Ring Theory: Proceedings of a Conference held in Granada, by Pere Ara (auth.), Jose Luis Bueso, Pascual Jara, Blas

By Pere Ara (auth.), Jose Luis Bueso, Pascual Jara, Blas Torrecillas (eds.)

The papers during this complaints quantity are chosen examine papers in numerous parts of ring concept, together with graded earrings, differential operator earrings, K-theory of noetherian jewelry, torsion conception, ordinary jewelry, cohomology of algebras, neighborhood cohomology of noncommutative earrings. The booklet might be very important for mathematicians lively in study in ring theory.

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Extra resources for Ring Theory: Proceedings of a Conference held in Granada, Spain, Sept. 1–6, 1986

Example text

R is a commutative ~-gradedring. g(R), Pic(R), F P(R) are respectively the categories of finitely generated projective modules, invertible modules and faithfullly projective modules. B y _ ~ ( R ) , Picg(R), ~ ( R ) we mean the cor- responding categories consisting of graded objects and where the morphisms involved have to be graded morphisms. Pg(R), Picg(R), FPg(R) denote the categories with the same objects, but with arbitrary morphisms. The space of dr-maximal ideals of R is denoted maxg(R), and its dimension by d.

I be any such right two-sided that I ~ uR e uR ideal some J of R. of R and Then R-modules. for and h e n c e implications ideals = r i(I). as r i g h t (ReR)R ~e URR is f a i t h f u l of the case. ring, conditions (b) Proof. ~(R) same. e is f a i t h f u l . 10]. are p r o v e d injectlve (a) ~ ring [2,Proposition lemmas R be a c o m p l e t e in R. The [Proposition of S. [] is s u r j e c t i v e Proposition situation and by [5, it is not So t a k i n g central u = 1. follow let u be as in [ 2 ] .

6]. (Other The following result was proved in [3]. 1. For R a G-graded ring, G any group, J(R#G*) = JG(R)#G*. 2. J(R e) = JG(R)nRe. There is a connection between groups G for which J(S#G) = 0 if J(S) = 0 and groups for which the Jacobson radical of a graded ring R is a graded ideal. For suppose we wish to show that J(R) ~ JG(R) for R a G-graded ring. This is equivalent to showing that J(R/JG(R)) = 0. Denote R/JG(R) by R'. 1, J(R'#G*) = JG(R')#G* = 0. implies J((R'#G*)#G) = 0. 3, J(R') = 0 and J(R) ~ JG(R), Conversely if S has G-action and if J(S#G) ~ JG(S#G), then J(S) = 0 implies J(S#G) = O.

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