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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2023; 63(2): 175-186

Published online June 30, 2023 https://doi.org/10.5666/KMJ.2023.63.2.175

### Weakly Right IQNN Rings

Yang Lee, Sang Bok Nam, Zhelin Piao∗

Department of Mathematics, Yanbian University, Yanji 133002, P. R. China and Institute for Applied Mathematics and Optics, Hanbat National University, Daejeon 34158, Korea
e-mail : ylee@pusan.ac.kr

Department of Computer Engineering, Kyungdong University, Geseong 24764, Korea
e-mail : k1sbnam@kduniv.ac.kr

Department of Mathematics, Yanbian University, Yanji 133002, P. R. China
e-mail : zlpiao@ybu.edu.cn

Received: June 11, 2022; Accepted: February 7, 2023

### Abstract

In this article we look at the property of a 2 by 2 full matrix ring over the ring of integers, of being weakly right IQNN. This generalisation of the property of being right IQNN arises from products of idempotents and nilpotents. We shown that it is, indeed, a proper generalization of right IQNN. We consider the property of beign weakly right IQNN in relation to several kinds of factorizations of a free algebra in two indeterminates over the ring of integers modulo 2.

Keywords: weakly right IQNN ring, idempotent, nilpotent, 2 by 2 full matrix ring, 2 by 2 upper triangular matrix ring, right IQNN ring

### 1. Prerequisites

Throughout this article every ring is an associative ring with identity unless otherwise stated. Let R be a ring. I(R) is used to denote the set of all idempotents of R, and I(R)=I(R)\{0,1}. We use N(R), and N*(R) to denote the set of all nilpotent elements, and upper nilradical (i.e., the sum of all nil ideals) of R, respectively. It is evident that N*(R)N(R). A nilpotent element is also called a nilpotent for simplicity. For n2, denote the full and upper triangular matrix rings over R by Matn(R) and Tn(R) respectively. Let , n, and denote the ring of integers, ring of integers modulo n, and the field of rational numbers, respectively. For m,n, gcd(m,n) is the greatest common divisor of m, n. The characteristic of R is denoted by ch(R).

Following Kwak et al. [5, Definition 1.2], a ring R is called right idempotent-quasi-normalizing on nilpotents, abbreviated to right IQNN, provided that I(R)' is empty, or else for every pair (e,a)I(R)×N(R) there exists (b,f)N(R)×I(R) such that ea=bf. A left IQNN ring is defined symmetrically. A ring is IQNN if it is both right and left IQNN. Abelian rings, in which every idempotent is central, are clearly IQNN but the converse is not true, as is shown in [5].

Define the sets

E1= 1 0 0 0E2= 0 0 0 1E3= 1 t 0 0E4= 1 0 u 0E5= 0 t 0 1E6= 0 0 u 1t0,u0B1= 0 0 0 0B2= 0 t 0 0B3= 0 0 u 0

The following is from [5, Lemma 2.3(2, 3)].

Lemma 1.1. If F is a commutative domain and R=Mat2(F), then

I(R)=E1E2E7 and N(R)=B1B4.

The following is from [1, Lemma 2.1(2)].

Lemma 1.2. Let F be a commutative domain, R=Mat2(F), and K be the quotient field of F. In Mat2(K), we have

(i) E7B4=sa+tc ba(sa+tc) us(sa+tc) buas(sa+tc) B4E7=as+bu ts(as+bu) ca(as+bu) ctas(as+bu) ;(ii) E4B4=a b ua ub =a aca ua uaca E5B4=tc ta c a =taba ta aba a ;(iii) B4E3=a ta c tc =a ta aba taba B4E6=ub b ua a =uaca aca ua a ,

from which it follows that if

ME7B4B4E7E4B4E5B4B4E3B4E6,

we have that if M is nonzero, then every entry of M is nonzero.

In the following, we see a practical application of Lemma 1.2 that may provide useful information to the studies related to products of idempotents and nilpotents.

Remark 1.3. Let F= and R=Mat2(F).

(1) Let p, q be any nonzero integers. Let C=0p0qR be such that C=EA for some EI(R) and AN(R). Then, by Lemma 1.1, we have the cases that A=B2=0v00N(R), and

E=E7=pmq1 p I(R), where p(1p)=qm0

or

E=E4=10u0I(R), where u0.

That is, EA is 0pv0qv with p=p'v and q=q'v, or 0v0uv with p=v and q=uv.

We will find BN(R) and EI(R) such that the left ideal RBE' of R contains EA.

Case 1. Suppose that p and q do not divide each other, and gcd(p,q)1. Evidently |p|,|q|2. Letting p=pv1 and q=qv1 with gcd(p',q')=v1 (then gcd(p,q)=1), we also have |p|,|q|2 since p and q do not divide each other.

Let p=p1u1pfuf and q=q1v1qgvg, with ui,vj1, we the prime number decompositions of p and q respectively. Since p(1p)=qm and gcd(p,q)=1, q must divide 1-p'. Letting 1p=qm, we have p+qm=1 and this implies gcd(p,q)=1 (hence gcd(v1,q)=1).

Since q divides 1p as above, we have that 1pq and

BE=qp p 2 q 2qp01p q 01=0p 0q ,

noting B=qp p 2 q 2qpN(R) and E=01p q01I(R). From this, we also have

v00vBE=v00vqp p 2 q 2qp01p q 01=0pv0qv=0p0qv

and

vv100vv1BE=vv100vv1qp p 2 q 2qp01p q 01=0pvv10qvv1=0pv10q,

noting v00vqp p 2 q 2qp,vv100vv1qp p 2 q 2qpN(R).

Thus RBE' contains the matrices

10000p0qv=0p00  and 00010pv10q=000q,

entailing EARBE.

Case 2. The results in this case are obtained by the argument of [1, Lemma 2.1(3)].

(i) Suppose that p divides q. Then BE=qpq2pq0001=0p0q=EA, where B=qpq2pqN(R) and E=0001I(R).

(ii) Suppose that q divides p. Then pp2qqp01+pq01=0p0q=EA, where B=pp2qqpN(R) and E=01+pq01I(R).

(iii) Suppose that p and q do not divide each other and gcd(p,q)=1. Then we get BE=qpp2q2qp01pq01=0p0q=EA, where B=qpp2q2qpN(R) and E=01pq01I(R).

Thus there exist BN(R) and EI(R) such that EARBE in any case of (i), (ii) and (iii).

(2) Let p, q be any nonzero integers. Let C=q0p0R be such that C=EA for some EI(R) and AN(R). Then, by Lemma 1.1, we have the cases that 0A=00s0N(R), and

E=E7=1pqmpI(R) (where p(1p)=qm0)

or

E=E5=0t01I(R) (where t0);

that is, EA=qs0ps0 with p=p's and q=q's, or EA=st0s0 with p=s and q=st.

We will find BN(R) and EI(R) such that the left ideal RBE' of R contains EA.

Case 1. Suppose that p and q do not divide each other, and gcd(p,q)1.

By applying the argument and using the notation of (1), we have

BE=qp q 2 p 2qp101p q 0=q 0p 0,

noting B=qp q 2 p 2qpN(R) and E=101p q0I(R). From this, we also have

v00vqp q 2 p 2qp101p q 0=qv0pv0=qv0p0

and

vv100vv1qp q 2 p 2qp101p q 0=qvv10pvv10=q0pv10,

noting v00vqp q 2 p 2qp,vv100vv1qp q 2 p 2qpN(R).

Thus RBE' contains the matrices 1000q0pv10=q000  and 0001qv0p0=00p0,

entailing EARBE.

Case 2. The results in this case are obtained by the argument of [1, Lemma 2.1(3)].

(i) Suppose that p divides q. Then

BE=qq2ppq1000=q0p0=EA, where B=qq2ppqN(R) and E=1000I(R).

(ii) Suppose that q divides p. Then BE=pqp2qp101+pq0=q0p0=EA, where B=pqp2qpN(R) and E=101+pq0I(R).

(iii) Suppose that p and q do not divide each other and gcd(p,q)=1. Then we get BE=qpq2p2qp101pq0=q0p0=EA, where B=qpq2p2qpN(R) and E=101pq0I(R).

Thus there exist BN(R) and EI(R) such that EARBE in any case of (i), (ii) and (iii).

(3) Let p, q be any nonzero integers. Let C=pq00R be such that C=EA for some EI(R) and AN(R). Then we have the cases that E is E1=1000I(R) or E3=1t00I(R), and A=B4=abcaN(R); that is, EA is ab00 with p=a, q=b, or a+tcbta00 with p=a+tc, q=b-ta.

We will find BN(R) and EI(R) such that the right ideal BE'R of R contains EA.

Case 1. Suppose that q divides p.

We have

0q0000pq1=pq00=EA,

noting B=0q00N(R) and E=00pq1I(R).

Based on Case 1, we can assume that |q|2 in the cases below.

Case 2. Suppose that p divides q.

We have

01001qq(1q)ppq=pq00=EA,

noting B=0100N(R) and E=1qq(1q)ppqI(R).

Case 3. Suppose that p and q do not divide each other.

Let p=p'k and q=q'k with gcd(p,q)=k. Take B=0k00N(R) and E=0001I(R). Then BE'R contains BE=0k00, 0k000011=kk00 and, consequently, contains k000.

Thus BE'R contains k000p000=p000 and 0k00000q=0q00, and hence contains pq00.

(4) Let p, q be any nonzero integers. Let C=00qpR be such that C=EA for some EI(R) and AN(R).

Then we have the cases that E is E2=0001I(R) or E6=00u1I(R), and A=B4=abcaN(R); that is, EA is 00ca with q=c, p=-a, or 0000 with q=ua+c, q=ub-a.

We will find BN(R) and EI(R) such that the right ideal BE'R of R contains EA.

Case 1. Suppose that q divides p.

We have

00q01pq00=00qp=EA,

noting B=00q0N(R) and E=1pq00I(R).

Based on Case 1, we assume |q|2 in the cases below.

Case 2. Suppose that p divides q.

We have

0010qpq(1q)p1q=00qp=EA,

noting B=0010N(R) and E=qpq(1q)p1qI(R).

Case 3. Suppose that p and q do not divide each other.

Let p=p'k and q=q'k with gcd(p, q)=k. Take B=00k0N(R) and E=1000I(R). Then BE'R contains BE=00k0, 00k01100=00kk and, consequently, contains 000k.

Thus BE'R contains 00k0q000=00q0 and 000k000p=000p, and hence contains 00qp.

(5) Let p be any nonzero integer.

Let C=p000R be such that C=EA for some EI(R) and AN(R).

Then we have the case that E=E3=1t00I(R) and A=B3=00c0N(R); that is, EA=ct000 with p=ct. Take B=0p00N(R) and E=0001I(R). Then BE'R contains 0p00, 0p000011=pp00 and, consequently, contains EA=p000.

Let C=000pR be such that C=EA for some EI(R) and AN(R).

Then we have the case that E=E6=00u1I(R) and A=B2=0b00N(R); that is, EA=000bu with p=bu. Take B=00p0N(R) and E=1000I(R). Then BE'R contains 00p0, 00p01100=00pp and, consequently, contains EA=000p.

Let C=0p00R be such that C=EA for some EI(R) and AN(R).

Then we have the case that E=E1=1000I(R) and A=B2=0p00N(R). Take B=0p00N(R) and E=E2=0001I(R). Then BE'R contains EA=0p00.

Let C=00p0R be such that C=EA for some EI(R) and AN(R).

Then we have the case that E=E2=0001I(R) and A=B3=00p0N(R). Take B=00p0N(R) and E=E1=1000I(R). Then BE'R contains EA=00p01000=00p0.

(6) Let p, q, r, s be nonzero integers, and let C=pqrsR be such that C=EA for some EI(R) and AN(R). Then, by Lemma 1.1, BE' must be one of B4E3,B4E6, or B4E7.

By (1) and (2), there exist CiN(R) and FiI(R) such that 0q0sRC1F1 and p0r0RC2F2, from which we obtain CRC1F1+RC2F2.

By (3) and (4), there exist CiN(R) and FiI(R) such that pq00C1F1R and 00rsC2F2R, from which we obtain CC1F1R+C2F2R.

Similar arguments are available to the cases of pqr0, pq0s, p0rs, 0qrs, p00s and 0qr0.

Mat2() is shown to be not right IQNN by [1, Theorem 2.3(1)]. In the next section, we introduce a closely related property that Mat2() satisfies, based on Remark 1.3.

### 2. Weakly Right IQNN Rings

Motivated by the arguments of Remark 1.3., we consider the following new ring property as a generalization of right IQNN ring.

Definition 2.1. A ring R is said to be weakly right IQNN provided that I(R)' is empty, or else every pair (e,a)I(R)×N(R) satisfies one of the following:

(i) There exist bN(R) and fI(R) such that eabfR;

(ii) There exist biN(R) and fiI(R) (i=1, 2) such that eab1f1R+b2f2R.

R is called weakly left IQNN provided that I(R)' is empty, or else every pair (e,a)I(R)×N(R) satisfies one of the following:

(i) There exist bN(R) and fI(R) such that aeRfb;

(ii) There exist biN(R) and fiI(R) (i=1, 2) such that aeRf1b1+Rf2b2.

A ring is weakly IQNN if it is both weakly right IQNN and weakly left IQNN.

Right IQNN rings are clearly weakly right IQNN, but not conversely as we see in the arguments below.

Theorem 2.2. Mat2(A) is weakly IQNN over any ring A.

Proof. Let R=Mat1(A). We apply the argument of Remark 1.3. Let 0M=pqrsR. Take

E1=0001,E2=1000I(R)

and

B1=0100,B2=0010N(R).

Then we have

M=B1E100pq+B2E2rs00B1E1R+B2E2R.

Thus R is weakly right IQNN. The proof for the case of weakly left IQNN can be done symmetrically.

Mat2() is not right IQNN as mentioned above, and can be shown to be not left IQNN by a symmetrical method of the proof of [1][Theorem 2.3(1)]. Thus the concept of weakly right (resp., left) IQNN is a proper generalization of right (resp., left) IQNN.

In the following, we see another kind of weakly right IQNN rings but not right IQNN.

Example 2.3. Let K=2 and A=Ka,b be the free algebra with noncommuting indeterminates a, b over K.

(1) We use the ring of the ring of [4, Example 2.3(2)]. Let I be the ideal of A generated by a2- a, b2, ab and set R=A/I and identify the elements in A with their images in R1 for simplicity. Then a2 = a and ab = 0 = b2.

By applying the arguments of [4, Example 2.3(1)] and [5, Example 2.6], we have the following:

(i) every element r∈ R is of the form r=α0+α1a+α2b+α3ba, where α0,α1,α2,α3K;

(ii) I(R)=1+a+γba,a+γbaγ,γK and N(R)=αba+βbα,β K, that is an ideal of R (i.e., N(R)=N*(R));

(iii) S1={eneI(R),nN(R)}=N(R) and S2={nenN(R), eI(R)}={b+ba,ηbaδ,ηK}. So S1S2.

Since S1S2, R is (weakly) left IQNN. Next consider e=1+aI(R) and bN(R). Then eb=b. Since bS2, R is not right IQNN. But

b=(b+ba)+ba=(b+ba)(1+a)+(ba)ac1e1R+c2e2R,

where e1=1+a,e2=aI(R) and c1=b+ba,c2=baN(R). Thus R is weakly right IQNN.

(2) Let J be the ideal of A generated by a2a,b2,ba. and set R=A/J. Then R' is the opposite ring of R of (1). Then a similar argument shows that R' is weakly IQNN but not left IQNN.

(3) Let I' be the ideal of A generated by a2a,b2,abb and set R=A/I and identify the elements in A with their images in R for simplicity. Then a2=a, ab=b, and b2=0 in R. From this relation we obtain the following:

(i) Every element r∈ R is of the form r=α0+α1a+α2b+α3ba, where αiK;

(ii) I(R)=1+a+γb+γba,a+γb+γbaγ,γK and N(R)=αb+α baα,αK, that is an ideal of R;

(iii) S1={eneI(R),nN(R)}=N(R) and S2={nenN(R), eI(R)}={αba,b+ba}. So S1S2.

Since S1S2, R is (weakly) left IQNN. Next consider e=aI(R) and b∈ N(R). Then eb=ab=b. But bS2 and so R is not right IQNN. But

b=(b+ba)+ba=(b+ba)(1+a)+(ba)ac1e1R+c2e2R,

where e1=1+a,e2=aI(R) and c1=b+ba,c2=baN(R). Thus R is weakly right IQNN.

(4) Let J' be the ideal of A generated by a2a,b2,bab. and set R=A/J. Then R' is the opposite ring of R of (3). Then a similar argument shows that R' is weakly IQNN but not left IQNN.

The non-Abelian rings of Example 2.3 are all weakly IQNN. Next we provide a method by which one can construct non-Abelian rings that are neither weakly right nor weakly left IQNN.

Example 2.4. We use the ring of [3][Example 1.2(2)]. Let K=2 and A=Ka,b be the free algebra with noncommuting indeterminates a, b over K. Let I be the ideal of A generated by a2-a, b2 and set R=A/I. Identify the elements in A with their images in R for simplicity. Then a2 = a and b2=0. By help of the argument of [3][Example 1.2(2)], we can express r∈ R and cN(R) by

r=k0+k1a+k2b+af1a+af2b+bf3a+bf4b and c=kb+bfb

where k,kiK and f,fjR for all j.

Let e=aI(R) and c=bN(R). Assume that ec=ab=c'e'r for some c=kb+bfbN(R), eI(R) and r∈ R. Since ab0, c=kb+bfb=b(k+fb)0 and this yields

ab=b(k+fb)er and 0bab=bb(k+fb)er=0,

a contradiction. Next assume that ec=ab=c1e1r1+c2e2r2 for some 0ci=kib+bfibN(R), eiI(R) and riR (i=1, 2). This yields

ab=b((k1+f1b)e1r1+(k2+f2b)e2r2)

and

0bab=bb((k1+f1b)e1r1+(k2+f2b)e2r2)=0,

a contradiction. Thus R is not weakly right IQNN. It is also shown by a symmetrical argument that R is not weakly left IQNN.

Next we consider two kinds of rings R over which T2(R) may be weakly right IQNN.

Proposition 2.5. Let R be a ring.

• (1) If N(R)=N*(R) then T2(R) is weakly right IQNN.

• (2) If I(R)={0, 1} then T2(R) is weakly right IQNN.

Proof. Write T=T2(R). Note that

I(T)=eg0fTe,fI(R),(e,f){(0,0),(1,1)},eg+gf=g

and

N(T)=ac0bTa,bN(R) and cR.

(1) Assume N(R)=N*(R). Let E=eg0fI(T) and A=ac0bN(T). Then EA=eaec+gb0fbN(T), i.e., ea,fbN(R), by assumption. Take E1=1000, E2=0001 and B1=ea000, B2=0ec+gb0fb. Then EiI(T) and BiN(T) such that EA=B1E1+B2E2B1E1T+B2E2T. Thus T is weakly right IQNN.

(2) Assume I(R)={0,1}. Then

I(T)=eg0fT(e,f){(1,0),(0,1)},gR.

Let E=eg0fI(T) and A=ac0bN(T). Then EA=eaec+gb0fbN(T) since e,f{0,1}; in fact, ea is zero or a, and fb is also zero or b. Take E1=1000, E2=0001 and B1=ea000, B2=0ec+gb0fb. Then EiI(T) and BiN(T). Since EA=B1E1+B2E2B1E1T+B2E2T. Thus T is weakly right IQNN.

In the following argument we see a condition under which the weakly IQNN property is right-left symmetric. Let R be a ring. An involution on a ring R is a function *:RR which satisfies the properties that (x+y)*=x*+y*, (xy)*=y*x*, 1*=1, and (x*)*=x for all x,yR. It is easily checked that 0*=0, aN(R) implies a*N(R), and e*I(R) for eI(R). We use these facts without referring.

Proposition 2.6. Let R be a ring with an involution *. Then R is weakly right IQNN if and only if R is weakly left IQNN.

Proof. Assume that I(R)' is nonempty. Suppose that R is weakly right IQNN. Let aN(R) and e∈ I(R)'. Then a*N(R) and e*I(R). Since R is weakly right IQNN, we have the following four cases. We proceed our argument on a case-by-case computation.

(i) There exist bN(R), fI(R) and s∈ R such that e*a*=bfs. This implies that

ae=((ae)*)*=(e*a*)*=(bfs)*=s*f*b*Rf*b*.

(ii) There exist biN(R), fiI(R) and siR (i=1, 2) such that e*a*=b1f1s1+b2f2s2. This implies that

ae=((ae)*)*=(e*a*)*=(b1f1s1+b2f2s2)*=s1*f1*b1*+s2*f2*b2*Rf1*b1*+Rf2*b2*.

Since b*,bi*N(R) and f*,fi*I(R), we now conclude that R is weakly left IQNN by the results (i) and (ii).

Conversely suppose that R is weakly left IQNN. Then we have the following cases.

(iii) There exist bN(R), fI(R) and r∈ R such that a*e*=rfb. This implies that

ea=((ea)*)*=(a*e*)*=(rfb)*=b *f *r*b *f *R.

(iv) There exist biN(R), fiI(R) and riR (i=1, 2) such that a*e*=r1 f 1 b 1+r2 f 2 b 2. This implies that

ea=((ea)*)*=(a*e*)*=(r1 f 1 b 1+r2 f 2 b 2)*=b 1*f 1*r1*+b 2*f 2*r2*b 1*f 1*R+b 2*f 2*R.

Since b*, bi*N(R) and f*, fi*I(R), we now conclude that R is weakly right IQNN by the results (iii) and (iv).

### Footnote

The second named author was supported by Kyungdong University Research Fund, 2022.

The third named author was supported by the Science and Technology Research Project of Education Department of Jilin Province, China(JJKH20210563KJ).

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