On Generalized Simple Singular AP-Injective Rings

Raida D. Mahammod raida.1961@uomosul.edu.iq Husam Q. Mohammad husam_alsabawi@yahoo.com College of Computer Sciences and Mathematics University of Mosul Received on: 21/11/2011 Accepted on: 15/02/2012 ABSTRACT A ring R is said to be generalized right simple singular AP-injective, if for any maximal essential right ideal M of R and for any bM, bR/bM is AP-injective. We shall study the characterization and properties of this class of rings. Some interesting results on these rings are obtained. In particular, conditions under which generalized simple singular AP-injective rings are weakly regular rings, and Von Neumann regular rings. Key word: AP-injective Rings, weakly continuous rings, socle of R, Von Neumann regular rings. طمنلا نم ةرماغو ةممعملا ةدرفنملا ةطيسبلا تاقلحلا لوح – AP  دومحم دؤاد ةدئا ر .د.أ    دمحم مساق ماسح .د.م.أ  تايضايرلاو بوساحلا مولع ةيلك  لصوملا ةعماج


Introduction:
Throughout this paper, R is an associative ring with identity, and R-module is unital. For a  R, r(a) and l(a) denote the right annihilator and the left annihilator of a, respectively. We write J(R), Y(R)(Z(R)), N(R) and Soc(RR) for the Jacobson radical, the right ( left ) singular ideal, the set of nilpotent elements and right socle of R, respectively. X ≤ M denoted that X is a submodule of module M.
Recall that a ring R is called right MC2-ring if eRa=0 implies aRe=0, where a, e 2 = e R and eR is minimal right ideal of R [8]. A ring R is Von Neumann (weakly) regular provided that for every aR there exists bR (b RaR) such that a=aba (a=ab resp.). Recall that a ring R is right (left) weakly continuous if J(R)=Y(R) ( J(R) = Z (R)), R / J(R) is regular and idempotent can be left module J(R) [5]. Clearly every regular ring is right (left) weakly continuous. A ring R is called zero commutative (briefly ZC-ring )if ab=0 implies ba=0, a,b R [1]. A right R-module M is principally injective (briefly P-injective), if for any principal right ideal aR of R and any right Rhomomorphism of aR into M can be extended to one of R into M [11]. The ring R is called right P-injective if RR is P-injective.

Generalized Simple Singular AP-injective Rings
Recall that a module MR with S=End(MR) is said to be almost principally injective (briefly AP-injective), if for any aR, there exists an S-submodule Xa of M such that lM(rR(a))=Ma Xa as left S-module [6]. AP-injectivety has been studied by many authors (see [9,10]). Actually, Zhao Yu-e [12] investigated some properties of rings whose simple singular right R-module is AP-injective. Now, we give a generalized AP-injective.

Definition 2.1:
A ring R is called a generalized right (left) simple singular AP-injective, if for any maximal essential right (left) ideal M of R, any bM, bR/bM (Rb/Mb) is APinjective.
The following lemma which is due to Zhao Yu-e [12], plays a central role in several of our proofs

Lemma 2.2:
Suppose M is a right R-module with S=End(MR). If lMrR(a)=MaXa, where Xa is left S-submodule of MR. Set f: aR→M is a right R-homomorphis, then f(a)= ma+x with mM, xXa.

Lemma 2.3:
If M is a maximal right ideal of R and r(a)  M with a  M, then 1-aR ≠ aM 2-R/M  aR/aM .

Proof:
(1) If aR = aM, then a = ay for some y in M, which implies that 1-y  r(a)  M, whence 1 M, contradicting M ≠ R. (2) From (1) aR ≠ aM, then the right Rhomomorphism g:R/M→aR/aM is defined by g(r+M) = ar+aM for all rR implies that R/M  aR/aM. ■ We start this section with the following results.

Proposition 2.4:
Let R be generalized right simple singular AP-injective ring, then (2) Let k Soc(RR) Y(R). If k≠ 0, then kR is a minimal right ideal and r(k) is an essential right ideal of R. Since every minimal one -sided ideal of R is either nilpotent or direct summand of R [8]. Thus, if (kR) 2 ≠ 0, then kR is a direct summand and hence r(k) is also direct summand which is a contradiction. If (kR) 2 =0, then k 2 = 0 and k  r(k). But r(k) is maximal essential right ideal of R. Therefore, by Lemma 2.3 R/r(k)k(r(k)). Hence, R/r(k) is AP-injective, so there exists cR and xXa as a proof (1) such that 1-ck r(k). Since, ck RkRr(k), then 1r(k). This is also contradiction, therefore Soc(RR) Y(R) = 0. ■ Following [7], for a prime ideal P of a ring R, we put Op={aP: ab=0 for some bR\P}.
In general, OP not subset of a prime ideal P. as the following example shows.

Example [2]:
Let R be a ring of 2×2 matrices over a field F. Then, but. P a ■

Theorem 2.5:
Let P be a prime ideal of a generalized right simple singular AP-injective ring with OP  P, then P is maximal.

Proof :
We claim that RaR + P = R for aR/P. if not, there exists a maximal ideal M of R containing RaR+ P. Moreover, M is a maximal right ideal of R. Suppose not, then there exists a maximal right ideal K of R such that M  K. If K is not essential in R. Then K is a direct summand of R, so we can write K=r(e) for some 0≠e=e 2 R. Then, ea=0, since eP, then a  OP P. Therefore, K must be essential right ideal of R. Now, suppose that aR=aK, then a=ac for some c K that implies a(1-c)= 0. Since, aP, then 1-cOP  P  K which is a contradiction. If aR≠aK, the right Rhomomorphism g: R/K→ aR/aK is defined by g(b+K)=ab+aK for all bR which implies that R/KaR/ak. Therefore, R/K is AP-injective . Let f:aR→R/K be defined by f(ar)= r+K for all rR. So by Lemma 2.2 f(a)= ca+K+x, xXa. Hence, 1-ca+K=xR/K Xa =0 , so 1-ca K whence 1K. Therefore, M is a maximal essential right ideal of R. So by the same method in the above proof P is a maximal of R. ■ Recall that R is called 2-Primal if its prime radical P(R) concedes with the set N(R) [7]. Kim and Kwak [3] showed that if R is a 2-primal, then OP  P for each prime ideal of R.

Corollary 2.6:
Let R be 2-primal generalized right simple singular AP-injective ring, then every prime ideal of R is maximal. ■

Proposition 2.7:
Let R be ZC-generalized simple singular AP-injective rings, then for any a,b  R with ab=0, then r(a) + r(b) = R.

Proof:
Suppose that ab=0 and r(a) + r(b)  R. Then, there exists a maximal right ideal M containing r(a) + r(b). If M not essential, then there exists 0 e = e 2  R such that M=r(e). Since br(a)  M= r(e)=l(e), then be=0 which implies that er(b)  M = r(e), so that e = e 2 = 0 which is a contradiction. Therefore, M must be essential.
Since, r(a)  M and a M , then by Lemma 2.3 R/ M  aR / aM. Therefore R / M is AP-injective. Let f:aR→R/M is defined by f( ar ) = r+M for all r  R. Note that f is well-defined and by Lemma 2.2 1+ M = f( a ) = ca + M +x, c  R, x  Xa. Hence, 1ca + M = xR/M  Xa=0, so 1-ca  M. Since, a  r(b) and R is ZC-ring, then ca  r(b)  M whence 1M which is a contradiction. Therefore, r(a) + r(b) = R. ■

The Connection between Generalized Simple Singular AP-injective and Other Rings
In this section, we give the connection between Von Neumann regular rings and generalized simple singular AP-injective rings.

Theorem 3.1:
Let R be right MC2-generalized right simple singular AP-injective, then R is right weakly regular ring.

Proof:
We will show that RaR + r(a) = R for any a  R. Suppose that there exists b  R such that RbR + r(b) ≠ R. Then, there exists a maximal right ideal M of R containing RbR + r(b). If M not essential, then M is a direct summand of R. So, we can write M=eR for some 0 ≠ e =e 2 R. Thus, (1-e)Rb= 0, since R is MC2 and (1-e)R is minimal, then bR(1-e) = 0. Hence, (1-e)  r(b)  M, so 1  M. It is a contradiction. Therefore, M must be essential right ideal of R.
Since, r(a)  M and a M, then by Lemma 2.3 R/ M  aR / aM. Therefore, R / M is AP-injective. Let f:bR→R/M defined by f( br ) = r +M for all r  R. Note that f is well-defined and by Lemma 2.2, 1+M=f(b) = cb + M +x, c  R, x  Xb. Hence, 1cb + M = x  R/M  Xb = 0, so 1-cb  M. Since, cb RbR  M, then 1  M which is a contradiction. Therefore, that RaR + r(a) = R for all a  R. Hence, R is a right weakly regular ring. ■ Now, we shall prove the main results of this section.

Theorem 3.2:
Let R be a ring, then the following statements are equivalent: (1) R is Von Neumann regular.
(2) R is generalized right simple singular AP-injective right weakly continuous.
(2)  (1) Suppose that Y(R) ≠ 0. Then, there exists a non-zero element a  Y(R) such that a 2 =0. We claim that Y(R) + r(a) =R. If not, there exists a maximal essential right ideal M containing Y(R) + r(a). Since, r(a)  M and a M , then by Lemma 2.3 R/ M  aR / aM. Therefore, R/M is AP-injective and lR/Mr(a)=(R/M)a  Xa, Xa ≤R/M. Let f:aR→R/M be defined by f( ar ) = r +M for all r  R. Note that f is well-defined and by Lemma 2.2, 1 + M = f( a ) = ba + M +x, b  R, x  Xa. Hence, 1ba + M = x  R/M  Xa = 0, so 1-ba  M. Since, a  Y(R) = J(R) implies that ca  J(R)  M and 1  M, which is a contradiction. Therefore, Y(R) + r(a) =R. Thus, we can write 1= c + d, for some cY(R) and d  r(a) . Thus, a=ca and so (1-c)a = 0. Since c  Y(R) = J(R), 1-c is invertible. Thus a=0 contradicting a ≠ 0. Therefore, Y(R)=0. ■

Lemma 3.3: [4]
For any a  Cent( R ), if a=ara for some r  R, then there exists bCent( R ) such that a=aba ( where Cent(R) is the center of R).