. Moreover, since R has right RM C, it is easy to check that for each nonzero submodule Vi of Ui , Ui /Vi is artinian. Hence U has Krull dimension at most 1. Thus R has right Krull dimension. 8, [180]). Therefore R has ACC on right annihilators. Hence R is quasi-Frobenius (see [70]).

### Cyclic Modules and the Structure of Rings by S.K. Jain

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