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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(2): 307-318

Published online June 30, 2020

### Rings which satisfy the Property of Inserting Regular Elements at Zero Products

Hong Kee Kim, Tai Keun Kwak*, Yang Lee, Yeonsook Seo

Department of Mathematics and RINS, Gyeongsang National University, Jinju 52828, Korea
e-mail : hkkim@gsnu.ac.kr
Department of Mathematics, Daejin University, Pocheon 11159, Korea
e-mail : tkkwak@daejin.ac.kr
Department of Mathematics, Yanbian University, Yanji 133002, China and Institute of Basic Science, Daejin University, Pocheon 11159, Korea
e-mail : ylee@pusan.ac.kr
Department of Mathematics, Pusan National University, Busan 46241, Korea
e-mail : ysseo0305@pusan.ac.kr

Received: October 22, 2019; Revised: January 21, 2020; Accepted: February 10, 2020

### Abstract

This article concerns the class of rings which satisfy the property of inserting regular elements at zero products, and rings with such property are called regular-IFP. We study the structure of regular-IFP rings in relation to various ring properties that play roles in noncommutative ring theory. We investigate conditions under which the regular-IFPness pass to polynomial rings, and equivalent conditions to the regular-IFPness.

Keywords: regular-IFP ring, regular element, IFP ring, polynomial ring, generalized reduced ring

### 1. Introduction

Throughout this article every ring is an associative ring with identity. Let R be a ring. An element u of R is right regular if ur = 0 implies r = 0 for rR. A left regular element is defined similarly. An element is regular if it is both left and right regular (and hence not a zero divisor). We use C(R) and U(R) to denote the monoid of regular elements and the group of units in R, respectively. The Wedderburn radical (i.e., sum of all nilpotent ideals), the upper nilradical (i.e., the sum of all nil ideals), the lower nilradical (i.e., the intersection of all prime ideals), the Jacobson radical, and the set of all nilpotent elements in R are denoted by N0(R), N*(R), N*(R), J(R), and N(R), respectively. It is well-known that N0(R) ⊆ N*(R) ⊆ N*(R) ⊆ N(R) and N*(R) ⊆ J(R). The polynomial ring with an indeterminate x over R is denoted by R[x] and Cf(x) denotes the set of all coefficients of f(x) for f(x) ∈ R[x]. Denote the n by n full (resp., upper triangular) matrix ring over R by Mn(R) (resp., Tn(R)). Dn(R) denotes the subring {(aij) ∈ Tn(R)|a11 = ⋯ = ann} of Tn(R). Let In and Eij be the identity matrix and the matrix, with (i, j)-entry 1 and elsewhere 0, in Mn(R), respectively. Let ℤ (resp., ℤn) denote the ring of integers (modulo n), and ℝ denote the field of real numbers.

### 2. Regular-IFP Rings

In this section we study the properties of regular-IFP rings as well as the relations between regular-IFP rings and ring properties that play important roles in noncommutative ring theory. Due to Bell [4], a ring is called IFP if ab = 0 for a, bR implies aRb = 0. Following Kim et al. [15], a ring R is called unit-IFP if ab = 0 for a, bR implies aU(R)b = 0. IFP rings are clearly unit-IFP, but not conversely by [15, Example 1.1]. A ring R is usually called reduced if N(R) = 0. Commutative rings are clearly IFP; and reduced rings are easily shown to be IFP, but not conversely because there exist many non-reduced commutative ring. A ring is usually called Abelian if every idempotent is central. Unit-IFP rings are Abelian by [15, Lemma 1.2(2)].

### Definition 2.1

A ring R is called regular-IFP if ab = 0 for a, bR implies aC(R)b = 0.

Regular-IFP rings are clearly unit-IFP (hence Abelian), but not conversely by the following example.

### Example 2.2

There exists a unit-IFP ring that is not regular-IFP. Let K be a field and A = Ka, b⟩ be the free algebra generated by the noncommuting indeterminates a, b over K. Let I be the ideal of A generated by b2 and set R = A/I. Identify a, b with their images in R for simplicity. Then R is unit-IFP by [15, Example 1.1]. But R is not a regular-IFP ring because b2 = 0 and bab ≠ 0 where aC(R).

### Remark 2.3

• (1) The following conditions are equivalent, which can be proved by applying the regular-IFPness iteratively:

• (i) A ring R is regular-IFP;

• (ii) a1C(R)a2C(R)a3an−1C(R)an = 0 whenever a1a2an = 0 for a1, a2, …, anR.

• (2) D3(R) is (regular-)IFP over a reduced ring R by [15, Proposition 2.1]. However Mn(R) and Tn(R), over any ring R for n ≥ 2, cannot be regular-IFP since they are not Abelian, noting that unit-IFP (or regular-IFP) rings are Abelian.

• (3) There exists an Abelian ring that is not regular-IFP. Set R = Dn(S) for n ≥ 4 over an Abelian ring S. Then R is Abelian by [10, Lemma 2]. Let A = E12, B = E34R. Then AB = 0. Consider C = In + E23R. Then CC(R) clearly. But ACB = E14 ≠ 0, so that R is not regular-IFP.

• (4) Let R be a regular-IFP ring such that R = C(R) ∪ N(R). Let ab = 0 for a, bR. Then aC(R)b = 0 since R is regular-IFP. Moreover aN(R)b = 0 by [15, Lemma 1.2(2)]. Thus R is IFP.

Based on Armendariz [3, Lemma 1], a ring R is called Armendariz if ab = 0 for all aCf(x) and bCg(x) whenever f(x)g(x) = 0 for f(x), g(x) ∈ R[x], by Rege and Chhawchharia [19]. Reduced rings are Armendariz by [3, Lemma 1]. The concepts of Armendariz rings and commutative rings are independent of each other by Example 2.2 and [19, Example 3.2], noting that the ring R in Example 2.2 is Armendariz by [2, Example 4.8].

By Goodearl [7], a ring R is called (von Neumann) regular if for every aR there exists bR such that a = aba, and a ring R is called strongly regular if aa2R for every aR. It is easily checked that J(R) = 0 for every regular ring R, and note that a ring is strongly regular if and only if it is Abelian regular, by [7, Theorems 3.2 and 3.5]. Recall that unit-IFP rings are Abelian, and Armendariz rings are also Abelian by [12, Corollary 8]. So for a regular ring R, we have that R is reduced if and only if R is Armendariz if and only if R is IFP if and only if R is regular-IFP if and only if unit-IFP if and only if R is Abelian, by [7, Theorem 3.2].

Following [11], a ring is called locally finite if every finite subset generates a finite multiplicative semigroup. Finite rings are clearly locally finite, but not conversely by the existence of algebraic closures of finite fields. It is shown that a ring is locally finite if and only if every finite subset generates a finite subring, in [11, Theorem 2.2(1)].

### Proposition 2.4

• (1) Let R be a locally finite ring.

• (i) If R is an Armendariz ring, then it is regular-IFP.

• (ii) If R is a regular-IFP ring, then R/J(R) is strongly regular with J(R) = N(R).

• (2) Let R be a right or left Artinian ring. If R is regular-IFP, then R/J(R) is a strongly regular ring with J(R) = N(R).

Proof

(1)–(i) Let R be an Armendariz ring and suppose ab = 0 for a, bR. Let cC(R). Since R is locally finite, cnI(R) for some n ≥ 1 by the proof of [12, Proposition 16]. But cnC(R), forcing cn = 1. This yields acnb = 0, and so acb = 0 by [12, Lemma 7]. This implies aC(R)b = 0, and hence R is regular-IFP.

(1)–(ii) Since regular-IFP rings are Abelian, we obtain the result by [11, Proposition 2.5].

(2) It is well-known that J(R) is nilpotent for the right (or left) Artinian ring R. Since R/J(R) is semisimple Artinian and Abelian, R/J(R) is a finite direct product of division rings. This completes the proof.

The class of regular-IFP rings is not closed under homomorphic images as can be seen by the ring R in Example 2.2. But the following constructions preserve the regular-IFPness. We use ⊕ and Π to denote the direct sum and the direct product of rings, respectively.

### Proposition 2.5

• (1) Let Rλ ∈ Λ) be Abelian rings. Then Rλ is regular-IFP for each λ ∈ Λ if and only if Πλ∈ΛRλ is regular-IFP if and only if the subring of Πλ∈ΛRλ generated byλ∈ΛRλ and 1Πλ∈ΛRλis regular-IFP.

• (2) Let R be an Abelian ring and e2 = eR. Then R is regular-IFP if and only if both eR and (1 − e)R are regular-IFP.

Proof

(1) Suppose that the subring of Πλ∈ΛRλ generated by ⊕λ∈ΛRλ and 1Πλ∈ΛRλ, say S, is regular-IFP. Let ab = 0 for a, bRλ, and cC(Rλ). Let α = (xi) and β = (yj) be sequences in S such that xλ = a, xi = 0 for all iλ, and yλ = b, yj = 0 for all jλ. Then αβ = 0. Consider a sequence δ = (zk) ∈ S in which zλ = c and zk = 1Rk for all kλ. Then δC(S). Since S is regular-IFP, we have αδβ = 0. This yields acb = 0, and so Rλ is regular-IFP. The remainder of the proof is routine.

(2) The proof is obtained from (1) since R = eR ⊕ (1 − e)R for e2 = eR.

For a given ring R, recall that R is called local if R/J(R) is a division ring; R is called semilocal if R/J(R) is semisimple Artinian; and R is called semiperfect if R is semilocal and idempotents can be lifted modulo J(R). Local rings are clearly Abelian and semilocal.

### Corollary 2.6

A ring R is semiperfect regular-IFP if and only if R is a finite direct product of local regular-IFP rings.

Proof

Suppose that R is regular-IFP and semiperfect. Since R is semiperfect, R has a finite orthogonal set {e1, e2, …, en} of local idempotents whose sum is 1 by [18, Proposition 3.7.2], i.e., each eiRei is a local ring. Since R is regular-IFP, R is Abelian and so eiR = eiRei for each i. This implies $R=∑i=1neiR$. Then each eiR is also a regular-IFP ring by Proposition 2.5.

Conversely assume that R is a finite direct product of local regular-IFP rings. Then R is semiperfect since local rings are semiperfect by [18, Corollary 3.7.1], and moreover R is regular-IFP by Proposition 2.5.

Recall that homomorphic images of regular-IFP rings need not be regular-IFP. Considering this fact, one may ask whether a ring R is regular-IFP when every homomorphic image of R is regular-IFP. But the following provides a negative answer.

### Example 2.7

There exists a non-regular-IFP ring R whose factor rings are regular-IFP. Consider R = T2(D) over a division ring D. Then every non-trivial factor ring is one of R/J(R) ≅ DD, R/ID and R/KD, where $J(R)=(0D00),I=(DD00),K=(0D0D)$. These factor rings are reduced and so (regular-)IFP. But R cannot be regular-IFP because R is non-Abelian.

### 3. Extensions of Regular-IFP Rings

In this section we examine the regular-IFP property of ring extensions that play roles in noncommutative ring theory.

Regarding Remark 2.3(3), we have the following.

### Proposition 3.1

For a ring R the following conditions are equivalent:

• (1) R is a reduced ring;

• (2) D3(R) is an IFP ring;

• (3) D3(R) is a regular-IFP ting;

• (4) D3(R) is a unit-IFP ring;

• (5) AN(D3(R))B = 0 whenever AB = 0 for A, BD3(R).

Proof

The equivalences of the conditions (1), (2), and (4) are proved by [15, Proposition 2.1], and so they are equivalent to (3).

(4) ⇒ (5): Suppose that (4) holds and let CN(D3(R)). Then I3CU(D3(R)), where I3 denotes the identity matrix in D3(R). If AB = 0 for A, BD3(R), then A(I3C)B = 0 by assumption since I3CU(D3(R)), implying that ACB = 0.

(5) ⇒ (1): Suppose that (5) holds. Assume on the contrary that there exists 0 ≠ aR with a2 = 0. We refer to the argument in the proof of [14, Proposition 2.8]. Consider two matrices

$A=(aa-10a-100a) and B=(a0a0a100a)$

in D3(R). Then AB = 0, but AE12B = aE13 ≠ 0 for E12N(D3(R)), which contradicts (5). Thus R is reduced.

Following Cohn [6], a ring R is called reversible if ab = 0 for a, bR implies ba = 0. It is easily checked that reduced rings are reversible and reversible rings are IFP. The condition “R is a reduced ring” in Proposition 3.1 cannot be weaken by the condition “R is a reversible ring” by next example.

### Example 3.2

We refer to the construction and argument in [16, Example 2.1]. Let

$A=ℤ2⟨a0,a1,a2,b0,b1,b2,c⟩$

be the free algebra generated by noncommuting indeterminates a0, a1, a2, b0, b1, b2, c over ℤ2. Next, let I be the ideal of A generated by

$a0b0,a0b1+a1b0,a0b2+a1b1+a2b0,a1b2+a2b1,a2b2,a0rb0,a2rb2,b0a0,b0a1+b1a0,b0a2+b1a1+b2a0,b1a2+b2a1,b2a2,b0ra0,b2ra2,(a0+a1+a2)r(b0+b1+b2),(b0+b1+b2)r(a0+a1+a2),and r1r2r3r4,$

where the constant terms of r, r1, r2, r3, r4A are zero. Now set R = A/I. We identity

$a0,a1,a2,b0,b1,b2,c$

with their images in R for simplicity. Then R is reversible by [16, Example 2.1] but not reduced clearly.

Now, consider

$A=(a0a100a0000a0),B=(b0b100b0000b0)∈D3(R).$

Then AB = 0. But

$ACB=(a0a100a0000a0) (c000c000c) (b0b100b0000b0)=(0a0cb1+a1cb00000000)≠0$

because a0cb1 + a1cb0I, noting $C=(c000c000c)∈N(D3(R))$. Thus D3(R) does not satisfy the condition (5) of Proposition 3.1.

### Remark 3.3

• (1) Note that D2(R) over a reduced ring R is IFP by [16, Proposition 1.6] and so it is regular-IFP. Moreover, there exists a non-reduced non-commutative reversible ring R over which D2(R) is regular-IFP by [15, Example 2.2]. However, the ring S is always regular-IFP when D2(S) is regular-IFP. For, suppose that D2(S) is regular-IFP and let ab = 0 for a, bS. For $A=(a00a),B=(b00b)∈D2(S)$, we have AB = 0 and so AC(D2(S))B = 0 by assumption. Set $C=(c00c)$ for any cC(S). Then CC(D2(S)) and ACB = 0, entailing acb = 0. Thus S is regular-IFP.

• (2) Related to (1) above, there exists a reversible ring R such that D2(R) is not regular-IFP. Let ℍ be the Hamilton quaternions over ℝ and R = D2(ℍ). Then R is reversible [16, Proposition 1.6]. We refer to the argument in [16, Example 1.7]. Consider

$A=((0i00)(j00j)(0000)(0i00)) and B=((0100)(k00k)(0000)(0100))$

in D2(R). Then AB = 0.

Note that

$C=((j00j)(0000)(0000)(j00j))∈C(D2(R))$

by [13, Lemma 2.1] because $(j00j)∈C(D2(ℍ))$. But ACB ≠ 0, hence D2(R) is not regular-IFP.

• (3) For a ring R and n ≥ 2, let Vn(R) be the ring of all matrices (aij) in Dn(R) such that ast = a(s+1)(t+1) for s = 1, …, n − 2 and t = 2, …, n − 1. Note that $Vn(R)≅R[x]xnR[x]$. If R is a reduced ring, then Vn(R) is (regular)-IFP by [17, Lemma 2.3 and Proposition 3.3], but the converse does not hold in general as can be seen by the commutative ring Vn(R) over a non-reduced commutative ring (e.g., ℤnl for n, l ≥ 2) R for n ≥ 2.

### Proposition 3.4

• Let M be a multiplicatively closed subset of a ring R consisting of central regular elements. Then R is regular-IFP if and only if M−1R is regular-IFP.

• Let R be a ring. Then R[x] is regular-IFP if and only if R[x, x−1] is regular-IFP.

Proof

(1) It comes from the fact that C(M−1R) = M−1C(R).

(2) Recall the ring of Laurent polynomials in x, written by R[x, x−1]. Letting M = {1, x, x2, x3, …}, M is clearly a multiplicatively closed subset of central regular elements in R[x] such that R[x, x−1] = M−1R[x]. By (1), the proof is completed.

In [15, Example 2.7], we see an IFP ring R over which R[x] is not unit-IFP, where the ring R is constructed in [12, Example 2]. So the regular-IFPness does not pass to polynomial rings since regular-IFP rings are unit-IFP. In the following we see a condition under which the regular-IFPness is preserved by polynomial rings.

### Proposition 3.5

Let R be a ring.

• {c + c1x + … + ctxtR[x] | cC(R) for t ≥ 1} ⊆ C(R[x]) and {d0 + d1x + … + ds−1xs−1 + dxsR[x] | dC(R) for s ≥ 1} ⊆ C(R[x]).

• If R[x] is regular-IFP, then so is R.

• Let R be a regular-IFP ring such that C(R[x]) = {c + xN(R)[x] | cC(R)}. If R is Armendariz then R[x] is regular-IFP.

Proof

(1) Consider h(x) = c + c1x + … + ctxtR[x] with cC(R). Suppose that h(x)g(x) = 0 for any g(x) = b0 + b1x + ⋯ + bnxnR[x]. Then h(x)g(x) = 0 implies cb0 = 0 and so b0 = 0 since cC(R). From 0 = h(x)g(x) = (c + c1x + ⋯ + ctxt)(b1x + ⋯ + bnxn), we have cb1 = 0 and hence b1 = 0. Continuing this process, we get b2 = 0, …, bn = 0, showing that g(x) = 0. Thus h(x) ∈ C(R[x]) and so {c + c1x1 + … + ctxtR[x] | cC(R) for t ≥ 1} ⊆ C(R[x]). The proof of the latter part is similar.

(2) It is routine.

(3) Suppose that R is Armendariz. Let f(x)g(x) = 0 for f(x), g(x) ∈ R[x]. Then ab = 0 for all aCf(x) and bCg(x) since R is Armendariz. Hence aC(R)b = 0 by hypothesis. Moreover, aN(R)b = 0 by help of [15, Lemma 1.2(1)]. This implies that f(x)C(R[x])g(x) = 0, showing that R[x] is regular-IFP.

The next example shows that the condition “R is an Armendariz ring” cannot be dropped in Proposition 3.5(3).

### Example 3.6

We use the ring and argument in [12, Example 2]. Let A = ℤ2a0, a1, a2, b0, b1, b2, c⟩ be the free algebra with noncommuting indeterminates a0, a1, a2, b0, b1, b2, c over ℤ2. Let B be the set of all polynomials with zero constant terms in A. Next, consider the ideal I of A generated by

$a0b0,a1b2+a2b1,a0b1+a1b0,a0b2+a1b1+a2b0,a2b2,a0rb0,a2rb2,(a0+a1+a2)r(b0+b1+b2),r1r2r3r4$

where rA and r1, r2, r3, r4B. Set R = A/I, and identify a0, a1, a2, b0, b1, b2, c with their images in R for simplicity. Then R is (regular-)IFP but not Armendariz, by [12, Example 2] and [19, Proposition 4.6].

Notice that (a0 + a1x + a2x2)(b0 + b1x + b2x2) = 0. But

$(a0+a1x+a2x2)(1+c)(b0+b1x+b2x2)=(a0+a1x+a2x2)c(b0+b1x+b2x2)≠0$

because a0cb1 + a1cb0I, noting 1 + cC(R[x]) as in [15, Example 2.7]. Thus R[x] is not regular-IFP.

Considering Proposition 3.5, it is natural to ask whether Cf(x)C(R) ≠ ∅︀ when f(x) ∈ C(R[x]), where R is a given ring. But the answer is negative by the following.

### Example 3.7

Let A be any ring and R = A × A. Consider a polynomial f(x) = (1, 0) + (0, 1)x in R[x]. Suppose f(x)g(x) = 0 for $g(x)=∑i=0maixi∈R[x]$ with ai = (bi, ci). Then from f(x)g(x) = 0, we obtain f1(x)g1(x) = 0 and f2(x)g2(x) = 0, where

$f1(x)=1+0x,f2(x)=0+x, and g1(x)=∑i=0mbixi,g2(x)=∑i=0mcixi.$

This implies $∑i=0mbixi=0$ and $∑i=0mcixi+1=0$, so that bi = 0 and ci = 0 for all i. Thus ai = 0 for all i, and hence f(x) ∈ C(R[x]). But (1, 0), (0, 1) ∉ C(R).

We consider next some equivalent conditions to the regular-IFP property in relation to the sum of coefficients of polynomials which satisfy some property of inserting regular polynomials. For f(x) ∈ R[x], let f(1) denote the sum of all coefficients of f(x).

### Proposition 3.8

For a ring R the following conditions are equivalent:

• (1) R is regular-IFP;

• (2) If f1(x)f2(x) ⋯ fn(x) = 0 for f1(x), f2(x), …, fn(x) ∈ R[x], then the sum of all coefficients of every polynomial in

$f1(x)C(R)[x]f2(x)C(R)[x]⋯fn-1(x)C(R)[x]fn(x)$

is zero;

• (3) If f(x)g(x) = 0 for f(x), g(x) ∈ R[x], then the sum of all coefficients of every polynomial in f(x)C(R)[x]g(x) is zero;

• (4) If f(x)g(x) = 0 for linear polynomials f(x), g(x) in R[x], then the sum of all coefficients of every polynomial in f(x)C(R)[x]g(x) is zero;

• (5) f(x)g(x) = 0 implies f(x)C(R)[x]g(x) = 0 for linear polynomials f(x), g(x) in R[x].

Proof

The procedure of the proof is almost similar to one of [15, Proposition 2.8], but we write it here for completeness. (1) ⇒ (2): Assume that the condition (1) holds. Let f1(x)f2(x) ⋯ fn(x) = 0 for f1(x), f2(x), …, fn(x) ∈ R[x]. Then we have

$f1(1)f2(1)⋯fn-1(1)fn(1)=0.$

By Remark 2.3(1), we have f1(1)C(R)f2(1)C(R) ⋯ fn−1(1)C(R)fn(1) = 0. This yields that the sum of all coefficients of every polynomial in

$f1(x)C(R)[x]f2(x)C(R)[x]⋯fn-1(x)C(R)[x]fn(x)$

is zero.

(2) ⇒ (3), (3) ⇒ (4), and (5) ⇒ (1) are obvious.

(4) ⇒ (5): Assume that the condition (4) holds. Let f(x) = a0 + a1x, g(x) = b0 + b1xR[x] such that f(x)g(x) = 0. Then a0b0 = 0, a0b1 + a1b0 = 0 and a1b1 = 0. From a0b0 = 0 and a1b1 = 0, we get (a0x)(b0x) = 0 and (a1x)(b1x) = 0; hence, by (4), we have

$a0C(R)b0=0 and a1C(R)b1=0.$

From f(x)g(x) = 0, we have

$0=f(1)cg(1)=(a0+a1)c(b0+b1)=a0cb0+a0cb1+a1cb0+a1cb1=a0cb1+a1cb0$

for all cC(R) by (4) and the equalities (3.1). Therefore f(x)C(R)[x]g(x) = 0.

### 4. Related Topic

Based on Proposition 3.1(5), a ring R is called nilpotent-IFP [8] if aN(R)b = 0 whenever ab = 0 for a, bR. Every unit-IFP ring is nilpotent-IFP by the proof of Proposition 3.1 and this direction is irreversible by [15, Example 2.5]. For a unit-IFP (or regular-IFP) ring R, we have N0(R) = N*(R) = N*(R) by [15, Theorem 1.3(1)]. But there exists a unit-IFP (hence nilpotent-IFP) ring R such that N0(R) ⊊ N(R) and N(R) ⊈ J(R), by [15, Example 1.1]. The following partially extends the result of [15, Theorem 1.3].

### Theorem 4.1

For a nilpotent-IFP ring R, we have the following.

• (1) N0(R) = N*(R) = N*(R).

• (2) J(R[x]) = N0(R[x]) = N*(R[x]) = N*(R[x]) = N0(R)[x] = N*(R)[x] = N*(R)[x] = N0(R)[x] ⊆ J(R)[x]. Moreover, J(R[x]) = J(R)[x] when J(R) is nil.

Proof

(1) Let aN*(R). Then an = 0 for some n ≥ 1 and RaRN*(R) ⊆ N(R). Since R is nilpotent-IFP,

$a(RaR)a(RaR)a⋯a(RaR)a(RaR)a=0,$

and hence (RaR)2n−1 = 0 and so aN0(R). Thus we have N0(R) = N*(R) = N*(R).

(2) By help of [1, Theorem 1] and [5, Corollary 4], we have J(R[x]) ⊆ N*(R)[x] and N0(R)[x] = N0(R[x]), respectively. By (1) and the facts that N*(R[x]) ⊆ N*(R[x]), N*(R[x]) ⊆ J(R[x]), and N*(R)[x] ⊆ J(R)[x]. Thus we get

$J(R[x])⊆N*(R)[x]=N*(R)[x]=N0(R)[x]=N0(R[x])⊆N*(R[x])=N*(R[x])⊆J(R[x]),$

and so

$J(R[x])=N*(R)[x]=N*(R[x])=N0(R[x])=N*(R[x])=N*(R)[x]=N0(R)[x]⊆J(R)[x].$

Moreover, J(R[x]) = J(R)[x] since J(R) = N*(R) when J(R) is nil.

Notice that J(R[x]) is always nil in any nilpotent-IFP ring R by Theorem 4.1(2).

On the other hand, Hong et al. [9] consider the duo property on the monoid of regular elements as follows. They call a ring R right (resp., left) duo on regularity (simply, right (resp., left) DR) if C(R)aaC(R) (resp., aC(R) ⊆ C(R)a) for all aR; and a ring is called DR if it is both left and right DR. Thus it is clear that a ring R is DR if and only if C(R)a = aC(R) for all aR. A ring R is clearly DR when C(R) is contained in the center of R. Division rings are clearly DR.

### Proposition 4.2

Every one-sided DR ring is regular-IFP.

Proof

Suppose that R is a right DR ring and let ab = 0 for a, bR. Then a(C(R)b) ⊆ a(bC(R)) = 0 since R is right DR. The proof is done as desired. The proof for the case of left DR is similar to the above.

Notice that the converse of Proposition 4.2 does not hold in general by the next example.

### Example 4.3

Let R be the Hamilton quaternions over ℤ. Then R is a domain and so regular-IFP. But R is not right DR by [9, Example 2.5(1)]. Moreover R is also not left DR by a similar argument to [9, Example 2.5(1)].

Consider the group ring KQ8, where K is a field and Q8 denotes the quaternion group.

### Corollary 4.4

For a field K of characteristic 0 and R = KQ8, the following conditions are equivalent:

• (1) R is DR;

• (2) R is regular-IFP;

• (3) R is unit-IFP;

• (4) R is nilpotent-IFP;

• (5) R is Abelian;

• (6) The equation 1 + x2 + y2 = 0 has no solutions in K;

• (7) R is isomorphic to a finite direct product of division rings;

• (8) R is reduced.

Proof

The equivalences of (1), (5), (6), (7) and (8) are shown in [9, Proposition 2.13]. (1) ⇒ (2) comes from Proposition 4.2. (2)⇒ (3) is obvious, and (3) ⇒ (4) is noted above. (4) ⇒ (5) is proved by [8, Proposition 1.5(1)].

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