Almost JGP-Injective Rings and Semipritive Rings

Available online at www.sciencedirect.com Available online at www.sciencedirect.com

Procedia Engineering

Procedia Engineering 00 (2011) 000–000 Procedia Engineering 29 (2012) 1697 – 1701 www.elsevier.com/locate/procedia

2012 International Workshop on Information and Electronics Engineering (IWIEE)

Almost JGP-Injective Rings and Semipritive Rings Zhanmin Zhu* Department of mathematics,Jiaxing University,J iaxing,314001,China

Abstract

A ring R is called right almost JGP-injective or right AJGP-injective for short, if for any 0≠a∈J(R), there exist a positive integer n such that an≠0 and R an is a direct summand of lr(an). A right R-module M is called right JGP-injective if for any 0≠a∈J(R), there exists a positive integer n such that an≠0 and any Rhomomorphism from anR to M extends to a homomorphism of R to M. In this Article, we investigate the properties of right AJGP-injective rings, and characterize semiprimitive rings by AJGP-injective rings and JGP-injective modules. © 2011 Published by Elsevier Ltd. Selection and/or peer-review under responsibility of Harbin University of Science and Technology Keywords: AJGP-injective rings; semiprimitive rings; JGP-injective modules.

In this paper, we generalize the concepts of right AGPinjective rings and right JGP-injective rings to right almost JGP-injective rings (or right AJGP-injective rings for short), some interesting properties of right AJGP-injective rings are obtained. Moreover, we definite the concept of JGP-injective modules, and semiprimitive rings are characterized by AJGP-injective rings and JGP-injective modules. Concepts not explained can be found in papers [1-11]. Definition 1. A ring R is called right almost JGP-injective or right AJGP-injective for short, if for any

0 ≠ a ∈ J ( R) , there exist a positive integer n such that a n ≠ 0 and Ra n is a direct summand of lr (a n ) .a module M R is called minimal C2 if every minimal submodule that is isomorphic to a direct summand of M is itself a direct summand, and we call a ring R right minimal C2 if R R is minimal C2. *

* Corresponding author. Tel.: +86-0573-82623480 E-mail address: [email protected].

1877-7058 © 2011 Published by Elsevier Ltd. doi:10.1016/j.proeng.2012.01.197

1698 2

Zhanmin / Procedia Engineering (2012) 1697 – 1701 Author nameZhu / Procedia Engineering 00 29 (2011) 000–000

Theorem 2. If R is a right AJGP-injective ring. Then (1) J(R) ⊆ Zr. (2) R is right minimal C2. (3) S r ⊆ r(J(R)). Proof. (1). Let a ∈ J(R), we will show that a ∈ Z r . If not, then, there is 0 ≠ b ∈ J ( R ) such that r(a) I bR = 0. Clearly, ab ≠ 0. Since R is right AJGP-injective, there is a positive integer n and a left ideal X such that ( ab) ≠ 0 and lr ((ab) ) = R ( ab) ⊕ X . Let s = ( ab) , t = b( ab) n

then ( ab)

n −1

n

n

n

n −1

. If x ∈ r ( s ) ,

n −1

x ∈ r(ab) = r(b), so tx = b(ab) x = 0 , i.e, x ∈ r (t ) . This follows that r(s) = r(t). Now since t ∈ lr(t) = lr(s) = Rs ⊕ X, so t = us + y = uat + y for some u ∈ R and y ∈ X . Note that 1 - ua is invertible for a ∈ J(R), so t ∈ X and whence s = at ∈ Rs ∩ X = 0 . A contradiction. 2 (2) Let I be a minimal right ideal of R with I ≅ eR , where e = e . Then I = aR for some a ∈ R 2 such that a = ae and r(e) is a maximal right ideal. If ( aR ) ≠ 0 , then exists k ∈ aR such that kaR ≠ 0 . Since aR is minimal, kaR = aR . Thus k = k f for some 0 ≠ f ∈ aR , this shows that f 2 − f ∈ raR (k ) . But raR (k ) ≠ aR because kf ≠ 0 , and note that aR is simple, we have

raR (k ) = 0 , and so f 2 = f and aR = fR is a summand of R R . If (aR) 2 ≠ 0 , then a ∈ J (R) . Since R is right AJGP-injective, there exist a positive integer n and a left ideal X a n such that

a n ≠ 0 and

lr (a n ) = Ra n ⊕ X a n . Noting that 0 ≠ a n = a n e and r (e) is a maximal right ideal, we have

r (a n ) = r (e) , and so Re = l ((1 − e) R ) = lr (e) = lr (a n ) = Ra n ⊕ X a n . Write e = ba n + x , where b ∈ R, x ∈ X a n , then

a = ae = aba n + ax , so a n − a n ba n ∈ Ra n ∩ X a n =0, and hence

a n = a n ba n . Let g = a n b , then g 2 = g ≠ 0 and aR = a n R = gR . Thus, I = aR = gR is a direct summand of R R . (3) Let tR ⊆ R be simple. Suppose jt ≠ 0 for some j ∈ J (R ) . Then r ( jt ) = r (t ) . Since R is n right AJGP-injective, there exist a positive integer n and a left ideal X such that ( jt ) ≠ 0 and lr (( jt ) n ) = R( jt ) n ⊕ X . Note that r (( jt ) n ) = r (t ) , we have t = a( jt ) n + x ，where a ∈ R and x ∈ X . Then (1 − a( jt ) n −1 j )t = x , so (1 − a( jt ) n −1 j ) −1 x ∈ X . It follows that ( jt ) n ∈ R( jt ) n ∩ X , and then ( jt ) n = 0 , a contradition. _Corollary 3 Let R be a semiperfect and right AJGP-injective ring. Then S r ⊆ S l and Z l ⊆ Z r = J (R ) . Proof. Since R is semiperfect, it is semilocal and semiregular. So r(J(R)) = S l by [1,Proposition 15.17], and Z l + Z r ⊆ J (R ) by [9, Theorem B.58 and Theorem B.59]. So the result follows from Theorem 2. Corollary 4 If R is a right AJGP-injective and right GC2 ring, then J ( R ) = Z r . In particular, if R is a right AJGP-injective left Kasch ring, then J ( R ) = Z r .Proof. Since R is right GC2, by [8, Proposition 2.6], J ( R ) ⊇ Z r . Since R is right AJGP-injective,by Theorem 2, J ( R ) ⊆ Z r . So J ( R ) = Z r . If R is a left Kasch ring, then by [7,Proposition 1.46], R is right C2, and hence right GC2, so the last assertion follows. Theorem 5 Let R be a left noetherian and right AJGP-injective ring. Then J(R) is nilpotent and

1699 3

Zhanmin / Procedia Engineering 29 00 (2012) 1697 – 1701 AuthorZhu name / Procedia Engineering (2011) 000–000

r(J(R)) is essential both as a left and right ideal. Proof. Let J = J(R). First we prove that r ( J ) ⊆

ess

RR .

Let 0 ≠ x ∈ R . Since R is left noetherian, the non-empty set F = {l (( xa) ) : a ∈ R, k > 0 such that k

( xa) k ≠ 0} has a maximal element, say l (( xy ) n ) . n n We claim that J ( xy ) = 0 = 0. If not, then there exists t ∈ J such that t ( xy ) ≠ 0. Since R is n right AJGP-injective and t ( xy ) ∈ J , there exists a positive integer m and a left ideal X such that (t ( xy ) n ) m ≠ 0 and lr ((t ( xy ) n ) m ) = R(t ( xy ) n ) m ⊕ X . Write (t ( xy ) n ) m = s( xy ) n , where s = (t ( xy ) n ) m −1 t ∈ J . We proceed with the following two cases. n n n n n n Case 1. r (( xy ) ) = r ( s ( xy ) ). Then ( xy ) ∈ lr (( xy ) ) = lr ( s ( xy ) ) = Rs ( xy ) ⊕ X , and n n n hence ( xy ) = cs ( xy ) + z for some c ∈ R and z ∈ X , i.e., (1 − cs )( xy ) = z . Since s ∈ J ,1 − cs is invertible, we have s ( xy) n ∈ Rs ( xy ) n ∩ X , and thus s ( xy ) n = 0 . This is a contradiction. Case 2. r (( xy ) ) ≠ r ( s ( xy ) ). Then there exists u ∈ r ( s ( xy ) ) but u ∉ r (( xy ) ) . Thus, n

n

n

n

s ( xy ) n u = 0 and ( xy ) n u ≠ 0 . It shows that s ∈ l (( xy ) n u ) and l (( xy ) n u ) ∈ F . Noting that s ∉ l (( xy ) n ) , so the inclusion l (( xy ) n ) ⊂ l (( xy ) n u ) is strict. This contracts the maximality of l (( xy ) n ) in F. n n Thus we have proved that J ( xy ) = 0 , and so 0 ≠ ( xy ) ∈ xR ∩ r ( J ) . Therefore, r ( J ) is an essential right ideal of R. Next we prove that J is nilpotent. Since R is left noetherian, there exists k > 0 such that r ( J ) = r ( J

k +n

) for all n > 0. Suppose J is not nilpotent. Then J k ≠ 0 and so k R M = R / r ( J ) is a nonzero left R-module. Since R is left noetherian, the set {l R ( m) : 0 ≠ m ∈ M } k k has a maximal element, l R ( m1 ) say. Write m1 = x + r ( J ) , where x ∈ R . Then J x ≠ 0 . Since r ( J k ) = r ( J 2k ) , J k x ⊄ r ( J k ). So, there exists b ∈ J k such that bx ∉ r ( J k ) . Since r ( J )⊆ ess RR , bxR ∩ r ( J k ) ≠ 0 , so we have 0 ≠ bxa ∈ r ( J k ) for some a ∈ R . Let m2 = xa + r ( J k ) . Then m 2 ≠ 0 and b ∈ l R (m2 ) , but b ∉ l R (m1 ) . So, the inclusion l R (m1 ) ⊂ l R (m2 ) is proper. This contradicts the choice of m1 . n n +1 Finally, for any 0 ≠ x ∈ R , we have Jx = 0 , or J x ≠ 0 and J x = 0 for some n > 0. This follows that Rx ∩ r ( J ) ≠ 0 . So r(J) is an essential left ideal of R. k

Corollary 6 Let R be a left noetherian right JGP-injective ring. Then the following statements are equivalent: (1) R is right Kasch. (2) R is left C2. (3) R is left GC2. (4) R is semilocal . (5) R is left artinian. Proof. (1) ⇒ (2). By [9, Proposition 1.46]. (2) ⇒ (3) is obvious. (3) ⇒ (4). Since left noetherian ring is left finite dimensional, and left finite dimensional left GC2 ring is

1700 4

Zhanmin / Procedia Engineering (2012) 1697 – 1701 Author nameZhu / Procedia Engineering 00 29 (2011) 000–000

semilocal [11, Corollary 2.5], so (4) follows from (3). (4) ⇒ (5). Since R is left noetherian right AJGP-injective, By Theorem 5, J is nilpotent. Hence R is left noetherian and semiprimary by hypothesis. And therefore R is left artinian. (5) ⇒ (1). Since R is right JGP-injective, by [11, Proposition 3.4], R is right mininjective. Then R is semiperfect right mininjective ring and S r ⊆ R R , and so R is a right minfull ring. By [9, Theorem 3.12], R is right Kasch. Definition 7 A right R-module M is called right JGP-injective if for any 0 ≠ a ∈ J ( R ) , there exists a ess

positive integer n such that a ≠ 0 and any R-homomorphism from a R to M extends to a homomorphism of R to M. Theorem 8. The following conditions are equivalent for a module M R : n

n

(1) M R is JGP-injective. (2) For any 0 ≠ a ∈ J ( R ) , there exists n > 0 such that a ≠ 0 and l M rR ( a ) = Ma . Proof. Trivial. We call an element x ∈ R right generalized π -regular if there exists a positive integer n such that n

n

n

x n = xyx n for some y ∈ R . Let S be a subset of R. Then S is called regular if every element in S is

regular. S is called right generalized π -regular if every element in S is right generalized π -regular. It is easy to see that an element a ∈ J (R ) is regular if and only if a = 0, and so a subset S of J(R) is regular if and only if S = 0. We call a ring R right JPP, if every right principal ideal contained in J(R) is projective. Lemma 9 Let R be a ring and a ∈ R . If a − ara is regular for some positive integer n and n

n

a ∈ R , then there exists y ∈ R such that a n = aya n , whence a is right generalized

π -regular.

Proof. Let d = a − ara . Since d is regular, d = dud for some u ∈ R . Hence n

n

a n = d + ara n = (a n − ara n )u (a n − ara n ) + ara n = a (a n −1 − ra n )u (1 − ar )a n + ara n = aya n n −1 , where y = ( a − ra n )u (1 − ar ) + r . Lemma 10 Let R be a ring and I an ideal of R. Then the following statements are equivalent. (1) I is regular.

(2) N ( I ) = {a ∈ I : a = 0} is regular and I is right generalized π -regular. Proof. (1) ⇒ (2) is clear. (2) ⇒ (1). Let a ∈ I . Since I is right generalized π -regular, there exist a positive integer n 2

a n = ara n . Next we shall show that a is regular. In fact, if n = 1, we are n −1 − ara n −1 , then d ∈ I , da = 0 , and so d 2 = d (a n −1 − ara n −1 ) = 0 . done. Let n > 1. Put d = a n −1 Since N ( I ) is regular, d is regular. Hence a = ay1 a n −1 for some y1 ∈ R by Lemma 9 . If n − 1 > 1 , then there exists y 2 ∈ R such that a n − 2 = ay1 a n − 2 by the preceding proof. Continuous in this way, we will get b ∈ R such that a = aba, i.e., a is regular. and an element r in R such that

Corollary 11 Let R be a ring. Then the following statements are equivalent. (1) R is semiprimitive. 2

(2) N ( J ) = {a ∈ J : a = 0} = 0 and J is right generalized Theorem 12 The following are equivalent for a ring R.

π -regular.

Zhanmin / Procedia Engineering 29 (2012) 1697 – 1701 AuthorZhu name / Procedia Engineering 00 (2011) 000–000

(1) R is semiprimitive. (2) R is right JPP and right AJGP-injective. (3) Every principally right ideal of R is JGP-injective. (4) Every principally right ideal of R is JGP-injective. Proof. (1) ⇒ (2), (3) and (3) ⇒ (4) are obvious. (2) ⇒ (1). Suppose R is right JPP and right AJGP-injective. If J ( R ) ≠ 0 . Then for any

0 ≠ a ∈ J ( R) , by the right AJGP-injectivity of R, there exist a positive integer n and a left ideal X a n such that

a n ≠ 0 and lr (a n ) = Ra n ⊕ X a n . By the projectivity of a n R , there exists e 2 = e ∈ R

such that

r (a n ) = eR . Then we have R (1 − e) = l (eR ) = lr (a n ) = Ra n ⊕ X a n and so

1 − e = ba n + x for some b ∈ R and x ∈ X a n . Thus, a n = a n (1 − e) = a n ba n + a n x , this follows that a = a contradiction. (4) ) ⇒ (1). Let n

n

ba n . So a n R is a direct summand of R. But a n R is small in R, so a n = 0 , a

a ∈ J ( R) . Write M = aR. Since M is JGP-injective, by Theorem 8, there exists a n n n n n positive integer n such that l M rR ( a ) = Ma , so a = a ba for some b ∈ R . Hence, J is right generalized π -regular. Next we prove that N(J)=0. If not, Let 0 ≠ b ∈ N ( J ) . Since bR is JGPinjective, any homomorphism from bR to bR extends to one from R R to bR. Thus bR is a direct summand of R R , and so b = 0 since b ∈ J ( R ) , a contradiction. So N(J)=0. Therefore, R is semiprimitive by Corollary 11. References [1] Anderson, F. W., Fuller, K. R., Rings and Categories of Modules, Springer, New York, 1974. [2] Azumaya, G., Finite splitness and finite projectivity, J. Algebra 106 (1) (1987), 114-134. [3] Camillo, V., Yousif, M. F., Continuous rings with ACC on annihilators, Canad. Math.Bull., 34 (4) (1991), 462-464. [4] Chen, J. L., Ding, N. Q., On general principally injective rings, Comm.Algebra, 27(5) (1999), 2097-2116. [5] Chen, J. L., Ding, N. Q., On regularity of rings, Algebra colloq. 8 (3) (2001), 267-274. [6] Chen, J. L., Zhou, Y. Q., Zhu, Z. M., GP-injective rings need not be P-injective,Comm. Algebra, 33 (7) (2005), 2395-2402. [7] Nam, S. B., Kim, N. K., Kim, J. Y., On simple GP-injective modules, Comm. Algebra, 23 (14) (1995), 5437-5444. [8] Nicholson, W. K., Yousif, M. F., Principally injective rings, J. Algebra, 174 (1) (1995), 77-93. [9] Nicholson, W. K., Yousif, M. F., Quasi-Frobenius Rings, Cambridge University Press, Cambridge (2003). [10] Page, S. S., Zhou, Y. Q., Generalizations of Principally injective rings, J. Algebra, 206 (2) (1998), 706-721. [11] Yousif, M. F., Zhou, Y. Q., Rings for which certain elements have the principal extension property, Algebra Colloq, 10 (4) (2003), 501-512.

1701 5