https://doi.org/10.4134/JKMS.j210180 pISSN: 0304-9914 / eISSN: 2234-3008
ON φ-EXACT SEQUENCES AND φ-PROJECTIVE MODULES
Wei Zhao
Abstract. Let R be a commutative ring with prime nilradical N il(R) and M an R-module. Define the map φ : R → RN il(R) by φ(r) = r1 for r ∈ R and ψ : M → MN il(R) by ψ(x) = x1 for x ∈ M . Then ψ(M ) is a φ(R)-module. An R-module P is said to be φ-projective if ψ(P ) is projective as a φ(R)-module. In this paper, φ-exact sequences and φ-projective R-modules are introduced and the rings over which all R-modules are φ-projective are investigated.
1. Introduction
Throughout this paper, it is assumed that all rings are commutative and associative with non-zero identity. We use N il(R) to denote the set of nilpotent elements of R, and Z(R) the set of zero-divisors of R. It is shown in [4] that if a ring R admits a strongly prime ideal, then N il(R) is a divided prime ideal.
A ring with N il(R) being divided prime is called a φ-ring. We recommend [1,2,4–21] for the study of the ring-theoretic characterizations on φ-rings. Their main ideas are investigating the nonnil ideals I of R (i.e., I * N il(R)). For example, a φ-ring is called a nonnil-Noetherian ring if all nonnil ideals are finitely generated and a φ-Noetherian rings if all nonnil ideals I are φ-finitely generated (see [8]).
In order to investigate module-theoretic characterizations on φ-rings, the authors [3, 24, 26, 27] introduce nonnil-injective modules and nonnil-flat mod- ules, and characterize nonnil-Noetherian rings, φ-von Neumann regular rings, nonnil-coherent rings, φ-coherent rings and φ-Pr¨ufer rings. Recall from [24]
and [27], an R-module E is said to be nonnil-injective (resp., nonnil-flat) if Ext1R(R/I, E) = 0 (resp., TorR1(R/I, M ) = 0) for all nonnil ideals I of R.
In order to generalize the concept of projective modules in the context of φ- rings, the authors in [28] consider the module-homomorphism ψ : M → MN il(R)
Received March 14, 2021; Accepted July 6, 2021.
2010 Mathematics Subject Classification. Primary 13C05, 13C10, 13C12.
Key words and phrases. φ-exact sequence, nonnil-divisible module, φ-projective module.
This work was financially supported by the National Natural Science Foundation of China 12061001, 11861001, the China Postdoctoral Science Foundation 2021M691526, the Science and Technology Plan Project of Aba Prefecture 20RKX0001, and Aba Teachers University ASB20-02, ASA20-02, ASC20-02, 201901011, 201907019, 201910107, 201910108.
c
2021 Korean Mathematical Society 1513
such that ψ(x) = x1 for x ∈ M . We observe that prime nilradical N il(R) of R is necessary in defining the homomorphism ψ. This condition justifies our focus on rings R where N il(R) is a prime ideal.
In this paper, if the nilradical N il(R) of a ring R is prime, then R is called a P N -ring (abbreviation for “prime nilradical”). If Z(R) = N il(R), then R is called a ZN -ring. If R is a φ-ring and Z(R) = N il(R), then R is called a strongly φ-ring. From the definitions, we see that a strongly φ-ring is a φ-ring, a strongly φ-ring is a ZN -ring, a φ-ring is a P N -ring, and a ZN -ring is a P N -ring, but the converses are not true (see [25]).
We know that the exact sequence is an important tool to study projective modules. In this paper, our main purpose is to generalize the concept of the exact sequence in the context of P N -rings and characterize φ-projective mod- ules.
In this paper, let R be a P N -ring and M be an R-module. Set φ : R → RN il(R), φ(r) = r1 for r ∈ R. Then φ(R) is a ZN -ring. Define the map ψ : M → MN il(R), ψ(x) = x1 for x ∈ M . Then ψ(M ) is a φ(R)-module. If f : M → N is a homomorphism of R-modules, then f induces naturally a φ(R)-homomorphism ef : ψ(M ) → ψ(N ) such that ef (x1) = f (x)1 for x ∈ M . For a sequence of R-modules and homomorphisms A → Bf → C, we wantg to characterize the exactness of ψ(A) → ψ(B)fe → ψ(C). In this paper, weeg introduce the nonnil-kernel and the nonnil-image of an R-homomorphism.
An exact sequence may be not φ-exact. In Section 2, we give some examples of a short exact sequence which is φ-exact. It is clear that a split short exact sequence is φ-exact. If 0 → A → Bf → C → 0 is an exact sequence of R-g modules, where C is nonnil-torsion-free, then it is φ-exact. The short exact sequence 0 → A→ Bf → C → 0 of R-modules, where ker(g) is nonnil-divisibleg (that is, rM = M for any r ∈ R\N il(R)), is φ-exact. In particular, if R is a φ-ring, then the short exact sequence
0 → nM ,→ Mi → M/nM → 0π
is φ-exact, where n = N il(R). At the end of Section 2, we show the existence of φ-exact sequences.
In Section 3, we introduce the concept of φ-projective modules as a general- ization of projective modules, and give some properties of φ-projective modules.
We give an example of φ-projective module which is not projective.
In Section 4, we investigate the φ-rings over which all R-modules are φ- projective. A φ-ring R is called a φ-domain if φ(R) is a domain. If R is a φ-domain and φ(R) is a field, then R is field. If R is a φ-ring over which all R-modules are φ-projective, then R is a φ-domain. If R is a strong φ-ring over which all R-modules are φ-projective, then R is a field. If R is a φ-ring over which all R-modules are φ-projective, then R is a field. This result identifies with that all semisimple domains are field.
In this paper, R always denotes a P N -ring.
2. On φ-exact sequences Let M be an R-module. We set
N N (R) = {I | I is a nonnil ideal of R, that is, I * N il(R)}
and
Ntor(M ) = {x ∈ M | Ix = 0 for some I ∈ N N (R)}.
Then Ntor(M ) is a submodule of M . If Ntor(M ) = M (resp., Ntor(M ) = 0), M is called a nonnil-torsion R-module (resp., a nonnil-torsion-free R-module).
Set
φ : R → RN il(R), φ(r) = r 1
for r ∈ R. Then φ(R) is a ZN -ring and M/Ntor(M ) is a φ(R)-module by Lemma 3.4 in [28].
Observe that if R is nonnil-torsion-free as an R-module, then R is a ZN - ring. In this case, φ(R) = R and an R-module M is also a φ(R)-module and if M is a nonnil-torsion-free R-module, then M is also a nonnil-torsion-free φ(R)-module. Moreover, if M is a nonnil-torsion-free φ(R)-module, then M is also a nonnil-torsion-free R-module by rx = r1x for x ∈ M, r ∈ R.
Define the map
ψ : M → MN il(R), ψ(x) = x 1
for x ∈ M . Then ψ is a homomorphism of R-modules. In this case, kerψ = Ntor(M ) and ψ(M ) ∼= M/Ntor(M ) is a φ(R)-module. If M is a nonnil-torsion- free R-module, then ψ(M ) ∼= M . If f : M → N is a homomorphism of R- modules, then f induces naturally a φ(R)-homomorphism ef : ψ(M ) → ψ(N ) of φ(R)-modules such that
f (ex
1) =f (x) 1 for x ∈ M .
It is easy to show that this induced homomorphism ef is a homomorphism of nonnil-torsion-free R-modules. If f : A → B and g : B → C are two homomorphisms of R-modules, then
gf =f g eef .
If f, g : A → B are homomorphisms of R-modules, then g + f =] g + ee f .
Therefore, ψ is an additive covariant functor from the category of R-modules to the category of nonnil-torsion-free φ(R)-modules. Moreover, ψ is an additive covariant functor from the category of R-modules to the category of nonnil- torsion-free R-modules. We have ψ(ψ(M )) = ψ(M ) and the following.
Proposition 2.1. Let M, N be R-modules. Then ψ(M ⊕ N ) = ψ(M ) ⊕ ψ(N ).
Proof. Observe that ψ(M ⊕ N ) = ψ(M ) ⊕ ψ(N ) if and only if ψ(M ⊕ N ) = ψ(M ) + ψ(N ) and ψ(M ) ∩ ψ(N ) = {0}.
For any x+y1 ∈ ψ(M ⊕ N ), we have x + y
1 =x 1 +y
1 ∈ ψ(M ) + ψ(N ).
On the other side, if x1 ∈ ψ(M )T ψ(N ), then x1 = a1 = 1b for some a ∈ M and b ∈ N . Hence sa = sb ∈ MT N = {0} for some s ∈ R\N il(R). So we have x1 = a1 = 0. Therefore,
ψ(M ⊕ N ) = ψ(M ) ⊕ ψ(N ).
Moreover, if Mi(i ∈ Γ) are R-modules, then ψ(⊕i∈ΓMi) = ⊕i∈Γψ(Mi).
We know that a sequence of R-modules and homomorphisms A→ Bf → Cg is called a complex (resp., an exact sequence) if gf = 0, that is, Imf ⊆ kerg (resp., Imf = kerg). We can consider the following φ-exact sequence with the help of the functor ψ.
Definition 2.2. A sequence of R-modules and homomorphisms A→ Bf → Cg
is called a φ-complex (resp., a φ-exact sequence) if ψ(A)→ ψ(B)fe → ψ(C)eg is a complex (resp., an exact sequence) of φ(R)-modules.
Note that an exact sequence may be not φ-exact, for example, 0 → I→ Ri → R/I → 0,π
where I ∈ N N (R) and I 6= R. By Proposition 2.1, a split short exact sequence is φ-exact.
In order to investigate the characterizations of a φ-exact sequence, we in- troduce nonnil-kernels and the nonnil-images of an R-homomorphism. Let f : A → B be a homomorphism of R-modules. Set
NKer(f ) = {a ∈ A | sf (a) = 0 for some s ∈ R\N il(R)}, NIm(f ) = {b ∈ B | sb = sf (a) for some a ∈ A and s ∈ R\N il(R)}.
It is clear that NKer(f ) is a submodule of A and NIm(f ) is a submodule of B and set NCoker(f ) = B/NIm(f ). We have
ker(f ) + Ntor(A) ⊆ NKer(f ) ⊆ A
and
Im(f ) + Ntor(B) ⊆ NIm(f ) ⊆ B.
Definition 2.3. Let f : A → B be a homomorphism of R-modules. The submodule NKer(f ) of A is called the nonnil-kernel of f and the submodule NIm(f ) of B is called the nonnil-image of f .
The following results indicate that nonnil-kernels and nonnil-images play an important role in the portrayal of the φ-exact sequences which identifies with the exact sequences.
Lemma 2.4. Let f ∈ HomR(A, B). If ef ∈ Homφ(R)(ψ(A), ψ(B)) is the in- duced naturally homomorphism, then
(a) ker( ef ) = ψ(NKer(f ));
(b) Im( ef ) = ψ(NIm(f )) = ψ(Im(f )).
Proof. (a) Suppose that ef (a1) = 0, a ∈ A. Hence sf (a) = 0 for some s ∈ R\N il(R) and so a ∈ NKer(f ). Thus ker( ef ) ⊆ ψ(NKer(f )).
Conversely, suppose that a ∈ NKer(f ). Hence sf (a) = 0 for some s ∈ R\N il(R) and so ef (a1) = 0. Thus ψ(NKer(f )) ⊆ ker( ef ). Therefore, ker( ef ) = ψ(NKer(f )).
(b) Suppose that b1 ∈ Im( ef ). we have 1b = ef (a1) = f (a)1 for some a ∈ A.
Thus sb = sf (a) for some s ∈ R\N il(R), so b ∈ NIm(f ). Therefore, Im( ef ) ⊆ ψ(NIm(f )).
Conversely, suppose that b1 ∈ ψ(NIm(f )). There is an element b0 ∈ NIm(f ) such that b1 = b10, so s0b = s0b0 for some s0 ∈ R\N il(R). There is an element a ∈ A such that sb0 = sf (a) for some s ∈ R\N il(R), so
ss0b = ss0b0= ss0f (a).
Thus
b
1 =f (a) 1 = ef (a
1) ∈ Im( ef ).
Therefore, ψ(NIm(f )) ⊆ Im( ef ). In this sense, Im( ef ) = ψ(NIm(f )).
By the fact that Im(f ) + Ntor(B) = NIm(f ), we have Im( ef ) = ψ(NIm(f )) =
ψ(Im(f )).
Lemma 2.5. Let A → Bf → C be a sequence of R-modules and homomor-g phisms. Then
(a) ψ(NIm(f )) ⊆ ψ(NKer(g)) if and only if NIm(f ) ⊆ NKer(g);
(b) ψ(NKer(g)) ⊆ ψ(NIm(f )) if and only if NKer(g) ⊆ NIm(f ).
Proof. It is clear that if NIm(f ) ⊆ NKer(g), then ψ(NIm(f )) ⊆ ψ(NKer(g)) and if NKer(g) ⊆ NIm(f ), then ψ(NKer(g)) ⊆ ψ(NIm(f )).
(a) Suppose that ψ(NIm(f )) ⊆ ψ(NKer(g)), that is Im( ef ) ⊆ ker(eg) by Lemma 2.4. If b ∈ NIm(f ), then there exist a ∈ A and s ∈ R\N il(R) such that
sb = sf (a). We have b
1 = f (a) 1 = ef (a
1) ∈ Im( ef ) ⊆ ker(g),e
thus eg(1b) = g(b)1 = 01. In this sense, s1g(b) = 0 for some s1 ∈ R\N il(R) and b ∈ NKer(g).
(b) If b ∈ NKer(g), then sg(b) = 0 for some s ∈ R\N il(R). Thus we have g(e b1) = g(b)1 =01, and so
b
1 ∈ ker(g) ⊆ Im( ee f ) = ψ(NIm(f )).
There is an element b0 ∈ NIm(f ) such that 1b = b10 and sb0 = sf (a), s0b = s0b0 for some s, s0 ∈ R\N il(R). We have
ss0b = ss0b0= ss0f (a),
so b ∈ NIm(f ).
Theorem 2.6. Let A → Bf → C be a sequence of R-modules and homomor-g phisms. Then
(a) A→ Bf → C is a φ-complex if and only ifg NIm(f ) ⊆ NKer(g);
(b) A→ Bf → C is a φ-exact sequence if and only ifg NIm(f ) = NKer(g).
Proof. Since A→ Bf → C is a φ-complex (resp., a φ-exact sequence) if and onlyg if ψ(A)→ ψ(B)fe → ψ(C) is a complex (resp., an exact sequence), if and only ifeg Im( ef ) ⊆ ker(eg) (resp. Im( ef ) = ker(eg)), if and only if ψ(NIm(f )) ⊆ ψ(NKer(g)) (resp., ψ(NIm(f )) = ψ(NKer(g))), if and only if NIm(f ) ⊆ NKer(g) (resp.,
NIm(f ) = NKer(g)).
A homomorphism f : A → B is called a φ-monomorphism (resp., a φ- epimorphism; a φ-isomorphism) if the induced homomorphism ef is a monomor- phism (resp., an epimorphism; an isomorphism).
Note that a φ-monomorphism is not always a monomorphism, a φ-epi- morphism is not always an epimorphism. Let K be a field and M be an R-module. Set R = K × M be the trivial extensions of K by [22]. Then f : R → φ(R) with f (r) = r/1 induces a φ-monomorphism ef , but f is not a monomorphism and g : M → R with g(x) = (0, x) induces a φ-epimorphismeg, but g is not an epimorphism.
We have the following characterizations of a φ-monomorphism and a φ- epimorphism with the help of the nonnil-kernel and the nonnil-image of an R-homomorphism.
Theorem 2.7. Let f : A → B be a homomorphism of R-modules. Then (a) f is a φ-monomorphism if and only if 0 → A→ B is φ-exact if and onlyf if
NKer(f ) = Ntor(A);
(b) f is a φ-epimorphism if and only if A→ B → 0 is φ-exact if and only iff NIm(f ) = B,
if and only if NCoker(f ) = 0;
(c) f is a φ-isomorphism if and only if 0 → A→ B → 0 is φ-exact.f
Proof. (a) It is clear that f is a φ-monomorphism if and only if 0 → A→ B isf φ-exact. If NKer(f ) = Ntor(A), then
ψ(NKer(f )) = ψ(Ntor(A)) = 0, and thus f is φ-monomorphic.
Conversely, if f is φ-monomorphic, then NKer(f ) = NIm(0 → A) = Ntor(A).
(b) We only show that A → B → 0 is φ-exact if and only if NIm(f ) = B.f This is because NKer(B → 0) = B.
(c) It is natural.
By Theorem 2.7, we have f : A → B is a φ-monomorphism if and only if ker(f ) is a nonnil-torsion R-module. If f is a monomorphism, then f is a φ- monomorphism. If N is a nonnil-torsion-free R-module, then ker( ef ) = ψ(kerf ), in this case, if f is a monomorphism, then f is a φ-monomorphism. In fact, if
ψ(x) = x
1 ∈ ker( ef ) for some x ∈ A, then
f (ψ(x)) = ee f (x
1) = f (x) 1 = 0.
Thus there is J ∈ N N (R) such that J f (x) = 0. If B is nonnil-torsion-free, then f (x) = 0. So ψ(x) ∈ ψ(kerf ) and thus ker( ef ) = ψ(kerf ).
If f : A → B is an epimorphism, then f is a φ-epimorphism. If f is an isomorphism, then f is a φ-isomorphism. If B is a nonnil-torsion-free R-module, then Im( ef ) ∼= Imf . In this case, φ-epimorphism identifies with epimorphism.
It is a fact that if 0 → A→ Bf → C → 0 is an exact sequence of R-modulesg with C being a nonnil-torsion-free R-module, then it is a φ-exact sequence.
Next, we give another example that a short exact sequence is a φ-exact sequence. For this purpose, we introduce the nonnil-divisible modules. Recall from [23] that an R-module M is said to be divisible if rM = M for any r ∈ R.
We have:
Definition 2.8. An R-module M is said to be nonnil-divisible if rM = M for any r ∈ R\N il(R).
If R is a φ-ring, then it is clear that n = N il(R) is nonnil-divisible due to nJ = n for any J ∈ N N (R). Moreover, if R is a φ-ring and M is an R-module, then nM is nonnil-divisible.
Lemma 2.9. If f : A → B is a φ-epimorphism, then cokerf = B/Im(f ) is a nonnil-torsion R-module, that is, ψ(cokerf ) = 0. If Im(f ) is nonnil-divisible and B/Im(f ) is a nonnil-torsion R-module, then f is a φ-epimorphism.
Proof. If f is a φ-epimorphism, then there is an element r ∈ R\N il(R) such that ry = rf (x) = f (rx) ∈ Im(f ) for any y ∈ B. Therefore, B/Im(f ) is nonnil-torsion and ψ(cokerf ) = 0.
Suppose that B/Im(f ) is a nonnil-torsion R-module. For any y ∈ B, there is an element r ∈ R\N il(R) such that ry ∈ Im(f ). If Im(f ) is nonnil-divisible, then there is an element x ∈ A such that ry = rf (x), thus y1 = f (x)1 ∈ Im( ef ).
Therefore, Im( ef ) = ψ(Im(f )) and f is a φ-epimorphism. Let f ∈ HomR(A, B). Recall Lemma 2.4, it is a fact that Im( ef ) = ψ(Im(f )).
We want to know when ker( ef ) = ψ(ker(f )) holds. With the help of nonnil- divisible modules, we have a situation such that ker( ef ) = ψ(ker(f )) holds.
Lemma 2.10. Let f ∈ HomR(A, B). If ker(f ) is nonnil-divisible, then ker( ef ) = ψ(ker(f )).
Proof. Suppose that x1 ∈ ker( ef ), x ∈ A. Then there exists some s ∈ R\N il(R) such that sx = z in ker(f ). Because ker(f ) is nonnil-divisible, so z = sy for some y ∈ ker(f ). Thus x1 =y1 ∈ ψ(ker(f )). Therefore,
ker( ef ) = ψ(ker(f )).
Corollary 2.11. Let N be a submodule of an R-module M . If N is nonnil- divisible, then ψ(M/N ) ∼= ψ(M )/ψ(N ). Moreover, if N ⊆ Ntor(M ),
ψ(M/N ) ∼= ψ(M ).
Proof. Consider the following commutative diagram with exact rows 0 // N i //
ψ
M π //
ψ
M/N //
ψ
0
0 // ψ(N) ei // ψ(M) eπ// ψ(M/N) // 0,
where π is the natural homomorphism. Thus ψ(M/N ) ∼= ψ(M )/ψ(N ). Theorem 2.12. The short exact sequence 0 → A → Bf → C → 0 of R-g modules, where ker(g) is nonnil-divisible, is φ-exact. In particular, if R is a φ-ring, then the short exact sequence
0 → nM ,→ Mi → M/nM → 0π is φ-exact, where n = N il(R).
At the end of this section, we show the existence of φ-exact sequences.
Let M be an R-module. By Theorem 1.6.7 in [23], M is the image of a free module, that is, there exist a free module F0and an epimorphism d0: F0→ M , or, there exist a free module F0 and an exact sequence F0
d0
→ M → 0. Set K0 = NKer(d0), we have 0 → K0 ,→ F0
d0
→ M → 0 is a φ-exact sequence.
Similarly there are a free module F1 and a φ-exact sequence F1 → Kd1 0 → 0.
Thus F1 d1
→ F0 d0
→ M → 0 is a φ-exact sequence. Continuing in this way, we get a φ-exact sequence
· · · −→ Fn dn
−→ Fn−1−→ · · · −→ F1 d1
−→ F0 d0
−→ M → 0,
where each Fi is a free module. Such a φ-exact sequence is called a free φ- resolution of M . Thus we have:
Proposition 2.13. Every R-module has a free φ-resolution.
3. On φ-projective modules Let M be an R-module. Set the mapping
ψ : M → MN il(R), ψ(x) =x 1,
for x ∈ M . We have ψ(M ) is a φ(R)-module. By the ideas of Badawi in [4], we call an R-module F be φ-free (resp., φ-projective ) if ψ(F ) is free (resp., projective) as a φ(R)-module.
Any free R-module is φ-free; any φ-free R-module is φ-projective; any pro- jective R-module P is φ-projective. If R is a ZN -ring, then a nonnil-torsion-free R-module is φ-free (resp., φ-projective) if and only if it is free (resp., projec- tive). The following example shows that a φ-free (resp., a φ-projective) module may be not free (resp., projective).
Example 3.1 (See Remark 1 in [2]). Let D be an integral domain and M a D-module. Then R = D(+)M has N il(R) = {0}(+)M , and N il(R) is a prime ideal of R. It is easily verified that N il(R) is a divided prime ideal of R if and only if M is divisible as a D-module. Moreover, N il(R) is a divided prime ideal and N il(R) = Z(R) if and only if M is torsion-free and divisible as a D-module. Taking M be not torsion-free, then φ(R) is a φ-free R-module but not free, also not projective.
We want to know whether φ-projective modules have the same properties as well as projective modules. Note that an R-module P is φ-projective if and only if ψ(P ) is projective as a φ(R)-module and φ(R) is a nonnil-torsion-free φ(R)-module. By Theorem 2.3.3 in [23], it is easy to show the following results.
Proposition 3.2. Let P be an R-module. Then the following conditions are equivalent.
(a) P is a φ-projective R-module.
(b) Ext1φ(R)(ψ(P ), A) = 0 for any φ(R)-module A (or any nonnil-torsion-free φ(R)-module A).
(c) Homφ(R)(ψ(P ), −) is an exact functor.
(d) If given any diagram of φ(R)-module and homomorphisms ψ(P )
h
|| f
B g // C // 0
with the bottom row exact (or ker(g) is nonnil-torsion-free), then there is a homomorphism h : ψ(P ) → B making the diagram commute.
(e) Every exact sequence of φ(R)-modules such as 0 → A → B → ψ(P ) → 0 (or A is nonnil-torsion-free) is split.
(f) ψ(P ) ⊕ K is a free φ(R)-module for some φ(R)-module K.
Recall Theorem 2.3.3 from [23] that an R-module P is projective if and only if every exact sequence 0 → A → B → P → 0 is split.
Theorem 3.3. Let P be an R-module. Then P is φ-projective if and only if every φ-exact sequence 0 → A → B → P → 0 is φ-split, that is, the sequence of φ(R)-module 0 → ψ(A) → ψ(B) → ψ(P ) → 0 is split.
Proof. Given any φ-exact sequence 0 → A → B → P → 0, we have 0 → ψ(A) → ψ(B) → ψ(P ) → 0 is an exact sequence. If P is φ-projective, then 0 → ψ(A) → ψ(B) → ψ(P ) → 0 is split.
Conversely, by Proposition 2.13, there exists a φ-exact sequence 0 → K → F → P → 0. If it is φ-split, then 0 → ψ(K) → ψ(F ) → ψ(P ) → 0 is split and so ψ(K)⊕ψ(P ) ∼= ψ(F ) is φ(R)-free. Thus ψ(P ) is φ(R)-projective. Therefore,
P is φ-projective.
Let M, N be R-modules. If ψ(M ) ∼= ψ(N ), then M is φ-projective if and only if N is φ-projective. We have M is φ-free if and only if N is φ-free. An R-module P is projective if and only if P is a direct summand of some free R-module. The following theorem gives the relationship between φ-projective modules and φ-free R-modules.
Theorem 3.4. Let P be an R-module. Then P is φ-projective if and only if P is a direct summand of some φ-free module.
Proof. Suppose that P is a φ-projective R-module. There is a φ-exact sequence 0 → A → F → P → 0, where F is free, and which is φ-split. So
ψ(F ) ∼= ψ(P ) ⊕ ψ(A) ∼= ψ(P ⊕ A).
Set F0= P ⊕ A. We have F0is φ-free and then P is a direct summand of some φ-free module.
Conversely, if P is a direct summand of some φ-free module, then P is
φ-projective by Proposition 3.2.
It is clear that a P N -ring R is φ-free and so any direct sum RΓ is φ-free.
Here is a problem whether a φ-free R-module is a direct sum of some R. The following theorem gives us the answer.
Lemma 3.5. Let F0 be a free φ(R)-module. Then there exists a free R-module F such that F0∼= ψ(F ).
Proof. Set L = ker(φ). We have
F0∼= ⊕i∈Γφ(R) ∼= ⊕i∈ΓR/L ∼= F/T = ψ(F ),
where F = ⊕i∈ΓR is free and T = Ntor(F ) = ⊕i∈ΓL. Theorem 3.6. Let F0 be an R-module. Then F0 is φ-free if and only if F0∼= (F ⊕ L)/K, where F is free and K = Ntor(F ), L = Ntor(F0).
Proof. Suppose that F0 is φ-free. Then there exists a free R-module F such that ψ(F0) ∼= ψ(F ) by Lemma 3.5. Consider the following diagram with exact rows
0 // K //
σ
F f //
h
ψ(F ) // 0
0 // L // F0 g // ψ(F0) // 0,
where K = Ntor(F ), L = Ntor(F0). By Schanuel’s Lemma, there exists an exact sequence 0 → K → F ⊕ L → F0→ 0. Therefore, F0∼= (F ⊕ L)/K. Next, we give another property of φ-projective modules which is in contrast to projective modules. Recall from [23], an R-module P is said to be projective if given any diagram of module homomorphisms
P
h
~~ f
B g // C // 0
with the bottom row exact (that is, g is an epimorphism), then there is a homomorphism h : P → B making this diagram commute.
Lemma 3.7. Let F0 be an R-module. If F0 is φ-free, then there is a homo- morphism α : F0→ B making any given diagram commute
F0
α
β
0 // A f // B g // C // 0,
where the bottom row is an exact sequence of nonnil-torsion-free R-modules.
Proof. Suppose that F0is φ-free. Then there is a φ(R)-module homomorphism δ : ψ(F0) → B making the following diagram commute.
ψ(F0)
δ
|| eβ
0 // A f // B g // C // 0.
By Proposition 3.2, the φ(R)-module homomorphism δ is also a R-module homomorphism. Consider the following diagram of R-module homomorphisms:
F0
β ##
α
ψ // ψ(F0)
δ
{{ eβ
0 // A f // B g // C // 0.
Set α = δψ, we have gα = gδψ = eβψ = β.
Lemma 3.8. Let T be a nonnil-torsion R-module. Then there exists a homo- morphism α : T → B making any given diagram commute
T
α
β
0 // A f // B g // C // 0,
where the bottom row is an exact sequence of nonnil-torsion-free R-modules.
Proof. Note that if β ∈ Hom(T, C), where T is nonnil-torsion and C is nonnil-
torsion-free, then β = 0.
Theorem 3.9. An R-module P is φ-projective if and only if there is an R- module homomorphism α : P → B making any diagram commute
P
α
β
0 // A f // B g // C // 0,
where the bottom row is an exact sequence of nonnil-torsion-free R-modules.
Proof. Suppose that P is φ-projective. Then P ⊕ Q ∼= F0 for some R-module Q and φ-free R-module F0. Let 0 → Q → F0→ P → 0 and 0 → Pπ → Fi 0 → Q → 0 be exact sequences such that πi = 1P. Consider the following diagram of R-module homomorphisms:
0 // Q // F0
δ
π // P
α
~~ β
// 0
0 // A f // B g // C // 0.
There exists a homomorphism δ : F0 → B such that gδ = βπ by Lemma 3.7.
Set α = δi, we have gα = gδi = βπi = β.
Conversely, consider the following commutative diagram of R-module homo- morphisms:
0 // K
ψK
// F
ψF
// P
α
zz ψP
// 0
0 // ψ(K) f // ψ(F) g // ψ(P) // 0,
where the above row is φ-exact and the bottom row is an exact sequence of nonnil-torsion-free R-modules. Suppose that gδ = ψP. We have βα = 1e P. Therefore, ψ(P ) is projective and so P is φ-projective.
4. On φ-rings over which all R-modules are φ-projective In this section, we investigate the φ-rings over which all R-modules are φ- projective. Set n = N il(R), the map φ : R → Rn such that φ(r) = r1 for r ∈ R.
A φ-ring R is called a φ-domain if φ(R) is a domain.
Theorem 4.1. Let R be a φ-domain. If φ(R) is a field, then R is a field.
Proof. Set L = ker(φ), we have φ(R) ∼= R/L. If φ(R) is a field, then there is an element s in R such that ¯r¯s = ¯1 for each r in R\L. Hence there is an element q in L such that 1 − q = rs. Because q is nilpotent, 1 − q is invertible in R and r is invertible in R. If r ∈ L, then r1 = 01, so rt = 0 for some t ∈ R\N il(R) and
we have r = 0. Therefore, R is a field.
Theorem 4.2. Let R be a φ-ring. If all R-modules are φ-projective, then R is a φ-domain.
Proof. We have the short exact sequence 0 → n → R → R/n → 0 is φ-exact by Theorem 2.12 and hence it is φ-split by Theorem 3.3. We have φ(R) ∼= φ(n) ⊕ R/n, so φ(n) is projective as a φ(R)-module. We have φ(n) = 0 by Proposition 6.7.12 in [23]. Therefore, R is a φ-domain. Lemma 4.3. Let R be a strongly φ-ring. Then
(a) A nonnil-torsion-free R-module P is φ-projective if and only if P is projective;
(b) If P is a φ-projective R-module, then P ∼= T ⊕ψ(P ), where T = Ntor(P );
(c) 0 → n → R → R/n → 0 is a short exact sequence of nonnil-torsion-free R-modules.
Proof. (a) Consider the φ-exact sequence 0 → A → F → P → 0, where F is free. Because R is a strongly φ-ring, 0 → A → F → P → 0 is an exact sequence of φ-torsion free R-modules. Suppose that P is a φ-projective R-module, we have 0 → A → F → P → 0 is split and hence F ∼= P ⊕ A. Therefore, P is projective.
(b) Consider the short exact sequence of R-modules 0 → T → P → ψ(P ) → 0, where T = Ntor(P ). Because R is a strongly φ-ring, φ(R) = R and hence 0 → T → P → ψ(P ) → 0 is an exact sequence of φ(R)-modules. If P is a φ-projective R-module, then 0 → T → P → ψ(P ) → 0 is split. Therefore, P ∼= T ⊕ ψ(P ).
(c) Note that R is nonnil-torsion-free if and only if R is a strongly φ-ring. Theorem 4.4. Let R be a strongly φ-ring. Then the following statements are equivalent:
(a) All R-modules are φ-projective;
(b) All short φ-exact sequences are φ-split;
(c) All short exact sequences of nonnil-torsion-free R-modules are split;
(d) All short exact sequences of torsion free R-modules are split;
(e) All torsion free R-modules are projective.
Proof. (a)⇒(b) It is clear by Theorem 3.3.
(b)⇒(c) All short exact sequences of nonnil-torsion-free R-modules are φ- exact, so is an exact sequence of φ(R)-modules.
(c)⇒(d) Because R is a strongly φ-ring, we have all torsion free R-modules are nonnil-torsion-free.
(d)⇒(e) If all short exact sequences of torsion free R-modules are split, then there is a short exact sequence 0 → K → F → M → 0 for any torsion free R-module M , where F is free. Because R is a strongly φ-ring, 0 → K → F → M → 0 is a short exact sequence of torsion free R-modules, hence it is split.
Therefore, M is projective.
(e)⇒(a) If all torsion free R-modules are projective, then we have M ∼= T ⊕ ψ(M ), where T = Ntor(M ) for any R-module M . Because T and ψ(M )
are φ-projective, we have M is φ-projective.
Theorem 4.5. Let R be a strong φ-ring. If all R-modules are φ-projective, then R is a field.
Proof. Note that the direct product of torsion free R-modules is torsion free and the direct product of projective R-modules is projective if and only if R is an Artinian ring. If all R-module are φ-projective, then R is a φ-domain.
Since R is a strong φ-ring, we have R is a domain. By Theorem 2.8.14 in [23],
every Artinian domain is necessarily a field.
Corollary 4.6. Let R be a φ-ring. If all R-modules are φ-projective, then R is a field.
Proof. If all R-modules are φ-projective, then R is a φ-domain, that is, φ(R) is a domain. Because every φ(R)-module is also a R-module, all φ(R)-modules are φ-projective, thus φ(R) is a field. Therefore, R is a field.
This result identifies with that all semisimple domains are fields.
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2020.1729362
Wei Zhao
Department of Mathematics Nanjing University
Nanjing 210093, P. R. China and
School of Mathematics ABa Teachers University
Wenchuan, Sichuan 623002, P. R. China Email address: zw9c248@163.com
