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##  eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2021; 61(1): 11-21

Published online March 31, 2021

Copyright © Kyungpook Mathematical Journal.

### Structures Related to Right Duo Factor Rings

Hongying Chen, Yang Lee, Zhelin Piao∗

Department of Mathematics, Pusan National University, Pusan 46241, Korea
e-mail : 1058695917@qq.com

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, Yanbian University, Yanji 133002, China
e-mail : zlpiao@ybu.edu.cn

Received: December 9, 2019; Revised: June 30, 2020; Accepted: June 4, 2020

We study the structure of rings whose factor rings modulo nonzero proper ideals are right duo; such rings are called right FD. We first see that this new ring property is not left-right symmetric. We prove for a non-prime right FD ring R that R is a subdirect product of subdirectly irreducible right FD rings; and that $R/N*(R)$ is a subdirect product of right duo domains, and $R/J(R)$ is a subdirect product of division rings, where $N*(R)$ (J(R)) is the prime (Jacobson) radical of R. We study the relation among right FD rings, division rings, commutative rings, right duo rings and simple rings, in relation to matrix rings, polynomial rings and direct products. We prove that if a ring R is right FD and $0≠e2=e∈R$ then eRe is also right FD, examining that the class of right FD rings is not closed under subrings.

Keywords: right FD ring, right duo ring, division ring, commutative ring, simple ring, non-prime right FD ring, matrix ring, polynomial ring, subring, idempotent

### 1. Introduction

Throughout this note every ring is an associative ring with identity unless otherwise stated. Let R be a ring. We use N(R), J(R), N*(R), N*(R), and W(R) to denote the set of all nilpotent elements, Jacobson radical, lower nilradical (i.e., prime radical), upper nilradical (i.e., the sum of all nil ideals), and the Wedderburn radical (i.e., the sum of all nilpotent ideals) of R, respectively. The center of R is denoted by Z(R). It is well-known that $W(R)⊆N*(R)⊆N*(R)⊆N(R)$ and $N*(R)⊆J(R)$. U(R) denotes the group of all units in R. The polynomial (resp., power series) ring with an indeterminate x over R is denoted by R[x] (resp., R[[x]]). $ℤ(ℤn)$ denotes the ring of integers (modulo n). Denote the n by n ($n≥2$) full (resp., upper triangular) matrix ring over R by $Matn(R)$ (resp., $Tn(R)$). Write $Dn(R)={(aij)∈Tn(R)∣a11=⋯=ann}$. Use $Eij$ for the matrix with (i, j)-entry 1 and zeros elsewhere. In denotes the identity matrix in $Matn(R)$. $∏$ means the direct product. Use |S| to denote the cardinality of a given set S. The characteristic of R is written by ch(R). An element u of R is called right (resp., left) regular if ur=0 (resp., ru=0) for $r∈R$ implies r=0. An element is regular if it is both left and right regular. The monoid of all regular elements in R is denoted by C(R).

This article is motivated by the results in . In Section 2 we study the structure of right FD rings, focusing on the relation among right FD rings, commutative rings and simple rings. We investigate that in several kinds of ring extensions that play important roles in ring theory. In Section 3 we examine the right FD property of polynomial rings, subrings and direct products for given right FD rings.

A ring is called Abelian if every idempotent is central. Following Feller , a ring is called right duo if every right ideal is two-sided. Left duo rings are defined similarly. A ring is called duo if it is both left and right duo. Right (left) duo rings are easily shown to be Abelian. A ring is usually called reduced if it has no nonzero nilpotents. It is easily checked that a ring R is reduced if and only if a2=0 for a ∈ R implies a=0. Reduced rings are clearly Abelian, but not conversely by [6, Lemma 2].

### 2. When Factor Rings are Right Duo

In this section we are concerned with the class of rings whose factor rings modulo nonzero proper ideals are right duo. A ring R shall be called right FD if R is simple, or else R/I is a right duo ring for every nonzero proper ideal I of R. A left FD ring is can be defined similarly. A ring is called FD if it is both right and left FD. There exist many non-simple FD rings as we will see. We first examine the following basic results about right duo rings.

### Lemma 2.1.

• (1) Every simple (or right primitive) right duo ring is a division ring.

• (2) Every prime right (left) duo ring is a domain.

• (3) The class of right (left) duo rings is closed under factor rings and direct products.

• (4) If R is a division ring then D2(R) is a duo ring; but Dn(A) is neither right no left duo for all n ≥ 3 over any ring A.

• (5) Let A be any ring and n ≥ 3. Then Tn(A) is neither right nor left FD.

• (6) Let A be any ring and n ≥ 4. Then Dn(A) is neither right nor left FD.

• (7) The class of right (left) FD rings is closed under factor rings.

Proof. (1) Let R be a simple right duo ring and 0 ≠ a ∈ R. Then aR=RaR=R, so that a ∈ U(R). Thus R is a division ring. Every right primitive right duo ring is a division ring through a simple computation.

(2) Let R be a prime right duo ring and suppose that ab=0 for a, b ∈ R. Then aRb ⊐ abR=0, so that a=0 or b=0. Thus R is a domain.

(3) is obvious.

(4) Take $0≠(aij)∈D2(R)$. If $aii≠0$then $(aij)∈U(D2(R))$, so that $D2(R)(aij)=D2(R)=(aij)D2(R)$. Assume $aii=0$. Then $a12≠0$ and so $D2(R)(aij)=0R00=D2(R)(aij)D2(R)=(aij)D2(R)$. So $D2(R)$ is duo. Next consider $Dn(A)$ for $n≥3$ over any ring A. Then $AE(n−1)n$ is a right ideal of $Dn(A)$, but not two-sided; and $AE12$ is a left ideal of $Dn(A)$, but not two-sided. So $Dn(A)$ is neither right nor left duo.

(5) Note that $Tn(A)$ is not simple. Consider the proper ideal $I=AE1n+AE2n+⋯+AE(n−1)n+AEnn$ of $Tn(A)$. Then $Tn(A)/I$ is isomorphic to $Tn−1(A)$ that is non-Abelian (hence not right duo) since $n≥3$. Thus $Tn(A)$ is not right FD.

(6) Note that $Dn(A)$ is not simple. Consider the proper ideal $J=AE1n+AE2n+⋯+AE(n−1)n$ of $Dn(A)$. Then $Dn(A)/J$ is isomorphic to $Dn−1(A)$ that is not right duo by (4) since $n−1≥3$. Thus $Dn(A)$ is not right FD.

(7) Let R be a right FD ring and I be a proper ideal of R. Consider R/I. If I=0 then R/0 is right FD. So assume I ≠ 0. Let J/I be a proper nonzero ideal of R/I. Then J is a nonzero proper ideal of R. So $(R/I)/(J/I)≅R/J$ is right duo. Thus R/I is right FD.

The proofs for the left cases of (1)-(7) are similar.

Right duo rings are right FD by Lemma 2.1(3); but the converse is not true in general by the following. Note that Matn(A) cannot be Abelian (hence neither right nor left duo) for n ≥ 2 over any ring A. If A is simple then Matn(A) is simple (hence FD). In the following we consider the right FD property of Tn(R) and Dn(R) for n=2 and n=3, respectively, based on Lemma 2.1(5, 6).

### Theorem 2.2.

Let R be a ring and n ≥ 2.

• (1) R is simple if and only if Matn(R) is right FD if and only if Matn(R) is simple.

• (2) The following conditions are equivalent:

• (i) R is a division ring;

• (ii) T2(R) is a right (left) FD ring;

• (iii) D3(R) is a right (left) FD ring.

• (3)] Let R be simple. Then R is a division ring if and only if D2(R) is right (left) FD.

Proof. (1) It suffices to show that if Matn(R) is right FD then R is simple. Let R be non-simple. Consider a nonzero proper ideal I of R. Then $Matn(R)/Matn(I)$ is isomorphic to Matn(R/I) that is not right duo. So Matn(R) is not right FD.

(2) We apply the proof of [9, Theorem 1.10(3)]. (i) $⇒$ (ii). Let F be a division ring and R=T2(F). A nonzero proper ideal of R is one of the following: $I1=0F00$, $I2=FF00$, and $I3=0F0F$. Then $R/I1≅F×F$ and $R/I2≅F≅R/I3$, hence they are duo. So R is FD.

(ii) $⇒$ (i). Suppose that T2(R) is right FD. Assume that R is not simple. Then $T2(R)/T2(M)$ is isomorphic to the non-Abelian (hence not right duo) ring T2(R/M) for each maximal ideal M of R, entailing that T2(R) is not right FD. Thus R must be simple. Next consider the proper ideal $I=0R0R$ of T2(R). Then T2(R)/I is isomorphic to R, and hence R is right duo because T2(R) is right FD. Summarizing, R is a division ring by Lemma 2.1(1). The proof for the left case is similar.

(i) $⇒$ (iii). Let R be a division ring. Then it is easy to check that each nonzero proper ideal of D3(R) is one of the following: $00R000000$, $0RR000000$, $00R00R000$ and $0RR00R000$. So the factor rings modulo by these ideals are isomorphic to

respectively. Note that D2(R) is duo by Lemma 2.1(4) and R is clearly duo. Next R' is isomorphic to the subring

$(ab00a000a,a000ac00a)∣a,b,c∈R$

of $D3(R)×D3(R)$, which is also isomorphic to the subring

$(ab0a,ac0a)∣a,b,c∈R$

of $D2(R)×D2(R)$. This ring is duo by the proof of Lemma 2.1(4), noting that every non-invertible element is of the form $(0b00,0c00)$. Therefore $D3(R)$ is FD.

(iii) $⇒$ (i). Suppose that D3(R) is right FD. Assume that R is not simple. Then $D3(R)/D3(M)$ is isomorphic to the noncommutative ring $D3(R/M)$, that is not right duo by Lemma 2.1(4), for a maximal ideal M of R. So D3(R) is not right FD. Thus R must be simple. Next consider the proper ideal $I=0RR00R000$ of D3(R). Then D3(R)/I is isomorphic to R, and hence R is right duo because D3(R) is right FD. Summarizing, R is a division ring by Lemma 2.1(1). The proof for the left case is similar.

(3) It suffices to show the sufficiency by Lemma 2.1(4). Let D2(R) is right FD. Then $R≅D2(R)I$ is right duo, and hence R is a division ring by Lemma 2.1(1), where I=RE12. The proof for the left case is similar.

Following , a ring R is called FC if R is simple, or else R/I is a commutative ring for every nonzero proper ideal I of R. FC rings are clearly FD, but the converse need not hold by Theorem 2.2(2). Indeed, letting R be a noncommutative division ring, T2(R) is right FD by Theorem 2.2(2), but not FC by [9, Theorem 1.10(3)].

Following Birkhoff , a ring R is called subdirectly irreducible if the intersection of all nonzero ideals in R is nonzero. It is obvious that a ring R is subdirectly irreducible if and only if for every set of nonzero proper ideals of R, ${Kl∣l∈L}$ say, we have $∩l∈LKl≠0$. We will use this fact freely.

### Lemma 2.3.

Let R be a non-prime right (resp., left) FD ring. Then each of the following holds.

• (1) R is a subdirect product of subdirectly irreducible right (resp., left) FD rings.

• (2) R/N*(R) is a subdirect product of right (resp., left) duo domains, and R/J(R) is a subdirect product of division rings.

• (3) If R is semiprime then R is a subdirect product of right (resp., left) duo domains (hence reduced).

Proof. (1) It is proved by Birkhoff  that any ring is a subdirect product of subdirectly irreducible rings. We apply the proof of [10, Theorem 4.12.3]. For any 0 ≠ a ∈ R, there exists a proper ideal Ma that is maximal with respect to the property that $a∉Ma$. Then $∩0≠a∈RMa=0$, and $R/Ma$ is subdirectly irreducible since every nonzero ideal of $R/Ma$ contains a+Ma by the maximality of Ma. Moreover $R/Ma$ is right FD by Lemma 2.1(7). Therefore R is a subdirect product of subdirectly irreducible right FD rings $R/Ma$. The left case can be proved similarly.

(2) Let Pi (i ∈ I) be all prime ideals of R. Then every Pi is nonzero because R is not prime. So R/Pi is right duo since R is right FD. Moreover R/Pi is a right duo domain by Lemma 2.1(2). Thus R/N*(R) is a subdirect product of right duo domains. The remainder is proved similarly by Lemma 2.1(1).

(3) is an immediate consequence of (2). The proofs of (1), (2) and (3) for the left case are similar.

There exist non-prime FD rings which are not subdirectly irreducible. In fact, each of $ℤ×ℤ$ and $ℤpq$ is not subdirectly irreducible because $J(ℤ×ℤ)=0$ and $pℤpq∩qℤpq=0$, where p and q are distinct prime numbers. This elaborates on Lemma 1.3(1). The condition "non-prime" is not superfluous in Lemma 1.3(3) as can be seen by the simple ring $Matn(A)$ for $n≥2$ over a simple ring A.

Based on Lemma 1.3, one may ask whether a ring R is right FD if every right primitive factor ring of R is a simple domain. But the answer is negative as follows. There exists a semiprime ring R for which $J(R)≠0$ and $R/J(R)$ is a simple domain, but R is neither right nor left FD.

### Example 2.4.

We refer to the construction and argument in [7, Example 1.2] and [8, Theorem 2.2(2)]. Let K be a simple domain that is neither right nor left duo (e.g., the first Weyl algebra over a field of characteristic zero). Let $Rn=D2n(K)$ for $n≥1$ with the function $σ:Rn→Rn+1$ by $A↦A00A$. Next set $R=∪ n=1∞Rn$, noting that Rn can be considered as a subring of Rn+1 via σ. Then R is a semiprime ring by [8, Theorem 2.2(2)]. But

This implies that J(R) is maximal (hence primitive), entailing that every primitive factor ring of R is a simple domain. But R is neither right nor left FD since R/J(R) is neither right nor left duo.

Let $R′=R×R$. Then $J(R′)=J(R)×J(R)$ and all right (left) primitive factor ring of R' is $M1=R×J(R)$ and $M2=J(R)×R$. Note$R′/J(R′)≅K×K$ and $R′/Mi≅K$. But R' is also neither right nor left FD by Lemma 2.1(7).

Following Neumann , a ring R is said to be regular if for each a ∈ R there exists b∈ R such that a=aba. Such a ring is also called von Neumann regular by Goodearl . It is shown that R is regular if and only if every principal right (left) ideal of R is generated by an idempotent in [3, Theorem 1.1]. From this fact we can easily conclude that every regular ring R is clearly semiprimitive (i.e., J(R)=0).

### Proposition 2.5.

Let R be a non-prime regular ring. Then the following conditions are equivalent:

• (1) R is right FD;

• (2) R is reduced;

• (3) R is right duo;

• (4) R is left duo;

• (5) R is left FD;

• (6) Every right primitive factor ring of R is a division ring;

• (7) R is a subdirect product of division ring;

• (8) R is a subdirect product of domains.

Proof. Since R is regular, J(R)=0 and hence N*(R)=0. Thus R is reduced by Lemma 2.3(3), showing (1) ⇒ (2) and (5) ⇒ (2). (2) ⇒ (3) and (2) ⇒ (4) are proved by [3, Theorem 3.2]. (3) ⇒ (1), (4) ⇒ (5), (7) ⇒ (8) and (8) ⇒ (2) are obvious.

(1) ⇒ (6) is obtained from Lemma 2.1(1) because R is right FD. (6) ⇒ (7) is obvious since R is semiprimitive.

The condition "non-prime" is not superfluous in Proposition 2.5 as can be seen by the regular ring Matn(A) for n≥ 2 over a division ring A (refer to [3, Lemma 1.6]). Indeed, Matn(A) is simple (hence FD) but not reduced.

Following , a ring R is called right (resp., left) quasi-duo if every maximal right (resp., left) ideal of R is two-sided. It is obvious that a ring R is right quasi-duo if and only if R/J(R) is right quasi-duo. Right duo rings are clearly right quasi-duo but not conversely. It is proved by [5, Proposition 1] that a ring R is right quasi-duo if and only if every right primitive factor ring of R is a division ring.

### Proposition 2.6.

• (1) Every non-prime right FD ring is right quasi-duo.

• (2) If R is a non-prime right FD ring then R/J(R) is a reduced right quasi-duo ring.

Proof. (1) Let R be a non-prime right FD ring. Since R is not prime, every right primitive ideal of R is nonzero. Then since R is right FD, R/P is right duo for every right primitive ideal P of R. Hence R/P is a division ring by Lemma 2.1(1). So R is right quasi-duo by [5, Proposition 1].

(2) is obtained from (1) and Lemma 2.3(2).

The following elaborates upon Proposition 2.6.

### Remark 2.7.

• (1) Simple (hence FD) rings need not be quasi-duo by the existence of simple domains which are not division rings (e.g., the first Weyl algebra over a field of characteristic zero), which is compared with Proposition 2.6(1). Indeed this domain is neither right nor left quasi-duo.

• (2) There exist non-prime noncommutative FD rings as can be seen by T2(R) and D3(R) over a division ring R (see Theorem 2.2(2)). This provides examples to Proposition 2.6.

• (3) Based on Proposition 2.6(1), one may ask whether a non-prime right quasi-duo ring is right FD. But the answer is negative. Let A be a right quasi-duo ring and R=Tn(A) for n ≥ 3. Then R is right quasi-duo by [14, Proposition 2.1]. Let I=AE1n. Then R/I is non-Abelian (hence not right duo), and so R is not right FD.

Next we will show that the FD property is not left-right symmetric.

### Example 2.8.

Consider a skewed trivial extension in [13, Definition 1.3] as follows. Let R be a commutative ring with an endomorphism σ and M be an R-module. For $R⊕M$, the addition and multiplication are given by $(r1,m1)+(r2,m2)=(r1+r2,m1+m2)$ and $(r1,m1)(r2,m2)=(r1r2,σ(r1)m2+m1r2)$. Then this construction forms a ring. Following the literature, this extension is called the skew-trivial extension of R by M, denoted by $R∝M$. Note that $R∝R$ is isomorphic to $R[x;σ]/(x2)$ via the corresponding $(r,m)↦r+x¯m$, where $R[x;σ]$ is the skew polynomial ring, with the coefficients written on the right, only subject to $ax=xσ(a)$ for $a∈R$ and (x2) is the ideal of $R[x;σ]$ generated by x2.

Now let K be a field with a monomorphism σ and M be a K-module. Suppose that σ is not surjective. Then $K∝M$ is a right duo ring that is not left duo by [11, Theorem 2.5]. Next set $E=R1×R2$ with R1=K and $R2=K∝M$. Then E is right duo (hence right FD) but not left duo, by Lemma 2.1(3). Let $I=K×0$. Then E/I is isomorphic to the ring $K∝M$ that is not left duo. Thus E is not left FD.

### 3. Subrings, Polynomial Rings and Direct Products

In this section we study the right FD property of polynomial rings, subrings and direct products of given right FD rings. We consider first the polynomial ring case.

### Theorem 3.1.

The following conditions are equivalent for a given ring R:

• (1) R[x] is right (left) FD;

• (2) R is commutative;

• (3) R[x] is commutative.

Proof. It suffices to prove that if R[x] is right FD then R is commutative. Let n ≥ 3 and suppose that R[x] is right FD. We first obtain

$R+Rx+⋯+Rxn−1≅R[x]xnR[x]$

from the nonzero proper ideal $xnR[x]$ of R[x]. Then since R[x] is right FD, the ring $R+Rx+⋯+Rxn−1$ is right duo. This implies that for any $a,b∈R$, $b(a+x)=(a+x)f(x)$ for some $f(x)∈R$. Comparing the degrees of both sides, we must get $f(x)∈R$. c say. It then follows that b=c and ba=ac=ab. The commuting of a and b can be shown also by [4, Lemma 3]. Thus R is commutative. The proof of the left case is similar.

We can show, by help of Theorem 3.1, that the right FD property does not pass to polynomial rings.

We can write the following by help of Theorem 3.1 and Lemma 2.1(3): For a ring R, R[x] is right (resp., left) FD if and only if R[x] is right (resp., left) duo.

We next argue about subrings of right (left) FD rings.

### Example 3.2.

• (1) Let R be the first Weyl algebra over a field of characteristic zero. Consider R[x]. Since R[x] is right Noetherian domain, there exists the quotient division ring, Q say. Q is clearly FD. But R[x] is noncommutative, hence R[x] is neither right nor left FD by Theorem 3.1.

• (2) We extend (1). Let D be any right Noetherian domain that is not a division ring. Let Q be the quotient division ring. Then T2(Q) is FD by Theorem 2.2(2). But the subring T2(D) is neither right nor left FD by Theorem 2.2(2) because D is not a division ring.

In the following we find a kind of subring which inherits the right FD property.

### Theorem 3.3.

• (1) Let R be a ring and 0 ≠ e2=e ∈ R. If R is right (resp., left) duo then eRe is right (resp., left) duo.

• (2) Let R be a ring and 0 ≠ e2=e ∈ R. If R is right (resp., left) FD then eRe is right (resp., left) FD.

Proof. (1) Let R be a right duo ring and $0≠e2=e∈R$. Consider $eae,ebe∈eRe$. Since R is right duo, ebeae=eaec for some c∈ R. This yields eaec=eaece=eaeece. Thus eRe is right duo. The proof for the left case is similar.

(2) We apply the proof of [9, Theorem 1.12]. Suppose that R is simple. Then eRe is simple (hence FD) by the proof of [9, Theorem 1.12].

Suppose that R is FC and eRe is non-simple. Then R is non-simple by the preceding argument. Let J be a nonzero proper ideal of eRe. Then, by the proof of [9, Theorem 1.12], J=eReJeRe=eIe where I=ReJeR is a nonzero proper ideal of R. Since R is right FD, R/I is right duo.

Write $R¯=R/I$ and $r¯=r+I$ for $r∈R$. Note that $e∉J$ implies $e∉I$. So $e¯≠0$ in $R¯$. Next consider the epimorphism $f:eRe→e¯R¯e¯$ defined by $f(ere)=e¯r¯e¯$.

Since $R¯$ is right duo, the subring $e¯R¯e¯$ of $R¯$ is also right duo by (1). So $eReKer(f)(≅e¯R¯e¯)$ is right duo, where Ker(f) is the kernel of f. But Ker(f)=J by the proof of [9, Theorem 1.12], so that (eRe)/J is right duo. Therefore eRe is right FD. The proof for the left case is similar.

Let A be a simple ring, B be a noncommutative ring and set $R=A×B[x]$. Then letting e=(1,0), we get $eRe≅A$ is simple (hence FD); but $R/(A×0)≅B[x]$ is not right FD by Theorem 3.1. Whence R is not right FD, showing that the converse of Theorem 3.3(2) need not hold.

Next let A be a simple ring and $R=A×A$. Then for $e=(1,0)∈R$, $eRe≅A$ is simple, but R is not simple; which shows that the converse of the first part of the proof of Theorem 3.3(2) need not hold.

Recall that right duo rings are right FD. In contrast to Lemma 2.1(3), one may ask whether the direct product of right FD rings is also right FD. But the answer is negative as follows. Let A be a simple ring and $n≥2$. Then $Matn(A)$ is simple (hence FD). Set $R=Matn(A)×Matn(A)×Matn(A)$ and $I=Matn(A)×0×0$. Then I is a nonzero proper ideal of R and R/I is isomorphic to $Matn(A)×Matn(A)$ that is not right duo. Thus R is not right FD.

In the following we see an equivalent condition for direct products of right FD rings to be right FD.

### Theorem 3.4.

Let Ri be rings for all $i∈I$, and $R=∏ i∈IRi$, where $|I|≥2$. The following conditions are equivalent:

• (1) R is right FD;

• (2) Ri is right duo for all $i∈I$;

• (3) R is right duo.

Proof. (1) ⇒ (2). Suppose R is right FD. Let j ∈ I and $Ij={(ai)i∈I∈R∣aj=0}$. Then Ij is a nonzero proper ideal of R such that R/Ij is isomorphic to Rj. Since R is right FD, Rj is right duo.

(3) ⇒ (1) is obvious, and (2) ⇒ (3) is shown by Lemma 2.1(3).

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