On Nil-Symmetric Rings

The concept of nil-symmetric rings has been introduced as a generalization of symmetric rings and a particular case of nilsemicommutative rings. A ring R is called right (left) nil-symmetric if, for a, b, c ∈ R, where a, b are nilpotent elements, abc = 0 (cab = 0) implies acb = 0. 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 R is right (left) nil-symmetric, then the polynomial ring R[x] is a nil-Armendariz ring.


Introduction
Throughout this paper, all rings are associative with unity. Given a ring , nil( ) 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 = 0 implies = 0 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 = 0 implies = 0 for all , ∈ ; is called semicommutative [4] if = 0 implies = 0 for all , ∈ . In [5], Rege-Chhawchharia introduced the concept of an Armendariz ring. A ring is called Armendariz if whenever polynomials ( ) = 0 + 1 + ⋅ ⋅ ⋅ + , ( ) = 0 + 1 + ⋅ ⋅ ⋅ + ∈ [ ] satisfy ( ) ( ) = 0, then = 0 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 ( ) = 0 + 1 + ⋅ ⋅ ⋅ + , ( ) = 0 + 1 + ⋅ ⋅ ⋅ + ∈ [ ] satisfy ( ) ( ) = 0, then ∈ nil( ) for each , . A ring is called nil-Armendariz if whenever ( ) = 0 + 1 + ⋅ ⋅ ⋅ + , ( ) = 0 + 1 + ⋅ ⋅ ⋅ + ∈ [ ] satisfy ( ) ( ) ∈ nil( )[ ], then ∈ nil( ) 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 = 0 implies = 0 for all , ∈ nil( ). In their paper it is shown that, in a nil-semicommutative ring , nil( ) 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) nilsymmetric 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]).

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 nil ( ( )) = ( ) .
(1) 2 Journal of Mathematics Definition 1. A ring is said to be right (left) nil-symmetric if whenever, for every , ∈ nil( ) and for every ∈ , = 0 ( = 0), then = 0. 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 2 = 0. Let 1 and 2 be the paths of length 0 at vertices 1 and 2, 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 ( 1 + ℎ 2 + + + ) be an arbitrary element. We have Thus is a right nil-symmetric ring. However, we have that Similarly by considering the opposite ring of , one can have a left nil-symmetric ring which is not right nilsymmetric.
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 , 2 ( ) is a nil-symmetric ring which is not symmetric. This can be verified as follows. Let Then Also Thus 2 ( ) is a right nil-symmetric ring. Similarly it can be shown that 2 ( ) is a left nil-symmetric ring. But whereas Thus 2 ( ) is not symmetric.
From the above example we observe that a nil-symmetric ring need not be Abelian, as ( 1 1 0 0 ) is an idempotent in 2 ( ), 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.
is an Abelian ring as ( 0 0 0 0 ) and ( 1 0 0 1 ) are the only idempotents. Again we have but Hence, is neither right nil-symmetric nor left nilsymmetric.

Proposition 6. Let be a reduced ring. Then
is a nil-symmetric ring.
Proof. Let be such that This implies Since is reduced, 1 3 2 = 0. 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 ≥ 2. Based on Proposition 6, one may think that may also be nil-symmetric for ≥ 4, but the following example nullifies that possibility.
Then ( ) forms a subring of 4 .    We also observe that every right (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 , 3 ( ) is a nilsemicommutative ring which is neither a right nil-symmetric ring nor a left nil-symmetric ring. This can be verified as follows.
We have By [6, Example 2.4], 4 is weak-Armendariz. By Example 7, 4 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 nil(∏ =1 ) = ∏ =1 nil( ) Since the class of right (left) nil-symmetric rings is closed under subrings, therefore, for any right (left) nil-symmetric ring and for any 2 = ∈ , is a right (left) nilsymmetric ring. The converse is, however, not true, in general as shown by the following example. 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
= 0 implies = 0 ( = 0 implies = 0) for any two nonempty subsets , of ( ) and any subset of .
Proof. Let be a right nil-symmetric ring and let , be nonempty subsets of nil( ); let be a nonempty subset of such that = 0. Then = 0 for all ∈ , ∈ , ∈ . Right nil-symmetric property of gives = 0 for all ∈ , ∈ , ∈ . Thus = 0. 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 ∈ , = 0 implies that = 0.
Proposition 22. For a semiprime ring , the following statements are equivalent.
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.
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 nilsymmetric. However, the following example eliminates the possibility.  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, = 3 ( ) is not nil-symmetric but nilsemicommutative. 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.

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 nilsemicommutative. 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.
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.