Research Article | Open Access

# On Nil-Symmetric Rings

**Academic Editor:**Li Guo

#### 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.

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#### Copyright

Copyright © 2014 Uday Shankar Chakraborty and Krishnendu Das. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.