On A Class of Lifting Modules
aa r X i v : . [ m a t h . R A ] D ec ON A CLASS OF LIFTING MODULES
HATICE INANKIL, SAIT HALICIO ˘GLU, AND ABDULLAH HARMANCI
Abstract.
In this paper, we introduce principally δ -lifting modules whichare analogous to δ -lifting modules and principally δ -semiperfect modules as ageneralization of δ -semiperfect modules and investigate their properties. Introduction
Throughout this paper all rings have an identity, all modules considered areunital right modules. Let M be a module and N, P be submodules of M . We call P a supplement of N in M if M = P + N and P ∩ N is small in P . A module M is called supplemented if every submodule of M has a supplement in M . A module M is called lifting if, for all N ≤ M , there exists a decomposition M = A ⊕ B suchthat A ≤ N and N ∩ B is small in M . Supplemented and lifting modules havebeen discussed by several authors (see [2, 4, 6]) and these modules are useful incharacterizing semiperfect and right perfect rings (see [4, 7]).In this note, we study and investigate principally δ -lifting modules and princi-pally δ -semiperfect modules. A module M is called principally δ -lifting if for eachcyclic submodule has the δ -lifting property , i.e., for each m ∈ M , M has a decom-position M = A ⊕ B with A ≤ mR and mR ∩ B is δ -small in B , where B is called a δ -supplement of mR . A module M is called principally δ -semiperfect if, for each m ∈ M , M/mR has a projective δ -cover. We prove that if M is semisimple, M is principally δ -lifting, M and M are relatively projective, then M = M ⊕ M isa principally δ -lifting module. Among others we also prove that for a principally δ -semiperfect module M , M is principally δ -supplemented, each factor module of M is principally δ -semiperfect, hence any homomorphic image and any direct sum-mand of M is principally δ -semiperfect. As an application, for a projective module M , it is shown that M is principally δ -semiperfect if and only if it is principally δ -lifting, and therefore a ring R is principally δ -semiperfect if and only if it isprincipally δ -lifting.In section 2, we give some properties of δ -small submodules that we use in thepaper, and in section 3, principally δ -lifting modules are introduced and variousproperties of principally δ -lifting and δ -supplemented modules are obtained. Insection 4, principally δ -semiperfect modules are defined and characterized in termsof principally δ -lifting modules.In what follows, by Z , Q , Z n and Z / Z n we denote, respectively, integers, ra-tional numbers, the ring of integers and the Z -module of integers modulo n . Forunexplained concepts and notations, we refer the reader to [1, 4]. Mathematics Subject Classification.
Key words and phrases. lifting modules, δ -lifting modules, semiperfect modules, δ -semiperfectmodules. δ -Small Submodules Following Zhou [9], a submodule N of a module M is called a δ -small submodule if, whenever M = N + X with M/X singular, we have M = X . We begin bystating the next lemma which is contained in [9, Lemma 1.2 and 1.3]. Lemma 2.1.
Let M be a module. Then we have the following. (1) If N is δ -small in M and M = X + N , then M = X ⊕ Y for a projectivesemisimple submodule Y with Y ⊆ N . (2) If K is δ -small in M and f : M → N is a homomorphism, then f ( K ) is δ -small in N . In particular, if K is δ -small in M ⊆ N , then K is δ -smallin N . (3) Let K ⊆ M ⊆ M , K ⊆ M ⊆ M and M = M ⊕ M . Then K ⊕ K is δ -small in M ⊕ M if and only if K is δ -small in M and K is δ -smallin M . (4) Let N , K be submodules of M with K is δ -small in M and N ≤ K . Then N is also δ -small in M . Lemma 2.2.
Let M be a module and m ∈ M . Then the following are equivalent. (1) mR is not δ - small in M . (2) There is a maximal submodule N of M such that m N and M/N singular.Proof. (1) ⇒ (2) Let Γ := { B ≤ M | B = M, mR + B = M , M/B singular } .Since mR is not δ -small in M , there exists a proper submodule B of M such that mR + B = M and M/B singular. So Γ is non empty. Let Λ be a nonempty totallyordered subset of Γ and B := ∪ B ∈ Λ B . If m is in B then there is a B ∈ Λ with m ∈ B . Then B = mR + B = M which is a contraction. So we have m / ∈ B and B = M . Since mR + B = M and M/B singular, B is upper bound in Γ. ByZorn’s Lemma, Γ has a maximal element, say N . If N is a maximal submodule of M there is nothing to do. Assume that there exists a submodule K containing N properly. Since N is maximal in Γ, K is not in Γ. Since M = mR + N and N ≤ K ,so M = mR + K . M/K is singular as a homomorphic image of singular module
M/N . Hence K must belong to the Γ. This is the required contradiction.(2) ⇒ (1) Let N be a maximal submodule with m ∈ M \ N and M/N singular.We have M = mR + N . Then mR is not δ -small in M . (cid:3) Let A and B be submodules of M with A ≤ B . A is called a δ -cosmall submodule of B in M if B/A is δ -small in M/A . Let A be a submodule of M . A is calleda δ -coclosed submodule in M if A has no proper δ -cosmall submodules in M . Asubmodule A is called δ - coclosure of B in M if A is δ -coclosed submodule of M andit is δ -cosmall submodule of B . Equivalently, for any submodule C ≤ A with A/C is δ -small in M/C implies C = A and B/A is δ -small in M/A . Note that δ -coclosedsubmodules need not always exist. N A CLASS OF LIFTING MODULES 3
Lemma 2.3.
Let A and B be submodules of M with A ≤ B . Then we have: (1) A is δ -cosmall submodule of B in M if and only if M = A + L for anysubmodule L of M with M = B + L and M/L singular. (2) If A is δ -small and B is δ -coclosed in M , then A is δ -small in B .Proof. (1) Necessity: Let M = B + L and M/L be singular. We have
M/A = B/A + ( L + A ) /A and M/ ( L + A ) is singular as homomorphic image of singularmodule M/L . Since
B/A is δ -small, M/A = ( L + A ) /A or M = L + A .Sufficiency: Let M/A = B/A + K/A and
M/K singular. Then M = B + K . Byhypothesis, M = A + K and so M = K . Hence A is δ -cosmall submodule of B in M .(2) Assume that A is δ -small submodule of M and B is δ -coclosed in M . Let B = A + K with B/K singular. Since B is δ -coclosed in M , to complete theproof, by part (1) it suffices to show that K is δ -small submodule of B in M. Let M = B + L with M/L singular. By assumption, M = A + K + L = K + L since M/ ( K + L ) is singular. By (1), K is δ -small submodule of B in M . (cid:3) Lemma 2.4.
Let A , B and C be submodules of M with M = A + C and A ⊆ B .If B ∩ C is a δ -small submodule of M , then A is a δ -cosmall submodule of B in M .Proof. Let
M/A = B/A + L/A with
M/L singular. We have M = B + L and B = A + ( B ∩ C ). Then M = A + ( B ∩ C ) + L = ( B ∩ C ) + L . Hence M = L since B ∩ K is δ -small in M and M/L is singular. Hence
B/A is δ -small in M/A . Thus A is δ -cosmall submodule of B in M . (cid:3) Principally δ -Lifting Modules In this section, we study and investigate some properties of principally δ -liftingmodules. The following definition is motivated by [9, Lemma 3.4] and Lemma 3.4. Definition 3.1.
A module M is called finitely δ -lifting if for any finitely generatedsubmodule A of M has the δ -lifting property , that is, there is a decomposition M = N ⊕ S with N ≤ A and A ∩ S is δ -small in S . In this case A ∩ S is δ -small in S if and only if A ∩ S is δ -small in M . A module M is called principally δ -lifting if foreach cyclic submodule has the principally δ -lifting property , i.e., for each m ∈ M , M has a decomposition M = A ⊕ B with A ≤ mR and mR ∩ B is δ -small in B . Example 3.2.
Every submodule of any semisimple module satisfies principally δ -lifting property. Example 3.3.
Let p be a prime integer and n any positive integer. Then the Z -module M = Z / Z p n is a principally δ -lifting module. HATICE INANKIL, SAIT HALICIO ˘GLU, AND ABDULLAH HARMANCI
Lemma 3.4 is proved in [7] and [9].
Lemma 3.4.
The following are equivalent for a module M . (1) M is finitely δ -lifting. (2) M is principally δ -lifting. Let M be a module and N a submodule of M . A submodule L is called a δ -supplement of N in M if M = N + L and N ∩ L is δ -small in L (therefore in M ). Proposition 3.5.
Let M be a principally δ -lifting module. The we have: (1) Every direct summand of M is a principally δ -lifting module. (2) Every cyclic submodule C of M has a δ -supplement S which is a directsummand, and C contains a complementary summand of S in M .Proof. (1) Let K be a direct summand of M and k ∈ K . Then M has a decom-position M = N ⊕ S with N ≤ kR and kR ∩ S is δ -small in M . It follows that K = N ⊕ ( K ∩ S ), and kR ∩ ( K ∩ S ) ≤ kR ∩ S is δ -small in M and so kR ∩ ( K ∩ S )is δ -small in K . Therefore K is a principally δ -lifting module.(2) Assume that M is a principally δ -lifting module and C is a cyclic submoduleof M . Then we have M = N ⊕ S , where N ≤ C and C ∩ S is δ -small in M . Hence M = N + S ≤ C + S ≤ M , we have M = C + S . Since S is direct summand and C ∩ S is δ -small in M , C ∩ S is δ -small in S . Therefore S is a δ -supplement of C in M . (cid:3) Theorem 3.6.
The following are equivalent for a module M . (1) M is a principally δ -lifting module. (2) Every cyclic submodule C of M can be written as C = N ⊕ S , where N isdirect summand and S is δ -small in M . (3) For every cyclic submodule C of M , there is a direct summand A of M with A ≤ C and C/A is δ -small in M/A . (4) Every cyclic submodule C of M has a δ -supplement K in M such that C ∩ K is a direct summand in C . (5) For every cyclic submodule C of M , there is an idempotent e ∈ End ( M ) with eM ≤ C and (1 − e ) C is δ -small in (1 − e ) M . (6) For each m ∈ M , there exist ideals I and J of R such that mR = mI ⊕ mJ ,where mI is direct summand of M and mJ is δ -small in M .Proof. (1) ⇒ (2) Let C be a cyclic submodule of M . By hypothesis there exist N and S submodules of M such that N ≤ C , C ∩ S is δ -small in M and M = N ⊕ S .Then we have C = N ⊕ ( C ∩ S ). N A CLASS OF LIFTING MODULES 5 (2) ⇒ (3) Let C be a cyclic submodule of M . By hypothesis, C = N ⊕ S , where N is direct summand and S is δ -small in M . Let π : M → M/N be the naturalprojection. Since S is δ -small in M , we have π ( S ) is δ -small in M/N . Since π ( S ) ∼ = S ∼ = C/N , C/N is δ -small in M/N .(3) ⇒ (4) Let C be a cyclic submodule of M . By hypothesis, there is a directsummand A ≤ M with A ≤ C and C/A is δ -small in M/A . Let M = A ⊕ A ′ .Hence C = A ⊕ ( A ′ ∩ C ). Let σ : M/A → A ′ denote the obvious isomorphism.Then σ ( C/A ) = A ′ ∩ C is δ -small in A ′ .(4) ⇒ (5) Let C be any cyclic submodule of M and K ≤ M such that C ∩ K is directsummand of C , M = C + K and C ∩ K is δ -small in K . So C = ( C ∩ K ) ⊕ X forsome X ≤ C . Then M = X +( C ∩ K )+ K = X ⊕ K . Let e : M → X ; e ( x + k ) = x and (1 − e ) : M → K ; e ( x + k ) = k are projection maps. e ( M ) ≤ X ≤ C and(1 − e ) C = C ∩ (1 − e ) M = C ∩ K is δ -small in (1 − e ) M .(5) ⇒ (6) Let mR be any cyclic submodule of M . By hypothesis, there exists anidempotent e ∈ End ( M ) such that eM ≤ mR , M = eM ⊕ (1 − e ) M and (1 − e ) mR is δ -small in (1 − e ) M . Note that ( mR ) ∩ ((1 − e ) M ) = (1 − e ) mR ( for if m = em + y ,where em ∈ eM , y ∈ ( mR ) ∩ ((1 − e ) M ). Then (1 − e ) m = em + (1 − e ) y = y and so (1 − e ) mR ≤ ( mR ) ∩ ((1 − e ) M ). Let mr = (1 − e ) m ′ ∈ ( mR ) ∩ ((1 − e ) M ).Then mr = (1 − e ) mr ∈ (1 − e ) mR . So ( mR ) ∩ ((1 − e ) M ) ≤ (1 − e ) mR . Thus( mR ) ∩ ((1 − e ) M ) = (1 − e ) mR ). So mR = eM ⊕ (1 − e ) mR . Let I = { r ∈ R : mr ∈ eM } and J = { t ∈ R : mt ∈ (1 − e ) mR } . Then mR = mI ⊕ mJ , mI = eM and mJ = (1 − e ) mR is δ -small in (1 − e ) M .(6) ⇒ (1) Let m ∈ M . By hypothesis, there exist ideals I and J of R such that mR = mI ⊕ mJ , where mI is direct summand and mJ is δ -small in M . Let M = mI ⊕ K for some submodule K . Since K ∩ mR ∼ = mJ and mJ is δ -small in M , M is principally δ -lifting. (cid:3) Note that every lifting module is principally δ -lifting. There are principally δ -lifting modules but not lifting. Example 3.7.
Let M be the Z -module Q and m ∈ M . It is well known thatevery cyclic submodule mR of M is small, therefore δ -small in M . Hence M is aprincipally δ -lifting Z -module. If N is a nonsmall proper submodule of M , then N is neither direct summand nor contains a direct summand of M . It follows that M is not a lifting Z -module.It is clear that every δ -lifting module is principally δ -lifting. However the converseis not true. Example 3.8.
Let R and T denote the rings in [9, Example 4.1], where HATICE INANKIL, SAIT HALICIO ˘GLU, AND ABDULLAH HARMANCI R = ∞ P i =1 L Z + Z . { ( f , f , . . . , f n , f, f, . . . ) ∈ ∞ Q i =1 Z } and T = (" x yo x : x ∈ R, y ∈ Soc ( R ) ) . Then Rad δ ( T ) = " Soc ( R )0 0 and T /
Rad δ ( T ) is not semisimple as isomorphic to R . So T is not δ -semiperfectby [9, Theorem 3.6]. Hence T is not a δ -lifting module over T . It is easy to showthat T /
Rad δ ( T ) lift to idempotents of T , so T is a semiregular ring. Since T isa δ -semiregular ring, every finitely generated right ideal H of T can be written as H = aT ⊕ S , where a = a ∈ T and S ≤ Rad δ ( T ) by [9, Theorem 3.5]. Hence T isa principally δ -lifting module. Proposition 3.9.
Let M be a principally δ -lifting module. If M = M + M suchthat M ∩ M is cyclic, then M contains a δ -supplement of M in M .Proof. Assume that M = M + M and M ∩ M is cyclic. Then we have M ∩ M = N ⊕ S , where N is direct summand of M and S is δ -small in M . Let M = N ⊕ N ′ and M = N ⊕ ( M ∩ N ′ ). It follows that M ∩ M = N ⊕ ( M ∩ M ∩ N ′ ) = N ⊕ S .Let π : M = N ⊕ ( M ∩ N ′ ) → N ′ be the natural projection. It follows that π ( M ∩ M ∩ N ′ ) = M ∩ M ∩ N ′ = π ( S ). Since S is δ -small in M , it is δ -small in N ′ by Lemma 2.1. Hence M = M + ( M ∩ N ′ ), M ∩ N ′ ≤ M and M ∩ ( M ∩ N ′ )is δ -small in M ∩ N ′ . M ∩ N ′ is contained in M and a δ -supplement of M in M . This completes the proof. (cid:3) Let M be a module. A submodule N is called fully invariant if for each endo-morphism f of M , f ( N ) ≤ N . Let S = End ( M R ), the ring of R -endomorphisms of M . Then M is a left S -, right R -bimodule and a principal submodule N of the right R -module M is fully invariant if and only if N is a sub-bimodule of M . Clearly0 and M are fully invariant submodules of M . The right R -module M is calleda duo module provided every submodule of M is fully invariant. For the readers’convenience we state and prove Lemma 3.10 which is proved in [5]. Lemma 3.10.
Let a module M = L i ∈ I M i be a direct sum of submodules M i ( i ∈ I ) and let N be a fully invariant submodule of M . Then N = L i ∈ I ( N ∩ M i ) .Proof. For each j ∈ I , let p j : M → M j denote the canonical projection and let i j : M j → M denote inclusion. Then i j p j is an endomorphism of M and hence i j p j ( N ) ⊆ N for each j ∈ I . It follows that N ⊆ L j ∈ I i j p j ( N ) ⊆ L j ∈ I ( N ∩ M j ) ⊆ N ,so that N = L j ∈ I ( N ∩ M j ). (cid:3) One may suspect that if M and M are principally δ -lifting modules, then M ⊕ M is also principally δ -lifting. But this is not the case. N A CLASS OF LIFTING MODULES 7
Example 3.11.
Consider the Z -modules M = Z / Z M = Z / Z
8. It isclear that M and M are principally δ -lifting. Let M = M ⊕ M . Then M isnot a principally δ -lifting Z -module. Let N = (1 , Z and N = (1 , Z . Then M = N + N , N is not a direct summand of M and does not contain any nonzerodirect summand of M . For any proper submodule N of M , M/N is singular Z -module. Hence the principal submodule does not satisfy δ -lifting property. Itfollows that M is not principally δ -lifting Z -module. By the same reasoning, for anyprime integer p , the Z -module M = ( Z / Z p ) ⊕ ( Z / Z p ) is not principally δ -lifting.We have already observed by the preceding example that the direct sum ofprincipally δ -lifting modules need not be principally δ -lifting. Note the followingfact. Proposition 3.12.
Let M = M ⊕ M be a decomposition of M with M and M principally δ -lifting modules. If M is a duo module, then M is principally δ -lifting.Proof. Let M = M ⊕ M be a duo module and mR be a submodule of M . ByLemma 3.10, mR = (( mR ) ∩ M ) ⊕ (( mR ) ∩ M ). Since ( mR ) ∩ M and ( mR ) ∩ M are principal submodules of M and M respectively, there exist A , B ≤ M suchthat A ≤ ( mR ) ∩ M ≤ M = A ⊕ B , B ∩ (( mR ) ∩ M ) = B ∩ ( mR ) is δ -small in B , and A , B ≤ M such that A ≤ ( mR ) ∩ M ≤ M = A ⊕ B , B ∩ (( mR ) ∩ M ) = B ∩ ( mR ) is δ -small in B . Then M = A ⊕ A ⊕ B ⊕ B , A ⊕ A ≤ N and ( mR ) ∩ ( B ⊕ B ) = (( mR ) ∩ B ) ⊕ (( mR ) ∩ B ) is δ -small in M ⊕ M . (cid:3) Lemma 3.13.
The following are equivalent for a module M = M ′ ⊕ M ′′ . (1) M ′ is M ′′ -projective. (2) For each submodule N of M with M = N + M ′′ , there exists a submodule N ′ ≤ N such that M = N ′ ⊕ M ′′ .Proof. See [7, 41.14] (cid:3)
Theorem 3.14.
Let M be a semisimple module and M a principally δ -liftingmodule. Assume that M and M are relatively projective. Then M = M ⊕ M isprincipally δ -lifting.Proof. Let 0 = m ∈ M and let K = M ∩ (( mR ) + M ). We divide the proof intotwo cases: Case (i): K = 0. Then M = K ⊕ K for some submodule K of M and so M = K ⊕ K ⊕ M = ( mR ) + ( M ⊕ K ). Hence K is M ⊕ K -projective.By Lemma 3.13, there exists a submodule N of mR such that M = N ⊕ ( M ⊕ HATICE INANKIL, SAIT HALICIO ˘GLU, AND ABDULLAH HARMANCI K ). We may assume ( mR ) ∩ ( M ⊕ K ) = 0. Note that for any submod-ule L of M , we have ( mR ) ∩ ( L + K ) = L ∩ (( mR ) + K ). In particular( mR ) ∩ ( M + K ) = M ∩ ( mR + K ). Then mR = N ⊕ ( mR ) ∩ ( K ⊕ M ). Thereexist n ∈ N and m ′ ∈ ( mR ) ∩ ( K ⊕ M ) such that m = n + m ′ . Then nR = N and m ′ R = ( mR ) ∩ ( K ⊕ M ). Since ( mR ) ∩ ( M + K ) = M ∩ (( mR ) + K ), M ∩ (( mR ) + K ) is a principal submodule of M and M is principally δ -lifting,there exists a submodule X of M ∩ (( mR ) + K ) = ( mR ) ∩ ( M ⊕ K ) suchthat M = X ⊕ Y and Y ∩ M ∩ (( mR ) + K ) = Y ∩ (( mR ) + K ) is δ -small in M ∩ (( mR ) + K ) and in M . Hence M = ( N ⊕ X ) ⊕ ( Y ⊕ K ). Since N ⊕ X ≤ mR and ( mR ) ∩ ( Y ⊕ K ) = Y ∩ (( mR ) + K ), ( mR ) ∩ ( Y ⊕ K ) = Y ∩ (( mR ) + K )is δ -small in Y ⊕ K . So M is δ -lifting. Case (ii): K = 0. Then mR ≤ M . Since M is δ -lifting, there exists a submodule X of mR such that M = X ⊕ Y and ( mR ) ∩ Y is δ -small in Y for some submodule Y of M . Hence M = X ⊕ ( M ⊕ Y ). Since ( mR ) ∩ ( M ⊕ Y ) = ( mR ) ∩ Y and( mR ) ∩ ( M ⊕ Y ) = ( mR ) ∩ Y is δ -small in Y . By Lemma 2.1 (3), ( mR ) ∩ ( M ⊕ Y )is δ -small in M ⊕ Y . It follows that M is δ -lifting. (cid:3) A module M is said to be a principally semisimple if every cyclic submodule isa direct summand of M . Tuganbayev calls a principally semisimple module as aregular module in [3]. Every semisimple module is principally semisimple. Everyprincipally semisimple module is principally δ -lifting. For a module M , we writeRad δ ( M ) = P { L | L is a δ -small submodule of M } . Lemma 3.15.
Let M be a principally δ -lifting module. Then M/ Rad δ ( M ) is aprincipally semisimple module.Proof. Let m ∈ M . There exists M ≤ mR such that M = M ⊕ M and ( mR ) ∩ M is δ -small in M . So( mR ) ∩ M is δ -small in M . Then M/ Rad δ ( M ) = [( mR + Rad δ ( M )) / Rad δ ( M )] ⊕ [( M + Rad δ ( M )) / Rad δ ( M )]because ( mR + Rad δ ( M )) ∩ ( M +Rad δ ( M )) =Rad δ ( M ). Hence every principalsubmodule of M/ Rad δ ( M ) is a direct summand. (cid:3) Proposition 3.16.
Let M be a principally δ -lifting module. Then M = M ⊕ M ,where M is a principally semisimple module and M is a module with Rad δ ( M ) essential in M .Proof. Let M be a submodule of M such that Rad δ ( M ) ⊕ M is essential in M and m ∈ M . Since M is principally δ -lifting, there exists a direct summand M of M such that M ≤ mR , M = M ⊕ M ′ and mR ∩ M ′ is δ -small in M . Hence mR ∩ M ′ is a submodule of Rad δ ( M ) and so mR ∩ M ′ = 0. Then m ∈ M and mR = M .Since M ∩ Rad δ ( M ) = 0, M is isomorphic to a submodule of M/ Rad δ ( M ). By N A CLASS OF LIFTING MODULES 9
Lemma 3.15, M/ Rad δ ( M ) is principally semisimple, M is principally semisimple.On the other hand, Rad δ ( M ) =Rad δ ( M ′ ) is essential in M that it is clear fromthe construction of M ′ . (cid:3) A nonzero module M is called δ -hollow if every proper submodule is δ -small in M , and M is principally δ -hollow if every proper cyclic submodule is δ -small in M ,and M is finitely δ -hollow if every proper finitely generated submodule is δ -smallin M . Since finite direct sum of δ -small submodules is δ -small, M is principally δ -hollow if and only if it is finitely δ -hollow. Lemma 3.17.
The following are equivalent for an indecomposable module M . (1) M is a principally δ -lifting module. (2) M is a principally δ -hollow module.Proof. (1) ⇒ (2) Let m ∈ M . Since M is a principally δ -lifting module, there exist N and S submodules of M such that N ≤ mR , mR ∩ S is δ -small in M and M = N ⊕ S . By hypothesis, N = 0 and S = M . So that mR ∩ S = mR is δ -smallin M .(2) ⇒ (1) Let m ∈ M . Then mR = ( mR ) ⊕ (0). By (2) mR is δ -small and (0) isdirect summand in M . Hence M is a principally δ -lifting module. (cid:3) Lemma 3.18.
Let M be a module, then we have (1) If M is principally δ -hollow, then every factor module is principally δ -hollow. (2) If K is δ -small submodule of M and M/K is principally δ -hollow, then M is principally δ -hollow. (3) M is principally δ -hollow if and only if M is local or Rad δ ( M ) = M .Proof. (1) Assume that M is principally δ -hollow and N a submodule of M . Let m + N ∈ M/N and ( mR + N ) /N + K/N = M/N . Suppose that
M/K is singular.We have mR + K = M . Since M/K is singular and M is principally δ -hollow, M = K .(2) Let m ∈ M . Assume that mR + N = M for some submodule N with M/N singular. Then ( m + K ) R = ( mR + K ) /K is a cyclic submodule of M/K and( mR + K ) /K +( N + K ) /K = M/K and M/ ( N + K ) is singular as an homomorphicimage of M/N . Hence ( N + K ) /K = M/K or N + K = M . By hypothesis N = M .(3) Suppose that M is principally δ -hollow and it is not local. Let N and K betwo distinct maximal submodules of M and k ∈ K \ N . Then M = kR + N and M/N is a simple module, and so
M/N is a singular or projective module. If
M/N is singular, then M = N since kR is δ -small. But this is not possible since N ismaximal. So M/N is projective. Hence N is direct summand. So M = N ⊕ N ′ for some nonzero submodule N ′ of M , that is, N and kR are proper submodules of M . Since every proper submodule of M is contained in Rad δ ( M ), M = Rad δ ( M ).The converse is clear. (cid:3) Proposition 3.19.
Let M be a module. Then the following are equivalent. (1) M is principally δ -hollow. (2) If N is submodule with M/N cyclic, then N is a δ -small submodule of M .Proof. (1) ⇒ (2) Assume that N is a submodule with M/N cyclic. Lemma 2.1implies that
M/N is principally δ -hollow since being δ -small is preserved underhomomorphisms. Since M/N has maximal submodules, and by Lemma 3.18,
M/N is local. There exists a unique maximal submodule N containing N . Hence N issmall, therefore it is δ -small.(2) ⇒ (1) We prove that every cyclic submodule is δ -small in M . So let m ∈ M and M = mR + N with M/N singular. Then
M/N is cyclic. By hypothesis, N is δ -small submodule of M . By Lemma 2.1, there exists a projective semisimplesubmodule Y of N such that M = ( mR ) ⊕ Y . Let Y = L i ∈ I N i where each N i issimple. Now we write M = (( mR ) L i = j N j ) ⊕ N i . Then M/ (( mR ) L i = j N j ) is cyclicmodule as it is isomorphic to simple module N i . By hypothesis, (( mR ) L i = j N j ) is δ -small in M . Again by Lemma 2.1, there exists a projective semisimple submodule Z of (( mR ) L i = j N j ) such that M = Z ⊕ N i . Hence M is projective semisimplemodule. So M = N ⊕ N ′ for some submodule N ′ . Then N ′ is projective. M/N is projective as it is isomorphic to N ′ . Hence M/N is both singular and projectivemodule. Thus M = N . (cid:3) Applications
In this section, we introduce and study some properties of principally δ -semiperfectmodules. By [9], a projective module P is called a projective δ -cover of a module M if there exists an epimorphism f : P −→ M with Kerf is δ -small in P , and a ringis called δ -perfect (or δ -semiperfect ) if every R -module (or every simple R -module)has a projective δ -cover. For more detailed discussion on δ -small submodules, δ -perfect and δ -semiperfect rings, we refer to [9]. A module M is called principally δ -semiperfect if every factor module of M by a cyclic submodule has a projective δ -cover. A ring R is called principally δ -semiperfect in case the right R -module R isprincipally δ -semiperfect. Every δ -semiperfect module is principally δ -semiperfect.In [9], a ring R is called δ -semiregular if every cyclically presented R-module has aprojective δ -cover. Theorem 4.1.
Let M be a projective module. Then the following are equivalent. (1) M is principally δ -semiperfect. N A CLASS OF LIFTING MODULES 11 (2) M is principally δ -lifting.Proof. (1) ⇒ (2) Let m ∈ M and P f → M/mR be a projective δ -cover and M π → M/mR the natural epimorphism.
MP M/mR ♣♣♣♣♣♣♣♣♣♣♣ ✠ g ❄ π ✲ f ✲ Then there exists a map M g → P such that f g = π . Then P = g ( M )+Ker( f ). SinceKer( f ) is δ -small, by Lemma 2.1, there exists a projective semisimple submodule Y of Ker( f ) such that P = g ( M ) ⊕ Y . So g ( M ) is projective. Hence M = K ⊕ Ker ( g )for some submodule K of M . It is easy to see that g ( K ∩ mR ) = g ( K ) ∩ Ker( f )and Ker( g ) ≤ mR . Hence M = K + mR . Next we prove K ∩ ( mR ) is δ -small in K . Since Ker( f ) is δ -small in P , g ( K ) ∩ Ker( f ) = g ( K ∩ mR ) is δ -small in P byLemma 2.1 (4). Hence K ∩ ( mR ) is δ -small in K since g − is an isomorphism from g ( M ) onto K .(2) ⇒ (1) Assume that M is a principally δ -lifting module. Let m ∈ M . Thereexist direct summands N and K of M such that M = N ⊕ K , N ≤ mR and mR ∩ K is δ -small in K . Let K π → M/mR denote the natural epimorphism de-fined by π ( k ) = k + mR where k ∈ K , k + mR ∈ M/mR . It is obvious thatKer( π ) = mR ∩ K . It follows that K is projective δ -cover of M/mR . So M isprincipally δ -semiperfect. (cid:3) Corollary 4.2.
Let R be a ring. Then the following are equivalent. (1) R is principally δ -semiperfect. (2) R is principally δ -lifting. (3) R is δ -semiregular.Proof. (1) ⇔ (2) Clear by Theorem 4.1.(2) ⇔ (3) By Theorem 3.6 (2), R is principally δ -lifting if and only if for everyprincipal right ideal I of R can be written as I = N ⊕ S , where N is directsummand and S is δ -small in R . This is equivalent to being R δ -semiregular sincefor any ring R , Rad δ ( R ) is δ -small in R and each submodule of a δ -small submoduleis δ -small. (cid:3) The module M is called principally δ -supplemented if every cyclic submodule of M has a δ -supplement in M . Clearly, every δ -supplemented module is principally δ -supplemented. Every principally δ -lifting module is principally δ -supplemented. In a subsequent paper we investigate principally δ -supplemented modules in detail.Now we prove: Theorem 4.3.
Let M be a principally δ -semiperfect module. Then (1) M is principally δ -supplemented. (2) Each factor module of M is principally δ -semiperfect, hence any homomor-phic image and any direct summand of M is principally δ -semiperfect.Proof. (1) Let m ∈ M . Then M/mR has a projective δ -cover P β → M/mR .There exists P α → M such that the following diagram is commutative, β = πα ,where M π → M/mR is the natural epimorphism.
PM M/mR ♣♣♣♣♣♣♣♣♣♣♣ ✠ α ❄ β ✲ π ✲ Then M = α ( P ) + mR , and α ( P ) ∩ mR is δ -small in α ( P ), by Lemma 2.1 (1).Hence M is principally δ -supplemented.(2) Let M f → N be an epimorphism and nR a cyclic submodule of N . Let m ∈ f − ( nR ) and P g → M/ ( mR ) be a projective δ -cover. Define M/ ( mR ) h → N/nR by h ( m ′ + mR ) = f ( m ′ ) + nR , where m ′ + mR ∈ M/ ( mR ). Then Ker( g ) is con-tained in Ker( hg ). By projectivity of P , there is a map α from P to N such that hg = πα . P M/mRN N/nR ✲ g ♣♣♣♣♣♣♣♣♣♣♣ ❄ α ❄ h ✲ π ✲ It is routine to check that ( nR ) ∩ α ( P ) = α (Ker( g )). By Lemma 2.1 (2), α ( Ker ( g ))is δ -small in N since Ker( g ) is δ -small. Let x ∈ Ker( πα ). Then hg ( x ) = ( πα )( x ) = 0or α ( x ) ∈ ( nR ) ∩ α ( P ). So Ker( πα ) is δ -small. Hence P is a projective δ -cover for N/ ( nR ). (cid:3) Theorem 4.4.
Let P be a projective module with Rad δ ( P ) is δ -small in P . Thenthe following are equivalent. (1) P is principally δ -lifting. (2) P/ Rad δ ( P ) is principally semisimple and, for any cyclic submodule xR of P/ Rad δ ( P ) that is a direct summand of P/ Rad δ ( P ) , there exists a cyclicdirect summand A of P such that xR = A . N A CLASS OF LIFTING MODULES 13
Proof. (1) ⇒ (2) Since P is a principally δ -lifting module, P/ Rad δ ( P ) is principallysemisimple by Lemma 3.15. Let xR be any cyclic submodule of P/ Rad δ ( P ). ByTheorem 3.6, there exists a direct summand A of P and a δ -small submodule B such that xR = A ⊕ B . Since B is contained in Rad δ ( R ), xR + Rad δ ( R ) = A +Rad δ ( R ). Hence xR = A .(2) ⇒ (1) Let xR be any cyclic submodule of P . Then we have P/ Rad δ ( P ) =[( xR +Rad δ ( P )) / Rad δ ( P )] ⊕ [ U/ Rad δ ( P )] for some U ≤ P . By (2), there existsa direct summand A of P such that P = A ⊕ B and U = B + Rad δ ( P ). Then P = A ⊕ B = A + U + Rad δ ( P ). Since Rad δ ( P ) is δ -small in P , there exists aprojective and semisimple submodule Y of P such that P = A ⊕ B = ( A + U ) ⊕ Y .Since P is projective, A + B is also projective and so by Lemma 3.13, we have A + B = V ⊕ B for some V ≤ A . Hence P = V ⊕ B ⊕ Y . On the other hand( xR ) ∩ ( B ⊕ Y ) = ( xR ) ∩ B ≤ ( xR ) ∩ U ≤ Rad δ ( R ). Since Rad δ ( R ) is δ -small in P , it is δ -small in B ⊕ Y by Lemma 2.1 (3). Thus P is principally δ -lifting. (cid:3) References [1] F.W. Anderson and K.R. Fuller, Rings and Categories of Modules, Springer-Verlag, NewYork, 1974.[2] J. Clark, C. Lomp, N. Vanaja and R. Wisbauer, Lifting modules, Brikhauser-Basel, 2006.[3] C. Lomp,
Regular and Biregular Module Algebras , Arab. J. Sci. and Eng., 33(2008), 351-363.[4] S. Mohamed and B. J. M¨uller, Continuous and discrete modules, Cambridge University Press,1990.[5] A. C. Ozcan , A. Harmanci and P. F. Smith,
Duo Modules , Glasgow Math. J., 48(3)(2006),533-545.[6] K. Oshiro,
Semiperfect modules and quasi-semiperfect modules , Osaka J. Math., 20(1983),337-372.[7] R. Wisbauer, Foundations of module and ring theory, Gordon and Breach , Reading, 1991.[8] M.F. Yousif and Y. Zhou,
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Department of Mathematics, Ankara University, 06100 Ankara, Turkey
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