On the relation between one-sided duoness and commutators

Nam Kyun Kim; Yang Lee

doi:10.1515/math-2024-0118

Article Open Access

On the relation between one-sided duoness and commutators

Nam Kyun Kim and Yang Lee

Published/Copyright: December 31, 2024

Published by

Become an author with De Gruyter Brill

Submit Manuscript Author Information Explore this Subject

$Open Mathematics$

From the journal Open Mathematics Volume 22 Issue 1

Abstract

This article studies the structure of rings R over which the 2 × 2 upper triangular matrix rings with the same diagonal are right duo, denoted by D 2 ( R ) . We prove that for any right regular element d of such a ring R , d R contains the ideal of R generated by all commutators. It is also proved that for a domain R , D 2 ( R ) is right duo if and only if R is either commutative or a division ring. Moreover, it is proved that if R is a ring of characteristic 2 such that D 2 ( R ) is right duo, then R has an ascending chain of nil ideals N R i ⊆ N R i + 1 ( i = 0 , 1 , … ) such that N R i + 1 ⁄ N R i is contained in the center of R ⁄ N R i . Furthermore, we give a simpler proof to the famous result that if R is a simple noncommutative ring then R coincides with its subring generated by all commutators (by Herstein). Finally, we show that if D 2 ( R ) is right duo over a ring R , then S a ⊆ a S for any a ∈ R , where S is any of the following: (i) the prime radical, (ii) the Jacobson radical, (iii) the group of all units in R , and (iv) the set of one-sided zero-divisors.

Keywords: right duo ring; D2(R) is right duo; commutator; right regular element

MSC 2010: 16U80; 16N40

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. It is well known that W ( R ) ⊆ N * ( R ) ⊆ N * ( R ) ⊆ N ( R ) and N * ( R ) ⊆ J ( R ) . Denote the n by n ( n ≥ 2 ) full (resp., upper triangular) matrix ring over R by M a t n ( R ) (resp., T n ( R ) ). Write D n ( R ) = { ( a i j ) ∈ T n ( R ) ∣ a 11 = … = a n n } .

Following Feller [1], a ring is called right duo if every right ideal is two-sided. Left duo rings are defined similarly. This concept is not left-right symmetric as we see in [2,3]. A ring is called duo if it is both left and right duo. It is evident that factor rings of right duo rings are right duo too. For a right (left) duo ring R , it is easily shown that W ( R ) = N * ( R ) = N * ( R ) = N ( R ) , and if R is prime, then R is a domain. A ring is called reduced if it has no nonzero nilpotent elements. It is evident that a ring R is reduced if and only if a 2 = 0 for a ∈ R implies a = 0 . A ring is called Abelian if every idempotent is central. It is easily checked that reduced rings and one-sided duo rings are Abelian. A ring R is Abelian if and only if D n ( R ) is Abelian by [4, Lemma 2]. Abelian rings are clearly directly finite (or Dedekind finite) (i.e., a b = 1 implies b a = 1 ).

Note that for any ring R , D n ( R ) is neither left nor right duo for n ≥ 3 . In fact, E 13 = E 12 E 23 ∉ E 23 D n ( R ) and E 12 E 23 ∉ D n ( R ) E 12 , where E i j stands for the matrix with ( i , j )-entry 1 and zeros elsewhere. So we can consider the right (left) duoness of D n ( R ) only when n = 2 .

In [5, Theorem 1.2 (1)], Hong et al. proved that if D 2 ( R ) is right duo, then [ a , b ] ∈ b 2 R for any a , b ∈ R , and it also implies that R is right duo, where [ a , b ] stands for the commutator a b − b a .

With this result as motivation, in this article, we continue the study of properties of rings R over which D 2 ( R ) is one-sided duo, in relation to commutators and the duo property of sets in R .

Actually, we prove that for any right regular element d of such a ring R , the right ideal d R contains the ideal of R generated by all commutators. It is proved that if R is a ring of characteristic 2 such that D 2 ( R ) is right duo, then R has an ascending chain of nil ideals N R i ⊆ N R i + 1 ( i = 0 , 1 , … ) such that N R i + 1 ⁄ N R i is contained in the center of R ⁄ N R i . Moreover, we provide a simpler proof for the well known result that if R is a simple noncommutative ring, then R coincides with its subring generated by all commutators (by Herstein).

Finally, we investigate the duo property of several types of sets in a ring R when D 2 ( R ) is right duo. In fact, we show that if D 2 ( R ) is right duo, then S a ⊆ a S for any a ∈ R , where S includes the prime radical (the set of all nilpotents), the Jacobson radical, the group of all units in R , and the set of all one-sided zero-divisors.

Here are some additional notations to be used in this study. The group of all units, the set of all regular elements, and the center of R are denoted by U ( R ) , C ( R ) , and Z ( R ) , respectively. For an integer k ≥ 2 , we write Nil k ( R ) = { a ∈ R ∣ a k = 0 } . Given an element a ∈ R , r R ( a ) (resp., l R ( a ) ) denotes the right (resp., left) annihilator of a in R .

2 Structure of commutators in R when D 2 ( R ) is right duo

In this section, we investigate the structure of rings R over which D 2 ( R ) is right duo, with a focus on their relation to commutators. Let R be a ring, and recall that [ a , b ] = a b − b a for a , b ∈ R . An element d in a ring R is called right (resp., left) regular if r R ( d ) = 0 (resp., l R ( d ) = 0 ). If d is both right and left regular, then d is called regular, meaning that d is a non-zero-divisor.

Lemma 2.1

Let R be a ring such that D 2 ( R ) is right duo. Then, the following hold.

[ a , b ] ∈ a 2 R ∩ b 2 R for any a , b ∈ R .
R is right duo.

Proof

Part (1) is proved by [5, Theorem 1.2 (1)]. Part (2) follows directly from (1).□

Remark 2.2

As demonstrated in [5, Example 1.3], there exists a right duo ring R for which D 2 ( R ) is not right duo.

We adopt the definitions and arguments presented by Herstein [6] and Lanski [7]. Let R be a ring and A ⊆ R . A is called a Lie subring of R if A is an additive subgroup such that [ a , b ] ∈ A for all a , b ∈ R . An additive subgroup U of a Lie subring A is called a Lie ideal of A if [ u , a ] ∈ U for all u ∈ U and a ∈ A . For subsets S , T of R , [ S , T ] denotes the additive subgroup of R generated by all [ s , t ] with s ∈ S and t ∈ T . For any Lie ideals A , B of R , [ A , B ] is also a Lie ideal of R due to the equality [ [ x , y ] , r ] = [ [ x , r ] , y ] + [ x , [ y , r ] ] ([7, p. 6]). Thus, the class of Lie ideals of R includes R , [ R , R ] , [ [ R , R ] , R ] ( = [ R , [ R , R ] ] ) , [ [ R , R ] , [ R , R ] ] , and so forth. Refer to [6, pp. 2,4] and [7, p. 6] for further details. For a given subset B of R , we will use ⟨ B ⟩ and ( B ) to denote the subring and the ideal of R generated by B , respectively. Note that ⟨ B ⟩ ⊆ B R , ⟨ B ⟩ ⊆ R B , and ( B ) = R B R .

Theorem 2.3

Let R be a noncommutative ring such that D 2 ( R ) is right duo. Then, the following assertions hold.

0 ≠ ( [ R , R ] ) ⊆ d R for every right regular element d in R .
⋂ d ∈ R d R ≠ 0 , where d runs over all right regular elements in R.

Proof

(1) Considering the fact that D 2 ( R ) is right duo, we can write: a c 0 a d b 0 d = d b 0 d s t 0 s , where s , t ∈ R . This yields a d = d s and a b + c d = d t + b s . By Lemma 2.1 (1), we have a d = d a + d 2 r 1 = d ( a + d r 1 ) for some r 1 ∈ R . Hence, from a d = d s , we can deduce d ( s − ( a + d r 1 ) ) = 0 . Since r R ( d ) = 0 , it implies s = a + d r 1 . From these results, we can then derive a b + c d = d t + b ( a + d r 1 ) and a b − b a = − c d + d t + b d r 1 . By applying Lemma 2.1 (2), we obtain c d = d r 2 and b d = d r 3 for some r 2 , r 3 ∈ R . Consequently, a b − b a = d ( − r 2 + t + r 3 r 1 ) ∈ d R . As a result, we can conclude that [ R , R ] ⊆ d R , which leads to ( [ R , R ] ) ⊆ d R . Furthermore, the noncommutativity of R ensures that ( [ R , R ] ) ≠ 0 .

(2) This is immediate from (1).□

Theorem 2.3 will play important role in the subsequent results. However, as we will soon observe, the converse of this theorem does not necessarily hold.

Given a ring R , consider the condition ( † ) :

For every right regular element d in R , it holds that a b − b a ∈ d R for any a , b ∈ R .

We can provide an example of a simple noncommutative ring that does not satisfy the condition ( † ) . Let A = Q ⟨ x , y ⟩ be the free algebra with noncommuting indeterminates x , y over the field Q of rational numbers. Consider the quotient ring R = A ⁄ I , where I is the ideal of A generated by y x − x y − 1 . Then, R is a simple noncommutative domain. Since x ¯ y ¯ − y ¯ x ¯ = − 1 ∉ x ¯ R , it follows that R does not satisfy the condition ( † ) . Note that R is neither left nor right duo; consequently, by Lemma 2.1 (2), D 2 ( R ) is also neither left nor right duo.

Furthermore, we observe that the condition ( † ) is equivalent to the necessity stated in Theorem 2.3 (1) by Proposition 2.6 (3). This implies that any ring R for which D 2 ( R ) is right duo satisfies the condition ( † ) . However, there exists a right duo ring that does not satisfy this condition. An example can be found by considering the Hamilton quaternions R over the field of real numbers. In this case, the power series ring R [ [ x ] ] is right duo by Remark 2.7 (2). Note that i j − j i = 2 k ∉ x R [ [ x ] ] .

Remark 2.4

Considering Lemma 2.1 (2) and Theorem 2.3, a natural question arises: if a ring R satisfies the condition ( † ) , is R necessarily a right duo? The answer is negative by the property of E = D n ( K ) that is not right duo, as previously explained, where K is a field and n ≥ 3 . We note that every regular element M of E is of the form ( a i j ) with a i i ≠ 0 . This implies that M ∈ U ( E ) , from which we see that E satisfies the condition ( † ) . This provides a counterexample that contradicts the converse of Theorem 2.3. However, if R is a domain, then the condition ( † ) implies the assertion in Lemma 2.1 (1) (hence, R is right duo).

Lemma 2.5

Let R be a ring and L be a Lie ideal of R. Then, the following hold.

( L ) = R L = L R .
Let I be a one-sided ideal of R. If L ⊆ I , and then ( L ) ⊆ I .

Proof

(1) For any a ∈ L and r ∈ R , the fact that L is a Lie ideal implies a r − r a ∈ L . Moreover, it follows from a r ∈ L R that r a ∈ L R , and vice versa, which completes the proof. (2) is an immediate consequence of (1).□

Proposition 2.6

Let R be a noncommutative ring that satisfies the condition ( † ) . Then, the following hold.

R is directly finite.
If ⟨ [ R , R ] ⟩ contains a right regular element of R, then every right regular element is a unit in R.
0 ≠ R [ R , R ] = [ R , R ] R = ( [ R , R ] ) ⊆ d R for every right regular element d of R.

Proof

(1) Let x y = 1 for x , y ∈ R . Then, we have r R ( y ) = 0 . By hypothesis, it follows that x y − y x ∈ y R , so that x y − y x = y z for some z ∈ R . This leads to the equation 1 = x y = y ( x + z ) , consequently, yielding x + z = x y ( x + z ) = x . This implies z = 0 , resulting in y x = 1 .

(2) Let A = ⟨ [ R , R ] ⟩ , and let u ∈ A such that r R ( u ) = 0 . Since u 2 is also right regular in R , by hypothesis, we have u ∈ u 2 R . Thus, u = u 2 g for some g ∈ R , which implies u ( 1 − u g ) = 0 and consequently leads to u g = 1 . This demonstrates that u ∈ U ( R ) since R is directly finite by (1). Moreover, by hypothesis once again, we have u ∈ d R for every right regular element d in R , which consequently implies d R = R .

(3) This is shown by hypothesis and Lemma 2.5.□

Hirano et al. [8, Lemma 3] proved that if R [ x ] is right duo over a ring R , then R must be commutative. From this result, we can deduce that when R is noncommutative, D 2 ( R ) [ x ] cannot be right duo. In the following, we present an alternative proof of this assertion that uses Theorem 2.3. For the sake of convenience, we will represent the identity matrix in M a t n ( R ) as I n .

Remark 2.7

(1) The ring D 2 ( R ) [ x ] ( ≅ D 2 ( R [ x ] ) ) is not right duo over any noncommutative ring R . To prove this, assume, on the contrary, that D 2 ( R ) [ x ] is right duo. Then, for any a , b ∈ R , we have a b − b a ∈ x R [ x ] as stated in Theorem 2.3 (1). This implies a b − b a = 0 , leading to the conclusion R is commutative. However, this contradicts the noncommutativity of R . Hence, we established that D 2 ( R ) [ x ] is not right duo. Furthermore, the same argument applies to demonstrate that D 2 ( R ) [ [ x ] ] is also not right duo.

However, we note that D 2 ( R ) is right duo over any noncommutative division ring, by Theorem 2.8 (1). Consequently, the preceding argument provides a class of right duo rings over which polynomial (or power series) rings are not right duo.

(2) Let R be a division ring (possibly noncommutative), and let I be a nonzero right ideal of R [ [ x ] ] . Consider the set S = { m ∣ a m x m + a m + 1 x m + 1 + … ∈ I and a m ≠ 0 } ⊆ { 0 , 1 , 2 , … } . Take f ( x ) = a k x k + a k + 1 x k + 1 + … ∈ I such that k is minimal in S . Then, I = x k R [ [ x ] ] = R [ [ x ] ] x k , which is a two-sided ideal of R [ [ x ] ] , since a k + a k + 1 x + … ∈ U ( R [ [ x ] ] ) . Thus, R [ [ x ] ] is right duo. From this, we infer that the validity of [8, Lemma 3] does not hold for power series rings.

(3) Consider the ring R = ∏ i ∈ I R i , which is the direct product of rings R i for i ∈ I . It can be easily proved that R is right duo if and only if each ring R i is also right duo. This result provides examples of noncommutative rings that are not domain (non-reduced) and are right duo. For example, let R 1 be a noncommutative right duo domain, and let R 2 be a commutative that is not reduced. Then, the direct product R 1 × R 2 forms a noncommutative ring that is not reduced and is also right duo, as described earlier.

Recall that Z ( D n ( R ) ) = Z ( R ) I n + Z ( R ) E 1 n for n ≥ 2 ([9, Proposition 1.7]). Note that a simple right duo ring is clearly a division ring.

Theorem 2.8

Let R be a ring.

Let R be a noncommutative domain. Then, the following conditions are equivalent:
1. D 2 ( R ) is right duo;
2. a b − b a ∈ d R for any a , b , d ∈ R \ { 0 } ;
3. R is a division ring;
4. D 2 ( R ) is left duo;
5. a b − b a ∈ R d for any a , b , d ∈ R \ { 0 } .
Let R be a domain. Then, D 2 ( R ) is right duo if and only if R is either commutative or a division ring.
D 2 ( D 2 ( R ) ) is right duo if and only if R is commutative.

Proof

(1) (i) ⇒ (ii) is proved by Theorem 2.3.

(ii) ⇒ (iii). Assuming condition (ii), since R is noncommutative, there exist e , f ∈ R such that e f − f e ≠ 0 . Since R is a domain, it is clear that e f − f e is regular. Consequently, it follows that every nonzero element is a unit in R by Proposition 2.6 (2). This implies that R is a division ring.

(iii) ⇒ (i). Suppose that R is a division ring. Let M 1 = a b 0 a and M 2 = c d 0 c be nonzero elements of D 2 ( R ) . If c ≠ 0 , then M 2 is invertible, so that M 2 D 2 ( R ) = D 2 ( R ) . Assume c = 0 , i.e., M 2 = 0 d 0 0 . Here, if a = 0 , then M 1 M 2 = 0 ∈ M 2 D 2 ( R ) . So consider the case a ≠ 0 . Then, M 1 M 2 = 0 a d 0 0 = 0 d 0 0 d − 1 a d 0 0 d − 1 a d ∈ M 2 D 2 ( R ) , noting that R is a division ring and a , d ∈ R \ { 0 } . Consequently, D 2 ( R ) M 2 ⊆ M 2 D 2 ( R ) , which implies that D 2 ( R ) is right duo.

The equivalences of conditions (iii), (iv), and (v) can be proved by a symmetrical procedure.

(2) This is obtained from condition (1).

(3) Consider the ring E = D 2 ( D 2 ( R ) ) . Assuming that E is right duo, we can establish Nil 2 ( E ) ⊆ Z ( E ) using Lemma 2.1 (1), where Nil 2 ( E ) = { A ∈ E ∣ A 2 = 0 } . However,

Z ( E ) = a b 0 a c d 0 c 0 0 0 0 a b 0 a ∣ a , b , c , d ∈ Z ( R )

by [9, Proposition 1.7]. Since 0 R I 2 0 0 ⊆ Nil 2 ( E ) (where 0 , I 2 ∈ D 2 ( R ) ), we see that R I 2 ⊆ Z ( D 2 ( R ) ) = Z ( R ) I 2 + Z ( R ) E 12 , so that R = Z ( R ) . Thus, R is commutative.□

Note that for any ring R , every (minimal) prime ideal Q of D 2 ( R ) has the form { ( a i j ) ∈ D 2 ( R ) ∣ a i i ∈ P } , where P is a (minimal) prime ideal of R . Thus, D 2 ( R ) ⁄ Q is isomorphic to R ⁄ P . Suppose that N ( R ) = N * ( R ) , and let P be minimal (thus, Q is also minimal). It follows from [10, Proposition 1.11] that R ⁄ P is a domain.

Next, let R be right duo. If R is prime, then by the argument earlier, R is a domain. Note that R ⁄ P is also a right duo ring, it follows that R ⁄ P is a domain for any prime ideal P of R .

Corollary 2.9

(1) Let R be a ring, and let P be a prime ideal of R. If D 2 ( R ) is right duo, then R ⁄ P is either a commutative domain or a noncommutative division ring.

(2) Let R be a ring such that D 2 ( R ) is right duo. Then, R ⁄ N * ( R ) is a subdirect product of one of the following types: (i) commutative domains, (ii) noncommutative division rings, and (iii) commutative domains and noncommutative division rings.

Proof

(1) Assuming that D 2 ( R ) is right duo, we can conclude that R is right duo by Lemma 2.1 (2). Consequently, R ⁄ P is a domain by the argument earlier. Since D 2 ( R ) ⁄ D 2 ( P ) is also right duo and D 2 ( R ) ⁄ D 2 ( P ) ≅ D 2 ( R ⁄ P ) , we have that R ⁄ P is a division ring when R ⁄ P is noncommutative, by Theorem 2.8 (1).

(2) is obtained from condition (1).□

Consider a right duo ring R . If there exists a prime ideal P of R such that R ⁄ P is not a division ring, it follows that P cannot be a right primitive ideal, as implied by the subsequent remark. However, when D 2 ( R ) is right duo, the quotient ring R ⁄ P is a commutative domain by Corollary 2.9 (1). For example, in the ring of integers Z (or in Z [ x ] ), the zero ideal (or x Z [ x ] ) is a prime ideal that is not primitive.

The following example illustrates two points: (i) the condition that R is a domain in Theorem 2.8 (1) is essential, and (ii) the converse of Corollary 2.9 (1) is not generally true. For any ring R and n ≥ 2 , we write N n ( R ) = { ( a i j ) ∈ T n ( R ) ∣ a i i = 0 for all i } .

Example 2.10

Consider R = D n ( K ) as described earlier, where K is a field and n ≥ 3 .

(1) By Lemma 2.1 (2), it is evident that D 2 ( R ) is not right duo. Note that every regular element C of R = D n ( K ) has the form ( a i j ) with a i i ≠ 0 , thus implying C ∈ U ( R ) and leading to [ R , R ] ⊆ C R = R . From this, we see that the condition (ii) need not necessarily imply condition (i) in Theorem 2.8 (1) when R is not a domain.

(2) Let P = N n ( K ) . Then, P forms a prime ideal of R , and R ⁄ P ≅ K .

The following arguments elaborate upon Theorem 2.8 (1).

Remark 2.11

(1) Remark 2.7 (2) establishes that R [ [ x ] ] is a right duo domain over a noncommutative division ring R , but D 2 ( R [ [ x ] ] ) is not right duo by Theorem 2.8 (1). This example underlines the well known fact that the class of right duo rings is not closed under subrings. Note that D 2 ( K ) is indeed right duo due to Theorem 2.8 (1), where K denotes the quotient division ring of R [ [ x ] ] .

(2) Let R be a noncommutative domain such that the right quotient division ring of R [ x ] exists (e.g., R is right Noetherian). Let F be a noncommutative domain containing R [ x ] . Suppose that D 2 ( F ) is right duo. Then, F is a division ring by Theorem 2.8 (1), from which we see R ( x ) ⊆ F , where R ( x ) is the right quotient division ring of R [ x ] .

(3) There exists a domain that does not satisfy the condition of Theorem 2.8 (2). Consider a field K and let R = K ⟨ x , y ⟩ be the free algebra with noncommuting indeterminates x , y over K . It is evident that x m R ∩ y n R = 0 , x y − y x ∉ x m R , and x y − y x ∉ y n R for any m , n ≥ 1 . Thus, R is not right duo, leading to the conclusion that D 2 ( R ) is also not right duo by Lemma 2.1 (2).

A ring R is called semicommutative if for a , b ∈ R , a b = 0 implies a R b = 0 . It is obvious that a right or left duo ring is semicommutative; moreover, reduced rings are semicommutative.

Example 2.12

Considering Remark 2.11, it is clear that D 2 ( R ) is generally not right duo over a right duo ring R . This naturally raises the question of whether D 2 ( R ) is semicommutative when R is right duo. However, this assertion does not hold by the following example. Let R be the Hamilton quaternions over the field of real numbers. Then, Theorem 2.8 (1) guarantees that D 2 ( R ) is right duo. Nonetheless, D 2 ( D 2 ( R ) ) is not semicommutative by Example 1.7 in [11].

Recall the Eroǧlu’s result that if R is a noncommutative ring and 1 ∈ [ R , R ] , then R = ⟨ [ R , R ] ⟩ . In the following, we see another condition under which R = ⟨ [ R , R ] ⟩ holds.

Theorem 2.13

(1) Let R be a noncommutative ring that satisfies the condition ( † ) . Then, ⟨ [ R , R ] ⟩ contains a right regular element of R if and only if R = ⟨ [ R , R ] ⟩ .

(2) Let R be a noncommutative ring such that D 2 ( R ) is right duo. Then, ⟨ [ R , R ] ⟩ contains a right regular element of R if and only if R = ⟨ [ R , R ] ⟩ .

Proof

(1) Note first that ( [ [ R , R ] , R ] ) ⊆ ⟨ [ R , R ] ⟩ by Lemma 4.1 to follow. Set I = ( [ [ R , R ] , R ] ) . Suppose that there exists u ∈ ⟨ [ R , R ] ⟩ such that r R ( u ) = 0 . Then, due to the hypothesis and Proposition 2.6 (2), we deduce that u ∈ U ( R ) . Now, assume, on the contrary, that ⟨ [ R , R ] ⟩ ⊊ R . This implies I ⊊ R , and as a result, u ∉ I . However, since u ∈ ⟨ [ R , R ] ⟩ , it follows that ⟨ [ R , R ] ⟩ ⁄ I is not nil. This contradicts Lemma 4.1 (4) to follow. Therefore, we conclude thta R = ⟨ [ R , R ] ⟩ . The converse is obvious.

(2) This is proved by Theorem 2.3 and condition (1).□

Finally, we introduce a few more of properties of rings R when D 2 ( R ) is right duo, especially, in relation to Nil k ( R ) for k ≥ 2 .

Proposition 2.14

Let R be a ring such that D 2 ( R ) is right duo. Then, the following hold.

Nil 2 ( R ) ⊆ Z ( R ) .
[ a , r ] ∈ Nil 2 ( R ) for any a ∈ Nil 3 ( R ) and r ∈ R .
If Nil 3 ( R ) ⊈ Z ( R ) , then there exist a ∈ Nil 3 ( R ) and b ∈ R such that [ a , b ] R is a nonzero nilpotent ideal of R satisfying ( [ a , b ] R ) 2 = 0 .
For any a ∈ N ( R ) and r ∈ R , there exists k = k ( a ) ≥ 1 such that [ a k , r ] R is a nilpotent ideal of R satisfying ( [ a k , r ] R ) 2 = 0 .

Proof

(1) Let a ∈ Nil 2 ( R ) and r ∈ R . Then, by Lemma 2.1 (1), [ a , r ] ∈ a 2 R = 0 ; hence, a r = r a .

(2) Let a ∈ Nil 3 ( R ) and r ∈ R . Then, [ a , r ] ∈ a 2 R . However, since a 2 ∈ Nil 2 ( R ) , it follows from condition (1) that a 2 ∈ Z ( R ) . Say [ a , r ] = a 2 s with s ∈ R . Then, [ a , r ] 2 = ( a 2 s ) 2 = ( a 2 ) 2 s 2 = 0 .

(3) Suppose Nil 3 ( R ) ⊈ Z ( R ) . In such a case, there exist a ∈ Nil 3 ( R ) and b ∈ R such that [ a , b ] ≠ 0 . Additionally, using (1) and (2), we deduce that [ a , b ] ∈ Nil 2 ( R ) ⊆ Z ( R ) . Consequently, this leads to [ a , b ] R = R [ a , b ] R and ( R [ a , b ] R ) 2 = 0 .

(4) Since a ∈ N ( R ) , it follows that a k ∈ Nil 3 ( R ) for some k = k ( a ) ≥ 1 . Thus, the conclusion can be derived from (3).□

As previously mentioned, for a right (left) duo ring R , we have that W ( R ) = N * ( R ) = N * ( R ) = N ( R ) . Additionally, we also note that Nil i ( R ) ⊆ Nil i + 1 ( R ) for all i ≥ 1 and N ( R ) = ∪ i = 1 ∞ Nil i ( R ) . Given these considerations, a natural question arises: if D 2 ( R ) is right duo over a noncommutative ring R , does it follows that N ( R ) = Nil 2 ( R ) ? However, it is generally not true, even when D 2 ( R ) is commutative, as illustrated by Example 3.2.

3 When D 2 ( R ) is right duo and c h ( R ) = 2

In this section, we study the structure of rings R with c h ( R ) = 2 over which D 2 ( R ) is right duo.

Lemma 3.1

Let R be a ring with c h ( R ) = 2 . If D 2 ( R ) is right duo, then the following hold.

Nil 2 ( R ) is an ideal of R.
For any ideal I of R , Nil 2 ( R ⁄ I ) is an ideal of R ⁄ I .

Proof

(1) Consider a , b ∈ Nil 2 ( R ) and r ∈ R . According to Proposition 2.14 (1), we deduce that a , b ∈ Z ( R ) . This implies ( a − b ) 2 = a 2 + b 2 − 2 a b = 0 when c h ( R ) = 2 , leading to a − b ∈ Nil 2 ( R ) . Furthermore, we also establish that r a , a r ∈ Nil 2 ( R ) . Consequently, it follows that Nil 2 ( R ) forms an ideal of R .

(2) Note that D 2 ( R ) ⁄ D 2 ( I ) ( ≅ D 2 ( R ⁄ I ) ) is also right duo and c h ( R ⁄ I ) = 2 . By applying the conclusion of (1), we derive the desired result.□

In Lemma 3.1 (1), the condition c h ( R ) = 2 is not superfluous by the ring below.

Example 3.2

Let A be any commutative ring of c h ( A ) ≠ 2 , and consider the quotient ring R = A [ x , y ] ⁄ ( x 2 , y 2 ) , where A [ x , y ] is the polynomial ring with indeterminates x and y over A and ( x 2 , y 2 ) is the ideal of A [ x , y ] generated by x 2 and y 2 . Clearly, D 2 ( R ) is commutative (hence duo); however, ( x ¯ − y ¯ ) 2 = − 2 x ¯ y ¯ ≠ 0 (i.e., x ¯ − y ¯ ∉ Nil 2 ( R ) ) in spite of x ¯ , y ¯ ∈ Nil 2 ( R ) .

Next, we consider a case of right duo ring R that possesses an ascending chain of nil ideals in R .

Theorem 3.3

Let R be a ring with c h ( R ) = 2 such that D 2 ( R ) is right duo. Then, there exist nil ideals N R i of R for i = 0 , 1 , … such that

N R i ⊆ N R i + 1 and N R i + 1 ⁄ N R i = Nil 2 ( R ⁄ N R i ) ⊆ Z ( R ⁄ N R i ) , for a l l i .

Proof

Since D 2 ( R ) is right duo and c h ( R ) = 2 , Nil 2 ( R ) is an ideal of R by Lemma 3.1 (1). Write N R 0 = Nil 2 ( R ) . In the following argument, we will use Lemma 3.1 (2) without mention. Consider the quotient ring

D 2 ( R ) ⁄ D 2 ( N R 0 ) ≅ D 2 ( R ⁄ N R 0 )

that is also right duo. We will denote this quotient ring as R 1 = R ⁄ N R 0 . Since c h ( R 1 ) = 2 , it implies that Nil 2 ( R 1 ) is an ideal of R 1 and Nil 2 ( R 1 ) ⊆ Z ( R 1 ) due to Lemma 3.1 (1) and Proposition 2.14 (1). Note that Nil 2 ( R 1 ) = N R 1 ⁄ N R 0 for some ideal N R 1 of R containing N R 0 . Next, consider the right duo ring D 2 ( R ) ⁄ D 2 ( N R 1 ) ≅ D 2 ( R ⁄ N R 1 ) . Applying the same process, we obtain an ideal N R 2 of R containing N R 1 , and such that Nil 2 ( R 2 ) = N R 2 ⁄ N R 1 , where R 2 = R ⁄ N R 1 . Proceeding in this manner, we can construct an ascending chain N R 0 ⊆ N R 1 ⊆ … ⊆ N R i ⊆ N R i + 1 ⊆ … for i ≥ 0 . Since N R 0 is a nil ideal of R , and N R i + 1 ⁄ N R i is a nil ideal of R ⁄ N R i , by an inductive argument, we conclude that N R i is also nil for all i ≥ 1 .□

Note that N ( R ) = ∪ i = 1 ∞ Nil i ( R ) holds for any ring R . In the case where c h ( R ) = 2 and D 2 ( R ) is right duo, the proof of Theorem 3.3 establishes that Nil 2 i ( R ) = N R i − 1 for i ≥ 2 .

Corollary 3.4

Let R be a ring with c h ( R ) = 2 such that D 2 ( R ) is right duo. If R satisfies the ascending chain condition on ideals, then R has a bounded index (of nilpotency).

Proof

According to Theorem 3.3, we have the ascending chain N R 0 ⊆ N R 1 ⊆ … ⊆ N R i ⊆ N R i + 1 ⊆ … . By the assumption, there exists an integer k ≥ 0 such that N R k = N R k + 1 = … . Then, we can deduce from the earlier observation that N R k = Nil 2 k + 1 ( R ) . From Theorem 3.3 and the notations in it, we have that N ( R ) = ∪ i = 0 ∞ N R i . Then, we further obtain

N ( R ) = ∪ i = 0 ∞ N R i = Nil 2 k + 1 ( R ) .

This implies that R has a bounded index of 2 k + 1 .□

Note that W ( A ) = N ( A ) holds for every right duo ring A . In particular, for the right duo ring R of Corollary 3.4 (2), we have that ( R a R ) 2 k + 1 = 0 for every a ∈ N ( R ) .

Remark 3.5

If Nil 2 ( R ) ⊆ Z ( R ) , then for any a , b ∈ Nil 2 ( R ) , we have ( a + b ) 2 = 0 when c h ( R ) = 2 , which implies that a + b ∈ Nil 2 ( R ) . In the case where c h ( R ) ≠ 2 , it is possible that ( a + b ) 2 ≠ 0 , but ( a + b ) 3 = 0 , leading to a + b ∈ Nil 3 ( R ) .

In general, if Nil i ( R ) ⊆ Z ( R ) for some i ≥ 2 , then for any a , b ∈ Nil i ( R ) , we can deduce ( a + b ) 2 i − 1 = 0 , resulting in a + b ∈ Nil 2 i − 1 ( R ) .

When c h ( R ) = p with p being a prime number, it follows that a + b ∈ Nil p ( R ) for any a , b ∈ Nil p ( R ) .

In the following, we present a non-stationary ascending chain of N R i ’s, illustrating an example for Theorem 3.3.

Example 3.6

Let R = ⊕ i = 1 ∞ Z 2 [ x ] ( x 2 i ) , 1 ∏ i = 1 ∞ Z 2 [ x ] ( x 2 i ) be the Z 2 -subalgebra of ∏ i = 1 ∞ Z 2 [ x ] ( x 2 i ) generated by ⊕ i = 1 ∞ Z 2 [ x ] ( x 2 i ) and 1 ∏ i = 1 ∞ Z 2 [ x ] ( x 2 i ) . Then, c h ( R ) = 2 and D 2 ( R ) is commutative (hence duo). Set S i = Z 2 [ x ] ( x 2 i ) . Then,

Nil 2 ( S i ) = Z 2 [ x ] ( x 2 i ) x 2 i − 1 and Nil 2 ( R ) = ⊕ i = 1 ∞ Z 2 [ x ] ( x 2 i ) x 2 i − 1 .

Hence, R 1 = R Nil 2 ( R ) ≅ Z 2 ⊕ Z 2 [ x ] ( x 2 ) ⊕ Z 2 [ x ] ( x 2 2 ) ⊕ … , 1 ∏ i = 0 ∞ Z 2 [ x ] ( x 2 i ) ≅ Z 2 × R , noting Z 2 ≅ Z 2 [ x ] ( x 2 0 ) .

We next note that Nil 2 ( R 1 ) = Nil 2 2 ( R ) ⁄ Nil 2 ( R ) , and hence, R ⁄ Nil 2 2 ( R ) ≅ Z 2 × Z 2 × R . Continuing this process, we have an ascending chain Nil 2 ( R ) ⊊ Nil 2 2 ( R ) ⊊ … ⊊ Nil 2 i ( R ) ⊊ … .

4 Some results for commutators in R

Consider a ring R and a subset A ⊆ R . Recall that A is a Lie subring of R if A is an additive subgroup such that [ a , b ] ∈ A for all a , b ∈ R . Furthermore, an additive subgroup B of a Lie subring A is a Lie ideal of A if [ u , a ] ∈ B for all u ∈ B and a ∈ A .

Lemma 4.1

(1) [ a , b ] a = [ a , b a ] , [ a , b ] b = [ a b , b ] , a [ a , b ] = [ a , a b ] , b [ a , b ] = [ b a , b ] , and r [ a , b ] = [ r a , b ] − [ r , b ] a for all r , a , b ∈ R . Particularly, r [ [ x , y ] , b ] = [ r [ x , y ] , b ] − [ r , b ] [ x , y ] for all x , y ∈ R .

(2) ( [ [ R , R ] , [ R , R ] ] ) ⊆ ( [ [ R , R ] , R ] ) ⊆ ⟨ [ R , R ] ⟩ .

(3) Both ⟨ [ R , R ] ⟩ ⁄ ( [ [ R , R ] , [ R , R ] ] ) and ⟨ [ R , R ] ⟩ ⁄ ( [ [ R , R ] , R ] ) are commutative nil rings that are prime radical.

Proof

(1) By referring to the proof of [12, Theorem 1.1], we can establish the equalities.

(2) The result immediately follows from Lemma 2.5 and the final equality in (1).

(3) Based on (2), consider the subring S = ⟨ [ R , R ] ⟩ ⁄ ( [ [ R , R ] , [ R , R ] ] ) of the quotient ring R ¯ = R ⁄ ( [ [ R , R ] , [ R , R ] ] ) . Then, S is clearly a commutative ring, meaning [ a ¯ , b ¯ ] [ c ¯ , d ¯ ] = [ c ¯ , d ¯ ] [ a ¯ , b ¯ ] in R ¯ for all a , b , c , d ∈ R . For the sake of simplicity, elements of R are identified with their images in R ¯ . In the subsequent computation, we freely use the equalities from (1) and the commutativity of commutators. Given any s , t ∈ R , we observe that

[ s , t ] 3 = [ s , t ] ( [ s , t ] [ s , t ] ) = [ s , t ] ( [ [ s , t ] s , t ] − [ [ s , t ] , t ] s ) = [ [ s , t ] s , t ] [ s , t ] − [ [ s , t ] , t ] [ s , t ] s = ( [ s , t ] s t − t [ s , t ] s ) [ s , t ] − [ [ s , t ] , t [ s , t ] ] s = ( [ s , t ] s t [ s , t ] − t [ s , t ] s [ s , t ] ) − ( [ [ s , t ] , [ t s , t ] ] s ) = ( [ s , t ] s t [ s , t ] − t [ s , t s ] [ s , t ] ) − ( [ [ s , t ] , [ t s , t ] ] s ) = ( [ s , t ] s t [ s , t ] − t [ s , t ] [ s , t s ] ) − ( [ [ s , t ] , [ t s , t ] ] s ) = ( [ s , t s ] [ t s , t ] − [ t s , t ] [ s , t s ] ) − ( [ [ s , t ] , [ t s , t ] ] s ) = [ [ s , t s ] , [ t s , t ] ] − [ [ s , t ] , [ t s , t ] ] s ∈ ( [ [ R , R ] , [ R , R ] ] ) .

Now, let x be a nonzero element of S = ⟨ [ R ¯ , R ¯ ] ⟩ . This implies that there exists an integer n ≥ 1 such that the expression of x involves n distinct [ a ¯ , b ¯ ] terms. As x ∈ Z ( R ¯ ) and [ s ¯ , t ¯ ] 3 = 0 for all s , t ∈ R , it follows that ( x R ¯ ) n + 2 = 0 . This further leads to ⟨ [ R ¯ , R ¯ ] ⟩ ⊆ W ( R ¯ ) , entailing S = W ( S ) = N * ( S ) . The result for ⟨ [ R , R ] ⟩ ⁄ ( [ [ R , R ] , R ] ) follows directly from the aforementioned result.□

In the following, we provide a kind of Lie ideals and Lie subrings, subsequently analyzing their interrelation.

Example 4.2

We will use the facts: (i) Z ( D n ( R ) ) = Z ( R ) I n + Z ( R ) E 1 n for n ≥ 2 [9, Proposition 1.7], (ii) for a given ring A , if a = a 1 + a 2 and b = b 1 + b 2 ∈ A with a 1 , b 1 ∈ A and a 2 , b 2 ∈ Z ( A ) , then [ a , b ] = ( a 1 + a 2 ) ( b 1 + b 2 ) − ( b 1 + b 2 ) ( a 1 + a 2 ) = a 1 b 1 − b 1 a 1 = [ a 1 , b 1 ] .

(1) Let Li ( n ) k = [ … [ [ [ R n , R n ] , R n ] , R n ] , … , R n ] , where k instances of R n are used. Let F be a commutative ring, and let R n = D n ( F ) for n ≥ 2 . Then, we observe that Li ( 2 ) 2 = 0 , meaning that R 2 is commutative. Through the computation of [ A , B ] = A B − B A = f E 13 for A = E 12 , B = f E 23 ∈ D 3 ( F ) (where f represents any element of F ), we deduce Li ( 3 ) 2 = F E 13 and Li ( 3 ) 3 = 0 . Furthermore, ⟨ Li ( 3 ) 2 ⟩ ⁄ ( Li ( 3 ) 3 ) ≅ F E 13 .

Next, take A 1 = E 12 = A 3 , B 1 = f E 23 , A 2 = E 23 , B 2 = f E 34 , and B 3 = f E 24 in R 4 , where f is any element of F . Then, by help of the computation of R 2 and R 3 , we obtain that [ A 1 , B 1 ] = f E 13 , [ A 2 , B 2 ] = f E 24 , and [ A 3 , B 3 ] = f E 14 . Consequently, we have

Li ( 4 ) 2 = ( F E 13 + F E 14 ) + F E 24 , Li ( 4 ) 3 = F E 14 and Li ( 4 ) 4 = 0 .

Additionally, ⟨ Li ( 4 ) 2 ⟩ ⁄ ( Li ( 4 ) 3 ) ≅ F E 13 + F E 24 .

Inductively, we have

Li ( n ) 2 = ( F E 13 + … + F E 1 n ) + ( F E 24 + … + F E 2 n ) + … + ( F E n − 3 , n − 1 + F E n − 3 , n ) + F E n − 2 , n , Li ( n ) 3 = ( F E 14 + … + F E 1 n ) + ( F E 25 + … + F E 2 n ) + … + ( F E n − 4 , n − 1 + F E n − 4 , n ) + ( F E n − 3 , n ) , Li ( n ) 4 = ( F E 15 + … + F E 1 n ) + ( F E 26 + … + F E 2 n ) + … + ( F E n − 5 , n − 1 + F E n − 5 , n ) + ( F E n − 4 , n ) , … … Li ( n ) n − 1 = ( F E 1 , n − 1 + F E 1 n ) + F E 2 , n , Li ( n ) n − 1 = F E 1 n and Li ( n ) n = 0 .

Note that Li ( n ) k = ⟨ Li ( n ) k ⟩ = ( Li ( n ) k ) and

⟨ Li ( n ) 2 ⟩ ⁄ ( Li ( n ) 3 ) ≅ F E 13 + F E 24 + … + F E n − 3 , n − 1 + F E n − 2 , n , for n ≥ 5 .

(2) Let R = ∏ j = 2 ∞ R j be the direct product of R j ’s, where R j = D j ( F ) as in (1). Write L i m = [ … [ [ [ R , R ] , R ] , R ] , … , R ] when m -number of R ’s occur. Then, by the argument of (1), we have that

L i m = ∏ j = 2 ∞ Li ( j ) m and L i m ≠ 0 , for all m ≥ 2 .

Therefore, R has a non-stationary descending chain

L i 2 ⊋ L i 3 ⊋ … ⊋ L i m ⊋ L i m + 1 ⊋ …

of (Lie) ideals.

(3) Let R = Z Z Q Q 0 Z Z Q 0 0 Z Q 0 0 0 Z be a subring of T 4 ( Q ) , where Q is the field of rational numbers. Then, ⟨ [ R , R ] ⟩ = 0 0 Z Q 0 0 0 Q 0 0 0 0 0 0 0 0 , ( [ R , R ] ) = 0 0 Q Q 0 0 0 Q 0 0 0 0 0 0 0 0 , and [ [ R , R ] , R ] = 0 0 0 Q 0 0 0 0 0 0 0 0 0 0 0 0 , from which we see that ⟨ [ R , R ] ⟩ ⁄ ( [ [ R , R ] , R ] ) ≅ Z × Q and ⟨ [ R , R ] ⟩ ⊊ ( [ R , R ] ) .

(4) There exists a ring R such that [ R , R ] ⊊ ⟨ [ R , R ] ⟩ . Let S be a commutative ring with c h ( S ) = 2 , and let A = ⟨ x , y ⟩ be the free algebra generated by noncommuting indeterminates x , y over S . Form the ideal I in A generated by y x + x y + 1 , x 2 , and y 2 . Define R = A ⁄ I , and identify every element of A with its image in R . Note that R = S + S x + S y + S y x .

For f = a 0 + b 0 x + a 1 y + b 1 y x and g = c 0 + d 0 x + c 1 y + d 1 y x in R , their commutator is

[ f , g ] = ( b 0 c 1 + d 0 a 1 ) + ( b 0 d 1 + d 0 b 1 ) x + ( b 1 c 1 + d 1 a 1 ) y ,

resulting in [ R , R ] = S + S x + S y . Interestingly, ⟨ [ R , R ] ⟩ = S + S x + S y + S y x = R , showing [ R , R ] ⊊ ⟨ [ R , R ] ⟩ .

(5) Let R 1 be the ring R from (3), and let R 2 denote the ring R from (4). Define R = R 1 × R 2 . Applying the arguments from (3) and (4), it follows that [ R , R ] ⊊ ⟨ [ R , R ] ⟩ ⊊ ( [ R , R ] ) .

Lemma 4.1 will play a crucial role in this section. Herstein proved that if R is a simple noncommutative ring (possibly without identity), then R = ⟨ [ R , R ] ⟩ ([6, Corollary of Theorem 1.5]). When the ring R has an identity, a simpler proof of this result can be provided as follows.

Theorem 4.3

Let R be a ring.

For a noncommutative ring R, the following conditions are equivalent:
1. ( [ [ R , R ] , [ R , R ] ] ) is a proper ideal of R ;
2. ( [ [ R , R ] , R ] ) is a proper ideal of R ;
3. ⟨ [ R , R ] ⟩ is a proper subring of R .
If R is a simple noncommutative ring, then
R = ⟨ [ R , R ] ⟩ = ( [ R , R ] ) = ( [ [ R , R ] , R ] ) = ( [ [ R , R ] , [ R , R ] ] ) .

Proof

(1) (i) ⇒ (iii) Suppose ( [ [ R , R ] , [ R , R ] ] ) ⊊ R . Then, R ¯ = R ⁄ ( [ [ R , R ] , [ R , R ] ] ) is a ring with identity. Because ⟨ [ R , R ] ⟩ ⁄ ( [ [ R , R ] , [ R , R ] ] ) is a nil subring of R ¯ as Lemma 4.1 (3), ⟨ [ R , R ] ⟩ cannot have invertible elements due to ( [ [ R , R ] , [ R , R ] ] ) ⊊ R . In particular, 1 ∉ ⟨ [ R , R ] ⟩ , so ⟨ [ R , R ] ⟩ ⊊ R . The other directions are evident from Lemma 4.1 (2).

(2) Let R be a simple noncommutative ring. Assume [ [ R , R ] , [ R , R ] ] = 0 . Then, [ [ R , R ] , R ] = 0 (equivalently, [ R , R ] ⊆ Z ( R ) ) using (1). Also, ⟨ [ R , R ] ⟩ becomes a nonzero nil subring of R by Lemma 4.1 (3). However, since R is simple, [ a , b ] R = R for any 0 ≠ [ a , b ] ∈ [ R , R ] , implying [ a , b ] ∈ U ( R ) . This contradicts [ a , b ] ∈ N ( R ) . Consequently, [ [ R , R ] , [ R , R ] ] ≠ 0 , i.e., ( [ [ R , R ] , [ R , R ] ] ) = R , and we establish the result using (1) and Lemma 4.1 (2).□

Eroǧlu [12, Theorem 1.1] proved that if R is a noncommutative ring with 1 ∈ [ R , R ] , then R = ⟨ [ R , R ] ⟩ . In the following, we present an alternative, more straightforward proof of Eroǧlu’s result using Theorem 4.3 (1).

Remark 4.4

(1) [12, Theorem 1.1] If R is a noncommutative ring with 1 ∈ [ R , R ] , then R = ⟨ [ R , R ] ⟩ . To prove this, assume ⟨ [ R , R ] ⟩ ⊊ R . Then, we have ( [ [ R , R ] , [ R , R ] ] ) ⊊ R by Theorem 4.3 (1). Consequently, ⟨ [ R , R ] ⟩ ⁄ ( [ [ R , R ] , [ R , R ] ] ) forms a non-nil subring in R ⁄ ( [ [ R , R ] , [ R , R ] ] ) , as 1 ∉ ( [ [ R , R ] , [ R , R ] ] ) and 1 ∈ ⟨ [ R , R ] ⟩ . However, this contradicts Lemma 4.1 (3). Hence, we conclude that ⟨ [ R , R ] ⟩ = R .

(2) Lee [13, p. 3, Fact 1] stated that for a simple ring R and a ∈ R , if [ a , R ] ⊆ Z ( R ) or [ a , [ R , R ] ] = 0 , then a ∈ Z ( R ) . We include the proof here. It suffices to consider the case where R is noncommutative. Then, R = ⟨ [ R , R ] ⟩ by Theorem 4.3 (2). Therefore, if [ a , [ R , R ] ] = 0 , then 0 = [ a , ⟨ [ R , R ] ⟩ ] = [ a , R ] , implying a ∈ Z ( R ) . Furthermore, if [ a , R ] ⊆ Z ( R ) , then [ a , [ R , R ] ] = 0 due to the fact that [ a , [ x , y ] ] = [ y , [ x , a ] ] + [ [ y , a ] , x ] = 0 for any x , y ∈ R . As a consequence, for a simple ring R and a ∈ R , we have that a ∈ Z ( R ) if and only if [ a , R ] ⊆ Z ( R ) if and only if [ a , [ R , R ] ] = 0 .

5 Duo property of sets in R when D 2 ( R ) is right duo

We begin with the following lemma.

Lemma 5.1

Let R be a ring such that D 2 ( R ) is right duo, and let I be an ideal of R. Then, the following hold.

For any b ∈ N ( R ) , R b = b R .
If R ⁄ I is reduced, then I a ⊆ a I for any a ∈ R .

Proof

(1) Using Lemma 2.1 (1), we have [ b , a ] ∈ a 2 R ∩ b 2 R for any a , b ∈ R , which means b a − a b = b 2 c 1 for some c 1 ∈ R . Applying by Lemma 2.1 (1) again, we obtain [ b 2 , c 1 ] = b 2 2 c 2 for some c 2 ∈ R . This implies [ b , a ] = [ b 2 , c 1 ] + c 1 b 2 and b a = a b + c 1 b 2 + b 2 2 c 2 . Continuing this process, we have

b a = a b + b 2 c 1 = a b + c 1 b 2 + b 2 2 c 2 = a b + c 1 b 2 + c 2 b 2 2 + … + c k − 1 b 2 k − 1 + b 2 k c k ,

for every k ≥ 1 .

Now, assuming b ∈ N ( R ) , we can choose k ≥ 1 such that b 2 k = 0 , leading to b a = a b + c 1 b 2 + c 2 b 2 2 + … + c k − 1 b 2 k − 1 = ( a + c 1 b + c 2 b 2 2 − 1 + … + c k − 1 b 2 k − 1 − 1 ) b ∈ R b . This implies b R ⊆ R b . However, since R is right duo by Lemma 2.1 (2), R b ⊆ b R is already established.

(2) For b ∈ I and a ∈ R , we have a b − b a = a 2 c with some c ∈ R . As a b − b a ∈ I and R ⁄ I is reduced, a 2 c ∈ I implies a c ∈ I . This leads to b a = a b − a 2 c = a ( b − a c ) ∈ a I , implying I a ⊆ a I .□

Proposition 5.2

(1) Let R be a right duo ring. For any additive subgroup A of R and a ∈ R , there exists an additive subgroup B of R such that A a = a B .

(2) Suppose D 2 ( R ) is right duo. For any ideal I of R and a ∈ N ( R ) , there exists an ideal J of R such that I a = a J .

Proof

(1) Let α ∈ A and a ∈ R . As R is right duo, α a = a β for some β ∈ R . Define

B = { β ∣ α a = a β for some α ∈ A } .

Clearly, 0 ∈ B . If β 1 , β 2 ∈ B , the α 1 a = a β 1 and α 2 a = a β 2 for some α 1 , α 2 ∈ A . Hence, ( α 1 − α 2 ) a = a ( β 1 − β 2 ) , implying β 1 − β 2 ∈ B . Thus, B forms an additive subgroup of R .

(2) Consider J = { β ∣ α a = a β for some α ∈ I } , as defined in (1). Then, by (1), I a = a J holds, with J being an additive subgroup of R . For β ∈ J and r ∈ R , there exists α ∈ I such that α a = a β . Using Lemma 5.1 (1), we have a ( r β ) = ( a r ) β = ( s a ) β = s ( a β ) = s ( α a ) = ( s α ) a ∈ I a , and a ( β r ) = ( a β ) r = ( α a ) r = α ( a r ) = α ( s a ) = ( α s ) a ∈ I a for some s ∈ R . Consequently, r J and J r are both contained in J , confirming that J is an ideal of R .□

According to Lemma 2.1 (2), a ring R is right duo when D 2 ( R ) is right duo. In [5], a ring R is called right nilpotent-duo if N ( R ) a ⊆ a N ( R ) for every a ∈ R . Left nilpotent-duo rings are defined similarly. A ring is called nilpotent-duo if it is both left and right nilpotent duo. In [5, Example 1.7 (1)], an example of a right nilpotent-duo ring R , which is not right duo is provided (thus, D 2 ( R ) is also not right duo). However, we are unaware of any example of a right duo ring that is not right nilpotent-duo.

In the following, we demonstrate that a given ring R can indeed be nilpotent-duo when D 2 ( R ) is right duo.

Theorem 5.3

Let R be a ring. If D 2 ( R ) is right duo, then R is nilpotent-duo.

Proof

By Lemma 5.1 (2), we have N ( R ) a ⊆ a N ( R ) .

To prove the reverse inclusion, letting b ∈ N ( R ) , then a b − b a = a 2 c for some c ∈ R . Consider the commutator [ a , a c ] in ( a c ) 2 R . This gives a b = b a + a ( a c ) = b a + a c a + ( a c ) 2 d 1 for some d 1 ∈ R . Consider the commutator [ a , c a c d 1 ] in ( c a c d 1 ) 2 R . This leads to a b = b a + a c a + ( c a c d 1 ) a + ( c a c d 1 ) 2 d 2 for some d 2 ∈ R . Since R ⁄ N ( R ) is reduced and a 2 c ∈ N ( R ) , we have c a ∈ N ( R ) , and consequently, c d 1 c a ∈ N ( R ) . This implies ( c a c d 1 ) 2 d 2 = c a ( c d 1 c a ) c d 1 d 2 = c a d 3 ( c d 1 c a ) for some d 3 ∈ R using Lemma 5.1 (1). As a result, we have

a b = b a + a c a + ( c a c d 1 ) a + c a d 3 ( c d 1 c a ) = ( b + a c + c a c d 1 + c a d 3 c d 1 c ) a ∈ N ( R ) a .

This shows that a N ( R ) ⊆ N ( R ) a , concluding that N ( R ) a = a N ( R ) .□

Let R be the right duo ring constructed by Courter as shown in [3, Example 3.4]. Consider a field K and the row-finite infinite matrix ring E over K . Define the K -subspace S in E generated by the matrices { a 0 , a 1 , … , a j , … } , where a 0 = E 12 , a 1 = E 13 , a 2 = E 14 + E 32 , … , a j = E 1 ( j + 2 ) + E ( j + 1 ) 2 for j ≥ 2 . Let I be the identity matrix in E , and let R = K I ⊕ S , where R constitutes the K -subspace of E with a basis { I , a 0 , a 1 , … , a j , … } . Note that a i a i + 1 = a 0 for all i ≥ 1 and J ( R ) = N ( R ) = S hold. However, Theorem 5.3 does not apply to this right duo ring R , as evidenced by a 0 ∈ a 1 N ( R ) ⊊ N ( R ) a 1 = 0 . Moreover, D 2 ( R ) is not right duo due to Lemma 2.1 (1), since a 0 = [ a 1 , a 2 ] ∉ a 1 2 R = 0 .

Proposition 5.4

If D 2 ( R ) is right duo, then the following hold.

For any a ∈ R , J ( R ) a ⊆ a J ( R ) .
For any a ∈ R , J ( R [ x ] ) a ⊆ a J ( R [ x ] ) .

Proof

(1) We observe that R ⁄ J ( R ) is reduced according to [14, Corollary 2.4]. Hence, it follows from Lemma 5.1 (2) that J ( R ) a ⊆ a J ( R ) for any a ∈ R .

(2) We observe that J ( R [ x ] ) = N [ x ] for some nil ideal N of R , a consequence of Amitsur’s theorem. Because R is right duo, we have J ( R [ x ] ) = N ∗ ( R [ x ] ) = W ( R [ x ] ) = N ( R ) [ x ] . By using Theorem 5.3, we obtain the desired result.□

Consider any right (or left) duo ring A , and assume a b = 1 for a , b ∈ A . As A is Abelian, it is straightforward to deduce b a = 1 . This property will be used freely.

A ring R is called right unit-duo if U ( R ) a ⊆ a U ( R ) for every a ∈ R . Left unit-duo rings are defined similarly. A ring will be called unit-duo if it is both left and right unit-duo. While [15, Example 1.7] provides an example of a unit-duo ring that is not left or right duo, the existence of a right duo ring that is not right unit-duo remains unknown. However, Han et al. proved in [15, Theorem 1.3 (3)] that in a domain R , if R is right duo, then R is right unit-duo.

Proposition 5.5

If D 2 ( R ) is right duo, then the following hold.

R is right unit-duo.
For any a ∈ N ( R ) , U ( R ) a = a U ( R ) .
If R ⁄ N ( R ) is a commutative ring or a noncommutative domain, then R is right unit-duo.

Proof

(1) Let b ∈ U ( R ) and a ∈ R . Given that D 2 ( R ) is right duo, we have b 0 0 b a 1 0 a = a 1 0 a a 1 c 1 0 a 1 for some a 1 , c 1 ∈ R . This yields

b a = a a 1 and b = a c 1 + a 1 .

Now, the fact that b ∈ U ( R ) leads to a = b − 1 a a 1 , and thus, b = b − 1 a a 1 c 1 + a 1 . As R is right duo, ( b − 1 a ) a 1 = a 1 c 2 for some c 2 ∈ R , implying b = a 1 c 2 c 1 + a 1 = a 1 ( c 2 c 1 + 1 ) . However, since b ∈ U ( R ) , we conclude that a 1 ∈ U ( R ) . Hence, U ( R ) a ⊆ a U ( R ) .

(2) Let b ∈ U ( R ) and a ∈ N ( R ) . Using the same argument as in the proof of Lemma 5.1 (1), we have a b = b a + c 1 a 2 + … + c s − 1 a s for some s ≥ 2 . Thus, a b = b ( 1 + b − 1 c 1 a + … + b − 1 c s − 1 a s − 1 ) a . Since b − 1 c 1 a + … + b − 1 c s − 1 a s − 1 ∈ N ( R ) , 1 + b − 1 c 1 a + … + b − 1 c s − 1 a s − 1 ∈ U ( R ) , which implies a b = b u a for some u ∈ U ( R ) . Furthermore, b u ∈ U ( R ) . Consequently, a U ( R ) ⊆ U ( R ) a , leading to the conclusion that U ( R ) a = a U ( R ) for any a ∈ N ( R ) by help of Theorem 5.3.

(3) Suppose that R ⁄ N ( R ) is commutative. Take b ∈ U ( R ) and a ∈ R . We have a b − b a ∈ N ( R ) and a b − b a = a 2 c for some c ∈ R . Since R ⁄ N ( R ) is reduced, we deduce a c ∈ N ( R ) . Moreover, by using the fact that N ( R ) r = r N ( R ) for any r ∈ R according to Theorem 5.3, we also have a b − b a = a 2 c = m a for some m ∈ N ( R ) . Consequently, a b = b a + m a = ( b + m ) a .

Let n = b − 1 ( − m ) ∈ N ( R ) , and note that n k = 0 for some k ≥ 1 . Since ( b + m ) u = b ( 1 − b − 1 ( − m ) ) u = 1 , where u = ( 1 + n + … + n k − 1 ) b − 1 , we conclude that a b = ( b + m ) a ∈ U ( R ) a . Consequently, a U ( R ) ⊆ U ( R ) a , and thus, we can establish a U ( R ) = U ( R ) a using (1).

Now suppose that R ⁄ N ( R ) is a noncommutative domain. Observe that D 2 ( R ⁄ N ( R ) ) is right duo. This fact implies that R ⁄ N ( R ) is a division ring by Theorem 2.8 (1). When we take u ∉ N ( R ) , we can find 1 − u v ∈ N ( R ) for some v ∈ R , leading to u v ∈ U ( R ) and consequently, u ∈ U ( R ) . This conclusion shows that R = U ( R ) ∪ N ( R ) , entailing that U ( R ) a = a U ( R ) using (1).□

z d l ( R ) and z d r ( R ) stand for the set of all left and right zero-divisors of R , respectively.

Theorem 5.6

Let D 2 ( R ) be a right duo ring and a ∈ C ( R ) . Then, the following hold.

z d l ( R ) a ⊆ a z d r ( R ) .
z d r ( R ) a ⊆ a z d l ( R ) .

Proof

(1) Let b ∈ z d l ( R ) and a ∈ C ( R ) . We have b b 1 = 0 for some 0 ≠ b 1 ∈ R . Due to R being right duo, b a = a b ′ for some b ′ ∈ R . Assume b ′ ∉ z d r ( R ) . Then, b 1 a b ′ ≠ 0 and b 1 b a ≠ 0 follow.

Now consider the commutator [ b 1 , b 1 b ] , which leads to b 1 2 b − b 1 b b 1 = ( b 1 b ) 2 c for some c ∈ R . Using the fact that b b 1 = 0 , we have

b 1 2 b = b 1 b b 1 + ( b 1 b ) 2 c = 0 .

From this, we find that b 1 2 b a = 0 , implying b 1 2 = 0 since b a = a b ′ ∉ z d r ( R ) . Additionally, by noting that [ b 1 , b ] = b 1 2 d for some d ∈ R , we obtain b 1 b = b b 1 + b 1 2 d = 0 , a contradiction. Consequently, we must have b ′ ∈ z d r ( R ) , proving that z d l ( R ) a ⊆ a z d r ( R ) .

(2) Consider the case where b ∈ z d r ( R ) and a ∈ C ( R ) . This implies that b 1 b = 0 for some nonzero b 1 ∈ R . Similarly, we also have b a = a b ′ for some b ′ ∈ R . Suppose b ′ ∉ z d l ( R ) . From the commutator [ b a b 1 , b 1 ] , we obtain

b a b 1 2 = ( b a b 1 ) b 1 = b 1 b a b 1 + ( b a b 1 ) 2 c = 0 ,

for some c ∈ R . Since b a = a b ′ ∉ z d l ( R ) , we deduce that b 1 2 = 0 . Additionally, given the fact that [ b a , b 1 ] = b 1 2 d for some d ∈ R , we have b a b 1 = b 1 b a + b 1 2 d = 0 , which leads to a contradiction. Therefore, we conclude that b ′ ∈ z d l ( R ) .□

Corollary 5.7

If D 2 ( R ) is right duo, then z d l ( R ) = z d r ( R ) .

Proof

This is a consequence of Theorem 5.6 when we set a = 1 .□

A ring R is called right regular-duo if C ( R ) a ⊆ a C ( R ) for every a ∈ R . Left regular-duo rings are defined similarly. A ring will be called regular-duo if it is both left and right regular-duo. While there exists an example of a right duo ring that is not right regular-duo, as shown in [16, Example 2.4 (2)], there is currently no known example of a right regular-duo ring that is not right duo.

Proposition 5.8

If D 2 ( R ) is right duo, then R is right regular-duo.

Proof

Let b ∈ C ( R ) and a ∈ R . Since D 2 ( R ) is right duo, the same calculation as in the proof of Proposition 5.5 (1) yields

b a = a a 1 and b = a c 1 + a 1 .

Furthermore, b 2 = b a c 1 + b a 1 = a a 1 c 1 + b a 1 . Since R is right duo, we have a a 1 = a 1 c 2 and b a 1 = a 1 b 2 for some b 2 , c 2 ∈ R . This implies b 2 = a 1 c 2 c 1 + a 1 b 2 = a 1 ( c 2 c 1 + b 2 ) . However, since b 2 ∈ C ( R ) , it follows that a 1 is left regular.

We want to show that a 1 is also right regular. Suppose that a 1 d = 0 for some nonzero d ∈ R . Then, b 2 d = ( a a 1 c 1 + b a 1 ) d = a a 1 c 1 d = a a 1 d c 3 = 0 , where c 1 d = d c 3 for some c 3 ∈ R due to a right duoness of R . As b 2 ∈ C ( R ) , we conclude that d = 0 , which implies that a 1 is right regular. Consequently, C ( R ) a ⊆ a C ( R ) .□

Acknowledgement

The authors are grateful for the reviewer’s valuable comments that improved the manuscript.

Funding information: This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No. 2022R1F1A1059494).
Author contributions: Both the authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results, and approved the final version of the manuscript. This manuscript was written as a collaborative research project during YL’s three-month visit to NKK University, which was conducted in a self-initiated, question-generating, collaborative manner. NKK prepared the manuscript based on the results of the collaboration.
Conflict of interest: The authors state no conflicts of interest.
Data availability statement: All data generated or analysed during this study are included in this published article.

References

[1] E. H. Feller, Properties of primary noncommutative rings, Trans. Amer. Math. Soc. 89 (1958), 79–91, DOI: https://doi.org/10.2307/1993133. 10.1090/S0002-9947-1958-0098763-0Search in Google Scholar

[2] A. W. Chatters and W. Xue, On right duo p.p. rings, Glasg. Math. J. 32 (1990), 221–225, DOI: https://doi.org/10.1017/S0017089500009253. 10.1017/S0017089500009253Search in Google Scholar

[3] R. C. Courter, Finite dimensional right duo algebras are duo, Proc. Amer. Math. Soc. 84 (1982), 157–161, DOI: https://doi.org/10.1090/S0002-9939-1982-0637159-6. 10.1090/S0002-9939-1982-0637159-6Search in Google Scholar

[4] C. Huh, H. K. Kim, and Y. Lee, p.p. rings and generalized p.p. rings, J. Pure Appl. Algebra 167 (2002), 37–52, DOI: https://doi.org/10.1016/S0022-4049(01)00149-9. 10.1016/S0022-4049(01)00149-9Search in Google Scholar

[5] C. Y. Hong, H. K. Kim, N. K. Kim, T. K. Kwak, and Y. Lee, One-sided duo property on nilpotents, Hacet. J. Math. Stat. 49 (2020), 1974–1987, DOI: https://doi.org/10.15672/hujms.571016. 10.15672/hujms.571016Search in Google Scholar

[6] I. N. Herstein, Topics in Ring Theory, University of Chicago Press, Chicago, 1969. Search in Google Scholar

[7] C. Lanski, Higher commutators, ideals and cardinality. Bull. Aust. Math. Soc. 54 (1996), 41–54, DOI: https://doi.org/10.1017/S0004972700015069. 10.1017/S0004972700015069Search in Google Scholar

[8] Y. Hirano, C. Y. Hong, J. Y. Kim, and J. K. Park, On strongly bounded rings and duo rings, Comm. Algebra 23 (1995), 2199–2214, DOI: https://doi.org/10.1080/00927879508825341. 10.1080/00927879508825341Search in Google Scholar

[9] K. J. Choi, T. K. Kwak, and Y. Lee, Reversibility and symmetry over centers, J. Korean Math. Soc. 56 (2019), 723–738, DOI: https://doi.org/10.4134/JKMS.j180364. Search in Google Scholar

[10] G. Shin, Prime ideals and sheaf representation of a pseudo symmetric rings, Trans. Amer. Math. Soc. 184 (1973), 43–60, DOI: https://doi.org/10.1090/S0002-9947-1973-0338058-9. 10.1090/S0002-9947-1973-0338058-9Search in Google Scholar

[11] N. K. Kim and Y. Lee, Extensions of reversible rings, J. Pure Appl. Algebra 185 (2003), 207–223, DOI: https://doi.org/10.1016/S0022-4049(03)00109-9. 10.1016/S0022-4049(03)00109-9Search in Google Scholar

[12] M. P. Eroǧlu, On the subring generated by commutators, J. Algebra Appl. 21 (2022), 2250059, DOI: https://doi.org/10.1142/S0219498822500591. 10.1142/S0219498822500591Search in Google Scholar

[13] T.-K. Lee, On higher commutators of rings, J. Algebra Appl. 21 (2022), 2250118, DOI: https://doi.org/10.1142/S0219498822501183. 10.1142/S0219498822501183Search in Google Scholar

[14] H.-P. Yu, On quasi-duo rings, Glasg. Math. J. 37 (1995), 21–31, DOI: https://doi.org/10.1017/S0017089500030342. 10.1017/S0017089500030342Search in Google Scholar

[15] J. Han, Y. Lee, and S. Park, Duo property restricted to groups of units, J. Korean Math. Soc. 52 (2015), 489–501, DOI: https://doi.org/10.4134/JKMS.2015.52.3.489. 10.4134/JKMS.2015.52.3.489Search in Google Scholar

[16] C. Y. Hong, H. K. Kim, N. K. Kim, T. K. Kwak, and Y. Lee, Duo property on the monoid of regular elements, Algebra Colloq. 29 (2022), 203–216, DOI: https://doi.org/10.1142/S1005386722000165. 10.1142/S1005386722000165Search in Google Scholar

Received: 2023-08-28

Revised: 2024-12-05

Accepted: 2024-12-06

Published Online: 2024-12-31

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2024-0118

Keywords for this article

right duo ring; D2(R) is right duo; commutator; right regular element

Creative Commons

BY 4.0