ON REGULARITY OF RINGS 1

ON REGULARITY OF RINGS 1 Jianlong Chen Department of Mathematics, Harbin Institute of Technology Harbin 150001, P. R. China and Department of Applied Mathematics, Southeast University Nanjing 210096, P. R. China e-mail: jlchen@seu.edu.cn Nanqing Ding Department of Mathematics, Nanjing University Nanjing 210093, P. R. China e-mail: nqding@netra.nju.edu.cn Abstract. It is shown that a ring R is von Neumann regular if and only if every cyclic left R-module is GP -injective if and only if R is left P P and left GP -injective. In addition, two examples are given to answer two open questions in the negative. 2000 Mathematics Subject Classification: 16E50, 16D60 Keywords: von Neumann regular ring, P -injective module, GP -injective module. 1. Introduction Throughout R is an associative ring with identity and all modules are unitary. We write M R to indicate a right R-module and N M to mean N is a submodule of M. For a subset X of R, the left right annihilator of X in R is denoted by lx rx. If X = {a}, we usually abbreviate it to lara. Let M be a right R-module. For each subset A of R, the left annihilator of A in M is denoted by l M A. For a set I, M I M I is the direct product sum of copies of M indexed by I. As usual, SocR R denotes the right socle of R. 1 Algebra Colloquium 8:3 2001 267-274 1

Recall that a right R-module M is called right P -injective if every right R-homomorphism from a principal right ideal ar to M extends to one from R R to M. M is said to be GP -injective [7] if, for any 0 a R, there exists a positive integer n with a n 0 such that any right R-homomorphism from a n R to M extends to one from R R to M. The ring R is called right GP - injective if the right R-module R R is GP -injective. Note that GP -injective modules defined here are also called Y J-injective modules in [18, 19], but differ from the GP -injective modules defined in [17]. An element a R is regular if a ara. A subset U of R is regular in case every element in U is regular. Von Neumann regular rings have been studied extensively by many authors See, e.g., [5, 11, 12, 14-20]. It is well known [14] that a ring R is regular if and only if every left R-module is P -injective. Therefore if R is regular then every left R-module is GP -injective. Is the converse true? This question was raised by R. Yue Chi Ming See [15, 16 18, 19]. Zhang and Wu [20] answered this question in the positive. Recently, the Third Korea-China- Japan International Symposium on Ring Theory was held in South Korea. During the conference, Professor J. Y. Kim asked whether R is regular if R is a ring whose every cyclic left R-module is GP -injective. In this paper, we first prove that a ring R is regular if and only if every cyclic left R-module is GP -injective if and only if every proper principal left ideal is GP -injective, which gives an affirmative answer to the question above. Then it is shown that R is regular if and only if every essential left ideal is GP -injective. Finally we answer two open questions in the negative. 2. Results Recall that an element a R is called π-regular if there exists a positive integer m such that a m = a m ba m for some b R. We shall call an element x R generalized π-regular if there exists a positive integer n such that x n = x n yx for some y R. R is called generalized π-regular if every element in R is generalized π-regular. Clearly, every π-regular element is generalized π-regular. We start with the following. Lemma 2.1 Let R be a ring and a R. If a n a n ra is regular for some positive integer n and r R, then there exists y R such that a n = a n ya, whence a is generalized π-regular. Proof. Let d = a n a n ra. Since d is regular, d = dud for some u R. 2

Hence a n = d + a n ra = a n a n raua n a n ra + a n ra = a n 1 raua n 1 a n ra + a n ra = a n ya, where y = 1 raua n 1 a n r + r R. Theorem 2.2 The following are equivalent for a ring R. 1 R is regular. 2 N 1 R = {0 a R a 2 = 0} is regular and R is generalized π-regular. Proof. 1 2 is clear. 2 1. Let 0 a R. Then there exists a positive integer n such that a n = a n ra for some r R since R is generalized π-regular. Next we shall show that a is regular. In fact, if n = 1, we are done. Let n > 1. Put d = a n 1 a n 1 ra, then ad = 0. If d = 0, then a n 1 = a n 1 ra. If d 0, then d 2 = a n 1 a n 1 rad = 0. Since N 1 R is regular, d is regular. Hence a n 1 = a n 1 ya for some y R by Lemma 2.1. Therefore we always have a n 1 = a n 1 z 1 a for some z 1 R. If n 1 > 1, then there exists z 2 R such that a n 2 = a n 2 z 2 a by the preceding proof. Repeating the above-mentioned process, we will get b R such that a = aba, i.e., a is regular. Theorem 2.3 The following are equivalent for a ring R. 1 R is regular. 2 Every cyclic left R-module is GP -injective. 3 Every principal left ideal is GP -injective. 4 Every proper principal left ideal is GP -injective. Proof. 1 2 3 4 is clear. 4 1. Let a N 1 R, i.e., 0 a R and a 2 = 0, then Ra R. Since Ra is GP -injective, any homomorphism from Ra to Ra extends to one from R R to Ra. Thus Ra is a direct summand of R R, and so a is regular. Hence N 1 R is regular. Next let 0 b R. If Rb = R, then b is regular, and hence b is generalized π-regular. If Rb R, then Rb is GP -injective by 4, and so there exists a positive integer n such that b n 0 and any homomorphism Rb n Rb extends to R R Rb. Now define f : Rb n Rb such that frb n = rb n for r R. Clearly f is a left R-homomorphism. Then there exists r R such that b n = b n rb, which shows that b is generalized π-regular. Consequently R is generalized π-regular, and so R is regular by Theorem 2.2. Remark 1. From Theorem 2.3 we know that the equivalences 1 4 5 in [18, Theorem 1] and 1 4 in [18, Theorem 2] are clear. Note that the direct summand of a GP -injective module is GP -injective. We obtain the following result of [20] immediately from Theorem 2.3 above. 3

Corollary 2.4 [20, Theorem 9]. The following are equivalent for a ring R. 1 R is regular. 2 Every left R-module is GP -injective. 3 Every direct product or sum of cyclic left R-modules is GP -injective. The following result of Zhang and Wu [20] will be needed in the sequel. Lemma 2.5 [20, Proposition 1]. Let M be a right R-module. Then M is GP -injective if and only if, for any 0 a R, there exists a positive integer n such that a n 0 and l M ra n = Ma n. It is known that the direct sum of P -injective modules is P -injective. We don t know whether the direct sum of two GP -injective modules is still GP -injective. However, for two right R-modules M 1 and M 2, if M 1 M 2 is GP -injective, then there exists a positive integer n such that a n 0 and l M1 M 2 ra n = M 1 M 2 a n by Lemma 2.5. In the meantime, it is easy to verify that l M1 ra n = M 1 a n and l M2 ra n = M 2 a n. Lemma 2.6 Let I 1 and I 2 be right ideals of a ring R with I 1 I 2 = 0. If R R/I 1 I 2 is GP -injective, the R/I i is GP -injective, i = 1, 2. Proof. We use an argument similar to that in [5, Lemma 32]. Let 0 a R. Since R R/I 1 I 2 is a GP -injective right R-module, there exists a positive integer n with a n 0 such that lra n = Ra n and l R/I1 I 2 ra n = R/I 1 I 2 a n by the remark preceding Lemma 2.6. Next we show that R/I 1 is right GP -injective. In fact, let b+i 1 l R/I1 ra n, where b R. Then bra n I 1 I 1 I 2, and so b + I 1 I 2 l R/I1 I 2 ra n = R/I 1 I 2 a n. Thus b + I 1 I 2 = xa n + I 1 I 2 for some x R, and hence b = xa n + r 1 + r 2 for some r 1 I 1 and r 2 I 2. Note that r i ra n I i, i = 1, 2, and r 2 ra n = b xa n r 1 ra n bra n +r 1 ra n I 1. Therefore r 2 ra n I 1 I 2 = {0}, i.e., r 2 lra n = Ra n, and so r 2 = ya n for some y R. Thus b = xa n + r 1 + ya n = x + ya n + r 1, and then b + I 1 R/I 1 a n. Consequently, l R/I1 ra n = R/I 1 a n, and so R/I 1 is right GP -injective by Lemma 2.5. Similarly, R/I 2 is GP -injective. Theorem 2.7 The following are equivalent for a ring R. 1 R is regular. 2 Every essential right ideal of R is GP -injective. 3 R R/L is GP -injective for every essential right ideal of R. Proof. Obviously, 1 implies 2 and 3. 2 1. Let a R. Then there exists a right ideal K of R such that ar K is essential in R R, and so ar K is GP -injective by 2. Therefore ar is GP -injective, and hence R is regular by Theorem 2.3. 3 1. Let a R. Then ra I is essential in R R for some right ideal I of R, and so R R/ra I is GP -injective. It follows that ar = R/ra 4

is GP -injective by Lemma 2.6. So R is regular by Theorem 2.3. Corollary 2.8 The following are equivalent for a ring R. 1 R is regular. 2 Every essential right ideal of R is P -injective. 3 R is right P -injective and cyclic singular right R-modules are P -injective. Recall that a ring R is called left P P in case each principal left ideal of R is projective. It is known that a ring R is regular if and only if R is left P P and left P -injective [12, Theorem 3]. This result can be generalized to the following. Theorem 2.9 A ring R is regular if and only if R is left P P and left GP -injective. Proof. is clear.. Let 0 a R. Since R is left GP -injective, there exists a positive integer n with a n 0 such that rla n = a n R. By hypothesis, Ra n is projective. Thus 0 la n R Ra n 0 splits, and hence la n = Re, where e 2 = e R. Hence a n R = rla n = rre = 1 er. It follows that a n is regular. Therefore R is π-regular, whence it is generalized π-regular. Let 0 a R with a 2 = 0. Then a is regular by the preceding proof. So N 1 R is regular. Consequently R is regular by Theorem 2.2. In relation with the preceding theorem, R. Yue Chi Ming [17] raised the following question: Is a left P P right P -injective ring regular? Here we answer this question in the negative. The ring R in the next example is a left hereditary and right P -injective ring, but it is not regular. This ring R appeared in [13], and it was essentially given by J. Clark [2]. Example 1. Let Z 2 = {0, 1} be the field of two elements and N the set of positive integers. Let A be the subring of Z 2 N consisting of elements of the form a 1, a 2,, a n, a, a, a,, i.e., A is obtained by adjoining the identity of Z 2 N to its ideal Z 2 N. Then A is a commutative countable regular ring with each element an idempotent, and so A is hereditary by [6, Theorem 1]. If k Z 2 and a 1, a 2,, a n, a, a, a, A, let k a 1, a 2,, a n, a, a, a, = ka, then Z 2 is a right A-module. With the obvious left Z 2 -module multiplication, Z 2 is then a Z 2 -A-bimodule. Thus we may form the formal triangular matrix ring R = Z2 Z 2 0 A 5.

, where Z 2 is a left A-module by the multipli- Z2 0 First, let S = Z 2 A cation a 1, a 2,, a n, a, a, a, k = ak for a 1, a 2,, a n, a, a, a, A and k Z 2. It is easy to check that R and S are anti-isomorphic. By [4, Theorem 4.7], S is right hereditary, and so R is left hereditary. Secondly, we shall prove that R is a right P -injective ring. It is sufficient to show that lrɛ = Rɛ for every ɛ R by [8, Lemma 1.1]. Let ɛ = k l, where x = a 1, a 2,, a n, a, a, a, A and k, l Z 2. Case 1. k = 0. In this case, ɛ = or ɛ =. 1.1 Let ɛ =. Then ɛ 2 = ɛ, and hence lrɛ = Rɛ. 1.2 Let ɛ =. If a = 1, then it is easy to see that ɛ 2 = ɛ, and so lrɛ = Rɛ. If a = 0, set e = x = a 1, a 2,, a n, 0, 0, 0, Z N 2, Z2 Z 2 then ɛ =. It is easily seen that rɛ = N and 0 e 0 1 ez 2 0 Z2 0 l l 1 lrɛ = Rɛ for =. Hence 0 Ae 0 ye 0 y 0 e lrɛ = Rɛ. we take this opportunity to point out that there is a gap in [13]. By 2 Z2 Z of [13, Example 1], rɛ = 2, but this is not true by the foregoing discussion. 1 0 1 1 Case 2. k = 1. In this case, ɛ = or ɛ =. 1 0 2.1 Let ɛ =. Then ɛ 2 = ɛ, and so lrɛ = Rɛ. 1 1 2.2 Let ɛ =. If a = 0, then ɛ 2 = ɛ, whence lrɛ = Rɛ. If 1 1 a = 1, put f = x = a 1, a 2,, a n, 1, 1, 1,, then ɛ =. It is 0 f easily checked that rɛ = = R and lrɛ = 0 1 fa f 1 0 1 1 1 1 l = R Rɛ for =. f 0 f 0 f 0 f So lrɛ = Rɛ. Consequently lrɛ = Rɛ for every ɛ R, and hence R is right P -injective. 6

Note that Xue s claim [13, p. 32] that R is not right P -injective is not true. Finally we claim that R is not regular. 0 Z2 Let σ =. Then Rσ = is a minimal left ideal of R. It is easy to check that lσ = R and rlσ = r = 1 0 R σr. Thus R is not left mininjective by [9, Lemma 1.1], whence R is not left GP -injective by [13, p. 35]. So R is not regular. By the way, we shall show that SocR R is not projective. Note that 0 Z2 σr = is a minimal right ideal of R. Let L = rσ. Then L = Z2 Z 2 N. We claim that R/L is not projective. In fact, if R/L is 0 Z 2 k l projective, then there exists u = L, where k, l Z 0 y 2 and y Z N 2, such that ξ = uξ, ξ L. Thus, for any x Z N 2, = k l, and so x = yx. Let x = e 0 y i = 0, 0,, 0, 1, 0, with 1 in the ith position and 0 s in all other positions, i = 1, 2,. Then e i = ye i, i = 1, 2,, and hence y = 1, 1, 1, is the identity of Z N 2, which contradicts y Z N 2. Therefore σr = R/L is not projective, and so SocR R is not projective. Remark 2. The right P -injective ring R in Example 1 is neither left GP - injective nor left mininjective. This shows that the notion of GP -injectivity and that of mininjectivity are not left-right symmetric. Recall that a ring R is a 2-primal ring if its prime radical coincides with the set of all nilpotent elements of R. R is right left weakly regular if I 2 = I for each right left ideal I of R. R is weakly regular if it is both right and left weakly regular. R is reduced if R contains no nonzero nilpotent element. Nam and Kim [7] proved that: if R is a 2-primal ring such that every simple right R-module is GP -injective, then R is a reduced weakly regular ring. They also asked whether the converse of this result is true [7, Question 2]. The following example gives a negative answer to this question. Example 2. Let R be a simple principal right ideal domain which is not a right V -ring this kind of rings can be found in [10]. Then it is clear that R is a reduced weakly regular ring. But not every simple right R-module is GP - injective for R is not a right V -ring. In fact, in this case, GP -injectivity of 7

a right R-module M coincides with injectivity of M. First M R is P -injective if and only if M R is injective since R is a principal right ideal ring. Next let M R be GP -injective. Then for any 0 a R, there exists a positive integer n with a n 0 such that l M ra n = Ma n by Lemma 2.5. Since R is a domain, ra n = 0. Thus M = Ma n, and so M = Ma, whence l M ra = Ma for ra = 0. Clearly l M r0 = M0. Therefore l M ra = Ma for any a R. So M R is P -injective. This shows that M R is GP -injective if and only if M R is P -injective. ACKNOWLEDGMENT This work was partially supported by the National Natural Science Foundation of China No. 19701008 & 19771046. REFERENCES [1] F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, Springer, New York Heidelberg Berlin, 1973. [2] J. Clark, A note on rings with projective socle, C. R. Math. Rep. Acad. Sci. Canada 18 1996, 85-87. [3] N. Q. Ding and J. L. Chen, Rings whose simple singular modules are Y J-injective, Math. Japon. 40 1 1994, 191-195. [4] K. R. Goodearl, Ring Theory: Nonsingular rings and modules, Monographs Textbooks Pure Appl. Math. 33, 1976. [5] Y. Hirano and H. Tominaga, Regular rings, V-rings and their generalizations, Hiroshima Math. J. 9 1979, 137-149. [6] C. U. Jensen, On homological dimensions of rings with countably generated ideals, Math. Scand. 18 1966, 97-105. [7] S. B. Nam, N. K. Kim and J. Y. Kim, On simple GP -injective modules, Comm. Algebra 23 14 1995, 5437-5444. [8] W. K. Nicholson and M. F. Yousif, Principally injective rings, J. Algebra 174 1995, 77-93. [9] W. K. Nicholson and M. F. Yousif, Mininjective rings, J. Algebra 187 1997, 548-578. 8

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