On Nil-Symmetric Rings

Abstract

The concept of nil-symmetric rings has been introduced as a generalization of symmetric rings and a particular case of nil-semicommutative rings. A ring is called right (left) nil-symmetric if, for , where are nilpotent elements, implies . A ring is called nil-symmetric if it is both right and left nil-symmetric. It has been shown that the polynomial ring over a nil-symmetric ring may not be a right or a left nil-symmetric ring. Further, it is also proved that if is right (left) nil-symmetric, then the polynomial ring is a nil-Armendariz ring.

1. Introduction

Throughout this paper, all rings are associative with unity. Given a ring , and denote the set of all nilpotent elements of and the polynomial ring over , respectively. A ring is called reduced if it has no nonzero nilpotent elements; is said to be Abelian if all idempotents of are central; is symmetric [1] if implies for all . An equivalent condition for a ring to be symmetric is that whenever product of any number of elements of the ring is zero, any permutation of the factors still gives the product zero [2]. is reversible [3] if implies for all ; is called semicommutative [4] if implies for all . In [5], Rege-Chhawchharia introduced the concept of an Armendariz ring. A ring is called Armendariz if whenever polynomials , satisfy , then for each . Liu-Zhao [6] and Antoine [7] further generalize the concept of an Armendariz ring by defining a weak-Armendariz and a nil-Armendariz ring, respectively. A ring is called weak-Armendariz if whenever polynomials ,   satisfy , then for each . A ring is called nil-Armendariz if whenever ,   satisfy , then for each . Mohammadi et al. [8] initiated the notion of a nil-semicommutative ring as a generalization of a semicommutative ring. A ring is nil-semicommutative if implies for all . In their paper it is shown that, in a nil-semicommutative ring , forms an ideal of . Getting motivated by their paper we introduce the concept of a right (left) nil-symmetric ring which is a generalization of symmetric rings and a particular case of nil-semicommutative rings. Thus all the results valid for nil-semicommutative rings are valid for right (left) nil-symmetric rings also. We also prove that if a ring is right (left) nil-symmetric and Armendariz, then is right (left) nil-symmetric. In the context, there are also several other generalizations of symmetric rings (see [9, 10]).

2. Right (Left) Nil-Symmetric Rings

For a ring , and denote the full matrix ring and the upper triangular matrix ring over , respectively. We observe that if is a ring, then

Definition 1. A ring is said to be right (left) nil-symmetric if whenever, for every and for every , , then . A ring is nil-symmetric if it is both right and left nil-symmetric.

Example 2. let be a field, and let be the path algebra of the quiver over , modulo the relation . Let and be the paths of length at vertices and , respectively. Composing arrows from left to right, is a nonzero path, while is not.
Then any nilpotent element is a linear combination of , , and .
Let and be two such elements and let be an arbitrary element. We have Thus is a right nil-symmetric ring. However, we have that , while . Hence, is not a left nil-symmetric ring.
Similarly by considering the opposite ring of , one can have a left nil-symmetric ring which is not right nil-symmetric.
Clearly every symmetric ring is nil-symmetric but the converse is not true by Example 3 and that every subring of a right (left) nil-symmetric ring is right (left) nil-symmetric.

Example 3. For a reduced ring , is a nil-symmetric ring which is not symmetric. This can be verified as follows.
Let Then Also Thus is a right nil-symmetric ring. Similarly it can be shown that is a left nil-symmetric ring. But whereas Thus is not symmetric.

From the above example we observe that a nil-symmetric ring need not be Abelian, as is an idempotent in , but

Remark 4. An Abelian ring also need not be either a right nil-symmetric or a left nil-symmetric ring as shown by the following example.

Example 5. We consider the ring in [11, Example 2.2] is an Abelian ring as and are the only idempotents. Again we have but Hence, is neither right nil-symmetric nor left nil-symmetric.

Proposition 6. Let be a reduced ring. Then is a nil-symmetric ring.

Proof. Let be such that This implies Since is reduced, . Thus Hence, is a right nil-symmetric ring. Similarly it can be shown that is a left nil-symmetric ring.

Let be a reduced ring and we define a new ring as follows: where . Based on Proposition 6, one may think that may also be nil-symmetric for , but the following example nullifies that possibility.

Example 7. Let be a reduced ring and let Now but Thus is neither a right nil-symmetric ring nor a left nil-symmetric ring.
For a ring , let Then forms a subring of .

Example 8. For every reduced ring , is nil-symmetric.
Let and let be such that This gives Thus ,  . Since is reduced, we have ,  . Therefore, Hence, is a right nil-symmetric ring. Similarly, it can be shown that is a left nil-symmetric ring.

We also observe that every right (left) nil-symmetric ring is nil-semicommutative.

Proposition 9. Every right (left) nil-symmetric ring is nil-semicommutative.

Proof. Let be a right nil-symmetric ring and such that . Let be arbitrary; then . By right nil-symmetric property of , . Thus . Hence, is nil-semicommutative. Proceeding similarly one can show that every left nil-symmetric ring is nil-semicommutative.

Remark 10. The converse is however not true, as shown by the following example.

Example 11. For every reduced ring , is a nil-semicommutative ring which is neither a right nil-symmetric ring nor a left nil-symmetric ring. This can be verified as follows.
We have but Thus is neither a right nil-symmetric ring nor a left nil-symmetric ring. But is nil-semicommutative by [8, Example 2.2].

Remark 12. Semicommutativity and nil-symmetry do not follow each other. In Example 3, is a nil-symmetric ring but not Abelian (and so not semicommutative [12]). The following example [13, Example 2.8] shows that a semicommutative ring need not be a right or left nil-symmetric ring.

Example 13. Let be the quaternion group and let be the ring of integers modulo 2. Consider the group ring . By [14, Corollary 2.3], is reversible and so semicommutative. Let , , . Then and such that , but . Hence, is neither a right nil-symmetric ring nor a left nil-symmetric ring.

Proposition 14. For a reduced ring and for ,

Proof. Let be a reduced ring. Then by [9, Theorem 2.3], is a symmetric ring and hence a nil-symmetric ring, where is the ideal generated by for any positive integer . Also by [15], for . Hence, for is nil-symmetric.

Since the class of nil-symmetric rings is contained in the class of nil-semicommutative rings, the results which are valid for nil-semicommutative rings are also valid for nil-symmetric rings. Mohammadi et al. [8, Example 2.8] have shown that is not a nil-semicommutative ring, where is a reduced ring. Thus is not nil-symmetric. Now we give an example of a weak-Armendariz ring which is not nil-symmetric.

Example 15. Let be a reduced ring and let
By [6, Example 2.4], is weak-Armendariz. By Example 7, is neither a right nor a left nil-symmetric ring.

Proposition 16. Finite product of right (left) nil-symmetric rings is right (left) nil-symmetric.

Proof. It comes from the fact that [8, Proposition 2.13]. Let and such that . Thus, for each , . Since is right nil-symmetric, for each . So, we get . The result can be similarly proved for left nil-symmetric rings.

Proposition 17. Let be a ring and let be a multiplicatively closed subset of consisting of central nonzero-divisors. Then is right (left) nil-symmetric if and only if is right (left) nil-symmetric.

Proof. It suffices to prove the necessary condition because subrings of right (left) nil-symmetric rings are also right (left) nil-symmetric. Let with , , and ; then , , and . Since is contained in the center of , we have and so . It follows that , since is right nil-symmetric. Thus . Hence, is right nil-symmetric. Similarly, can be shown to be left nil-symmetric if itself is a left nil-symmetric ring.

Corollary 18. For a ring , is a right (left) nil-symmetric ring if and only if is a right (left) nil-symmetric ring.

Proof. It directly follows from Proposition 17. If , then is clearly a multiplicatively closed subset of and .

Proposition 19. Let be a ring. Then and are right (left) nil-symmetric for some central idempotent of if and only if is right (left) nil-symmetric.

Proof. It suffices to prove the necessary condition because subrings of right (left) nil-symmetric rings are also right (left) nil-symmetric. Let and be right (left) nil-symmetric rings for some central idempotent of . Since, , is right (left) nil-symmetric by Proposition 16.

Since the class of right (left) nil-symmetric rings is closed under subrings, therefore, for any right (left) nil-symmetric ring and for any , is a right (left) nil-symmetric ring. The converse is, however, not true, in general as shown by the following example.

Example 20. Let be any reduced ring. Then by Example 11, is neither a right nil-symmetric nor a left nil-symmetric ring.
But for is a reduced ring and so a nil-symmetric ring.

For any nonempty subsets of a ring , denotes the set of all finite sums of the elements of the type , where , , .

Proposition 21. A ring is right (left) nil-symmetric if and only if implies ( implies ) for any two nonempty subsets of and any subset of .

Proof. Let be a right nil-symmetric ring and let be nonempty subsets of ; let be a nonempty subset of such that . Then for all , , . Right nil-symmetric property of gives for all , , . Thus . Similar proof can be given for left nil-symmetric rings. The converse is straightforward.

The following result shows that, for a semiprime ring, the properties of reduced, symmetric, reversible, semicommutative, nil-semicommutative, and nil-symmetric rings coincide. Note that a ring is said to be semiprime if, for , implies that .

Proposition 22. For a semiprime ring , the following statements are equivalent.(1)is reduced.(2) is symmetric.(3) is reversible.(4) is semicommutative.(5) is nil-semicommutative.(6) is right (left) nil-symmetric.

Proof. (1)–(4) are equivalent by [16, Lemma 2.7]. (1)(5) by [8, Proposition 2.18]. (2)(6) is clear. (6) (1): let for . Then for any , and so , since is right nil-symmetric. Thus by semiprimeness of and, therefore, is reduced.

Given a ring and a bimodule  , the trivial extension of by is the ring with the usual addition and the following multiplication: This is isomorphic to the ring of all matrices: where and and the usual matrix operations are used.

Proposition 23. For a reduced ring , is a nil-symmetric ring.

Proof. Let be a reduced ring. Since is a subring of in Proposition 6 and the class of right(left) nil-symmetric rings is closed under subrings, thus is a nil-symmetric ring.

Considering the above proposition one may conjecture that if a ring is nil-symmetric, then is nil-symmetric. However, the following example eliminates the possibility.

Example 24. Let be the Hamilton quaternions over the real number field and let Then by Proposition 6, is a nil-symmetric ring. Let be the trivial extension of by itself. Then is not a right nil-symmetric ring. Note that However we have Thus is not a right nil-symmetric ring.

Example 25. Let be a ring and let be an ideal of such that is nil-symmetric. Then may not be nil-symmetric. This can be verified as follows. Let be any reduced ring. Then by Example 11, is not nil-symmetric but nil-semicommutative. Thus is an ideal of and is reduced, so nil-symmetric.

Homomorphic image of a right (left) nil-symmetric ring need not be a right (left) nil-symmetric ring. This is discussed after Example 26.

3. Polynomial Extension of Nil-Symmetric Rings

Anderson-Camillo [17] proved that a ring is Armendariz if and only if is Armendariz; Huh et al. [12] have shown that polynomial rings over semicommutative rings need not be semicommutative; Kim-Lee [16] showed that polynomial rings over reversible rings need not be reversible. Recently Mohammadi et al. [8] have given an example of a nil-semicommutative ring for which is not nil-semicommutative. Based on the above findings, it is natural to check whether the polynomial ring over a nil-symmetric ring is nil-symmetric. However, the answer is given in the negative through the following example.

Example 26. Let be the field of integers modulo 2 and let be the free algebra of polynomials with zero constant terms in noncommuting indeterminates ,  ,  ,  ,  ,  , and over . Consider an ideal of the ring , say , generated by the following elements: , , , , , , , , , , , , , , , and , where . Now is symmetric by [9, Example 3.1] and so a nil-symmetric ring. By [8, Example 3.6], we have , . Now , , but because . Hence is neither a right nil-symmetric ring nor a left nil-symmetric ring.

Remark 27. The above example also helps in showing that homomorphic image of a right (left) nil-symmetric ring need not be a right (left) nil-symmetric ring. This is verified as follows.

Example 28. In Example 26, is a domain [16] and so a nil-symmetric ring. But the quotient ring is neither a right nil-symmetric ring nor a left nil-symmetric ring.

Now we study some conditions under which the answer may be given positively. Since every right (left) nil-symmetric ring is nil-semicommutative by Proposition 9, therefore, by [8, Theorem 3.3] for each right (left) nil-symmetric ring , . The converse is, however, not true, in general. Now we give an example of a ring which satisfies , but is neither a right nil-symmetric ring nor a left nil-symmetric ring.

Example 29. We use the ring in [7, Example 4.8]. Let be a field, and . Then is not an ideal of . Thus is neither a right nil-symmetric nor a left nil-symmetric ring by Proposition 9 and [8, Theorem 2.5]. But is a nil-Armendariz ring and hence by [7, Corollary 5.2], .

Proposition 30. If is a right (left) nil-symmetric and Armendariz ring, then the polynomial ring is right (left) nil-symmetric.

Proof. Let be a right nil-symmetric and Armendariz ring and let and such that . Since is right nil-symmetric, by Proposition 9 and [8, Theorem 3.3]. Thus for ; . Since is Armendariz, therefore, by [17, Proposition 1]. Thus by right nil-symmetric property of , . Therefore, . Hence, is a right nil-symmetric ring. Similarly it can be shown that is a left nil-symmetric ring if is a left nil-symmetric and Armendariz ring.

Proposition 31. If is a right (left) nil-symmetric ring, then is nil-Armendariz.

Proof. Let be a right (left) nil-symmetric ring. Thus by Proposition 9, is nil-semicommutative. By [8, Corollary 2.9], is a nil-Armendariz ring. Again by [8, Theorem 3.3], . Thus by [7, Theorem 5.3], is nil-Armendariz.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors are deeply indebted to Dr. Pierre-Guy Plamondon, Laboratory of Mathematics, University of Paris, France, for providing Example 2 and Professor Mangesh B. Rege, Department of Mathematics, NEHU, Shillong, India, for his valuable suggestions.