A Generalization of Lifting Modules
We introduce the notion of $I$ -lifting modules as a proper generalization of the notion of lifting modules and present some properties of this class of modules. It is shown that if $M$ is an $I$ -lifting direct projective module, then $S/▽$ is regular and $▽ = \text{Jac} S$, where $S$ is the ring o...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860508192351453184 |
|---|---|
| author | Kalati, Amouzegar T. Калаті, Амузегар Т. |
| author_facet | Kalati, Amouzegar T. Калаті, Амузегар Т. |
| author_sort | Kalati, Amouzegar T. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T10:27:00Z |
| description | We introduce the notion of $I$ -lifting modules as a proper generalization of the notion of lifting modules and present some properties of this class of modules. It is shown that if $M$ is an $I$ -lifting direct projective module, then $S/▽$ is regular and $▽ = \text{Jac} S$, where $S$ is the ring of all $R$-endomorphisms of $M$ and $▽ = \{ϕ ∈ S | Im ϕ ≪ M\}$. Moreover, we prove that if $M$ is a projective $I$ -lifting module, then $M$ is a direct sum of cyclic modules. The connections between $I$ -lifting modules and dual Rickart modules are presented. |
| first_indexed | 2026-03-24T02:21:18Z |
| format | Article |
| fulltext |
UDC 512.5
T. Amouzegar Kalati (Quchan Univ. Advanced Technology, Iran)
A GENERALIZATION OF LIFTING MODULES
УЗАГАЛЬНЕНННЯ ПIДЙОМНИХ МОДУЛIВ
We introduce the notion of I-lifting modules as a proper generalization of the notion of lifting modules and give some
properties of this class of modules. It is shown that if M is an I-lifting direct projective module, then S/∇ is regular and
∇ = JacS, where S is the ring of all R-endomorphisms of M and ∇ = {φ ∈ S | Imφ � M}. Moreover, we prove
that if M is a projective I-lifting module, then M is a direct sum of cyclic modules. The connections between I-lifting
modules and dual Rickart modules are given.
Введено поняття I-пiдйомних модулiв як природне узагальнення пiдйомних модулiв. Наведено деякi властивостi
цього класу модулiв. Показано, що якщо M — прямий проективний модуль I-пiдйому, то S/∇ є регулярною i
∇ = JacS, де S — кiльце всiх R-ендоморфiзмiв M, а ∇ = {φ ∈ S | Imφ � M}. Бiльш того, доведено, що
якщо M — проективний I-пiдйомний модуль, то M є прямою сумою циклiчних модулiв. Встановлено зв’язки мiж
I-пiдйомними модулями та дуальними модулями Рiкарта.
1. Introduction. Throughout this paper, R will denote an arbitrary associative ring with identity, M
a unitary right R-module and S = EndR(M) the ring of all R-endomorphisms of M. We will use the
notationN �M to indicate thatN is small inM (i.e., L+N 6=M ∀L �M ). The notationN ≤⊕ M
denotes that N is a direct summand of M. N �M means that N is a fully invariant submodule of
M (i.e., φ(N) ⊆ N ∀φ ∈ EndR(M)). We denote DS(N) = {φ ∈ S | Imφ ⊆ N} for N ⊆M.
We recall that L is a cosmall submodule of K in M (denoted by L
cs
↪→ K in M ) if K/L�M/L.
Recall that a submodule L of M is called coclosed if L has no proper cosmall submodule. It is clear
that every direct summand of M is a coclosed submodule of M. A module M is called lifting if for
every A ≤M, there exists a direct summand B of M such that B ⊆ A and A/B �M/B [2].
A number of results concerning lifting modules have appeared in the literature in recent years and
many generalizations of the concept of lifting modules have been introduced and studied by several
authors (see [7 – 9, 17]). Motivated by the definition of a lifting module, we say that a module M
is I-lifting if for every φ ∈ EndR(M), there exists a direct summand N of M such that N ⊆ Imφ
and Imφ/N � M/N. It is obvious that every lifting module is I-lifting. In this note, we study
some properties of I-lifting modules. In Section 2, as we state in the abstract, we show that if M is
a direct projective module, then M is I-lifting if and only if SS is f -lifting and ∇ = JacS, where
∇ = {φ ∈ S | Imφ � M} (Corollary 2.3). Moreover, we prove that if M is a projective I-lifting
module, then M is a direct sum of cyclic modules (Theorem 2.6).
The notion of right Rickart rings (or right p.p. rings) initially appeared in Maeda [14, p. 510]
and was further studied by a number of authors [1, 3 – 5]. A ring R is called right Rickart if the
right annihilator of any single element of R is generated by an idempotent as a right ideal. The
notion of Rickart modules was introduced by Rizvi and Roman in [16], and was studied recently (see
[10, 12, 13]). A module M is said to be Rickart if, for every φ ∈ EndR(M), Kerφ ≤⊕ M. It is clear
that for M = RR, the notion of a Rickart module coincides with that of a right Rickart ring. Lee,
Rizvi and Roman investigate the dual notion of Rickart modules in [11]. A module M is called dual
Rickart if for every φ ∈ EndR(M), Imφ ≤⊕ M. It is easy to see that every dual Rickart module
is I-lifting. In Section 3, we investigate the connection between dual Rickart modules and I-lifting
modules. It is shown that M is dual Rickart if and only if M is I-lifting and T -noncosingular
(Corollary 3.1). We prove that if R is a right V -ring, then an R-module M is dual Rickart if and
only if M is I-lifting (Corollary 3.2).
c© T. AMOUZEGAR KALATI, 2014
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11 1477
1478 T. AMOUZEGAR KALATI
2. I-lifting modules.
Definition 2.1. A module M is called I-lifting if for every φ ∈ EndR(M), there exists a direct
summand N of M such that N ⊆ Imφ and Imφ/N �M/N.
It is clear that every lifting module is I-lifting, while the converse in not true (the Z-module Q
is I-lifting but it is not lifting).
Proposition 2.1. The following conditions are equivalent for a module M :
(1) M is a I-lifting module.
(2) For every φ ∈ S there exists a decomposition M = M1 ⊕M2 such that M1 ⊆ Imφ and
M2 ∩ Imφ�M2.
(3) For every φ ∈ S, Imφ can be written as Imφ = N ⊕ S such that N ≤⊕ M and S �M.
Proof. It follows from [2] (22.1).
Definition 2.2. A module M is called a N -I-lifting module if for every homomorphism φ :
M → N, there exists L ≤⊕ N such that L ⊆ Imφ and Imφ/L� N/L.
In view of the above definition, a right module M is I-lifting if and only if M is M -I-lifting.
Theorem 2.1. Let M and N be right R-modules. Then M is N -I-lifting if and only if for all
direct summands M ′ ≤⊕ M and coclosed submodule N ′ of N, M ′ is N ′-I-lifting.
Proof. Let M ′ = eM for some e2 = e ∈ EndR(M), and N ′ be a coclosed submodule
of N. Assume that ψ ∈ Hom(M ′, N ′). Since ψeM = ψM ′ ⊆ N ′ ⊆ N and M is N -I-lifting,
there exists a decomposition N = N1 ⊕ N2 such that N1 ⊆ Imψe and N2 ∩ Imψe � N2. As
N1 ⊆ Imψe ⊆ N ′, N ′ = N1 ⊕ (N2 ∩ N ′). By [2] (3.7(3)), N2 ∩ N ′ ∩ Imψ � N ′. Again by [2]
(3.7(3)), N ′ ∩N2 ∩ Imψ � N2 ∩N ′. Therefore M ′ is N ′-I-lifting. The converse is clear.
Corollary 2.1. The following conditions are equivalent for a module M :
(1) M is an I-lifting module.
(2) For any coclosed submodule N of M, every direct summand L of M is N -I-lifting.
Corollary 2.2. Every direct summand of an I-lifting module is I-lifting.
An R-module M is called T -noncosingular if, ∀φ ∈ EndR(M), Imφ � M implies that
φ = 0 [18].
Proposition 2.2. The following conditions are equivalent for a T -noncosingular module M :
(1) M is an indecomposable I-lifting module.
(2) Every nonzero endomorphism φ ∈ S is an epimorphism.
Proof. Let M be an indecomposable I-lifting module. Assume that 0 6= φ ∈ EndR(M). Then
there exists a decomposition M = M1 ⊕M2 with M1 ⊆ Imφ and M2 ∩ Imφ � M2. Since M is
indecomposable, M1 = 0 or M1 =M. If M1 = 0, then Imφ�M. By T - noncosingularity, φ = 0,
a contradiction. Thus M1 =M and so φ is an epimorphism. The converse follows easily.
Recall that a module M is said to be Hopfian if every epimorphism φ ∈ EndR(M) is an
isomorphism.
Proposition 2.3. Let M be a T -noncosingular Noetherian I-lifting module. Then there exists a
decomposition M = M1 ⊕ . . . ⊕Mn, where Mi is an indecomposable Noetherian I-lifting module
with EndR(Mi) a division ring.
Proof. Since M is Noetherian, it has a finite decomposition with indecomposable Noetherian
direct summands. By Corollary 2.2, every direct summand is I-lifting. By Proposition 2.2 and since
every Noetherian module is Hopfian, each indecomposable direct summand has a division ring.
An R-module M is called direct projective if, for every direct summand X of M, every epi-
morphism M → X splits. A module M is called finitely lifting, or f -lifting for short, if for every
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A GENERALIZATION OF LIFTING MODULES 1479
finitely generated submodule A of M, there exists a direct summand B of M such that B ⊆ A and
A/B �M/B.
Proposition 2.4. Let M be an I-lifting direct projective module. Then:
(1) S/∇ is regular and ∇ = JacS, where ∇ = {φ ∈ S | Imφ�M}.
(2) SS is f -lifting.
Proof. (1) Let f be an arbitrary element of S. As M is I-lifting, there exists a decomposition
M =M1⊕M2 with M1 ⊆ Im f and Im f ∩M2 �M. Let π denote the projection M1⊕M2 →M1.
Since πf : M → M1 is an epimorphism and M is direct projective, Kerπf ≤⊕ M. So M =
= Kerπf ⊕ U for some U ⊆ M. The restriction of πf to U is an isomorphism onto M1 and the
inverse isomorphism of M1 to U can be extended to an element g ∈ S. Note that gπf = 1U . Now
(f−fgπf)M = (f−fgπf)(Kerπf⊕U) = f(Kerπf) ≤ f(M)∩M2. Hence (f−fgπf)M �M
and so f − fgπf ∈ ∇. Therefore S/∇ is a regular ring. It follows that JacS ⊆ ∇. Now we want to
show that ∇ ⊆ JacS. Let f ∈ ∇. Since M = fM + (1 − f)M and Im f � M, (1 − f)M = M.
As M is direct projective, 1− f is right invertible. But ∇ is an ideal, so ∇ ⊆ JacS.
(2) Let f ∈ S and fS be an arbitrary cyclic right ideal of S. Consider the proof of (1). Set
h = fgπ ∈ S. It is clear that h2 = h and hS ⊆ fS. Since (f − fgπf)M �M, (f − fgπf) ∈ ∇ =
= JacS. So (1− h)fS � S. By [2] (22.7), SS is f -lifting.
Corollary 2.3. Let M be a direct projective module. Then M is I-lifting if and only if SS is
f -lifting and ∇ = JacS.
Proof. Let M be an I-lifting direct projective module. Then, by Proposition 2.4, SS is f -
lifting and ∇ = JacS. Conversely, let SS be f -lifting and ∇ = JacS. Assume that f ∈ S. Then
there exists an idempotent e ∈ S such that eS ⊆ fS and (1 − e)fS ⊆ JacS = ∇. Therefore
M = eM + (1− e)M, eM ⊆ fM and (1− e)fM �M.
Let K and N be submodules of M. K is called a supplement of N in M if M = K +N and K
is minimal with respect to this property, or equivalently, M = K +N and K ∩N � K. A module
M is called supplemented if every submodule of M has a supplement in M. We say a module M
is I-supplemented if for every f ∈ S, Im f has a supplement in M. Recall that a submodule U of
the R-module M has ample supplements in M if, for every V ⊆ M with M = U + V, there is a
supplement V ′ of U with V ′ ⊆ V. We call M amply I-supplemented if, for every f ∈ S, Im f has
ample supplements in M.
Proposition 2.5. Let M be an amply I-supplemented R-module. Then every direct summand of
M is amply I-supplemented.
Proof. Let V be a direct summand of M. Then M = V ⊕ U for some U ⊆ M. Assume
that f ∈ EndR(V ) and V = Im f + X. Thus M = Im f + X + U. Note that Im f = Im ιfπ,
where ι is the injection map from V to M and π is the projection map from M onto V. Since
M is amply I-supplemented, there exists a supplement Y of U + X with Y ⊆ Im f. We get
X ∩ Y ⊆ (U +X) ∩ Y � Y and M = Y +X + U. Thus X + Y = V and X ∩ Y � Y. Therefore
V is amply I-supplemented.
Proposition 2.6. Let M be an amply I-supplemented module and let for every supplement
submodule X of M we have X ≤⊕ M, then M is I-lifting.
Proof. Let f ∈ S and V be a supplement of Im f in M and X a supplement of V in M with
X ⊆ Im f. By hypothesis, M = X ⊕ X ′ for some X ′ ≤ M. Since Im f ∩ V � M, this X ′ is a
supplement of X + (Im f ∩ V ) = Im f (see [19] (41.1)). Hence Im f ∩X ′ � X ′.
Proposition 2.7. Let M be an I-lifting module and N be a submodule of M invariant under all
maps f ∈ EndR(M) with Im f a direct summand of M. Then N is a fully invariant submodule of M.
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1480 T. AMOUZEGAR KALATI
Proof. Let f ∈ EndR(M). Since M is I-lifting, M = M1 ⊕ M2 where M1 ⊆ Im f and
M2∩Im f �M2. We have Im f =M1⊕(M2∩Im f). Consider the projection maps πM1 : M →M
and πM2 : M → M of M onto M1 and M2, respectively. Note that Im(πM1f) = M1 is a direct
summand ofM. By hypothesis, πM1f(N) ⊆ N.As Im(πM2f)�M,we have Im(1M−πM2f) =M.
By assumption, (1M − πM2f)(N) ⊆ N. So πM2f(N) ⊆ N. Therefore f(N) ⊆ N.
An element a of the ring R is called (von Neumann) regular if axa = a for some x ∈ R.
Corollary 2.4. Suppose that M is a direct projective I-lifting module and N ⊆ M. Then the
following are equivalent:
(1) N is invariant under all (von Neumann) regular elements of EndR(M);
(2) N is invariant under all f ∈ EndR(M) with Im f a direct summand of M ;
(3) N is a fully invariant submodule of M.
Proof. (1) ⇒ (2). Let f : M → M be any homomorphism with Im f a direct summand of M.
Since M is direct projective, Ker f is also direct summand. By [19] (37.7), f is a (von Neumann)
regular element of EndR(M).
(2) ⇒ (3). By Proposition 2.7, N is a fully invariant submodule of M.
(3) ⇒ (1). It is clear.
A module M is called a N -I-supplemented module if for every homomorphism φ : M → N,
there exists L ≤ N such that Imφ+ L = N and Imφ ∩ L� L. It is clear that a right module M is
I-supplemented if and only if M is M -I-supplemented.
Theorem 2.2. Let M1, M2 and N be modules. If N is Mi-I-supplemented for i = 1, 2, then N
is M1 ⊕M2-I-supplemented. The converse is true if M1 ⊕M2 is a duo module.
Proof. Suppose N is Mi-I-supplemented for i = 1, 2. We will prove that N is M1 ⊕M2-I-
supplemented. Let φ = (π1φ, π2φ) be any homomorphism from N to M1 ⊕M2, where πi is the
projection map from M1 ⊕M2 to Mi for i = 1, 2. Since N is Mi-I-supplemented, there exists
a submodule Ki of Mi such that πiφN + Ki = Mi and πiφN ∩ Ki � Ki, for i = 1, 2. Let
K = K1⊕K2. Then M1⊕M2 = π1φN +π2φN +K1+K2 = φN +K. Since φN ∩ (K1+K2) ≤
≤ (φN+K1)∩K2+(φN+K2)∩K1,we have φN∩(K1+K2) ≤ (φN+M1)∩K2+(φN+M2)∩K1.
As φN+M1 = π2φN⊕M1 and φN+M2 = π1φN⊕M2, thus φN ∩K ⊆ (π2φN ∩K2)+(π1φN ∩
∩K1). Since πiφN ∩Ki � Ki for i = 1, 2, φN ∩K � K1 +K2 = K. Hence N is M1 ⊕M2-
I-supplemented. Conversely, let N be M1 ⊕M2-I-supplemented. Let φ be a homomorphism from
N to M1. Then Im ιφ = Imφ, where ι is the canonical inclusion from M1 to M1 ⊕M2. Since
N is M1 ⊕ M2-I-supplemented, there exists K ⊆ M1 ⊕ M2 such that M1 ⊕ M2 = Imφ + K
and Imφ ∩ K � K. Thus M1 = Imφ + (K ∩ M1) and Imφ ∩ K ∩ M1 = Imφ ∩ K � K.
As M1 ⊕M2 is a duo module, K �M1 ⊕M2 and so K ∩M1 is a direct summand of K. Hence
Imφ ∩K ∩M1 � (K ∩M1). Therefore N is M1-I-supplemented.
Corollary 2.5. Suppose M = M1 ⊕M2 and M is Mi-I-supplemented module for i = 1, 2.
Then:
(1) M is I-supplemented and for every f ∈ S, Im f has a supplement of the form K1+K2 with
K1 ⊆M1 and K2 ⊆M2.
(2) Let f ∈ S and Im f be a supplement submodule of M. Then K1 and Im f +K2 are mutual
supplements in M and the same is true for K2 and Im f +K1.
Proof. (1) By using the proof of Theorem 2.2.
(2) Let f ∈ S and Im f be a supplement submodule of M. Consider the proof of Theorem 2.2.
Then, by [2] (20.2(c)), since Im f ∩ (K1 + K2) � M we have Im f ∩ (K1 + K2) � Im f.
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A GENERALIZATION OF LIFTING MODULES 1481
Hence K1 ∩ (Im f +K2) ⊆ [Im f ∩ (K1 +K2)] + [K2 ∩ (Im f +K1)] � Im f +K2. Similarly,
K2 ∩ (Im f +K1)� Im f +K1.
A module K is said to be generalized M -projective if, for any epimorphism g : M → X and
morphism f : K → X, there exist decompositions K = K1 ⊕K2, M =M1 ⊕M2, a morphism h1 :
K1 →M1 and an epimorphism h2 : M2 → K2, such that h1g = f |K1 and h2f = g|M2 .
Lemma 2.1. Assume that M is a module, N ≤⊕ M and N = K ⊕ L. Let M be a L-I-lifting
module. If K is generalized L-projective, then for every homomorphism f : M → N such that
N = Im f +L and π(Im f) = Im f ∩ Imπ where π is an arbitrary projection map of N, there exist
X
cs
↪→ Im f in N, K ′ ⊆ K and L′ ⊆ L such that N = X ⊕K ′ ⊕ L′.
Proof. Let f : M → N be a homomorphism such that N = Im f + L. Consider the homomor-
phism πLf : M → L, where πL is the projection map from N onto L. Note that ImπLf = Im f ∩L
by hypothesis. Since M is L-I-lifting, there exists a decomposition L = L1 ⊕ L2 such that
L2 ⊆ L ∩ Im f and (L ∩ Im f) ∩ L1 = L1 ∩ Im f � L1. Thus we get N = Im f + L = Im f + L1
and L1 ∩ Im f � L1. As L2 ⊆ Im f, Im f = L2 ⊕ ((L1 ⊕ K) ∩ Im f). Set U = L1 ⊕ K. Since
N = Im f + L1, U = (U ∩ Im f) + L1. By [2] (4.43 and 4.42), there exists a decomposition
U = T ⊕K ′ ⊕ L′1 = T + L1 with T ⊆ U ∩ Im f, K ′ ⊆ K and L′1 ⊆ L1. As T ⊆ U ∩ Im f and
Im f = L2 + (U ∩ Im f), we have L2 ⊕ T ⊆ Im f. Since N = (L2 + T ) +L1 and Im f ∩L1 � N,
we have, by [2] (3.2(6)), that (L2 ⊕ T )
cs
↪→ Im f in N. As N = L2 ⊕ U.
Lemma 2.1 is proved.
Theorem 2.3. Suppose M = M1 ⊕M2 and M is Mi-I-lifting for i = 1, 2. Let M1 and M2 be
relatively generalized projective modules. Then for every f ∈ S, Im f is a direct summand of M if
Im f is a coclosed submodule of M and π(Im f) = Im f ∩ Imπ, where π is any projection map of
M. Moreover, M =M1 ⊕M2 is an exchange decomposition.
Proof. Let f ∈ S such that Im f is a coclosed submodule of M. Since M is Mi-I-supplemented,
M is a I-supplemented module and Im f has a supplement M ′1 ⊕ M ′2, where M ′1 ⊆ M1 and
M ′2 ⊆ M2 (see Corollary 2.5). As M is I-supplemented, the coclosed images of M are precisely
the supplement images and Im f + M ′1 and Im f + M ′2 are supplement submodules of M (see
Corollary 2.5 again). Since M is M2-I-lifting, M1 is generalized M2-projective and Im f +M ′1 is
a supplement, it follows that there exists a decomposition M = (Im f +M ′1) ⊕M ′′1 ⊕M ′′2 , with
M ′′1 ⊆ M1 and M ′′2 ⊆ M2 (see Lemma 2.1). Set U = M1 + Im f and N = M ′′1 ⊕M ′′2 . Then
M = U⊕N and M/ Im f = U/ Im f⊕ (N+Im f)/ Im f. Hence (N+Im f)/ Im f is a supplement
of M/ Im f. By [2] (20.5(2)), N + Im f is a supplement of M because Im f is a supplement of
M. Since N + Im f = Im f ⊕ M ′′1 ⊕ M ′′2 , by [2] (20.5(1)), Im f ⊕ M ′′2 is a supplement of M
and M = Im f +M ′′2 +M ′1 +M ′′1 ⊆ (Im f ⊕M ′′2 ) +M1. By using Lemma 2.1 again, we have
M = (Im f⊕M ′′2 )⊕M∗1⊕M∗2 ,withM∗1 ⊆M1 andM∗2 ⊆M2.HenceM = Im f⊕M∗1⊕(M∗2⊕M ′′2 ),
where M∗1 ⊆M1 and M∗2 ⊕M ′′2 ⊆M2. Therefore Im f ≤⊕ M and since any direct summand of M
is a coclosed epimorphic image of M, M =M1 ⊕M2 is an exchange decomposition.
Theorem 2.4. Let M1 and M2 be modules and M =M1 ⊕M2 such that M is Mi-I-lifting for
i = 1, 2 and let for every f ∈ S we have π(Im f) = Im f ∩ Imπ, where π is any projection map of
M. If any direct summand of M1 is generalized M2-projective and vice versa, then M is I-lifting.
Proof. Let f ∈ S. SinceM isM1-I-lifting, there exists a decompositionM =M ′1⊕M ′′1 such that
M ′1 ⊆ Im f ∩M1 = Im(πM1f) and Im(πM1f)∩M ′′1 �M ′′1 (π will denote the obvious projections).
Set K1 = Im f ∩ (M ′′1 ⊕M2). Note that Im(π(M ′′
1 ⊕M2)f) = K1. Since M is M2-I-lifting, there
exists a decomposition M2 =M ′2⊕M ′′2 such that M ′2 ⊆ πM2(K) and M ′′2 ∩πM2(K1)�M ′′2 . Setting
K2 = (M ′′1 ⊕M ′′2 )∩ Im f, we get that πM ′′
i
(K2)�M ′′i for i = 1, 2, M = πM1(Im f)+M ′′1 +M2 =
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1482 T. AMOUZEGAR KALATI
= Im f+M ′′1 +M2 = Im f+M ′′1 +πM2(K1)+M
′′
2 = Im f+M ′′1 +K1+M
′′
2 = Im f+(M ′′1 ⊕M ′′2 )
and K2 ⊆ πM ′′
1
(K2) ⊕ πM ′′
2
(K2) � M ′′1 ⊕M ′′2 . By hypothesis and [2] (4.43), direct summands of
M1 and M2 are relatively generalized projective and by Theorem 2.1, direct summands of M and Mi
are relatively I-lifting for i = 1, 2. Set N =M ′1⊕M ′′2 . Then N =M ′′2 +[(Im f +M ′′1 )∩N ]. Define
L = (Im f+M ′′1 )∩N. Then N = L+M ′′2 . Consider the endomorphism g : M ′1⊕M ′′1 ⊕M ′2⊕M ′′2 →
→M ′1⊕M ′′1 ⊕M ′2⊕M ′′2 defined by g(m′1+m
′′
1+m
′
2+m
′′
2) = f(m′1+m
′′
1+m
′
2+m
′′
2)+m
′′
1. Note
that Im g = Im f +M ′′1 and L = ImπNg. By using the Lemma 2.1, we have N = U ⊕ M̃ ′1 ⊕ M̃ ′′2 ,
where U
cs
↪→ L in N, M̃ ′1 ⊆ M ′1 and M̃ ′′2 ⊆ M ′′2 . By [2] (3.2(1)), we get N = U + M ′′2 . Let
M ′1 = M ′1 ⊕ M̃ ′1 and M ′′2 = M ′′2 ⊕ M̃ ′′2 . Now, we get M = M ′′1 ⊕M ′2 ⊕ U ⊕ M̃ ′1 ⊕ M̃ ′′2 . Next,
set T = U ⊕M ′′1 ⊕M ′2. Then T is a direct summand of M. Now M ′2 ⊆ πM2(K1) ⊆ Im f +M ′′1 .
Then M = N ⊕M ′′1 ⊕M ′2 and M ′′1 ⊕M ′2 ⊆ (Im f +M ′′1 ) imply Im f +M ′′1 = L ⊕M ′′1 ⊕M ′2.
Since U
cs
↪→ L in N and N is a direct summand of M, we have T
cs
↪→ (Im f +M ′′1 ) in M. Also
T = M ′′1 + (Im f ∩ T ) and M = T +M ′′2 . Hence M = (Im f ∩ T ) +M ′′1 +M ′′2 . As Im f =
= (Im f ∩ T ) + Im f ∩ (M ′′1 ⊕M ′′2 ) = (Im f ∩ T ) + K2 and K2 � M, by [2] (3.2(6)), we get
that (Im f ∩ T ) cs
↪→ Im f in M. Now, set A = M ′′1 ⊕M ′1 and B = M ′2 ⊕M ′′2 . Then A and B are
relatively generalized projective and M is A-I-lifting. Let φ = πA⊕B|T . Then φ : T → A⊕B is an
isomorphism and A ⊕ B = φ(T ) = φ(Im f ∩ T ) +M ′′1 = φ(Im f ∩ T ) + A = φ(πT f(M)) + A.
Using the Lemma 2.1 again, then there exists a direct summand T ′ ⊆ T such that T ′
cs
↪→ (Im f ∩ T )
in M. Since (Im f ∩ T ) cs
↪→ Im f in M, we have T ′
cs
↪→ Im f in M by [2] (3.2(2)). Therefore M is
an I-lifting module.
Corollary 2.6. LetM beMi-I-lifting for i ∈ {1, 2, . . . , n} and putM =M1⊕. . .⊕Mn. Assume
that M ′i and T are relatively generalized projective for any direct summand M ′i of Mi and any direct
summand T of ⊕j 6=iMj , for any 1 ≤ i ≤ n and let for every f ∈ S we have π(Im f) = Im f ∩ Imπ,
where π is any projection map of M. Then M is I-lifting.
Proof. It follows from Theorem 2.4 by induction on n.
Corollary 2.7. Let M be Mi-I-lifting for i ∈ {1, 2, . . . , n} and set M = M1 ⊕ . . . ⊕ Mn.
Suppose that Mi and Mj are relatively projective for each 1 ≤ i, j ≤ n, i 6= j and let for every
f ∈ S we have π(Im f) = Im f ∩ Imπ, where π is any projection map of M. Then M is I-lifting.
Theorem 2.5. Let M = ⊕n
i=1Mi be a module and Mi �M for all i ∈ {1, . . . , n}. Then M is
an I-lifting module if and only if Mi is I-lifting for all i ∈ {1, . . . , n}.
Proof. The necessity follows from Theorem 2.1. Conversely, let Mi be an I-lifting module for
all i ∈ {1, . . . , n}. Let φ = (φij)i,j∈{1,...,n} ∈ EndR(M) be arbitrary, where φij ∈ Hom(Mj ,Mi).
Since Mi � M for all i ∈ {1, . . . , n}, Imφ = ⊕n
i=1 Imφii. As Mi is I-lifting, there exists a
direct summand Xi of Mi and a submodule Yi of Mi with Xi ⊆ Imφii, Imφii = Xi + Yi and
Yi � Mi. Set X = ⊕n
i=1Xi, then X is a direct summand of M. Moreover, Imφ = ⊕n
i=1 Imφii =
=
∑n
i=1
Xi +
∑n
i=1
Yi and ⊕n
i=1Yi � ⊕n
i=1Mi =M. Therefore M is I-lifting.
Proposition 2.8. Let M be an I-lifting projective module. Then Rad(M)�M.
Proof. Let N ⊆M be any submodule with N +Rad(M) =M. Then Rad(M)→M →M/K
is epimorphism and there exists f : M → Rad(M) with M = Im f + N. Since M is I-lifting,
there exists a decomposition M = M1 ⊕ M2 with M1 ⊆ Im f and M2 ∩ Im f � M2. Note
that M1 ⊆ Rad(M). By [19] (22.3), M1 = 0 and so Im f � M. Hence N = M. Therefore
Rad(M)�M.
Theorem 2.6. LetM be a projective I-lifting module. ThenM is a direct sum of cyclic modules.
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A GENERALIZATION OF LIFTING MODULES 1483
Proof. If M is a projective module, then, by Kaplansky,s Theorem [19] (8.10), M is a direct sum
of countably generated module. Hence it is enough to prove the assertion for countably generated
modules. We prove by induction. First, consider M =
∑
i∈N
Rmi, mi ∈ M. Since the canonical
map f : ⊕NRmi →M splits, there exists g : M → ⊕NRmi with gf = idM . Consider the morphisms
gi = πig ∈ EndR(M), where πi is the canonical projection. Then M =
∑
N
gi(M). Since M is
I-lifting, there exists a decomposition M = P1 ⊕Q1 with P1 ⊆ Im g1 and K1 = Im g1 ∩Q1 �M.
So Im g1 = P1 +K1. Note that P1 is cyclic because P1 is a direct summand of Rm1. Suppose, for
k ∈ N, we have found cyclic modules Pi ⊆ M with M =
(∑
i≤k
Pi
)
⊕ Qk and
∑
i≤k
Im gi =
= (⊕i≤kPi)+Kk, Kk �M. Since M is I-lifting, by Theorem 2.1, M is Qk-I-lifting. Hence there
exists a decomposition Qk = Pk+1 ⊕ Qk+1, with Pk+1 ⊆ Im gk+1, Im gk+1 = Pk+1 +K ′k+1, and
K ′k+1 = Im gk+1 ∩ Qk+1 � M. Thus we have M = (⊕i≤k+1Pi) ⊕ Qk+1 and
∑
i≤k+1
Im gi =
= (⊕i≤k+1Pi) +Kk+1 with Kk+1 = K ′k+1 +Kk �M. By Proposition 2.8, Rad(M)�M and so∑
i∈N
Ki ⊆ Rad(M)�M. Therefore M =
∑
i∈N
Im gi = (⊕i∈NPi) +
∑
i∈N
Ki = ⊕i∈NPi.
A ring R is called f-semiperfect if, every finitely generated R-module has a projective cover. A
module M is said to be principally lifting if, for every cyclic submodule N of M, there exists a
decomposition M =M1 ⊕M2 such that M1 ⊆ N and N ∩M2 �M2.
The following theorem gives a characterization of f-semiperfect rings.
Theorem 2.7. The following are equivalent for a ring R :
(1) R is f-semiperfect;
(2) RR is finitely supplemented;
(3) every cyclic right ideal has a supplement in RR;
(4) RR is I-supplemented;
(5) RR is principally lifting;
(6) RR is I-lifting.
Proof. (1)⇔ (2) ⇔ (3)⇔ (5) by [19] (42.11).
(3) ⇒ (4) and (5) ⇒ (6) are clear because Imφ is cyclic for every φ ∈ EndR(RR).
(4)⇒ (3). Assume that I = aR is any cyclic right ideal of R. Consider the R-homomorphism φ :
RR → RR defined by φ(r) = ar, where r ∈ R. Then Imφ = I. By (4), Imφ = I has a supplement
in RR.
(6) ⇒ (5). It is similar to the proof of (4)⇒ (3).
3. Relation between dual Rickart modules and I-lifting modules. It is clear that if M is a
dual Rickart module, then M is I-lifting, while the converse is not true (the Z-module Z4 is I-lifting
but it is not dual Rickart).
Lemma 3.1. Let M be a module. Then M is a dual Rickart module if and only if for every
g ∈ S, there exists an idempotent e ∈ S such that DS(Im g) = eS.
Proof. Let M be a dual Rickart module and g ∈ S. Then there exists an idempotent e ∈ S
such that Im g = eM. Hence e ∈ DS(Im g) and so eS ⊆ DS(Im g). Now if f ∈ DS(Im g), then
Im f ⊆ Im g = eM. Moreover, since S = eS ⊕ (1 − e)S, we have f = es1 + (1 − e)s2 for some
s1, s2 ∈ S. Since Im f ⊆ eM, f = es1. Therefore f ∈ eS and so eS = DS(Im g). Conversely,
let for every g ∈ S, there exists an idempotent e ∈ S such that DS(Im g) = eS. Then for g ∈ S,
eM ⊆ Im g. On the other hand, we have g ∈ DS(Im g) = eS. Thus there exists s ∈ S such that
g = eS. It follows that Im g ⊆ eM. Therefore Im g = eM.
Lemma 3.2. An T -noncosingular I-lifting module M is a dual Rickart module.
ISSN 1027-3190. Укр. мат. журн., 2014, т. 66, № 11
1484 T. AMOUZEGAR KALATI
Proof. Let g ∈ S. Since M is I-lifting, Im g = eM ⊕ B, where e2 = e ∈ S and B � M.
Hence eS = DS(eM) ⊆ DS(Im g). Now, let φ ∈ DS(Im g). We want to show that φ ∈ eS. Note
that M = eM⊕(1−e)M and Im g∩(1−e)M = (eM⊕B)∩(1−e)M ⊆ (1−e)B. Since B �M,
we have (1− e)B �M. Thus Im g∩ (1− e)M �M. As S = eS⊕ (1− e)S, there exists s1 and s2
in S such that φ = es1 + (1− e)s2. Thus Im(1− e)s2 ≤ Im g ∩ (1− e)M �M. By hypothesis, we
have (1− e)s2 = 0 and so φ = es1 ∈ eS. Thus DS(Im g) = eS. By Lemma 3.1, M is dual Rickart.
The following results exhibits the connection between dual Rickart modules and I-lifting modules.
Corollary 3.1. Let M be a module. Then M is dual Rickart if and only if M is I-lifting and
T -noncosingular.
Proposition 3.1. Let M be a module with Rad(M) = 0. Then M is I-lifting if and only if M
is dual Rickart.
Proof. Let M be an I-lifting module and let φ ∈ S. Then there exists a direct summand X of
M and a submodule Y of M such that Imφ = X ⊕ Y and Y � M. Hence Y ⊆ Rad(M) = 0 and
so Imφ is a direct summand of M, this means that M is dual Rickart. The converse is clear.
Recall that a ring R is said to be a right V -ring if every simple right R-module is injective.
Corollary 3.2. Let R be a right V -ring and M be an R-module. Then M is dual Rickart if and
only if M is I-lifting.
Proof. By [19] (23.1), Rad(M) = 0 for right R-module M. Thus, by Proposition 3.1, every
I-lifting R-module is dual Rickart.
Corollary 3.3. Let R be a commutative regular ring andM be an R-module. ThenM is I-lifting
if and only if M is dual Rickart.
Proof. It is clear by Corollary 3.2 and [19] (23.5(2)).
1. Armendariz E. P. A note on extensions of Baer and P. P.-rings // J. Austral. Math. Soc. – 1974. – 18. – P. 470 – 473.
2. Clark J., Lomp C., Vanaja N., Wisbauer R. Lifting modules-supplements and projectivity in module theory // Front.
Math. – Birkhäuser, 2006.
3. Endo S. Note on p. p. rings // Nagoya Math. J. – 1960. – 17. – P. 167 – 170.
4. Evans M. W. On commutative P. P. rings // Pacif. J. Math. – 1972. – 41, № 3. – P. 687 – 697.
5. Kaplansky I. Rings of operators // Math. Lect. Note Ser. – New York: W. A. Benjamin, 1968.
6. Keskin D. On lifting modules // Communs Algebra. – 2000. – 28, № 7. – P. 3427 – 3440.
7. Keskin D., Xue W. Generalizations of lifting modules // Acta. Math. hung. – 2001. – 91. – P. 253 – 261.
8. Keskin D., Orhan N. CCSR-modules and weak lifting modules // East-West J. Math. – 2003. – 5, № 1. – P. 89 – 96.
9. Kosan M. T. δ-Lifting and δ-supplemented modules // Algebra Colloq. – 2007. – 14, № 1. – P. 53 – 60.
10. Lee G., Rizvi S. T., Roman C. S. Rickart modules // Communs Algebra. – 2010. – 38, № 11. – P. 4005 – 4027.
11. Lee G., Rizvi S. T., Roman C. S. Dual Rickart modules // Communs Algebra. – 2011. – 39. – P. 4036 – 4058.
12. Liu Q., Ouyang B. Y. Rickart modules // Nanjing Daxue Xuebao Shuxue Bannian Kan. – 2006. – 23, № 1. –
P. 157 – 166 (in Chinese).
13. Liu Q., Ouyang B. Y., Wu T. S. Principally quasi-Baer modules // J. Math. Res. Exposition. – 2009. – 29, № 5. –
P. 823 – 830.
14. Maeda S. On a ring whose principal right ideals generated by idempotents form a lattice // J. Sci. Hiroshima Univ.
Ser. A. – 1960. – 24. – P. 509 – 525.
15. Mohamed S. H., Müller B. J. Continuous and discrete modules // London Math. Soc. Lect. Not. Ser. 147. – Cambridge:
Cambridge Univ. Press, 1990.
16. Rizvi S. T., Roman C. S. Baer property of modules and applications // Adv. Ring Theory. – 2005. – P. 225 – 241.
17. Talebi Y., Amouzegar T. Pjective modules and a generalization of lifting modules // East-West J. Math. – 2010. – 12,
№ 1. – P. 9 – 15.
18. Tütüncü D. K., Tribak R. On T -noncosingular modules // Bull. Austral. Math. Soc. – 2009. – 80. – P. 462 – 471.
19. Wisbauer R. Foundations of module and ring theory. – Reading: Gordon and Breach, 1991.
Received 15.12.12
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| id | umjimathkievua-article-2239 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:21:18Z |
| publishDate | 2014 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/84/fdf560085597e89c6733e6f668c31284.pdf |
| spelling | umjimathkievua-article-22392019-12-05T10:27:00Z A Generalization of Lifting Modules Узагальненння підйомних модулів Kalati, Amouzegar T. Калаті, Амузегар Т. We introduce the notion of $I$ -lifting modules as a proper generalization of the notion of lifting modules and present some properties of this class of modules. It is shown that if $M$ is an $I$ -lifting direct projective module, then $S/▽$ is regular and $▽ = \text{Jac} S$, where $S$ is the ring of all $R$-endomorphisms of $M$ and $▽ = \{ϕ ∈ S | Im ϕ ≪ M\}$. Moreover, we prove that if $M$ is a projective $I$ -lifting module, then $M$ is a direct sum of cyclic modules. The connections between $I$ -lifting modules and dual Rickart modules are presented. Введено поняття $I$-підйомних модулів як природне узагальнення підйомних модулів. Наведено дєякі властивості цього класу модулів. Показано, що якщо $M$ — прямий проективний модуль $I$-підйому, то $S/▽$ є регулярною i $▽ = \text{Jac} S$, де $S$ — кільце всіх $R$-ендоморфізмів $M$, а $▽ = \{ϕ ∈ S | Im ϕ ≪ M\}$. Більш того, доведено, що якщо $M$ — проективний $I$-підйомний модуль, то $M$ є прямою сумою циклічних модулів. Встановлено зв'язки між $I$-підйомними модулями та дуальними модулями Рікарта. Institute of Mathematics, NAS of Ukraine 2014-11-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2239 Ukrains’kyi Matematychnyi Zhurnal; Vol. 66 No. 11 (2014); 1477–1484 Український математичний журнал; Том 66 № 11 (2014); 1477–1484 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2239/1479 https://umj.imath.kiev.ua/index.php/umj/article/view/2239/1480 Copyright (c) 2014 Kalati Amouzegar T. |
| spellingShingle | Kalati, Amouzegar T. Калаті, Амузегар Т. A Generalization of Lifting Modules |
| title | A Generalization of Lifting Modules |
| title_alt | Узагальненння підйомних модулів |
| title_full | A Generalization of Lifting Modules |
| title_fullStr | A Generalization of Lifting Modules |
| title_full_unstemmed | A Generalization of Lifting Modules |
| title_short | A Generalization of Lifting Modules |
| title_sort | generalization of lifting modules |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2239 |
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