ON CNZ RING PROPERTY VIA IDEMPOTENT ELEMENTS

. In this paper, the concept of e − CNZ rings is introduced as a generalization of symmetric rings and a particular case of e − reversible rings. Regarding the question of how idempotent elements affect CNZ property of rings. In this note, we show that e − CNZ is not left-right symmetric. We present examples of right e − CNZ rings that are not CNZ and basic properties of right e − CNZ are provided. Some subrings of matrix rings and some extensions of rings such as Jordan extension are investigated in terms of right e − CNZ.


INTRODUCTION
Throughout this paper, all rings are noncommutative with identity and associative unless otherwise stated. For a ring R, let Id(R), Z(R) and N(R) denote the set of all idempotents, the center and the set of all nilpotents of R, respectively. We denote the ring of integers (resp., modulo n) by Z(resp., Z n ). We write Mat n (R) (resp., U n (R)) for the ring of all n by n full matrix (resp., upper triangular matrix) over R, and D n (R)stands for the subring of the Laurent polynomial ring, the power series ring and the Laurent power series ring over R, respectively. We use E i j for the matrix with (i, j) − entry 1 and elsewhere 0.
A ring is called reduced if it has no nonzero nilpotent elements. A ring R is called CNZ [1] if ab = 0 implies ba = 0 for a, b ∈ N(R). The ring R with e ∈ Id(R) is called left e − reversible (resp., right e − reversible) if ab = 0 implies eba = 0 (resp., bae = 0) for any a, b ∈ R, and R is e − reversible [13] if it is both left and right e − reversible. The ring R is called left (resp., right) e − reduced [16] if eN(R) = 0 (resp., N(R)e = 0), and also R is called e − symmetric [16] if abc = 0 for all a, b, c ∈ R implies acbe = 0. The ring R is von Neumann regular [6] if for each a ∈ R, there is a b ∈ R such that a = aba. An element r of a ring R is central if ar = ra for all a ∈ R, and R is said to be abelian [2] if every idempotent is central. Also an idempotent e of R is called right (resp., left) semicentral [4] if for each a ∈ R, ea = eae (resp., ae = eae). R is called left (resp. right) e − semicommutative [14] if ab = 0 implies eaRb = 0 (resp. aRbe = 0) for any a, b ∈ R, and R is e − semicommutative if it is both left and right e − semicommutative.
The ring R is reflexive [10] ring if aRb = 0 then bRa = 0 for any a, b ∈ R.

PROPERTIES OF RIGHT e − CNZ RINGS
In this section we deal with the basic properties of right e − CNZ rings. Being with the following definition.
Definition 2.1. Let R be a ring and e ∈ Id(R) with e = 0. Then R is called right e − CNZ (resp., left e − CNZ) if for any a, b ∈ N(R), ab = 0 implies bae = 0 (resp., eba = 0). The ring R is e − CNZ if it is both left and right e − CNZ.
It is obvious that a ring R is CNZ if and only if R is 1 − CNZ. The following example shows that e − CNZ property is not left-right symmetric. Also, the CNZ property of a ring with respect to an idempotent e depends on e. There are rings R and idempotents e 1 and e 2 such that R is right e 1 − CNZ but not right e 2 − CNZ as the following example shows. (1) R is not CNZ.
Let A, B ∈ N(R) with AB = 0. Then BA is of the form In the next result, we give a characterization of right e − CNZ property in terms of subsets of rings and its proof is straightforward. (1) R is right e − CNZ.
(3) Every right e − reduced ring is right e − CNZ.
Proof. (1) Let R be a CNZ ring with e 2 = e ∈ R. Suppose that ab = 0 for a, b ∈ N(R) . Then ba = 0 and so bae = 0 and eba = 0.
(2) Let R be an e − reversible ring with e 2 = e ∈ R. Suppose that ab = 0 for a, b ∈ N(R).
Since N(R) ⊆ R and R is e − reversible then bae = 0 and eba = 0. (4) Assume that R is an e − symmetric ring. Let a, b ∈ N(R) with ab = 0. R is with identity 1ab = 0. By assumption 1bae = 0 implies bae = 0. Therefore R is right e − CNZ.
There are right e−CNZ rings R for some e ∈ Id(R) but not CNZ. This yields that the converse The following properties of right e − CNZ rings do an essential roles throughout this paper. Proposition 2.6.
(1) The class of right e − CNZ ring is closed under subrings.
(2) For a family {R λ : λ ∈ ∆} of rings, the following statements are equivalent: Thus S is right e − CNZ.
Lemma 2.7. The following are equivalent for a ring R and e ∈ Id(R): (i) R is a CNZ ring.  We get that ba = 0 as R is an e − CNZ ring. Hence R is a CNZ ring. Proof. Let R be a right e − CNZ ring with (exe)(eye) = 0, for x, y ∈ N(R). By assumption (eye)(exe) = 0 . Thus eRe is CNZ.
Similar to Theorem 2.8., we have the following result.  (1) R is right e − CNZ.
It is not necessary for the ring R to be right e − CNZ even if the ring R/I is rightē − CNZ.
We illustrate this by the following example. Proposition 2.12. Let R be a ring with an ideal I and e ∈ Id(R) with e is central. If R/I is a rightē − CNZ ring and Iis reduced as a ring without identity, then R is right e − CNZ.
Proposition 2.13. Let R be an e − symmetric ring and I an ideal of R with I = r R (S) for some subset S of N(R). Then R/I is rightē − CNZ.
Proposition 2.14. Every right e − semicommutative reflexive ring is right e − CNZ. Proof. Let a, b ∈ N(R) with ab = 0. There exists an c ∈ R such that a = aca. Multiply this equality by b from left, we get ba = b(aca). Also, ca is idempotent. Since R is abelian, we have bae = (ca) bae = c (ab) ae = 0. Therefore R is a right e − CNZ ring.

RELATIONS TO RIGHT e − CNZ RINGS
In this section we study the related rings and the extension rings of right e − CNZ rings, concentrating on matrix rings, polynomial rings, Jordan extension and some other kinds of extensions.
For a reduced ring R, we now show that Mat n (R) is neither right E − CNZ nor left E − CNZ for some E ∈ Id(Mat n (R)).
Example 3.1. Let R be a reduced and E i j denote the matrix unit in Mat n (R) whose (i, j) − th entry is 1 and the others are zero. Consider A = E 23 , B = E 12 in N(Mat n (R)) and E = E 11 + E 33 ∈ Id(Mat n (R)). Then AB = 0 and BA = 0. Also, BAE = 0 and EBA = 0. Thus Mat n (R) is If R is a reduced ring, then both U 2 (R) and D 2 (R) are e − CNZ for every idempotent e of R The next example shows that the ring U 2 (R) is not e − CNZ when we replace the condition "R is reduced" with condition "R is e − CNZ".
For a reduced ring R, D 3 (R) need not be CNZ by the same argument as in the proof as noted in [1, Remark 2.6(2)]. We note also the following.
In case n ≥ 3, the ring R being reduced and e ∈ Id(R) need not imply D n (R) being eI n −CNZ as illustrated below.  is right E 22 − CNZ ring.
For any ring R, U n (R), D n (R) need not be e − CNZ for n ≥ 3 by the following example. For any ring R and n ≥ 2, V n (R) is the subring of Mat n (R).
. . a n−1 a n . . . a n−2 a n−1 . . . a n−3 a n−2 . . . . . . . . .  The condition "R is reduced" is not superfluous. By the following example. A ring R is Armendariz [18] if whenever any polynomials And also a ring R is called power-serieswise   α]] with respect to the Ore set x j j≥0 , also when α is an automorphism of R, it consists elements of the form x s a s + x s+1 a s+1 + · · · + a 0 + a 1 x + · · · , for a i ∈ R and integers s ≤ 0 and i ≥ s, with usual addition and the multiplication is defined by xa = α(a)x for any a ∈ R.  (1) R is right e − CNZ for each e ∈ Id(R).
An element a of a ring R is called right regular if ac = 0 implies c = 0 for c ∈ R. In the same way, left regular is defined and it is regular if it is both left and right regular (and hence non-zero divisor).
Note that: Id(∆ −1 R) = {u −1 e : e ∈ Id(R) and u ∈ ∆} Theorem 3.13. Let R be a ring, ∆ be a multiplicatively closed subset of R consisting of central regular elements, 1 ∈ ∆ and e ∈ Id(R). Then R is right e − CNZ if and only if ∆ −1 R is right Note that: N(∆ −1 R) = ∆ −1 N(R).
Proof. Assume that R is right e − CNZ, and let a, b ∈ N(R), s,t ∈ ∆ such that (s  Let R be a ring and α a monomorphism of R. Now we consider the Jordan's construction of an over-ring of R by α [9]. Let A(R, α) be the subset x −i rx i : r ∈ R and i ≥ 0 of the skew Laurent with the natural operations that follow: x −i rx i +x − j sx j = x −(i+ j) (α j (r)+ α i (s))x i+ j and (x −i rx i )(x − j sx j ) =x −(i+ j) α j (r)α i (s)x i+ j for i, j ≥ 0 and r, s ∈ R.
Note that A(R, α) is an over-ring of R, andᾱ : A(R, α) → A(R, α) defined byᾱ(x −i rx i ) = x −i α(r)x i is an automorphism of A(R, α). Jordan demonstrated, with the use of left localization of the skew polynomial R[x; α] with respect to the set of powers of x, that for any pair (R, α), such an extension A(R, α) always exists in [9]. This ring A(R, α) is usually called the Jordan extension of R by α. Note that: Id(A) = x −i rx i : r ∈ Id(R) and i ≥ 0 . Proof. It is enough to show the necessity by Proposition 2.6 (1). Suppose that R is right e−CNZ and ab = 0 for a = x −i rx i , b = x − j sx j ∈ N(A) for i, j ≥ 0 and for x −k ex k ∈ Id(A), e ∈ Id(R).

Thus the Jordan extension
14 IMAN JALAL ALI, CHENAR ABDUL KAREEM AHMED Let R be an algebra over a commutative ring C. Due to Dorroh [5], consider the abelian group R ⊕ C with multiplication defined by (a, b)(c, d) = (ac + da + bc, bd) where a, c ∈ R, b, d ∈ S. By this operation R ⊕Cbecomes a ring called Dorroh extension of R by C and denoted by D(R,C). By definition, C is isomorphic to a subring of R. By this reason we may assume that C is contained in the center of R and use this fact in the sequel.  Thus bae = 0. Therefore R is right e − CNZ.

RELATED TOPICS
As consequences of our observation in our previous sections, we introduce further results related to our main concept.
Let T and S be any rings. Take M as (T, S) − bimodule and R the formal triangular matrix (1) T is a right e − CNZ ring,    (1) T [R, S] is a right E − CNZ ring.    The rings H (x,y) (R) [7]: Let R be a ring, and let x, y ∈ Z(R) be invertible in R.