QUASI-ARMENDARIZ PROPERTY ON POWERS OF COEFFICIENTS

The study of Armendariz rings was initiated by Rege and Chhawchharia, based on a result of Armendariz related to the structure of reduced rings. Armendariz rings were generalized to quasi-Armendariz rings by Hirano. We introduce the concept of power-quasi-Armendariz (simply, p.q.Armendariz) ring as a generalization of quasi-Armendariz, applying the role of quasi-Armendariz on the powers of coefficients of zero-dividing polynomials. In the process we investigate the power-quasi-Armendariz property of several ring extensions, e.g., matrix rings and polynomial rings, which have roles in ring theory. Mathematics Subject Classification (2010): 16U80


Introduction
Throughout this note every ring is associative with identity unless otherwise specified.Given a ring R, J(R), N * (R) and N (R) denote the Jacobson radical, the upper nilradical (i.e., sum of all nil ideals) and the set of all nilpotent elements in R, respectively.It is well-known that N * (R) ⊆ J(R) and N * (R) ⊆ N (R).We use R[x] to denote the polynomial ring with an indeterminate x over given a ring R. For f (x) ∈ R[x], let C f (x) denote the set of all coefficients of f (x).Z (resp., Z n ) denotes the ring of integers (resp., the ring of integers modulo n).Denote the n by n full (resp., upper triangular) matrix ring over a ring R by M at n (R) (resp., U n (R)) for n ≥ 2. Next let D n (R) be the subring {m ∈ U n (R) | the diagonal entries of m are all equal} of U n (R), N n (R) = {(a ij ) ∈ D n (R) | a ii = 0 for all i}, and a (i+1)(j+1) for i = 1, . . ., n−2 and j = 2, . . ., n−1}.
Note that V n (R) ∼ = R[x]/(x n ), where (x n ) is the ideal of R[x] generated by x n .Use E ij for the matrix with (i, j)-entry 1 and other entries 0.
For a ring R and an (R, R)-bimodule M , the trivial extension of R by M is the ring T (R, M ) = R ⊕ M with the usual addition and the following multiplication: . This is isomorphic to the ring of all matrices r m 0 r , where r ∈ R and m ∈ M and the usual matrix operations are used.
A ring is called reduced if it has no nonzero nilpotent elements.Rege and Chhawchharia [15] called a ring R (not necessarily with identity) Armendariz if based on [ Semiprime rings are quasi-Armendariz rings by [7,Corollary 3.8], but not conversely in general.
On the other hand, Han et al. [6] called a ring R (not necessarily with identity) .
The class of quasi-Armendariz rings and the class of power-Armendariz rings do not imply each other by Example 2.1 to follow.

Power-quasi-Armendariz rings
We first consider the following condition ( †): There exist m, n ≥ 1 such that where R is a ring, not necessarily with identity.
It is obvious that a m Rb n = 0 for some m, n ≥ 1 if and only if a Rb = 0 for some ≥ 1, in the condition ( †) above.Quasi-Armendariz rings clearly satisfy the condition ( †), but each part of the following example shows that the class of rings satisfying the condition ( †) need not be quasi-Armendariz or power-Armendariz.
(2) Consider a ring R = M at n (A) where A is a quasi-Armendariz ring and n ≥ 2. Then R is quasi-Armendariz by [7, Theorem 3.12] and so it satisfies the condition ( †), but not power-Armendariz by [6, Example 1.5 (1)].
Based on the above, we will call a ring R (not necessarily with identity) powerquasi-Armendariz (shortly, p.q.-Armendariz) if it satisfies the condition ( †).Hence, the concept of p.q.-Armendariz ring is a generalization of a quasi-Armendariz ring.
Due to Lambek [13], an ideal I of a ring R is called symmetric if abc ∈ I implies acb ∈ I for all a, b, c ∈ R. If the zero ideal of a ring R is symmetric then R is called symmetric.Following Bell [4], a ring R is called to satisfy the Insertion-of-Factors-Property (simply, an IFP ring) if ab = 0 implies aRb = 0 for a, b ∈ R. Note that N (R) = N * (R) for an IFP ring R by [16,Theorem 1.5].Reduced rings are symmetric and symmetric rings are IFP, and a simple computation yields that IFP rings are Abelian.We see that D 3 (R) is an IFP ring and D n (R) is not IFP for n ≥ 4 in [11], where R is a reduced ring.
Recall that a ring R is called almost symmetric [16] if R is IFP and satisfies the following condition: (2) The class of p.q.-Armendariz rings is closed under direct sum.
Then there exist m, n ≥ 1 such a m Rb n = 0 for any a (2) Let R u be p.q.-Armendariz rings for all u ∈ U and We apply the proof of [6, Proposition 1.1 (1)].Note that f (x) and g(x) can be rewritten by -Armendariz for all u ∈ U .Then there exists h ≥ 1 such that [a(i) u ] h [b(j) u ] h = 0 for all i, j, u.This implies that (a(i) u ) h E(b(j) u ) h = 0 for all i, j, showing that E is p.q.-Armendariz.
(3) Let R be an almost symmetric ring.Then N (R) = N * (R).Suppose that We use R and r to denote R/N (R) and r + N (R), respectively.Since R/N (R) is reduced (hence quasi-Armendariz) and ( āi x i ) R( bj x j ) = 0, we have aRb ⊆ N (R) for any a ∈ C f (x) and b ∈ C g(x) .Then (aRb) n = 0 and so (ab) n = 0 for some n ≥ 1.Since R is almost symmetric, a l b l = 0 and so a l Rb l = 0 for some l ≥ 1.Thus R is p.q.-Armendariz.
(4) is simply checked through a simple computation.Proof.We adapt the proof of [6,Theorem 1.11(4)].Let f (x)Rg(x) = 0 for Since R/I is p.q.-Armendariz, there exists s ≥ 1 such that . By the same computation as in the proof of [6, Theorem 1.11( 4)], we have aIb = 0 for any a ∈ C f (x) and b ∈ C g(x) , and thus and hence a s+1 Rb s+1 = 0. Therefore R is p.q.-Armendariz.
We actually do not know whether M at n (R) (resp., R is a p.q.-Armendariz ring. Question.If R is a p.q.-Armendariz ring, then is M at n (R) (resp., U n (R)) p.q.-Armendariz?
But we find the following kinds of subrings of M at n (R) which preserve the p.q.
Theorem 2.7.Let R be an IFP ring and n ≥ 2. The following conditions are equivalent: (1) R is p.q.-Armendariz.
(2) D n (R) is p.q.-Armendariz. ( , where A i = (a(i) cd ) and B j = (b(j) hk ) for 0 ≤ i ≤ s and 0 ≤ j ≤ t.The proof is similar to one of [6,Theorem 1.4(1)], but we write it here for completeness.
Note that f (x) and g(x) can be expressed by , where Since f (x)D n (R)g(x) = 0, f 11 (x)Rg 11 (x) = 0 and so there exist w ≥ 1 such that a(i) w 11 Rb(j) w 11 = 0 for all i, j since R is p.q.-Armendariz.Next note that every sum-factor of each entry of A wn i (resp., B wn j ) contains a(i) w 11 (resp., b(j) w 11 ) in its product by [9,Lemma 1.2(1)].Now since R is IFP, we get A wn i RB wn j = 0 because every sum-factor in each entry of A wn i RB wn j is of the form sa(i) w 11 tb(j) w 11 u = 0, for any s, t, u ∈ R. ( for any a ∈ C f (x) and b ∈ C g(x) .In particular, for any r ∈ R, we get (a . Therefore R is p.q.-Armendariz. (1)⇔( 3) and ( 1)⇔( 4) can be obtained by the same argument as in the proof of (1)⇔(2). The (S II) ab m c m = 0 for some positive integer m whenever a(bc) n = 0 for given n ≥ 1 and a, b, c ∈ R. Symmetric rings are almost symmetric, but not conversely by [16, Proposition 1.4 and Example 5.1], and almost symmetric rings are obviously IFP, however the class of IFP rings and the class of rings satisfying the condition (S II) are independent of each other by [16, Example 5.1(c) and Example 5.2(b)].Symmetric rings are power-Armendariz by [6, Proposition 1.1(4)].Proposition 2.2.(1) If R is a p.q.-Armendariz ring, then so is eRe for 0 = e 2 = e ∈ R.

Corollary 2 . 3 .Example 2 . 4 .Proposition 2 . 5 .
Let e be a central idempotent of a ring R. Then R is p.q.-Armendariz if and only if eR and (1 − e)R are both p.q.-Armendariz.Proof.It follows from Proposition 2.2(1,2), since R ∼ = eR ⊕ (1 − e)R.The ring R = U 2 (D) for a domain D is quasi-Armendariz by [7, Corollary 3.15] and hence R is p.q.-Armendariz, but not IFP.Let R be a ring and I be a proper ideal of R. If R/I is a p.q.-Armendariz ring and I is reduced as a ring without identity, then R is p.q.-Armendariz.

( 2 )
Let R be an IFP ring.If R is p.q.-Armendariz, then so is R[x].Proof.(1) Suppose that R[x] is a p.q.-Armendariz ring.Let f (x)Rg(x) = 0 for f (x), g(x) ∈ R[x].Let y be an indeterminate over R[x].Then f (y)Rg(y) = 0 and so f (y)R[x]g(y) = 0 for f (y), g(y) ∈ R[x][y], since x commutes with y.By hypothesis, there exist s, t ≥ 1 such that a s R[x]b t = 0 for any a ∈ C f (y) and following result comes from Theorem 2.7 and Proposition 2.2(3).Notice that a regular ring R is power-Armendariz if and only if R is Armendariz if and only if R is Abelian if and only if R is reduced by help of [6, Theorem 1.8].However, there exists a von Neumann regular p.q.-Armendariz ring but not reduced, by considering M at 2 (D) with D a division ring in Example 2.9.Theorem 2.10.(1) If R[x] is a p.q.-Armendariz ring, then so is R.
[5]ollary 2.8.If R is an almost symmetric ring, then D n (R) is p.q.-Armendariz for any n ≥ 2.Recall that a ring R is called directly finite if ba = 1 whenever ab = 1 for a, b ∈ R. Abelian rings are directly finite and power-Armendariz rings are Abelian by [6, Proposition 1.1(5)].However, there exists a p.q.-Armendariz ring which is not directly finite (hence non-Abelian) by the following.Example 2.9.There exists a domain (hence p.q.-Armendariz) D such that R = M at 2 (D) is not directly finite by [14, Theorem 1.0].Then R is quasi-Armendariz by [7, Theorem 3.12], and so it is p.q.-Armendariz.But R is non-Abelian obviously.A ring R is called (von Neumann) regular if for each a ∈ R there exists b ∈ R such that a = aba.in[5].