Studia Scientiarium Mathematicarum Hungarica 28. (1993)

1-2. szám - Conchigdorzh R.: Generalized p.p. rings and rings of pi-regular quotients

R. GONCHIGDORZH 1. Preliminaries Throughout this paper, all rings considered are associative and without assuming the existence of identity elements, unless otherwise indicated. Let R be a ring and S be a subset of R. Then rR(S) and £R(S) denote the right and left annihilator of 5 in R, respectively. ann^(5) will stand for rR(S) when rR(S) = £R(S). E(R) and B(R) denote the set of all idempotents and the set of all central idempotents of the ring R, respectively. A ring R is said to be a normal ring if E(R) — B(R), i.e. if every idempotent of R is central. DEFINITION 1.1. A ring R is said to be a right generalized p.p. ring if E(R)R = R and for any a £ R there exist a positive integer n and an idempotent e G E(R) with rR(an) = rR(e). A left generalized p.p. ring is defined analogously. A ring R is a generalized p.p. ring (abbreviated g.p.p. ring) if it is both right and left generalized p.p. ring. It is clear that if the ring R has an identity this definition coincides with the usual one given in [8], [10]. An extension of the concept of g.p.p. ring to rings without identity was made in [7]. There it was not assumed that E{R)R = R and of course, in such a ring some nil direct summand may occur. Nil rings, however, can be considered as a trivial case of g.p.p. rings. Therefore, in our definition of a g.p.p. ring R we have supposed the condition E(R)R— R for excluding such trivial cases. Lemma 1.2 (cf. [8, Corollary 4]). Let R be a normal ring. Then R is g.p.p. ring if and only if R ■ E(R) = R and for any a G R there exist a positive integer n and an idempotent e G E(R) such that for every K > n annR(aK ) = = ann/i(e) and aK e = aK . Proof. The necessity is obvious. Let £ be a g.p.p. ring and a be an arbitrary element of R. Then there are integers m,n> 0 and idempotents e, / G E(R) such that rR(an) — rR(e), £R{aTn) — rR(f). Then for all K ^ max{n,m} we have rR(aK) — annfi(e) and £R(aK) — ann/*(/) (cf. [8, Lemma 3]). For, let an+16 = 0, b G R. Then 0 = eab = aeb and hence aneb = 0, rf>(an+l) — rR(an). In a similar way we have lR(am+1) = £Ii{am). Now, let rR(ah ) = ann/?(e), £R(aK ) = annfi(/). Then 0 = ah (/ — ef) = = (/ — ef)aK, and hence /(/ - ef) = 0, f — ef. Similarly we have e = = ef and therefore e — f and rR(aK ) = ann«(e) = £R(aK ), i.e. ann/j(aA ) = = annR(e). Because E(R)- R= R, we have an idempotent e' G E(R) such that ah — eah = e'(aK — eah ) = (e' — e'e)aK . Since e(e' — ee') = 0, we have 0 = = (e' — ee')aK = aK — eaK and so, aK = eaK. Definition 1.3. An element r G R is called reduced if there is a central idempotent e G B(R) such that ann#(r) = ann/{(e) and r = er. In this case it is easy to verify that the central idempotent e is unique and we call it associated idempotent of the reduced element r.

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