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 n≥2, denote the full and upper triangular matrix rings over R by Mat_n(R) and T_n(R) respectively. Let ℤ, ℤn, and ℚ denote the ring of integers, ring of integers modulo n, and the field of rational numbers, respectively. For m,n∈ℤ, 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].

The following facts have essential role in this article.

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 1t≠0,u≠0B1= 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)′=E1∪E2∪…∪E7 and N(R)=B1∪…∪B4.

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

M∈E7B4∪B4E7∪E4B4∪E5B4∪B4E3∪B4E6,

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=ℤ and R=Mat2(F).

(1) Let p, q be any nonzero integers. Let C=0p0q∈R be such that C=EA for some E∈I(R)′ and A∈N(R). Then, by Lemma 1.1, we have the cases that A=B2=0v00∈N(R), and

E=E7=p′mq′1− p ′ ∈I(R)′, where p′(1−p′)=q′m≠0

or

E=E4=10u0∈I(R)′, where u≠0.

That is, EA is 0p′v0q′v with p=p'v and q=q'v, or 0v0uv with p=v and q=uv.

We will find B∈N(R) and E′∈I(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′=p″v1 and q′=q″v1 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″=p1u1⋯pfuf and q″=q1v1⋯qgvg, with ui,vj≥1, we the prime number decompositions of p″ and q″ respectively. Since p″(1−p′)=q″m and gcd(p″,q″)=1, q″ must divide 1-p'. Letting 1−p′=q″m′, we have p′+q″m′=1 and this implies gcd(p′,q″)=1 (hence gcd(v1,q″)=1).

Since q″ divides 1−p′ as above, we have that −1−p′q″∈ℤ and

BE′=−q″p′ p ′ 2− q ″ 2q″p′0−1−p′ q ″ 01=0p ′ 0q″ ,

noting B=−q″p′ p ′ 2− q ″ 2q″p′∈N(R) and E′=0−1−p′ q″01∈I(R)′. From this, we also have

v00vBE′=v00v−q″p′ p ′ 2− q ″ 2q″p′0−1−p′ q ″ 01=0p′v0q″v=0p0q″v

and

vv100vv1BE′=vv100vv1−q″p′ p ′ 2− q ″ 2q″p′0−1−p′ q ″ 01=0p′vv10q″vv1=0pv10q,

noting v00v−q″p′ p ′ 2− q ″ 2q″p′,vv100vv1−q″p′ p ′ 2− q ″ 2q″p′∈N(R).

Thus RBE' contains the matrices

10000p0q″v=0p00  and 00010pv10q=000q,

entailing EA∈RBE′.

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′=−qp−q2pq0001=0p0q=EA, where B=−qp−q2pq∈N(R) and E′=0001∈I(R)′.

(ii) Suppose that q divides p. Then p−p2qq−p01+pq01=0p0q=EA, where B=p−p2qq−p∈N(R) and E′=01+pq01∈I(R)′.

(iii) Suppose that p and q do not divide each other and gcd(p,q)=1. Then we get BE′=−qpp2−q2qp0−1−pq01=0p0q=EA, where B=−qpp2−q2qp∈N(R) and E′=0−1−pq01∈I(R)′.

Thus there exist B∈N(R) and E′∈I(R)′ such that EA∈RBE′ in any case of (i), (ii) and (iii).

(2) Let p, q be any nonzero integers. Let C=q0p0∈R be such that C=EA for some E∈I(R)′ and A∈N(R). Then, by Lemma 1.1, we have the cases that 0≠A=00s0∈N(R), and

E=E7=1−p′q′mp′∈I(R)′ (where p′(1−p′)=q′m≠0)

or

E=E5=0t01∈I(R)′ (where t≠0);

that is, EA=q′s0p′s0 with p=p's and q=q's, or EA=st0s0 with p=s and q=st.

We will find B∈N(R) and E′∈I(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′=q″p′− q ″ 2 p ′ 2−q″p′10−1−p′ q ″ 0=q″ 0p ′ 0,

noting B=q″p′− q ″ 2 p ′ 2−q″p′∈N(R) and E′=10−1−p′ q″0∈I(R)′. From this, we also have

v00vq″p′− q ″ 2 p ′ 2−q″p′10−1−p′ q ″ 0=q″v0p′v0=q″v0p0

and

vv100vv1q″p′− q ″ 2 p ′ 2−q″p′10−1−p′ q ″ 0=q″vv10p′vv10=q0pv10,

noting v00vq″p′− q ″ 2 p ′ 2−q″p′,vv100vv1q″p′− q ″ 2 p ′ 2−q″p′∈N(R).

Thus RBE' contains the matrices 1000q0pv10=q000  and 0001q″v0p0=00p0,

entailing EA∈RBE′.

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′=q−q2pp−q1000=q0p0=EA, where B=q−q2pp−q∈N(R) and E′=1000∈I(R)′.

(ii) Suppose that q divides p. Then BE′=−pq−p2qp101+pq0=q0p0=EA, where B=−pq−p2qp∈N(R) and E′=101+pq0∈I(R)′.

(iii) Suppose that p and q do not divide each other and gcd(p,q)=1. Then we get BE′=qp−q2p2−qp10−1−pq0=q0p0=EA, where B=qp−q2p2−qp∈N(R) and E′=10−1−pq0∈I(R)′.

Thus there exist B∈N(R) and E′∈I(R)′ such that EA∈RBE′ in any case of (i), (ii) and (iii).

(3) Let p, q be any nonzero integers. Let C=pq00∈R be such that C=EA for some E∈I(R)′ and A∈N(R). Then we have the cases that E is E1=1000∈I(R)′ or E3=1t00∈I(R)′, and A=B4=abc−a∈N(R); that is, EA is ab00 with p=a, q=b, or a+tcb−ta00 with p=a+tc, q=b-ta.

We will find B∈N(R) and E′∈I(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=0q00∈N(R) and E′=00pq1∈I(R)′.

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

Case 2. Suppose that p divides q.

We have

01001−qq(1−q)ppq=pq00=EA,

noting B=0100∈N(R) and E′=1−qq(1−q)ppq∈I(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=0k00∈N(R) and E′=0001∈I(R)′. Then BE'R contains BE′=0k00, 0k000011=kk00 and, consequently, contains k000.

Thus BE'R contains k000p′000=p000 and 0k00000q′=0q00, and hence contains pq00.

(4) Let p, q be any nonzero integers. Let C=00qp∈R be such that C=EA for some E∈I(R)′ and A∈N(R).

Then we have the cases that E is E2=0001∈I(R)′ or E6=00u1∈I(R)′, and A=B4=abc−a∈N(R); that is, EA is 00c−a with q=c, p=-a, or 0000 with q=ua+c, q=ub-a.

We will find B∈N(R) and E′∈I(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=00q0∈N(R) and E′=1pq00∈I(R)′.

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

Case 2. Suppose that p divides q.

We have

0010qpq(1−q)p1−q=00qp=EA,

noting B=0010∈N(R) and E′=qpq(1−q)p1−q∈I(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=00k0∈N(R) and E′=1000∈I(R)′. Then BE'R contains BE′=00k0, 00k01100=00kk and, consequently, contains 000k.

Thus BE'R contains 00k0q′000=00q0 and 000k000p′=000p, and hence contains 00qp.

(5) Let p be any nonzero integer.

Let C=p000∈R be such that C=EA for some E∈I(R)′ and A∈N(R).

Then we have the case that E=E3=1t00∈I(R)′ and A=B3=00c0∈N(R); that is, EA=ct000 with p=ct. Take B=0p00∈N(R) and E′=0001∈I(R)′. Then BE'R contains 0p00, 0p000011=pp00 and, consequently, contains EA=p000.

Let C=000p∈R be such that C=EA for some E∈I(R)′ and A∈N(R).

Then we have the case that E=E6=00u1∈I(R)′ and A=B2=0b00∈N(R); that is, EA=000bu with p=bu. Take B=00p0∈N(R) and E′=1000∈I(R)′. Then BE'R contains 00p0, 00p01100=00pp and, consequently, contains EA=000p.

Let C=0p00∈R be such that C=EA for some E∈I(R)′ and A∈N(R).

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

Let C=00p0∈R be such that C=EA for some E∈I(R)′ and A∈N(R).

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

(6) Let p, q, r, s be nonzero integers, and let C=pqrs∈R be such that C=EA for some E∈I(R)′ and A∈N(R). Then, by Lemma 1.1, BE' must be one of B4E3,B4E6, or B4E7.

By (1) and (2), there exist Ci∈N(R) and Fi∈I(R)′ such that 0q0s∈RC1F1 and p0r0∈RC2F2, from which we obtain C∈RC1F1+RC2F2.

By (3) and (4), there exist C′i∈N(R) and F′i∈I(R)′ such that pq00∈C′1F′1R and 00rs∈C′2F′2R, from which we obtain C∈C′1F′1R+C′2F′2R.

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

Mat2(ℤ) 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(ℤ) satisfies, based on Remark 1.3.

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 b∈N(R) and f∈I(R)′ such that ea∈bfR;

(ii) There exist bi∈N(R) and fi∈I(R)′ (i=1, 2) such that ea∈b1f1R+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 b′∈N(R) and f′∈I(R)′ such that ae∈Rf′b′;

(ii) There exist b′i∈N(R) and f′i∈I(R)′ (i=1, 2) such that ae∈Rf′1b′1+Rf′2b′2.

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 0≠M=pqrs∈R. Take

E1=0001,E2=1000∈I(R)′

and

B1=0100,B2=0010∈N(R).

Then we have

M=B1E100pq+B2E2rs00∈B1E1R+B2E2R.

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

Mat2(ℤ) 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=ℤ2 and A=K〈a,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,α3∈K;

(ii) I(R)′=1+a+γba,a+γ′ba∣γ,γ′∈K and N(R)=αba+βb∣α,β∈ K, that is an ideal of R (i.e., N(R)=N^*(R));

(iii) S1={en∣e∈I(R)′,n∈N(R)}=N(R) and S2={n′e′∣n′∈N(R), e′∈I(R)′}={b+ba,ηba∣δ,η∈K}. So S1⊋S2.

Since S1⊋S2, R is (weakly) left IQNN. Next consider e=1+a∈I(R)′ and b∈N(R). Then eb=b. Since b∉S2, R is not right IQNN. But

b=(b+ba)+ba=(b+ba)(1+a)+(ba)a∈c1e1R+c2e2R,

where e1=1+a,e2=a∈I(R)′ and c1=b+ba,c2=ba∈N(R). Thus R is weakly right IQNN.

(2) Let J be the ideal of A generated by a2−a,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 a2−a,b2,ab−b 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 αi∈K;

(ii) I(R)′=1+a+γb+γba,a+γ′b+γ′ba∣γ,γ′∈K and N(R)=αb+α′ ba∣α,α′∈K, that is an ideal of R;

(iii) S1={en∣e∈I(R)′,n∈N(R)}=N(R) and S2={n′e′∣n′∈N(R), e′∈I(R)′}={αba,b+ba}. So S1⊋S2.

Since S1⊋S2, R is (weakly) left IQNN. Next consider e=a∈I(R)′ and b∈ N(R). Then eb=ab=b. But b∉S2 and so R is not right IQNN. But

b=(b+ba)+ba=(b+ba)(1+a)+(ba)a∈c1e1R+c2e2R,

where e1=1+a,e2=a∈I(R)′ and c1=b+ba,c2=ba∈N(R). Thus R is weakly right IQNN.

(4) Let J' be the ideal of A generated by a2−a,b2,ba−b. 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=ℤ2 and A=K〈a,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 c∈N(R) by

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

where k,ki∈K and f,fj∈R for all j.

Let e=a∈I(R)′ and c=b∈N(R). Assume that ec=ab=c'e'r for some c′=kb+bfb∈N(R), e′∈I(R)′ and r∈ R. Since ab≠0, c′=kb+bfb=b(k+fb)≠0 and this yields

ab=b(k+fb)e′r and 0≠bab=bb(k+fb)e′r=0,

a contradiction. Next assume that ec=ab=c1e1r1+c2e2r2 for some 0≠ci=kib+bfib∈N(R), ei∈I(R)′ and ri∈R (i=1, 2). This yields

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

and

0≠bab=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)′=eg0f∈T∣e,f∈I(R),(e,f)∉{(0,0),(1,1)},eg+gf=g

and

N(T)=ac0b∈T∣a,b∈N(R) and c∈R.

(1) Assume N(R)=N*(R). Let E=eg0f∈I(T)′ and A=ac0b∈N(T). Then EA=eaec+gb0fb∈N(T), i.e., ea,fb∈N(R), by assumption. Take E1=1000, E2=0001 and B1=ea000, B2=0ec+gb0fb. Then Ei∈I(T)′ and Bi∈N(T) such that EA=B1E1+B2E2∈B1E1T+B2E2T. Thus T is weakly right IQNN.

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

I(T)′=eg0f∈T∣(e,f)∈{(1,0),(0,1)},g∈R.

Let E=eg0f∈I(T)′ and A=ac0b∈N(T). Then EA=eaec+gb0fb∈N(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 Ei∈I(T)′ and Bi∈N(T). Since EA=B1E1+B2E2∈B1E1T+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 *:R→R which satisfies the properties that (x+y)*=x*+y*, (xy)*=y*x*, 1*=1, and (x*)*=x for all x,y∈R. It is easily checked that 0^*=0, a∈N(R) implies a*∈N(R), and e*∈I(R)′ for e∈I(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 a∈N(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 b∈N(R), f∈I(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 bi∈N(R), fi∈I(R)′ and si∈R (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 b′∈N(R), f′∈I(R)′ and r∈ R such that a*e*=rf′b′. This implies that

ea=((ea)*)*=(a*e*)*=(rf′b′)*=b′ *f′ *r*∈b′ *f′ *R.

(iv) There exist b′i∈N(R), f′i∈I(R)′ and ri∈R (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′*, b′i*∈N(R) and f′*, f′i*∈I(R)′, we now conclude that R is weakly right IQNN by the results (iii) and (iv).