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 I⊈Nil(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:={R∣R be a commutative ring, and Nil(R) be a divided prime ideal of R}. If R∈H, 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):={x∈M∣sx=0 for some s∈R∖Nil(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:A⟶B with Coker(f) being a φ-torsion R-module, we have 1F⊗Rf:F⊗RA⟶F⊗RB 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:R→RNil(R) by φ(r)=r1 for every r∈R. Then φ(R) is a ZN-ring. In [17], Zhao defined the map ψ:M→MNil(R) by ψ(x)=x1 for every x∈M. This makes ψ(M) a φ(R)-module. If f:M→N 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 x∈M. A sequence of R-modules and homomorphisms A→fB→gC 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:A→B be a homomorphism of R-modules. Set

 NKer(f):={a∈A∣sf(a)=0 for some s∈R∖Nil(R)} and 
   NIm(f):={b∈B∣sb=sf(a) for some a∈A and s∈R∖Nil(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:A→B,g:B→D,h:A→C,k:C→D 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 A→fB→gC 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:A→B of R-modules is called a nonnil-monomorphism if NKer(f)=Ntor(A), equivalently 0→A→fB is a nonnil-exact sequence; f is called a nonnil-epimorphism if NIm(f)=B (i.e., NCoker(f)=0), equivalently A→fB→0 is a nonnil-exact sequence. Also f is called a nonnil-isomorphism if there exists a homomorphism g:B→A such that NIm1A-gf=Ntor(A) and NIm1B-fg=Ntor(B). If there exists a nonnil-isomorphism f:A→B, we say that A and B are nonnil-isomorphic, denoted by A≃NB. Note that if f:A→B 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:P→B 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.

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

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

• (1)

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

• (2)

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

Proof. (1) Let b∈NKer(β). Then there exists t1∈R∖Nil(R) such that t1β(b)=0. On the other hand, there exists t2∈R∖Nil(R) such that t2γ∘g(b)=t2g′∘β(b). Hence t1t2γ∘g(b)=t2g′(t1β(b))=0. Therefore, g(b)∈NKer(γ). Since γ is a nonnil-monomorphism, there exists t3∈R∖Nil(R) such that t3g(b)=0, and so b∈NKer(g)=NIm(f). Then t4b=t4f(a) for some a∈A and t4∈R∖Nil(R). Hence

t4(β∘f(a)-f′∘α(a))=t4(β(b)-f′(α(a)). Since a∈A, it follows that t5(f′∘α(a)-β∘f(a))=0 for some t5∈R∖Nil(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 t6∈R∖Nil(R) and x′∈D′. Since δ is a nonnil-epimorphism, there exist some x∈D and t7∈R∖Nil(R) such that t7δ(x)=t7x′. Hence

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

On the other hand, since x∈D, it follows that t8h′δ(x)=t8αh(x) for some t8∈R∖Nil(R). So t6t7t8α(a)=t6t7t8h′∘δ(x)=t6t7t8α∘h(x), and hence t6t7t8α(a-h(x))=0. Therefore, a-h(x)∈NKer(α)=Ntor(A), and hence there exists t9∈R∖Nil(R) such that t9a=t9h(x). Since h(x)∈Im(h)⊆NIm(h)=NKer(f), we get t10f∘h(x)=0 for some t10∈R∖Nil(R). Then

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

Therefore, tb=0 with t:= t4t9t10∈R∖Nil(R), and so b∈Ntor(B). Thus β is a nonnil-monomorphism.

(2) Let b′∈B′. Since γ is a nonnil-epimorphism, there exist c∈C and t1∈R∖Nil(R) such that t1γ(c)=t1g′(b′). Nonnil-commutativity of the right square gives t2μ∘k(c)=t2jk′∘γ(c) for some t2∈R∖Nil(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 t3∈R∖Nil(R) such that t3k′∘g′(b′)=0, and so t1t2t3μ∘k(c)=0. Therefore, k(c)∈NKer(γ)=Ntor(E). Consequently there exists t4∈R∖Nil(R) such that t4k(c)=0, and hence c∈NKer(k)=NIm(g), that is, t5c=t5g(b) for some t5∈R∖Nil(R) and b∈B. On the other hand, since b∈B, there exists t6∈R∖Nil(R) such that t6γ∘g(b)=t6g′∘β(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 t7∈R∖Nil(R) and a′∈A′ such that t7(b′-β(b))=t7f′(a′). Since α is a nonnil-epimorphism, there exist some a∈A and t8∈R∖Nil(R) such that t8α(a)=t8a′. Hence

t8t7(b′-β(b))=t8t7f′(a′)=t7t8f′∘α(a).

Since a∈A, there exists t9∈R∖Nil(R) such that t9f′∘α(a)=t9β∘f(a), and so t9t8t7(b′-β(b))=t7t8t9β∘f(a). Thus tb′=tβ(b+f(a) with t:=t7t8t9∈R∖Nil(R). Consequently β is a nonnil-epimorphism.

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

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

Proof. Let x,y∈A. Then we have:

f(x)1=f(y)1∈ψ(ℑ(f))⟺∃s∈(R∖Nil(R)):sf(x)=sf(y) ⟺∃s∈(R∖Nil(R)):sf(x−y)=0 ⟺x−y∈NKer(f) ⟺x¯=y¯∈A/NKer(f).

Hence the homomorphism:

g:A/NKer(f)→ψ(ℑ(f))x¯↦g(x¯)=f(x)1

is an isomorphism.

A nonempty subset S of R is said to be a multiplicative subset if 1∈S, 0∉S, and for each a,b∈S, we have ab∈S. Note that if there exists s∈S∩Nil(R), then there exists a positive integer n such that 0=sn∈S, a contradiction. Hence we always assume that S∩Nil(R)=∅.

It is well known that if M′→fM→gM′′ is an exact sequence of R-modules, then MS′→fSMS→gSMS′′ 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 M′→fM→gM′′ be a nonnil-exact sequence of R-modules. Then MS′→fSMS→gSMS′′ is a nonnil-exact sequence.

Proof. Let ys∈NIm(fS). Then there exist ts1∈RS∖Nil(RS) and x′s′∈MS such that ts1ys=ts1fS(x′s′)=tf(x′)s1s′. Thus there exists s2∈S such that s2ts′s1y=s2s1stf(x′)=s2tf(s1sx′). Hence s1sy∈NIm(f)=NKer(g) since s2t∈R∖Nil(R), and so t′g(s1sy)=0 for some t′∈R∖Nil(R). Therefore, t′g(y)s=0, whence t′1gS(ys)=0 and t′1∈RS∖Nil(RS). Thus ys∈NKer(gS). Conversely, let xs∈NKer(gS). Then ts1g(x)s=0 for some ts1∈RS∖Nil(RS). Thus there exists s2∈S such that ts2f(x)=0, whence s2x∈NKer(f)=NIm(g) since t∈R∖Nil(R), that is, t1s2x=t1f(x′) for some x′∈M′ and t1∈R∖Nil(R). Then

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

Thus xs∈NIm(fS) since t1s21∈RS∖Nil(RS).

Remark 2.4. If S:=R∖Nil(R), then M′→fM→gM′′ is a nonnil-exact sequence if and only if MS′→fSMS→gSMS′′ 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=K∝M be the trivial extension. Then the homomorphism g:M→R 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=Z∝Z/2Z and consider g:Z→R defined by g(a)=(a,0). Since 2(0∝Z/2Z)=0, it follows that (0∝Z/2Z)⊆Ntor(R). Then NIm(g)=ℑ(g)+Ntor(R)=R. Hence g is a nonnil-epimorphism, which is not an epimorphism.

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

• (1)

  f is a nonnil-monomorphism,

• (2)

  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: 0→M′→iM→fN. Thus 0→MS′→iSMS→fSNS is also a nonnil-exact sequence. Hence Ntor(MS)+ℑ(iS)=NIm(iS)=NKer(fS)=Ntor(MS), and so MS′⊆Ntor(MS). Now let x∈M′. Then x1∈Ntor(MS), whence ts1x1=0 for some ts1∈RS∖Nil(RS). Thus stx=0 for some s∈S. Since Nil(R) is a prime ideal of R, st∈R∖Nil(R), and so x∈Ntor(M). Therefore, NKer(f)=Ntor(M). Consequently f is a nonnil-monomorphism.

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

• (1)

  f is a nonnil-epimorphism,

• (2)

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

• (3)

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

Proof. (1)⇒(2) Assume that f is a nonnil-epimorphism. Then M→fN→0 is nonnil-exact. Let p be a prime ideal of R. Then for S:=R∖p, we have Mp→fpNp→0 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 y∈N. Then y1∈Nm=NIm(fm) for any maximal ideal m of R. Thus for every m∈Max(R), there exist tmαm∈Rm∖Nil(Rm), x∈M, and sm∈R∖m such that tmαmy1=tmαmfm(xsm). So sm′αmtmsmy=sm′αmtmf(x) for some sm′∈R∖m. Set S:={sm∣m is a maximal ideal of R}. Since S generates R, there exist finite elements sm1,…,smn of S and α1,…,αn∈R such that 1=α1sm1+⋯+αnsmn. For all i=1,…,n, we have smi′αmitmismiy=smi′αmitmif(x), and so sαmitmismiy=sαmitmif(x) with s:=sm1′sm2′⋯smn′. 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  sαmitmi∈R∖Nil(R). Therefore, y∈NIm(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:

• (1)

  Every nonnil-commutative diagram is commutative,

• (2)

  Every nonnil-exact sequence is exact,

• (3)

  Every nonnil-monomorphism is a monomorphism,

• (4)

  Every nonnil-epimorphism is an epimorphism,

• (5)

  Every nonnil-isomorphism is an isomorphism,

• (6)

  R is a φ-von Neumann regular ring.

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

(3)⇒(6) Let a∈R∖Nil(R) and consider the following homomorphism f:R/Ra→0. 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 a∈R∖Nil(R) and consider the following homomorphism f:0→R/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=ℑ(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 a∈R∖Nil(R). Since a(R/Ra)=0, it is easy to verify that R/Ra≃N0 (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.

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 A1≃NB1 and A2≃NB2, then A1⊕A2≃NB1⊕B2.

Proof. Let f1:A1→B1 and f2:A2→B2 be two nonnil-isomorphisms. Then there exist two homomorphisms g1:B1→A1 and g2:B2→A2 such that NIm(1A1-f1∘g1)=Ntor(A1), NIm(1B1-g1∘f1)=Ntor(B1), NIm(1A2-f2∘g2)=Ntor(A1), and NIm(1B2-g2∘f2)=Ntor(B2). Define

f:A1⊕A2→B1⊕B2by(x1,x2)↦f(x1,x2)=(f1(x1),f2(x2))

and

g:B1⊕B2→A1⊕A2by(x1,x2)↦g(x1,x2)=(g1(x1),g2(x2)).

Then it is easy to verify that:

NIm(1A1⊕A2−f°g)=NIm(1A1−f1°g1)⊕NIm(1A2−f2°g2) =Ntor(A1)⊕Ntor(A2) =Ntor(A1⊕A2)

and

NIm(1B1⊕B2−g°f)=NIm(1B1−g1°f1)⊕NIm(1B2−g2°f2) =Ntor(B1)⊕Ntor(B2) =Ntor(B1⊕B2).

Hence A1⊕A2≃NB1⊕B2.

Lemma 3.3. Let M be an R-module. Then M≃N0 if and only if M is a φ-torsion R-module.

Proof. Let f:M→0 be a nonnil-isomorphism. Then NIm(1M-f∘0)=Ntor(M). Since NIm(1M-f∘0)=NIm(1M)=M, we get M=Ntor(M). Conversely, assume that M=Ntor(M). Then f:M→0 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):={x∈M∣sx∈N for some s∈S},

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:=R∖Nil(R).

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

Proof. Let g:B→A such that NIm(1A-g∘f)=Ntor(A) and NIm(1B-f∘g)=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 n′∈N. Then there exists t1∈R∖Nil(R) such that t1n′∈N. On the other hand, since NIm(1A-g∘f)=Ntor(A), we get n′-(g∘f)(n′)∈Ntor(A). Then t2n′=t2(f∘g)(n′) for some t2∈R∖Nil(R), and hence t2t1g(y)=t2t1n′∈N. 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=N⊕L for some submodule L of A. Let x=n+l∈S(N) with n∈N and l∈L. Then tx=tn+tl∈N for some t∈R∖Nil(R). Then tl=tx-tn∈N∩L=0, and so tl=0, that is, t∈Ntor(L). Therefore, S(N)⊆N⊕Ntor(L).

Conversely, let x=n+l∈N⊕Ntor(L). Then tl=0 for some t∈R∖Nil(R). Hence tx=tn∈N, and so x∈S(N). Consequently S(N)=N⊕Ntor(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=P⊕L is nonnil-isomorphic to F. Let f:A→F be a nonnil-isomorphism. Our aim now is to show that F=f(P)⊕f(L). For this, let g:F→A such that NIm(1A-g∘f)=Ntor(A) and NIm(1F-f∘g)=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 ℑ(1F-f∘g)⊆NIm(1F-f∘g)=Ntor(F)=0. Therefore, f is an epimorphism, that is, F=f(A). Consequently F=f(P)+f(L). Let y∈f(P)∩f(L). Then there exist x∈P and l∈L such that y=f(x)=f(l). Thus f(x-l)=0, and so x-l∈Ker(f)⊆NKer(f)=Ntor(A). Then there exists a non-nilpotent element t∈R such that tx=tl. Since tx=tl∈P∩L=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, P≃NS(P), and then P≃Nf(S(P)) according to Lemma 3.4. Note that f(S(P))=f(P⊕Ntor(L))=f(P)+f(Ntor(L)). Since f(Ntor(L))⊆Ntor(F)=0, we get f(S(P))=f(P). Thus P≃Nf(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 P≃NK. Since K is projective, it is a direct summand of a free module F, and so F=K⊕L for some L. Since P≃NK, it follows from Lemma 3.2 that P⊕L≃NK⊕L=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 A1≃NB1 and A2≃NB2, then A1⊗A2≃NB1⊗B2.

Proof. Let f1:A1→B1 and f2:A2→B2 be two nonnil-isomorphisms. Then there exist two homomorphisms g1:B1→A1 and g2:B2→A2 such that NIm(1B1-f1∘g1)=Ntor(B1), NIm(1A1-g1∘f1)=Ntor(A1), NIm(1B2-f2∘g2)=Ntor(B2), and NIm(1A2-g2∘f2)=Ntor(A2). Set A:=A1⊗A2, B:=B1⊗B2, f:=f1⊗f2, and g:=g1⊗g2. Then for every (a1⊗a2)∈A1⊗A2 (resp., (b1⊗b2)∈B1⊗B2) we have f(a1⊗a2)=f1(a1)⊗f2(a2) (resp., g(b1⊗b2)=g1(b1)⊗g2(b2)). By [13, Example 2.2.10] we get that f∘g=(f1∘g1)⊗(f2∘g2). Our aim now is to show that NIm(1A-g∘f)=Ntor(A) and NIm(1B-f∘g)=Ntor(B). Since Im(1A-g∘f)+Ntor(A)=NIm(1A-g∘f), to show that NIm(1A-g∘f)=Ntor(A), it is enough to show that Im(1A-g∘f)⊆Ntor(A). Let a1⊗a2)∈A1⊗A2. Then there exist s1,s2∈R∖Nil(R) such that s1(g1∘f1(a1)-a1)=0 and s2(g2∘f2(a2)-a2)=0. Thus g∘f(a1⊗a2)-a1⊗a2=g1∘f1(a1)⊗g2∘f2(a2)-a1⊗a2, which implies that s(g∘f(a1⊗a2)-a1⊗a2)=0 with s=s1s2∈R∖Nil(R). Similarly, we can deduce that Im(1A-g∘f)⊆Ntor(A), since Ntor(A) is a submodule of A. Therefore NIm(1A-g∘f)=Ntor(A). Likewise, we can deduce that NIm(1B-f∘g)=Ntor(B).

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

Proof. Let P1′ and P2′ be projective modules such that P1≃NP1′ and P2≃NP2′. Then by Lemma 3.7, P1⊗P2≃NP1′⊗P2′. Since P1′ and P2′ are projective modules, P1′⊗P2′ is projective by [13, Theorem 2.3.8]. Hence P1⊗P2 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 P≃NP0. 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 xi∣i∈ Γ}⊆I and R-homomorphisms fi∣i∈Γ⊆HomR(I,R) such that:

• (1)

  If x∈I, then almost all fi(x)=0,

• (2)

  If x∈I, then there exists an element s∈R∖Nil(R) such that sx=s∑ifi(x)xi.

Let a∈I be a non-nilpotent element. Then there exists a finite subset K of Γ such that fi(a)=0 for all i∈Γ∖K. Now let x∈I. Then there exists an element s∈R∖Nil(R) such that sx=s∑ifi(x)xi. Hence asx=as∑ifi(x)xi=s∑ixfi(a)xi=s∑k∈Kxfk(a)xk=sa∑k∈Kfk(x)xk. Since sa is regular, we conclude that x=∑k∈Kfk(x)xk. Therefore, I=∑k∈KRxk 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, I⊆Nil(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/I≃N0.

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].