Article Search
eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2024; 64(1): 1-14

Published online March 31, 2024

### A Characterization of Nonnil-Projective Modules

Hwankoo Kim, Najib Mahdou and El Houssaine Oubouhou

Division of Computer Engineering, Hoseo University, Asan 31499, Republic of Korea
e-mail : hkkim@hoseo.edu

Department of Mathematics, Faculty of Science and Technology of Fez, Box 2202, University S. M. Ben Abdellah Fez, Morocco
e-mail : najib.mahdou@usmba.ac.ma and hossineoubouhou@gmail.com

Received: September 13, 2023; Revised: October 20, 2023; Accepted: October 23, 2023

Recently, Zhao, Wang, and Pu introduced and studied new concepts of nonnil-commutative diagrams and nonnil-projective modules. They proved that an R-module that is nonnil-isomorphic to a projective module is nonnil-projective, and they proposed the following problem: Is every nonnil-projective module nonnil-isomorphic to some projective module? In this paper, we delve into some new properties of nonnil-commutative diagrams and answer this problem in the affirmative.

Keywords: Nonnil-commutative diagram, Nonnil-exact sequence, Nonnil-projective module

In this paper, all rings are assumed to be commutative with non-zero identity and all modules are assumed to be unitary. For a ring R, we denote by Nil(R) and Z(R) the ideal of all nilpotent elements of R and the set of all zero-divisors of R, respectively. A ring R is called a PN-ring if Nil(R) is a prime ideal of R and a ZN-ring if Z(R)=Nil(R). An ideal I of R is said to be nonnil if INil(R).

Recall from [4] that a prime ideal P of R is said to be divided if it is comparable to every ideal of R. Let H:={RR be a commutative ring, and Nil(R) be a divided prime ideal of R}. If RH, then R is called a φ-ring. A φ-ring is called a strongly φ-ring if it is also a ZN-ring. Recall from [1] that for a φ-ring R with total quotient ring T(R), the map φ:T(R)RNil(R) such that φba=ba is a ring homomorphism, and the image of R, denoted by φ(R), is a strongly φ-ring. The classes of φ-rings and strongly φ-rings are good extensions of integral domains to commutative rings with zero-divisors. In 2002, Badawi [6] generalized the concept of Noetherian rings to that of nonnil-Noetherian rings in which all nonnil ideals are finitely generated. He showed that a φ-ring R is nonnil-Noetherian if and only if φ(R) is nonnil-Noetherian, if and only if R/Nil(R) is a Noetherian domain. Generalizations of Dedekind domains, Prüfer domains, Bézout domains, pseudo-valuation domains, Krull domains, valuation domains, Mori domains, piecewise Noetherian domains, and coherent domains to the context of rings that are in the class H are also introduced and studied. We recommend [2, 3, 5, 7, 8, 9, 10, 11] for studying the ring-theoretic characterizations on φ-rings.

To investigate module-theoretic characterizations on φ-rings, the authors [12, 14, 17, 18, 19, 21] introduce nonnil-injective modules, φ-projective, and φ-flat modules, and characterize nonnil-Noetherian rings, φ-von Neumann regular rings, nonnil-coherent rings, φ-coherent rings, φ-Dedekind rings, and φ-Prüfer rings. Let M be an R-module and set

Ntor(M):={xMsx=0 for some sRNil(R)}.

If Ntor(M)=M, then M is called a φ-torsion module, and if Ntor(M)=0, then M is called a φ-torsion-free module. Recall from [18] that an R-module F is said to be φ-flat if for every R-monomorphism f:AB with Coker(f) being a φ-torsion R-module, we have 1FRf:FRAFRB is an R-monomorphism; equivalently, Tor1R(F,M)=0 for every φ-torsion R-module M (see for instance [15, 16, 18]). If R is a PN-ring, define φ:RRNil(R) by φ(r)=r1 for every rR. Then φ(R) is a ZN-ring. In [17], Zhao defined the map ψ:MMNil(R) by ψ(x)=x1 for every xM. This makes ψ(M) a φ(R)-module. If f:MN is a homomorphism of R-modules, then f induces naturally a φ(R)-homomorphism f˜:ψ(M)ψ(N) such that f˜x1=f(x)1 for xM. A sequence of R-modules and homomorphisms AfBgC is called φ-exact if the φ(R)-sequence: ψ(A)f˜ψ(B)g˜ψ(C) is exact, and an R-module P is said to be φ-projective (resp., φ-free) if ψ(P) is projective (resp., free) as a φ(R)-module. Let R be a PN-ring and let f:AB be a homomorphism of R-modules. Set

NKer(f):={aAsf(a)=0 for some sRNil(R)} and
NIm(f):={bBsb=sf(a) for some aA and sRNil(R)}.

Because Nil(R) is prime, NKer(f) is a submodule of A, called the nonnil-kernel of f, and NIm(f) is a submodule of B, called the nonnil-image of f. We set NCoker(f):=B/NIm(f). It is easy to verify that Ker(f)+Ntor(A)NKer(f) and Im(f)+ Ntor(B)=NIm(f). Let A,B,C,D be R-modules and f:AB,g:BD,h:AC,k:CD be homomorphisms of R-modules. Then the following diagram:

is said to be nonnil-commutative if NIm(gf-kh)=Ntor(D); equivalently, NKer(gf-kh)=Ntor(A). A sequence of R-modules and homomorphisms AfBgC is called a nonnil-complex (resp., a nonnil-exact sequence) if it is φ-complex (resp., φ-exact); equivalently, NIm(f)NKer(g) (resp., NIm(f)=NKer(g)) according to [17, Theorem 2.6]. A homomorphism f:AB of R-modules is called a nonnil-monomorphism if NKer(f)=Ntor(A), equivalently 0AfB is a nonnil-exact sequence; f is called a nonnil-epimorphism if NIm(f)=B (i.e., NCoker(f)=0), equivalently AfB0 is a nonnil-exact sequence. Also f is called a nonnil-isomorphism if there exists a homomorphism g:BA such that NIm1A-gf=Ntor(A) and NIm1B-fg=Ntor(B). If there exists a nonnil-isomorphism f:AB, we say that A and B are nonnil-isomorphic, denoted by ANB. Note that if f:AB is a nonnil-isomorphism, then f is both a nonnil-monomorphism and a nonnil-epimorphism. Interestingly, a homomorphism f of R-modules is both a nonnil-monomorphism and a nonnil-epimorphism without being a nonnil-isomorphism (see [20]). Following [20], an R-module P is said to be nonnil-projective if given any diagram of module homomorphisms

with the bottom row nonnil-exact, there is a homomorphism h:PB making this diagram nonnil-commutative. Also an R-module F0 is said to be N-free if it is nonnil-isomorphic to a free module. Following [20, Theorem 3.7], an R-module is nonnil-projective if and only if it is a direct summand of an N-free module. If an R-module P is nonnil-isomorphic to a projective module, then P is nonnil-projective (cf. [20, Corollary 3.8]). Afterward, they proposed an interesting problem as follows.

Problem: Is every nonnil-projective module nonnil-isomorphic to some projective module?

One of the main aims of this paper is to answer this problem. Section 2 studies some new properties of nonnil-commutative diagrams and nonnil-exact sequences. In the last section, we solved the previous problem in the affirmative: An R-module is nonnil-projective if and only if it is nonnil-isomorphic to a projective module (Theorem 3.1 and Remark 3.6). In this paper, R always denotes a PN-ring.

### 2. On Nonnil-Commutative Diagrams

We start this section by providing a nonnil-analog of Five Lemma.

Theorem 2.1. Consider the following nonnil-commutative diagram with exact rows:

• If α and γ are nonnil-monomorphisms and δ is a nonnil-epimorphism, then β is a nonnil-monomorphism.

• If α and γ are nonnil-epimorphisms and μ is a nonnil-monomorphism, then β is a nonnil-epimorphism.

Proof. (1) Let bNKer(β). Then there exists t1RNil(R) such that t1β(b)=0. On the other hand, there exists t2RNil(R) such that t2γg(b)=t2gβ(b). Hence t1t2γg(b)=t2g(t1β(b))=0. Therefore, g(b)NKer(γ). Since γ is a nonnil-monomorphism, there exists t3RNil(R) such that t3g(b)=0, and so bNKer(g)=NIm(f). Then t4b=t4f(a) for some aA and t4RNil(R). Hence

t4(βf(a)-fα(a))=t4(β(b)-f(α(a)). Since aA, it follows that t5(fα(a)-βf(a))=0 for some t5RNil(R). Therefore

0=t1t4t5(f°α(a)β°f(a)) =t1t5t4(β(b)+f°α(a)) =t5t4β(t1b)+t1t4t5f°α(a) =t1t4t5f°α(a).

Hence α(a)NKer(f)=NIm(h), and so t6α(a)=t6h(x) for some t6RNil(R) and xD. Since δ is a nonnil-epimorphism, there exist some xD and t7RNil(R) such that t7δ(x)=t7x. Hence

t6t7α(a)=t7t6h(x) =t6h(t7x) =t6h(t7δ(x)) =t6t7h°δ(x).

On the other hand, since xD, it follows that t8hδ(x)=t8αh(x) for some t8RNil(R). So t6t7t8α(a)=t6t7t8hδ(x)=t6t7t8αh(x), and hence t6t7t8α(a-h(x))=0. Therefore, a-h(x)NKer(α)=Ntor(A), and hence there exists t9RNil(R) such that t9a=t9h(x). Since h(x)Im(h)NIm(h)=NKer(f), we get t10fh(x)=0 for some t10RNil(R). Then

t4t9t10b=t9t10t4f(a). =t10t4f(t9h(a)) =t4t9t10f°h(a)=0

Therefore, tb=0 with t:= t4t9t10RNil(R), and so bNtor(B). Thus β is a nonnil-monomorphism.

(2) Let bB. Since γ is a nonnil-epimorphism, there exist cC and t1RNil(R) such that t1γ(c)=t1g(b). Nonnil-commutativity of the right square gives t2μk(c)=t2jkγ(c) for some t2RNil(R). Then

t1t2μ°k(c)=t2k(t1γ(c)) =t2k(t1g(b)) =t1t2k°g(b).

Since g(b)Im(g)NIm(g)=NKer(k), there exists t3RNil(R) such that t3kg(b)=0, and so t1t2t3μk(c)=0. Therefore, k(c)NKer(γ)=Ntor(E). Consequently there exists t4RNil(R) such that t4k(c)=0, and hence cNKer(k)=NIm(g), that is, t5c=t5g(b) for some t5RNil(R) and bB. On the other hand, since bB, there exists t6RNil(R) such that t6γg(b)=t6gβ(b). Then

t1t5t6g(b)=t1t5t6γ(c) =t1t6γ(t5g(b) =t1t5t6g°β(b).

Thus t1t5t6g(b-β(b))=0, and so b-β(b)NKerg=NIm(f). Hence there exist t7RNil(R) and aA such that t7(b-β(b))=t7f(a). Since α is a nonnil-epimorphism, there exist some aA and t8RNil(R) such that t8α(a)=t8a. Hence

t8t7(b-β(b))=t8t7f(a)=t7t8fα(a).

Since aA, there exists t9RNil(R) such that t9fα(a)=t9βf(a), and so t9t8t7(b-β(b))=t7t8t9βf(a). Thus tb=tβ(b+f(a) with t:=t7t8t9RNil(R). Consequently β is a nonnil-epimorphism.

Let M be an R-module. Define ψ:MMNil(R) such that ψ(x)=x1 for every xM.

Proposition 2.2. Let f:AB be an R-module homomorphism. Then A/NKer(f)ψ(Im(f)).

Proof. Let x,yA. Then we have:

f(x)1=f(y)1ψ((f))s(RNil(R)):sf(x)=sf(y) s(RNil(R)):sf(xy)=0 xyNKer(f) x¯=y¯A/NKer(f).

Hence the homomorphism:

g:A/NKer(f)ψ((f))x¯g(x¯)=f(x)1

is an isomorphism.

A nonempty subset S of R is said to be a multiplicative subset if 1S, 0S, and for each a,bS, we have abS. Note that if there exists sSNil(R), then there exists a positive integer n such that 0=snS, a contradiction. Hence we always assume that SNil(R)=.

It is well known that if MfMgM is an exact sequence of R-modules, then MSfSMSgSMS is also exact. The following theorem gives the nonnil-version of this result.

Theorem 2.3. Let R be a ring, S be a multiplicative subset of R, and MfMgM be a nonnil-exact sequence of R-modules. Then MSfSMSgSMS is a nonnil-exact sequence.

Proof. Let ysNIm(fS). Then there exist ts1RSNil(RS) and xsMS such that ts1ys=ts1fS(xs)=tf(x)s1s. Thus there exists s2S such that s2tss1y=s2s1stf(x)=s2tf(s1sx). Hence s1syNIm(f)=NKer(g) since s2tRNil(R), and so tg(s1sy)=0 for some tRNil(R). Therefore, tg(y)s=0, whence t1gS(ys)=0 and t1RSNil(RS). Thus ysNKer(gS). Conversely, let xsNKer(gS). Then ts1g(x)s=0 for some ts1RSNil(RS). Thus there exists s2S such that ts2f(x)=0, whence s2xNKer(f)=NIm(g) since tRNil(R), that is, t1s2x=t1f(x) for some xM and t1RNil(R). Then

t1s21xs=t1f(x)s=t1s2f(x)s2s=t1s21fS(xs2s).

Thus xsNIm(fS) since t1s21RSNil(RS).

Remark 2.4. If S:=RNil(R), then MfMgM is a nonnil-exact sequence if and only if MSfSMSgSMS is exact.

Note that a nonnil-monomorphism is not always a monomorphism (see [17]). But if we consider K as a field and M as a K-vector space, and let R=KM be the trivial extension. Then the homomorphism g:MR defined by g(x)=(0,x) is not a nonnil-epimorphism; in fact, (1,0)NIm(g). Now we give an example of a nonnil-epimorphism which is not an epimorphism.

Example 2.5. Let R=ZZ/2Z and consider g:ZR defined by g(a)=(a,0). Since 2(0Z/2Z)=0, it follows that (0Z/2Z)Ntor(R). Then NIm(g)=(g)+Ntor(R)=R. Hence g is a nonnil-epimorphism, which is not an epimorphism.

Proposition 2.6. Let f:MN be an R-module homomorphism and S be a multiplicative subset of R. Then the following statements are equivalent:

• f is a nonnil-monomorphism,

• fS is a nonnil-monomorphism.

Proof. (1)(2) This is straightforward by Theorem 2.3.

(2)(1) Assume that fS is a nonnil-monomorphism. Set M:=NKer(f). Then we have the following nonnil-exact sequence: 0MiMfN. Thus 0MSiSMSfSNS is also a nonnil-exact sequence. Hence Ntor(MS)+(iS)=NIm(iS)=NKer(fS)=Ntor(MS), and so MSNtor(MS). Now let xM. Then x1Ntor(MS), whence ts1x1=0 for some ts1RSNil(RS). Thus stx=0 for some sS. Since Nil(R) is a prime ideal of R, stRNil(R), and so xNtor(M). Therefore, NKer(f)=Ntor(M). Consequently f is a nonnil-monomorphism.

Proposition 2.7. Let f:MN be an R-module homomorphism. Then the following statements are equivalent:

• f is a nonnil-epimorphism,

• fp is a nonnil-epimorphism for any prime ideal p of R,

• fm is a nonnil-epimorphism for any maximal ideal m of R.

Proof. (1)(2) Assume that f is a nonnil-epimorphism. Then MfN0 is nonnil-exact. Let p be a prime ideal of R. Then for S:=Rp, we have MpfpNp0 is nonnil-exact according to Theorem 2.3. Thus fp is a nonnil-epimorphism for any prime ideal p of R.

(2)(3) This is straightforward.

(3)(1) Let yN. Then y1Nm=NIm(fm) for any maximal ideal m of R. Thus for every mMax(R), there exist tmαmRmNil(Rm), xM, and smRm such that tmαmy1=tmαmfm(xsm). So smαmtmsmy=smαmtmf(x) for some smRm. Set S:={smm is a maximal ideal of R}. Since S generates R, there exist finite elements sm1,,smn of S and α1,,αnR such that 1=α1sm1++αnsmn. For all i=1,,n, we have smiαmitmismiy=smiαmitmif(x), and so sαmitmismiy=sαmitmif(x) with s:=sm1sm2smn. Then

sαmitmiy=sαmitmi(α1sm1++αnsmn)y =sαmitmiα1sm1y++sαmitmiαnsmny =sαmitmiα1f(x)++stαnf(x) =sαmitmif(α1x++stαnx).

Since Nil(R) is a prime ideal of R, it follows that  sαmitmiRNil(R). Therefore, yNIm(f).

Recall that a ring R is called a φ-von Neumann regular ring if R/Nil(R) is a field [18, Theorem 4.1]. Note that if R is a φ-von Neumann regular ring, then every non-nilpotent element of R is a unit. We end this section with the following theorem, which characterizes when each nonnil-commutative diagram (resp., nonnil-exact sequence, nonnil-monomorphism, nonnil-epimorphism, nonnil-isomorphism) is commutative (resp., exact, monomorphism, epimorphism, isomorphism).

Theorem 2.8. Let R be a ring. Then the following conditions are equivalent:

• Every nonnil-commutative diagram is commutative,

• Every nonnil-exact sequence is exact,

• Every nonnil-monomorphism is a monomorphism,

• Every nonnil-epimorphism is an epimorphism,

• Every nonnil-isomorphism is an isomorphism,

• R is a φ-von Neumann regular ring.

Proof. (1)(5), (2)(3) (5), and (6)(2) (3) are straightforward.

(3)(6) Let aRNil(R) and consider the following homomorphism f:R/Ra0. Since Ntor(R/Ra)=R/Ra, it follows that R/Ra=Ntor(R/Ra)NKer(f)R/Ra, and so NKer(f)=Ntor(R/Ra). Hence f is a nonnil-monomorphism, and so it is a monomorphism by (3). Then R/Ra=Ker(f)=0, and hence a is a unit. Consequently (R,Nil(R)) is a local ring. Hence Nil(R) is a divided prime ideal of R. Thus R is a φ-ring with R/Nil(R) being a field. Therefore, R is a φ-von Neumann regular ring by [18, Theorem 4.1].

(4)(6) Let aRNil(R) and consider the following homomorphism f:0R/Ra. Since Ntor(R/Ra)=R/Ra, it follows that NIm(f)=Im(f)+Ntor(R/I)=0+R/Ra=R/Ra. Hence f is a nonnil-epimorphism, and so it is an epimorphism. Consequently 0=(f)=R/Ra, and so a is a unit. Hence as in the above, R is a φ-von Neumann regular ring.

(5)(6) Let aRNil(R). Since a(R/Ra)=0, it is easy to verify that R/RaN0 (see Lemma 3.3), and so R/Ra=0 by (5). Therefore, a is a unit, and so as in the above, R is a φ-von Neumann regular ring.

### 3. Characterizing Nonnil-Projective Modules Using Projective Modules

The nonnil-projective module was studied in [20] using an N-free module, a right nonnil-split sequence, and a nonnil-projective basis. In particular, if an R-module P is nonnil-isomorphic to a projective module P0, then P is nonnil-projective, and they conclude their paper by proposing the following problem.

Problem: Is every nonnil-projective module nonnil-isomorphic to some projective module?

The following theorem solves this difficulty by stating that an R-module is nonnil-projective if and only if it is nonnil-isomorphic to a projective module.

Theorem 3.1. Let R be a ZN-ring. Then every nonnil-projective module is nonnil-isomorphic to some projective module.

We need simple but necessary lemmas to prove Theorem 3.1.

Lemma 3.2. If A1NB1 and A2NB2, then A1A2NB1B2.

Proof. Let f1:A1B1 and f2:A2B2 be two nonnil-isomorphisms. Then there exist two homomorphisms g1:B1A1 and g2:B2A2 such that NIm(1A1-f1g1)=Ntor(A1), NIm(1B1-g1f1)=Ntor(B1), NIm(1A2-f2g2)=Ntor(A1), and NIm(1B2-g2f2)=Ntor(B2). Define

f:A1A2B1B2by(x1,x2)f(x1,x2)=(f1(x1),f2(x2))

and

g:B1B2A1A2by(x1,x2)g(x1,x2)=(g1(x1),g2(x2)).

Then it is easy to verify that:

NIm(1A1A2f°g)=NIm(1A1f1°g1)NIm(1A2f2°g2) =Ntor(A1)Ntor(A2) =Ntor(A1A2)

and

NIm(1B1B2g°f)=NIm(1B1g1°f1)NIm(1B2g2°f2) =Ntor(B1)Ntor(B2) =Ntor(B1B2).

Hence A1A2NB1B2.

Lemma 3.3. Let M be an R-module. Then MN0 if and only if M is a φ-torsion R-module.

Proof. Let f:M0 be a nonnil-isomorphism. Then NIm(1M-f0)=Ntor(M). Since NIm(1M-f0)=NIm(1M)=M, we get M=Ntor(M). Conversely, assume that M=Ntor(M). Then f:M0 is a nonnil-isomorphism since NIm(1M)=M=Ntor(M).

For any submodule N of an R-module M and any multiplicative subset S of R, we define

SM(N):={xMsxN for some sS},

called the S-component of N in M. If no confusion can arise, we will also write S(N) instead of SM(N). From this point on, set S:=RNil(R).

Lemma 3.4. Let f:AB be a nonnil-isomorphism and N be a submodule of A. Then S(N)Nf(S(N)).

Proof. Let g:BA such that NIm(1A-gf)=Ntor(A) and NIm(1B-fg)=Ntor(B). Define fS(N):S(N)f(S(N)) as the restriction of f on S(N). Let y=f(n)f(S(N)) with nN. Then there exists t1RNil(R) such that t1nN. On the other hand, since NIm(1A-gf)=Ntor(A), we get n-(gf)(n)Ntor(A). Then t2n=t2(fg)(n) for some t2RNil(R), and hence t2t1g(y)=t2t1nN. Therefore, f(y)S(N) and it is easy to verify that NIm(1S(N)-gf(S(N))fS(N))=Ntor(S(N)) and NIm(1f(S(N))-fS(N)gf(S(N)))=Ntor(f(S(N))). Hence S(N)Nf(S(N)).

Lemma 3.5. If N is a direct summand of A, then S(N)NN.

Proof. Let A=NL for some submodule L of A. Let x=n+lS(N) with nN and lL. Then tx=tn+tlN for some tRNil(R). Then tl=tx-tnNL=0, and so tl=0, that is, tNtor(L). Therefore, S(N)NNtor(L).

Conversely, let x=n+lNNtor(L). Then tl=0 for some tRNil(R). Hence tx=tnN, and so xS(N). Consequently S(N)=NNtor(L). Since Ntor(L)N0 by Lemma 3.3, S(N)NN according to Lemma 3.2.

Proof of Theorem 3.1. Let P be a nonnil-projective module. Then by [20, Theorem 3.7], P is a direct summand of an N-free module. Hence there is a free R-module F such that A=PL is nonnil-isomorphic to F. Let f:AF be a nonnil-isomorphism. Our aim now is to show that F=f(P)f(L). For this, let g:FA such that NIm(1A-gf)=Ntor(A) and NIm(1F-fg)=Ntor(F). Since F is a free R-module and Z(R)=Nil(R), it follows from [13, Example 1.6.12 (1)] that Ntor(F)=tor(F)=0, and hence (1F-fg)NIm(1F-fg)=Ntor(F)=0. Therefore, f is an epimorphism, that is, F=f(A). Consequently F=f(P)+f(L). Let yf(P)f(L). Then there exist xP and lL such that y=f(x)=f(l). Thus f(x-l)=0, and so x-lKer(f)NKer(f)=Ntor(A). Then there exists a non-nilpotent element tR such that tx=tl. Since tx=tlPL=0, it follows that tx=0, whence ty=f(tx)=0. Then y=0 since F is a free R-module. Thus F=f(P)f(L). Therefore, f(P) is a projective R-module. By Lemma 3.5, PNS(P), and then PNf(S(P)) according to Lemma 3.4. Note that f(S(P))=f(PNtor(L))=f(P)+f(Ntor(L)). Since f(Ntor(L))Ntor(F)=0, we get f(S(P))=f(P). Thus PNf(P) and f(P) is a projective R-module.

Note that Lemma 3.2 can be used to provide another demonstration of [20, Corollary 3.8] as shown below.

Remark 3.6. If P is nonnil-isomorphic to a projective module, then P is nonnil-projective.

Proof. Let K be a projective module such that PNK. Since K is projective, it is a direct summand of a free module F, and so F=KL for some L. Since PNK, it follows from Lemma 3.2 that PLNKL=F. Hence P is a direct summand of an N-free module. Then P is a nonnil-projective module by [20, Theorem 3.7].

Lemma 3.7. Let R be a ring. If A1NB1 and A2NB2, then A1A2NB1B2.

Proof. Let f1:A1B1 and f2:A2B2 be two nonnil-isomorphisms. Then there exist two homomorphisms g1:B1A1 and g2:B2A2 such that NIm(1B1-f1g1)=Ntor(B1), NIm(1A1-g1f1)=Ntor(A1), NIm(1B2-f2g2)=Ntor(B2), and NIm(1A2-g2f2)=Ntor(A2). Set A:=A1A2, B:=B1B2, f:=f1f2, and g:=g1g2. Then for every (a1a2)A1A2 (resp., (b1b2)B1B2) we have f(a1a2)=f1(a1)f2(a2) (resp., g(b1b2)=g1(b1)g2(b2)). By [13, Example 2.2.10] we get that fg=(f1g1)(f2g2). Our aim now is to show that NIm(1A-gf)=Ntor(A) and NIm(1B-fg)=Ntor(B). Since Im(1A-gf)+Ntor(A)=NIm(1A-gf), to show that NIm(1A-gf)=Ntor(A), it is enough to show that Im(1A-gf)Ntor(A). Let a1a2)A1A2. Then there exist s1,s2RNil(R) such that s1(g1f1(a1)-a1)=0 and s2(g2f2(a2)-a2)=0. Thus gf(a1a2)-a1a2=g1f1(a1)g2f2(a2)-a1a2, which implies that s(gf(a1a2)-a1a2)=0 with s=s1s2RNil(R). Similarly, we can deduce that Im(1A-gf)Ntor(A), since Ntor(A) is a submodule of A. Therefore NIm(1A-gf)=Ntor(A). Likewise, we can deduce that NIm(1B-fg)=Ntor(B).

Corollary 3.8. Let R is a ZN-ring and let P1 and P2 be nonnil-projective R-modules. Then P1P2 is nonnil-projective.

Proof. Let P1 and P2 be projective modules such that P1NP1 and P2NP2. Then by Lemma 3.7, P1P2NP1P2. Since P1 and P2 are projective modules, P1P2 is projective by [13, Theorem 2.3.8]. Hence P1P2 is nonnil-projective.

Corollary 3.9. Let R be a local ring. Then every nonnil-projective module is N-free.

Proof. Let P be a nonnil-projective R-module. Then there exists a projective R-module P0 such that PNP0. Since R is a local ring, P is free by [13, Theorem 2.3.17]. Hence P is nonnil-isomorphic to a free R-module. Thus P is N-free.

Theorem 3.10. Let R be a ZN-ring and I be a nonnil-projective nonnil-ideal of R. Then I is finitely generated.

Proof. Let I be a nonnil-projective nonnil-ideal of R. Then by [20, Theorem 3.9], there exist elements xii Γ}I and R-homomorphisms fiiΓHomR(I,R) such that:

• If xI, then almost all fi(x)=0,

• If xI, then there exists an element sRNil(R) such that sx=sifi(x)xi.

Let aI be a non-nilpotent element. Then there exists a finite subset K of Γ such that fi(a)=0 for all iΓK. Now let xI. Then there exists an element sRNil(R) such that sx=sifi(x)xi. Hence asx=asifi(x)xi=sixfi(a)xi=skKxfk(a)xk=sakKfk(x)xk. Since sa is regular, we conclude that x=kKfk(x)xk. Therefore, I=kKRxk is finitely generated.

Let M be a nonnil-torsion-free R-module. Then M is nonnil-projective if and only if M is projective by [20, Lemma 4.1]. In particular, if R is a ZN-ring and I is an ideal of R, then I is nonnil-projective if and only if I is projective. Note that if I is a nil ideal (i.e, INil(R)), then I is not projective by [13, Proposition 6.7.12], and so it is not nonnil-projective. It is well known that in an integral domain every projective ideal is finitely generated according to [13, Corollary 5.2.7]. The following corollary gives a generalization of this fact.

Corollary 3.11. Let R be a ZN-ring. Then every projective ideal of R is finitely generated.

We know that every projective module is flat. So a natural question is whether a nonnil-projective module is φ-flat. The following example shows that a nonnil-projective module is not always φ-flat.

Example 3.12. Let R be a ring with w.gl.dim(R)2 (for example R=k[X,Y] with k a field). Then there exists a non-zero ideal I of R such that R/I is not flat. Hence R/I is not a φ-flat module, but it is nonnil-projective since R/IN0.

Remark 3.13. Note that a nonnil-projective module is not necessarily φ-flat. However, if every R-module is nonnil-projective, then every R-module is φ-flat by [20, Theorem 4.5].

The authors would like to thank the reviewer for his/her careful reading and comments.

1. D. F. Anderson and A. Badawi. On φ-Prüfer rings and φ-Bézout rings, Houston J. Math., 30(2)(2004), 331-343.
2. D. F. Anderson and A. Badawi. On φ-Dedekind rings and φ-Krull rings, Houston J. Math., 31(4)(2005), 1007-1022.
3. A. Badawi. On φ-pseudo-valuation rings, Lecture Notes in Pure and Appl. Math., Marcel Dekker, New York, 205(1999), 101-110.
4. A. Badawi. On divided commutative rings, Comm. Algebra, 27(3)(1999), 1465-1474.
5. A. Badawi. On φ-chained rings and φ-pseudo-valuation rings, Houston J. Math., 27(4)(2001), 725-736.
6. A. Badawi. On nonnil-Noetherian rings, Comm. Algebra, 31(4)(2003), 1669-1677.
7. A. Badawi and D. E. Dobbs. Strong ring extensions and φ-pseudo-valuation rings, Houston J. Math., 32(2)(2006), 379-398.
8. A. Badawi and T. Lucas. On φ-Mori rings, Houston J. Math., 32(1)(2006), 1-31.
9. A. El Khalfi, H. Kim and N. Mahdou. On φ-piecewise Noetherian rings, Comm. Algebra, 49(3)(2021), 1324-1337.
10. B. Khoualdia and A. Benhissi. Nonnil-coherent rings, Beitr. Algebra Geom., 57(2)(2016), 297-305.
11. H. Kim and F. Wang. On φ-strong Mori rings, Houston J. Math., 38(2)(2012), 359-371.
12. W. Qi and X. L. Zhang. Some remarks on Nonnil-coherent rings and φ-IF rings, J. Algebra Appl., 21(11)(2021), Paper No. 2250211.
13. F. Wang and H. Kim, Foundations of Commutative Rings and Their Modules, Algebr. Appl. 22, Springer, Singapore, 2016, xx+699 pp.
14. X. Y. Yang and Z. K. Liu. On nonnil-Noetherian rings, Southeast Asian Bull. Math., 33(6)(2009), 1215-1223.
15. X. L. Zhang and W. Zhao. On w-φ-flat modules and their homological dimensions, Bull. Korean Math. Soc., 58(4)(2021), 1039-1052.
16. W. Zhao. On φ-flat modules and φ-Prüfer rings, J. Korean Math. Soc., 55(5)(2018), 1221-1233.
17. W. Zhao. On φ-exact sequences and φ-projective modules, J. Korean Math. Soc., 58(6)(2021), 1513-1528.
18. W. Zhao, F. Wang and G. Tang. On φ-von Neumann regular rings, J. Korean Math. Soc., 50(1)(2013), 219-229.
19. W. Zhao, F. Wang and X. Zhang. On φ-projective modules and φ-Prüfer rings, Comm. Algebra, 48(7)(2020), 3079-3090.
20. W. Zhao, M. Wang and Y. Pu. On nonnil-commutative diagrams and nonnil-projective modules, Comm. Algebra, 50(7)(2022), 2854-2867.
21. W. Zhao and X. L. Zhang. On nonnil-injective modules, J. Sichuan Normal Univ., 42(6)(2019), 808-815.