δ ss -supplemented modules and rings

Abstract

In this paper, we introduce the concept of δss-supplemented modules and provide the various properties of these modules. In particular, we prove that a ring R is δss-supplemented as a left module if and only if R Soc(R R) is semisimple and idempotents lift to Soc(RR) if and only if every left R-module is δss-supplemented. We define projective δss-covers and prove the rings with the property that every (simple) module has a projective δss-cover are δss-supplemented. We also study on δss-supplement submodules.

Full text

DOI: 10.2478/auom-2020-0041
An. S¸t. Univ. Ovidius Constant¸a Vol. 28(3),2020,193–216
δss-supplemented modules and rings
Burcu Ni¸sancı T¨urkmen and Erg¨ul T¨urkmen
Abstract
In this paper, we introduce the concept of δss-supplemented modules
and provide the various properties of these modules. In particular, we
prove that a ring Ris δss-supplemented as a left module if and only
if R
Soc(RR)is semisimple and idempotents lift to Soc(RR) if and only
if every left R-module is δss-supplemented. We define projective δss -
covers and prove the rings with the property that every (simple) module
has a projective δss -cover are δss-supplemented. We also study on δss-
supplement submodules.
1 Introduction
Throughout this paper, all rings are associative with identity and all modules
are unitary left modules. Let Rbe such a ring and Mbe an R-module. The
notation N⊆Mmeans that Nis a submodule of M.Soc(M) and Rad(M)
will stand for the socle of Mand the radical of M. Let Mbe a module.
A submodule L⊆Mis said to be essential in M, denoted as LEM, if
L∩N6= 0 for every non-zero submodule N⊆M. A module Mis called
singular if M∼
=N
Lfor some module Nand an essential submodule LEN.
As a dual to the notion of an essential submodule, a submodule Nof Mis
said to be small in M, denoted by NM, if M6=N+Kfor every proper
submodule Kof M([13, 19.1]). A non-zero module Mis called hollow if every
proper submodule of Mis small in M, and it is called local if it is hollow and
finitely generated.
Key Words: semisimple module, strongly δ-local module, δss -supplemented module, left
δss-perfect ring, pro jective δss-cover
2010 Mathematics Subject Classification: Primary 16D10; 16D60 Secondary 16D99.
Received: 08.10.2019.
Accepted: 09.01.2020.
193

δss-supplemented modules and rings 194
Let Mbe a module and U, V be submodules of M. The submodule Vis
said to be supplement of Uin Mor Uis said to have a supplement Vin Mif
Vis minimal with respect to M=U+V. It is well known that a submodule
Vof Mis a supplement of Uin Mif and only if M=U+Vand U∩VV.
Mis called supplemented if every submodule Uof Mhas a supplement in M.
A submodule Uof Mhas ample supplements in Mif every submodule Lof
Msuch that M=U+Lcontains a supplement of Uin M. The module Mis
called amply supplemented if every submodule of Mhas ample supplements in
M. Semisimple modules and hollow modules are (amply) supplemented ([13,
41]).
Zhou [15] generalizes small submodules to δ-small submodules of a module
Mas follows. A submodule N⊆Mis said to be δ-small in Mand indicated
by NδMif M6=N+Kfor every proper submodule Kof Mwith M
K
singular. It is clear that every small submodule or projective semisimple sub-
module of Mis δ-small in M. By δ(M) we will denote the sum of all δ-small
submodules of Mas in [15, Lemma 1.5 (2)]. Since Rad(M) is the sum of all
small submodules of M, it follows that Rad(M)⊆δ(M) for a module M. For
an arbitrary ring R, let δ(R) = δ(RR).
Let Mbe a module. In [7], Mis said to be δ-supplemented if every
submodule Uof Mhas a δ-supplement Vin M, that is, M=U+Vand
U∩VδV. The module Mis called amply δ-supplemented if, whenever M=
U+V,Uhas a δ-supplement V0⊆V. Clearly, every (amply) supplemented
module is (amply) δ-supplemented. For characterizations of supplemented and
δ-supplemented modules we refer to [1], [7] and [13].
In [6], the authors define ss-supplemented modules as a proper general-
ization of semisimple modules. A module Mis said to be ss-supplemented if
every submodule Uof Mhas a supplement Vin Msuch that U∩Vis semisim-
ple. They give in the same paper the structure of ss-supplemented modules.
In particular, it is shown in [6, Theorem 41] that a ring Ris semiperfect and
Rad(R)⊆Soc(RR) if and only if every left R-module is ss-supplemented if
and only if RRis the finite sum of strongly local submodules. Here a module
Mis called strongly local if it is local and the radical is semisimple ([6]).
Motivated by these results, we introduce the concept of δss-supplemented
modules. In this paper, we study on δss -supplemented modules and we obtain
the various properties of these modules. We show that strongly δ-local (see
below) modules are δss-supplemented. Every direct sum of strongly δ-local
modules and projective semisimple modules is coatomic. The class of δss-
supplemented modules is closed under finite sums and factor modules. We
prove that a module Mwith δ-small δ(M) is δss-supplemented if and only
if it δ-supplemented and δ(M)⊆Soc(M). We study on the rings with the
property that every left module is δss-supplemented and call these rings left

δss-supplemented modules and rings 195
δss-perfect. We also show that a ring Ris left δss-perfect if and only if RRis
δss-supplemented if and only if R
Soc(RR)is semisimple and idempotents lift to
Soc(RR) if and only if for any module every maximal submodule has a δss -
supplement in the module. We define projective δss-covers and prove that a
ring is left δss-perfect if and only if every left module has a projective δss-cover
if and only if every semisimple left module has a projective δss -cover if and
only if every simple left R-module has a projective δss -cover. We also study
on δss-supplement submodules.
The following lemma follows from [15, Lemma 1.2] and we will use it
throughout the paper.
Lemma 1.1. Let Mbe a module. A submodule N⊆Mis δ-small in Mif and
only if whenever X+N=Mthere exists a projective semisimple submodule
N0of Nsuch that X⊕N0=M.
It is obvious that a module Mis projective semisimple if and only if Mδ
M. A ring Ris called local if RR(or RR) is a local module.
Remark 1.2.Let Rbe a commutative domain (which is not field) or a local
ring and Mbe a non-zero R-module. Suppose that a submodule Nof Mis
δ-small in M. Let M=N+Kfor some submodule Kof M. Then there
exists a projective semisimple submodule N0of Nsuch that M=N0⊕K. By
[12, Proposition 2.5], we get that N0= 0 and so K=M. It means that Nis
a small submodule of M.
2 Strongly δ-Local Modules
It is well known that Mis local if and only if Rad(M)Mand Rad(M) is
maximal. Using this characterization, δ-local modules are defined in [4]. A
module Mis called δ-local if δ(M)δMand δ(M) is maximal. Maybe, it
is expected that local modules are also δ-local. But unfortunately, it is not
the case. Let Sbe a simple module. Since Sis projective or singular, it is
δ(S) = Sor δ-local. It follows that a projective simple module is local but not
δ-local.
As we have mentioned in the introduction, a module Mis strongly local if
it is local and Rad(M) is semisimple ([6]). Note that every simple module is
strongly local.
We say that a module Mstrongly δ-local if it is δ-local and δ(M)⊆
Soc(M). It is clear that every strongly δ-local module is δ-local but the con-
verse is not true in general. For example, let Mbe the left Z-module Z8. Then
Mis δ-local but not strongly δ-local. Then we have the following implications
on modules:

δss-supplemented modules and rings 196
singular simple
vv))
strongly local

strongly δ −local

local δ −local
We start the next lemma which are taken from [15, Lemma 1.3 and Lemma
1.5]. Recall that a module Mcoatomic if every proper submodule of Mis
contained in a maximal submodule of M. Note that a coatomic module has
small radical.
Lemma 2.1. Let Mbe a module.
(1) For any submodules Nand Lof M,N+LδMif and only if NδM
and LδM.
(2) If KδMand f:M−→ Nis a homomorphism, then f(K)δN.
In particular, if M⊆N, then KδN.
(3) If f:M−→ Nis a homomorphism, then f(δ(M)) ⊆δ(N).
(4) If M=Li∈IMi, then δ(M) = Li∈Iδ(Mi).
(5) If Mis coatomic, then δ(M)is the unique largest δ-small submodule of
M.
It is well known that every (strongly) local module is indecomposable. On
the other hand, the following theorem gives a characterization of a semisimple
module which is strongly δ-local. Firstly we need the following facts.
Lemma 2.2. Let Mbe a module and let Nbe a semisimple submodule of M
such that N⊆δ(M). Then NδM.
Proof. Let Kbe a submodule such that M=N+K. Since Nis semisimple,
then there exists a semisimple submodule Xof Nsuch that N= (N∩K)⊕X.
Therefore M= [(N∩K)⊕X] + K=X⊕K.
Next we prove that Xis projective. Let X=⊕i∈ISi, where Iis some
index set and each Siis simple. Since X⊆N⊆δ(M), by the modular law,
we have δ(M) = δ(M)∩M=δ(M)∩(X⊕K) = X⊕(K∩δ(M)) = X⊕δ(K).
Note that, by Lemma 2.1 (4), δ(M) = δ(X)⊕δ(K) . Therefore X=δ(X).
Let πi:X−→ Sibe the canonical projection. It follows from Lemma 2.1 (3)

δss-supplemented modules and rings 197
that Si=πi(X) = πi(δ(X)) ⊆δ(Si) and so δ(Si) = Si, for all i∈I. This
implies that each Siis projective for all i∈I. Then X=⊕i∈ISiis projective
as the direct sum of projective submodules. Hence NδM.
Observe from Lemma 2.2 that a module Mis strongly δ-local if and only
if δ(M) is maximal and semisimple. It follows that a semisimple module is
strongly δ-local if and only if δ(M) is maximal. The following result is a direct
consequence of Lemma 2.2.
Corollary 2.3. Let Mbe a module. Then Mis semisimple and δ(M) = M
if and only if it is projective semisimple.
Theorem 2.4. Let Mbe a semisimple module. Then Mis strongly δ-local if
and only if Mhas the decomposition M=M1⊕M2, where M1is a projective
semisimple submodule and M2is a singular simple submodule.
Proof. (=⇒) Let Mbe a strongly δ-local module. Since δ(M) is maximal
and Mis semisimple, there exists a simple submodule M2of Msuch that
M=δ(M)⊕M2. Put M1=δ(M). Since M1=δ(M)δM, it follows
from Lemma 2.1 (2) that M1δM1and so M1is semisimple projective by
Corollary 2.3. Therefore δ(M2)⊆δ(M)∩M2= 0 and so δ(M2) = 0. It means
that M2is singular. Hence we get the decomposition M=M1⊕M2as desired.
(⇐=) Clearly, δ(M1) = M1δM1and δ(M2)=0δM. It follows from
Lemma 2.1 (2)-(4) that δ(M) = δ(M1)⊕δ(M2) = M1⊕0 is δ-small in M.
Since δ(M) is maximal, we deduce that Mis strongly δ-local.
Observe from Theorem 2.4 that any factor module (in particular, direct
summand) of a strongly δ-local module need not be strongly δ-local in general.
Proposition 2.5. Let Mbe an indecomposable module. If Mis strongly
δ-local, then it is strongly local.
Proof. If Mis simple, then it is singular simple because Mis strongly δ-
local. Suppose that Mis not singular simple. Since Mis indecomposable, we
get that Soc(M)⊆Rad(M). This implies that Soc(M)M. Since Mis
strongly δ-local, we have δ(M)⊆Soc(M) and so δ(M) = Soc(M) is maximal.
Therefore Soc(M) = Rad(M). Thus Mis strongly local.
Proposition 2.6. Let Rbe a local ring. If Mis a strongly δ-local R-module,
then it is a strongly local R-module.
Proof. By Remark 1.2.
Proposition 2.7. Let Mbe a module. Assume that M
δ(M)is semisimple. Then
Mis coatomic if and only if δ(M)is δ-small in M.

δss-supplemented modules and rings 198
Proof. (=⇒) By Lemma 2.1 (5).
(⇐=) If δ(M) = M, then clearly MδMand so Mis projective semisim-
ple. Let δ(M)6=Mand let Ube any submodule of M. If U+δ(M) = M,
then there exists a (projective) semisimple submodule Sof δ(M) such that
U⊕S=M. Let S=⊕i∈ISi, where (i∈I)Siis simple and Iis some index
set. For some i0∈I, put U0=U⊕(⊕i∈I\{i0}Si). Then clearly U⊆U0.
Therefore M
U0∼
=Si0and hence U0is a maximal submodule of M. Suppose
that U+δ(M)6=M. Then U+δ(M)
δ(M)is a proper submodule of M
δ(M). Since
M
δ(M)is semisimple, there exists a maximal submodule K
δ(M)of M
δ(M)such that
U+δ(M)
δ(M)⊆K
δ(M). So Kis a maximal submodule of Mwhich contains U. It
means that Mis coatomic.
Recall that a module Mis called radical if Mhas no maximal submodules,
that is, M=Rad(M). Let P(M) be the sum of all radical submodules of M.
It is easy to see that P(M) is the largest radical submodule of M. If P(M) = 0,
Mis called reduced.
Corollary 2.8. Any strongly δ-local module is reduced and coatomic.
Proof. Let Mbe a strongly δ-local module. Therefore δ(M)⊆Soc(M). Since
Rad(M)⊆δ(M), it follows that Mis reduced. Since M
δ(M)is simple, we get
Mis coatomic by Proposition 2.7.
Theorem 2.9. Let M=⊕i∈IMi, where each Miis either strongly δ-local or
projective semisimple. Then Mis coatomic.
Proof. Note that M
δ(M)=π(M)∼
=Li∈I
Mi
δ(Mi). Let i0∈I. If Mi0is projective
semisimple, then δ(Mi0) = Mi0and so the factor module Mi0
δ(Mi0)= 0. It
follows that we can consider the module M
δ(M)is the direct sum of simple
modules Mk
δ(Mk), where (k∈Λ) Mkis strongly δ-local and Λ ⊆I. Thus M
δ(M)
is semisimple.
By Proposition 2.7, it is enough to prove that δ(M) is δ-small in M. By
the hypothesis, we have δ(Mi)⊆Soc(Mi). Applying Lemma 2.1 (4) and [13,
21.2 (5)], we obtain that δ(M) = ⊕i∈Iδ(Mi)⊆ ⊕i∈ISoc(Mi) = Soc(M). That
is, δ(M) is semisimple. It follows from Lemma 2.2 that δ(M) is δ-small in M.
This completes the proof.
3δss-Supplement Submodules
Let Mbe a module. By Socs(M) we denote the sum of all simple submodules
of Mthat are small in Mas in [14]. Since every small submodule of M

δss-supplemented modules and rings 199
is δ-small in M, the notation motives us to introduction the sum all simple
submodules of Mthat are δ-small in M. For a module M, let
Socδ(M) = P{S⊆M|S is simple and S δM}.
The properties of Socδ(M) for a module Mare given in the next proposi-
tion.
Proposition 3.1. Let Rbe a ring and Mbe a left R-module. Then:
(1) Socδ(M) = Soc(M)∩δ(M),
(2) Socδ(M)δM,
(3) Rad(Socδ(M)) = 0,
(4) Socδ(M) = Mif and only if Mis projective semisimple,
(5) If M0is a left R-module and f:M−→ M0is a homomorphism, then
f(Socδ(M)) ⊆Socδ(f(M)).
Proof. (1) Let x∈δ(M)∩S oc(M). Then Rx δMand Rx is semisimple. So
there exist m∈Z+and simple submodules Siof Mfor every i∈ {1,2, . . . , m}
such that Rx =S1⊕S2⊕ · · · ⊕ Smby [10, Proposition 3.3]. Since Rx δM,
it follows from Lemma 2.1 (2) that each SiδM. Thus x∈Rx ⊆Socδ(M).
The converse is clear by the definition of Socδ(M).
(2) Clearly, Socδ(M) is semisimple. Then the proof follows from Lemma
2.2.
(3) Since semisimple modules have zero radical, it is clear.
(4) Let Socδ(M) = M. By (1), we get Mis semisimple and δ(M) = M.
Hence Mis projective semisimple by Corollary 2.3. The converse is clear.
(5) Let f:M−→ M0be a homomorphism of modules and x∈f(Socδ(M)).
Then x=f(m) for some element m∈Socδ(M). Applying (1), we obtain that
m∈Soc(M)∩δ(M). Therefore x=f(m)∈f(Rm)⊆S oc(f(M)) by [13, 21.2
(1)] and x=f(m)∈f(Rm)⊆δ(f(M)) by Lemma 2.1 (3). It means that
x∈Soc(f(M)) ∩δ(f(M)). Again applying (1), we have x∈Socδ(f(M)).
Let Mbe a module and Sbe a simple submodule of M. Then SMor
we have the decomposition M=S⊕Kfor some submodule Kof M. Using
this fact we have:
Corollary 3.2. Let Mbe a module and let Sbe a simple submodule of M.
Then SδMif and only if Sis projective or small in M.
Proof. Let SδM. Suppose that Sis not small in M. Then we get M=
S⊕K. By the assumption, Sis projective as desired. The converse is clear.

δss-supplemented modules and rings 200
Let Mbe a module and U, V be submodules of M. Following [6], Vis
called ss-supplement of Uin Mif M=U+Vand U∩V⊆Socs(V). For any
left module X, we have Socs(X)⊆Socδ(X) and so it is natural to introduce
another notion that we called δss-supplement. A submodule Vof Mis called
δss-supplement of Uin Mif M=U+Vand U∩V⊆Socδ(V). Under given
definitions we obtain the following diagram:
ss −supplement
vv))
supplement
((
δss −supplement
uu
δ−supplement
Modifying of [6, Lemma 3] we characterize δss-supplement submodules of
a module M. Note that we shall freely use the next lemma without reference
in this paper.
Lemma 3.3. Let Mbe a module and U, V be submodules of M. Then the
following statements are equivalent.
(1) Vis a δss-supplement of Uin M,
(2) M=U+V,U∩V⊆δ(V)and U∩Vis semisimple,
(3) M=U+V,U∩VδVand U∩Vis semisimple.
Proof. Using Proposition 3.1, we have clearly (1) ⇒(2) and (3) ⇒(1).
(2) ⇒(3) It follows from Lemma 2.2.
Proposition 3.4. Let Mbe a module and Ube a maximal submodule of M.
If Uhas a δss-supplement Vin M, then Vis strongly δ-local or projective
semisimple
Proof. Let Vbe a δss-supplement of Uin M. Then M=U+V,U∩V⊆δ(V)
and U∩Vis semisimple. Note that M
U∼
=V
U∩Vis simple and thus U∩Vis a
maximal submodule of V. Hence δ(V) = U∩Vor δ(V) = V. If δ(V) = U∩V,
then δ(V)⊆Soc(V). Therefore Vis strongly δ-local. Now suppose that
δ(V) = V. By [11, Lemma 2.22], we get that Vis projective semisimple.
Proposition 3.5. Let Mbe module and let V⊆Mbe a δss-supplement in
M.

δss-supplemented modules and rings 201
(1) If Lis a submodule of V, then V
Lis a δss-supplement in M
L,
(2) Whenever V⊆K⊆M,Vis also a δss -supplement in K,
(3) Socδ(V) = V∩Socδ(M).
Proof. Since Vis a δss -supplement in M, then there exists a submodule Uof
Msuch that M=U+V,U∩VδVand U∩Vis semisimple.
(1) Since M=U+V, we have M
L= (U+L
L) + V
L. Let π:V−→ V
L
be the canonical homomorphism. Then by Lemma 2.1 (2), we obtain that
π(U∩V) = (U∩V)+L
L=(U+L)∩V
L= (U+L
L)∩V
LδV
L. It follows from [5,
8.1.5 (2)] that π(U∩V)=(U+L
L)∩V
Lis semisimple. It means that V
Lis a
δss-supplement of U+L
Lin M
L.
(2) By the modular law, we have K=K∩M=K∩(U+V) = U∩K+V.
Therefore (U∩K)∩V=U∩V⊆Socδ(V).
(3) It follows from Proposition 3.1, [4, Corollary 2.5] and [13, 21.2 (2)] that
we can write V∩Socδ(M) = V∩[Soc(M)∩δ(M)] = [V∩Soc(M)]∩[V∩δ(M)] =
Soc(V)∩δ(V) = Socδ(V).
Lemma 3.6. Let Mbe a module and let Kbe a direct summand of M. Then a
submodule V⊆Kis a δss-supplement in Kif and only if it is a δss-supplement
in M.
Proof. (=⇒) By the hypothesis, we have M=K⊕Lwhere L⊆M. Since
Vis a δss-supplement in K, then there exists a submodule Uof Ksuch that
K=U+V,U∩V <<δVand U∩Vis semisimple. So M= (U+V)⊕L=
(U⊕L) + V. It can be seen that (U⊕L)∩V=U∩V. Hence Vis a
δss-supplement of U⊕Lin M.
(⇐=) By Proposition 3.5 (2).
Theorem 3.7. Let Mbe a module. Then Mis a δss-supplement in every
extension if and only if it is a δss-supplement in E(M), where E(M)is the
injective hull of M.
Proof. One direction is clear. Conversely, let M⊆N. Then we have E(M)⊆
E(N). So by [10, Theorem 2.15], E(N) = E(M)⊕Lfor some submodule L
of E(N). Since Mis a δss-supplement in E(M), it follows from Lemma 3.6
that it is a δss-supplement in E(N). Hence Mis a δss -supplement in Nby
Proposition 3.5 (2).
Proposition 3.8. Let Mbe a module with Socδ(M) = 0. Then Mis a
δss-supplement in E(M)if and only if it is injective.

δss-supplemented modules and rings 202
Proof. Let Mbe a δss-supplement in E(M). Then there exists a submodule
Nof E(M) such that E(M) = N+Mand N∩M⊆Socδ(M). Since
Socδ(M) = 0, we obtain that N∩M= 0. Thus E(M) = N⊕M. It means
that Mis injective. The converse is clear.
Let Rbe a commutative domain and Mbe an R-module. We denote
by T or(M) the set of all elements mof Mfor which there exists a non-zero
element rof Rsuch that rm = 0, i.e. Ann(m)6= 0. Then T or (M), which is a
submodule of M, called the torsion submodule of M. If M=T or(M), then M
is called a torsion module and Mis called torsion-free provided T or(M) = 0.
Let Rbe a commutative domain which is not field and Mbe an R-module.
Suppose that Sis a simple submodule of M. Let mbe a non-zero element
of S. Then Rm =Sand so we can write S∼
=R
Ann(m). Since Ris not field,
Ann(m)6= 0. Therefore, for some non-zero element r∈R, we get rm = 0.
So m∈T or(M). It means that Soc(M)⊆T or (M). Using this fact and
Proposition 3.8, we obtain that the next result. By Remark 1.2, we get that
δss-supplements are ss-supplements in this case.
Corollary 3.9. Let Rbe a commutative domain which is not field and Mbe
a torsion-free R-module. Then Mis a ss-supplement in E(M)if and only if
it is injective.
Proof. Since Mis torsion-free, we get that Socδ(M) = 0. It follows from
Proposition 3.8 that the proof is clear.
Let Rbe a commutative domain which is not field. Ris said to be one
dimensional if, for every non-zero ideal Iof R,R
Iis an artinian ring.
Corollary 3.10. Let Rbe a one dimensional domain and Mbe a torsion-free
R-module. Then the following statements are equivalent:
(1) Mis a ss-supplement in E(M),
(2) Mis injective,
(3) Mis radical, i.e. Mhas no maximal submodules.
Proof. By Corollary 3.9 and [2, Lemma 4.4]
Proposition 3.11. Let Rbe a Dedekind domain and Mbe an R-module.
Then Mis a ss-supplement in E(M)if and only if it is injective.
Proof. Let Mbe a ss-supplement of some submodule Nin E(M). For every
non-zero element r∈R, we can write E(M) = rE(M) = rN +rM =N+rM
and so, by the minimality of M, we obtain that Mis divisible. By [2, Lemma
4.4], Mis injective.

δss-supplemented modules and rings 203
A module Mis said to be π-projective if whenever Uand Vare submodules
of Msuch that M=U+V, there exists an endomorphism fof Msuch that
f(M)⊆Uand (1 −f)(M)⊆V. Hollow (local) modules and self-projective
modules are π-projective.
Lemma 3.12. Let Mbe a π-projective module and U, V be submodules of
M. If Uand Vare mutual δ-supplements in M, then they are mutual δss-
supplements in M.
Proof. It follows from [7, Lemma 2.15].
Recall from [13, 41.16 (1)] that every supplement submodule of a π-projective
supplemented module is a direct summand. Analogous to that we have:
Corollary 3.13. Let Mbe a π-projective and δ-supplemented module. Then
every δ-supplement in Mis δss-supplement in M.
Proof. Let Vbe a δ-supplement of some submodule Uin M. Then M=U+V
and U∩VδV. Since Mis π-projective and δ-supplemented, it follows from
[1, Theorem 4.4] that it is amply δ-supplemented. So Vhas a δss-supplement
U0⊆Uin M. Therefore Vand U0are mutual δ-supplements in M. Thus by
Lemma 3.12, Vis a δss-supplement in M.
Theorem 3.14. The following conditions are equivalent for a module Mwith
non-zero δ(M).
(1) every cyclic submodule of Mis a δss-supplement in M,
(2) every cyclic submodule of Mis a δ-supplement in M,
(3) Mis projective semisimple.
Proof. (3) =⇒(1) and (1) =⇒(2) are clear.
(2) =⇒(3) Let 0 6=m∈δ(M). By (2), there exists a submodule Uof
Msuch that M=U+Rm and U∩Rm δRm. Since Rm δM, we
can write M=X⊕Rm, where Xis a projective semisimple submodule of
U. Since Rm is a δ-supplement in M, it follows from [4, Corollary 2.5] that
δ(Rm) = Rm ∩δ(M) = Rm. By Lemma 2.1 (2), we get δ(Rm) = Rm δ
Rm and so Rm is projective semisimple. Hence M=X⊕Rm is projective
semisimple.
Theorem 3.15. The following conditions are equivalent for a module Mwith
zero δ(M).
(1) every (resp., cyclic) submodule of Mis a δss-supplement in M,

δss-supplemented modules and rings 204
(2) every (resp., cyclic) submodule of Mis a δ-supplement in M,
(3) Mis (resp., regular) semisimple.
Proof. (3) =⇒(1) and (1) =⇒(2) are clear.
(2) =⇒(3) Since δ(M) = 0, every (cyclic) submodule of Mis a direct
summand of Mand so Mis (regular) semisimple.
It is well known that a ring Ris semisimple if and only if, for every left
R-module, every submodule is direct summad (see [13, 20.7]). Using Theorem
3.14 and Theorem 3.15, we generalize this fact.
Corollary 3.16. Let Rbe a ring. Then Ris semisimple if and only if, for
every left R-module M, every submodule of Mis δss -supplement in M.
4δss-Supplemented Modules
In this section, we define the concept of δss-supplemented modules and obtain
the basic properties of such modules.
Let Mbe a module. We say that Maδss-supplemented module if ev-
ery submodule Uof Mhas a δss-supplement Vin M, and Mamply δss-
supplemented if in case M=U+Vimplies that Uhas a δss-supplement
V0⊆V. It is clear that every (amply) ss-supplemented module is (am-
ply) δss-supplemented, and (amply) δss -supplemented modules are (amply)
δ-supplemented.
Now we begin by giving some examples of module to seperate (amply) ss-
supplemented, (amply) δss-supplemented and (amply) δ-supplemented. Firstly
we need the following facts:
Lemma 4.1. Every strongly δ-local module is δss-supplemented.
Proof. Let Mbe a strongly δ-local module and Ube any submodule of M. If
U⊆δ(M), then Uis semisimple since δ(M) is semisimple. By Lemma 2.2,
we get UδM. Thus Mis the δss-supplement of Uin M. Let U*δ(M).
Since δ(M) is maximal, we can write the equality M=U+δ(M). Then there
exists a projective semisimple submodule Vof δ(M) such that M=U⊕V
because δ(M)δM. Hence Mis δss -supplemented.
π-projective supplemented modules are amply supplemented. Similarly, we
show that π-projective δss-supplemented modules are amply δss -supplemented.
The proof is virtually the same that of [13, 41.15], but we give it for complete-
ness.
Proposition 4.2. Let Mbe a π-projective and δss-supplemented module.
Then Mis amply δss-supplemented.

δss-supplemented modules and rings 205
Proof. Let Uand Vbe submodules of Msuch that M=U+V. Since Mis
π-projective, there exists an endomorphism fof Msuch that f(M)⊆Uand
(1−f)(M)⊆V. Note that (1−f)(U)⊆U. Let V0be a δss-supplement of Uin
M. Then M=f(M)+(1−f)(M) = f(M)+(1−f)(U+V0)⊆U+(1−f)(V0)⊆
M, so that M=U+ (1 −f)(V0). Here (1 −f)(V0) is a submodule of V. Let
y∈U∩(1 −f)(V0). Then, y∈Uand y= (1 −f)(x) = x−f(x) for some
x∈V0. We have x=y+f(x)∈Uso that y∈(1 −f)(U∩V0). Since
U∩V0δV0, we get U∩(1 −f)(V0) = (1 −f)(U∩V0)δ(1 −f)(V0) by
[15, Lemma 1.3 (2)]. Also U∩(1 −f)(V0) = (1 −f)(U∩V0) is semisimple.
Thus, (1 −f)(V0) is a δss -supplement of Uin M. Therefore Mis amply
δss-supplemented.
Combining Proposition 4.2 and Lemma 4.1, we obtain the next result:
Corollary 4.3. A projective strongly δ-local module is amply δss-supplemented.
Example 4.4. (1) Consider the non-noetherian commutative ring Swhich is
the direct product Q∞
i≥1Fi, where Fi=Z2. Suppose that Ris the subring of
Sgenerated by L∞
i=1 Fiand 1S. Let M=RR. Then Mis a regular module
which is not semisimple. Therefore Soc(M) is maximal. By [15, Example
4.1], we have Soc(M) = δ(M)δM. This means that Mis strongly δ-
local. Since Mis projective, it follows from Lemma 4.1 and Corollary 4.3
that Mis amply δss-supplemented. On the other hand, it is not (amply)
ss-supplemented because Rad(M) = 0.
(2) Let Mbe the local Z-module Zpk, for pis any prime integer and k≥3.
It is clearly that Mis amply δ-supplemented. Since Socδ(Zpk) = Soc(Zpk)∼
=
Zpand δ(M) = Rad(M) = pZpk,Mis not (amply) δss-supplemented.
It is well known that artinian modules are (amply) δ-supplemented. Ex-
ample 4.4 (2) also shows that in general artinian modules need not to be δss-
supplemented. Now, we have the following implications the classes of modules:
artinian =⇒supplemented =⇒δ-supplemented
and
ss −supplemented
uu**
supplemented
))
δss −supplemented
tt
δ−supplemented

δss-supplemented modules and rings 206
Now we study on the various properties of δss-supplemented modules.
Proposition 4.5. Let Mbe a δ-local module. Then Mis δss-supplemented if
and only if it is strongly δ-local.
Proof. (=⇒) Since Mis δ-local, it suffices to show that δ(M)⊆Soc(M). Let
m∈δ(M). Then Rm δM. Since Mis δss-supplemented, Rm has a δss-
supplement Vin M. Therefore M=Rm +Vand Rm ∩Vis semisimple. So
we can write M=S⊕V, where Sis a projective semisimple submodule of
Rm. Applying the modular law, we have Rm =Rm ∩M=Rm ∩(S⊕V) =
S⊕(Rm∩V). So Rm is semisimple as the sum of two semisimple submodules.
Hence Rm ⊆Soc(M). It means that δ(M)⊆Soc(M).
(⇐=) By Lemma 4.1.
Proposition 4.6. Let Mbe a δ-supplemented module with δ(M)⊆Soc(M).
Then Mis δss-supplemented.
Proof. Let U⊆M. Since Mis δ-supplemented, there exists a submodule V
of Msuch that M=U+Vand U∩VδV. Then U∩V⊆δ(V)⊆δ(M).
By the hypothesis, U∩V⊆Soc(M). Therefore Vis a δss -supplement of Uin
M. It means that Mis δss-supplemented.
Proposition 4.7. Let Mbe a δss-supplemented module. Then M
Socδ(M)is
semisimple.
Proof. Let S ocδ(M)⊆U⊆M. Since Mis δss -supplemented, there exists
a submodule Vof Msuch that M=U+Vand U∩V⊆Socδ(V). Then
U∩V⊆Socδ(M) and so the sum M
Socδ(M)=U
Socδ(M)+V+Socδ(M)
Socδ(M)is direct
sum. Hence M
Socδ(M)is semisimple.
In order to prove that every finite sum of δss-supplemented modules is
δss-supplemented, we use the following standard lemma (see, [13, 41.2]).
Lemma 4.8. Let Mbe a module and Ube a submodule of M. Suppose that
a submodule M1of Mis δss-supplemented. If M1+Uhas a δss-supplement
in M,Uhas also a δss-supplement in M.
Proof. Suppose that Xis a δss-supplement of M1+Uin Mand Yis a δss-
supplement of (X+U)∩M1in M1. So M=M1+U+X,M1= (X+U)∩M1+Y,
(M1+U)∩YδY, (X+U)∩YδY, (M1+U)∩Yand (X+U)∩Yis
semisimple. Then M= (X+U)∩M1+Y+U+X=U+X+Yand by [11,
Lemma 2.1 (2)] U∩(X+Y)⊆X∩(U+Y)+Y∩(U+X)⊆X∩(U+M1) + Y∩
(U+X)δX+Y. Moreover, X∩(Y+U) is semisimple as a submodule of
semisimple module X∩(Y+U). Note that Y∩[(X+U)∩M1] = Y∩(X+U)

δss-supplemented modules and rings 207
is semisimple. It follows from [5, 8.1.5] that (X+Y)∩Uis semisimple. Hence
X+Yis a δss-supplement of Uin M.
Proposition 4.9. The class of δss-supplemented modules is closed under finite
sums.
Proof. Let Mi,i= 1,2, . . . , n be any finite collection of δss-supplemented
modules and let M=M1+M2+· · ·+Mn. To prove that Mis δss -supplemented
by induction on n, it is sufficient to prove this in the case, where n= 2. Hence,
suppose n= 2. Let M1,M2be any submodules of a module Msuch that
M=M1+M2. If M1and M2are δss-supplemented, Mis δss -supplemented.
Let Ube any submodule of M. The trivial submodule 0 is δss -supplement
of M=M1+M2+Uin M. Since M1is δss -supplemented, M2+Uhas a
δss-supplement in Mby Lemma 4.8. Again applying Lemma 4.8, we have that
Uhas a δss-supplement in M. This shows that Mis δss-supplemented.
A submodule Uof a module Mis said to be cofinite if M
Uis finitely gen-
erated (see [2]). Note that maximal submodules of Mare cofinite.
Proposition 4.10. Let Mbe a module. Then the following conditions are
equivalent.
(1) Mis the sum of strongly δ-local or projective semisimple submodules,
(2) Mis coatomic and every cofinite submodule of Mhas a δss-supplement
in M,
(3) Mis coatomic and every maximal submodule of Mhas a δss-supplement
in M.
Proof. (1) =⇒(2) Let M=Pi∈IMi, where Iis some index set and each
Miis strongly δ-local submodules or projective semisimple submodules. Put
N=Li∈IMi. It follows from Theorem 2.9 that Nis coatomic. Consider the
epimorphism ψ:N−→ Mvia ψ((mi)i∈I) = Pi∈Imifor all (mi)i∈I∈N. By
[16, Lemma 1.5 (a)], we get Mis coatomic.
Let Ube any cofinite submodule of M. Then M
Uis finitely generated and
so there exists a finite subset Λ ⊆Isuch that M=U+Pi∈ΛMi. By Lemma
4.1 and Proposition 4.9, we obtain that Pi∈ΛMiis δss-supplemented as the
finite sum of δss-supplemented submodules. Hence Uhas a δss -supplement in
Maccording to Lemma 4.8.
(2) =⇒(3) Clear.
(3) =⇒(1) Let Xbe the sum of all strongly δ-local submodules or semisim-
ple projective submodules. Suppose that X6=M. Since Mis coatomic, there
exists a submodule Uof Msuch that X⊆U⊂M. By the assumption, U

δss-supplemented modules and rings 208
has a δss-supplement, say V, in M. It follows from Proposition 3.4 that V
is projective simple or Vis strongly δ-local. Then V⊆X⊆U. This is a
contradiction.
It is clear that every submodule of a finitely generated module is cofinite.
Using this fact and Proposition 4.10, we obtain the following result:
Corollary 4.11. Let Mbe a finitely generated module. Then the following
conditions are equivalent:
(1) M=Pn
i=1 Mi, where each Miis strongly δ-local or projective semisim-
ple,
(2) Mis δss-supplemented,
(3) every maximal submodule of Mhas a δss-supplement in M.
Theorem 4.12. Let Mbe a module. Then Mis δss -supplemented if and only
if every submodule Uof Mcontaining S oc(M)has a δss -supplement in M.
Proof. One direction is clear. Conversely, let U⊆M. By the assumption,
Soc(M) + Uhas a δss -supplement Vin M. Since Soc(M) is δss -supplemented,
it follows from Lemma 4.8 that Uhas a δss-supplement in M. Hence Mis
δss-supplemented.
It is trivial to show that:
Corollary 4.13. Let Rbe a ring and Mbe an R-module.
(1) Soc(M)has a δss -supplement in Mif and only if Soc(M)has a δ-
supplement in M.
(2) If Ris a commutative domain, then Soc(M)has a δss-supplement in
Mif and only if Soc(M)has a supplement in M.
Proposition 4.14. If Mis a (amply) δss-supplemented module, then every
factor module of Mis (amply) δss-supplemented.
Proof. Let Mbe a δss-supplemented module and M
Lbe a factor module of
M. By the assumption, for any submodule Uof Mwhich contains L, there
exists a submodule Vof Msuch that M=U+V,U∩VδVand U∩V
is semisimple. Let π:M−→ M
Lbe the canonical projection. Then we have
that M
L=U
L+V+L
Land U
L∩V+L
L=(U∩V)+L
L=π(U∩V)δπ(V) = V+L
L
by Lemma 2.1 (2). Since U∩Vis semisimple, it follows from [5, 8.1.5 (2)]
that π(U∩V) = (U∩V)+L
Lis semisimple. That is, V+L
Lis a δss-supplement of
U
Lin M
L, as required.
It can be proved similarly that if Mis amply δss-supplemented, then M
Lis
amply δss-supplemented for every submodule Lof M.

δss-supplemented modules and rings 209
Lemma 4.15. Let Mbe a δss-supplemented module and NδM. Then
N⊆Socδ(M).
Proof. Let Kbe a δss-supplement of Nin M. Then M=N+K,N∩KδK
and N∩Kis semisimple. Since NδM, there exists a semisimple projective
submodule N0of Nsuch that M=N0⊕K. By the modular law, we obtain
that N=N0⊕(N∩K). Hence Nis semisimple.
Corollary 4.16. Let Mbe a coatomic module and Mbe a δss-supplemented
module. Then Rad(M)⊆δ(M)⊆Soc(M).
The following result is a generalization of Corollary 2.3.
Proposition 4.17. Let Mbe a δss-supplemented module and δ(M) = M.
Then Mis projective semisimple.
Proof. Let mbe any element of M. It follows from δ(M) = Mthat Rm δM.
By the assumption and Lemma 4.15, we have Rm ⊆Socδ(M)⊆S oc(M) and
so m∈Soc(M). Therefore Mis semisimple. Hence it is projective semisimple
by Corollary 2.3.
Note that a hollow module is either radical or local. Observe from Propo-
sition 4.17 that a hollow-radical module is not δss-supplemented.
Proposition 4.18. Let Mbe a hollow module. If Mis δss -supplemented,
then it is strongly local.
Proof. Let Mbe a δss-supplemented module. If δ(M) = M, it follows from
Proposition 4.17 that Mis projective semisimple and so Mis projective simple
because Mis hollow. Assume that δ(M)6=M. Since Rad(M)⊆δ(M) and
Mis hollow, Mis local. Therefore we have Rad(M) = δ(M) is maximal and
small in M. It follows from Lemma 4.15 that δ(M)⊆Socδ(M)⊆Soc(M). It
means that Mis strongly local.
In the following next theorem we give the structure of a δss-supplemented
module Mwith δ-small δ(M) in terms of δ-supplemented modules.
Theorem 4.19. Let M be a module and δ(M)δM. Then the following
statements are equivalent:
(1) Mis δss-supplemented,
(2) Mis δ-supplemented and δ(M)has a δss-supplement in M,
(3) Mis δ-supplemented and δ(M)⊆Soc(M).
Proof. Clearly we have (1) =⇒(2), and (2) =⇒(3) follows from Lemma 4.15.
(3) =⇒(1) By Proposition 4.6.

δss-supplemented modules and rings 210
5 Rings whose modules are δss -supplemented
It follows from [15] that a projective module Pis called a projective δ-cover of
a module Mif there exists an epimorphism f:P−→ Mwith Ker(f)δP.
A ring Ris called δ-semiperfect if every simple R-module has a projective
δ-cover, and it is called δ-perfect if every left R-module has a projective δ-
cover. It is proven in [7, Theorem 3.3 and Theorem 3.4] that a ring Ris
δ-perfect (respectively, δ-semiperfect) if and only if every left (respectively,
finitely generated) R-module is δ-supplemented. Now we characterize the
rings the property that every left R-module is (amply) δss-supplemented.
Lemma 5.1. Let Mbe a module. If every submodule of Mis δss-supplemented,
then Mis amply δss-supplemented.
Proof. Let Uand Vbe submodules of Msuch that M=U+V. Since Vis δss-
supplemented, there exists a submodule V0of Vsuch that V= (U∩V) + V0,
U∩V0δV0and U∩V0is semisimple. Note that M=U+V=U+ (U∩
V) + V0=U+V0. It means that Uhas ample δss-supplements in M. Hence
Mis amply δss-supplemented.
A module Mis called locally projective in case whenever g:N−→ Kis
an epimorphism and f:M−→ Kis a homomorphism then for every finitely
generated submodule M0of Mthere exists a homomorphism h:M−→ N
such that gh|M0=f|M0. Every projective module is locally projective. Also,
a finitely generated locally projective module is projective.
Proposition 5.2. Let Mbe a locally projective module and N⊆Soc(M).
Then NδM.
Proof. Let M=N+Kfor some submodule Kof M. Since Nis semisimple,
we can write N= (N∩K)⊕Xwhere Xis a semisimple submodule of N.
Therefore the sum M=X+Kis direct sum. Since being locally projective
is inherited by direct summands, it follows that every direct summand of
Xis locally projective and so every simple submodule of Xis projective.
Therefore Xis projective as the direct sum projective simple submodules.
Hence NδM.
Theorem 5.3. Let Rbe a ring. Then the following statements are equivalent.
(1) RRis δss-supplemented,
(2) Ris a δ-semiperfect ring and δ(R) = S oc(RR),
(3) R
Soc(RR)is semisimple and idempotents lift to Soc(RR),

δss-supplemented modules and rings 211
(4) every projective left R-module is δss-supplemented,
(5) every left R-module is (amply) δss-supplemented,
(6) for every left R-module Mevery maximal submodule has δss-supplement
in M,
(7) every left maximal ideal of Rhas a δss-supplement in R.
Proof. (1) =⇒(2) By the hypothesis, RRis δ-supplemented and so it follows
from [7, Theorem 3.3] that Ris a δ-semiperfect ring. Since RRis coatomic,
it follows from Lemma 2.1 (5) that δ(R) is δ-small in RR. Applying Theorem
4.19, we get that δ(R)⊆Soc(RR). On the other hand, by Proposition 5.2,
Soc(RR)⊆δ(R) and so we obtain that the equality δ(R) = Soc(RR).
(2) =⇒(3) By [15, Theorem 3.6].
(3) =⇒(4) Let Pbe a projective left R-module. Since R
Soc(RR)is artinian
semisimple, it follows from [15, Corollary 1.7] that δ(R) = Soc(RR) and so
δ(P) = δ(R)P=Soc(RR)P⊆Soc(P) by [15, Theorem 1.8]. According to
Proposition 4.6, it suffices to prove that Pis δ-supplemented. Since semisimple
rings are perfect, it follows from assumption and [15, Theorem 3.8] that Ris
aδ-perfect ring. By [7, Theorem 3.4], we obtain that Pis δ-supplemented.
(4) =⇒(5) Let Mbe a left R-module. Since every left R-module is a
homomorphic image of a free left R-module, it follows from Proposition 4.14
that every submodule of Mis δss-supplemented. By Lemma 5.1, it is amply
δss-supplemented.
(5) =⇒(6) and (6) =⇒(7) Clear.
(7) =⇒(1) By Corollary 4.11.
Hence we have the following strict containments of classes of rings:
{rings in [6, Theorem 41]}⊂{rings in Theorem 5.3}⊂{δ-perfect rings}
Examples for showing these implications are not invertible can be found
[15, Example 4.1 and Example 4.3]. So we say that a ring Ris left δss-perfect
if the equal conditions satisfy in the above theorem. Right δss-perfect rings
are defined similarly. Ris said to be δss -perfect if it is both a right and a left
δss-perfect.
Proposition 5.4. Let Rbe a left δss -perfect ring. Then Rad(R)is semisimple.
In particular, (Rad(R))2= 0.
Proof. Since Ris a left δss -perfect ring, it follows from Theorem 5.3 that
Rad(R)⊆δ(R) = Soc(RR). It means that Rad(R) is semisimple. By [13,
21.12 (4)], we obtain that (Rad(R))2= 0.

δss-supplemented modules and rings 212
A ring Ris called a left max ring if every left R-module has a maximal
submodule. It is well known that a ring Ris left max if and only if every
non-zero left R-module is coatomic.
Proposition 5.5. Let Rbe a left δss-perfect ring. Then it is a left max ring.
Proof. Let Mbe a radical module, that is, Rad(M) = M. Then δ(M) =
M. Since Ris a left δss-perfect ring, by Theorem 5.3, R
Soc(RR)=R
δ(R)is a
semisimple ring. By [15, Theorem 1.8], we obtain that δ(M) = δ(R)M=
Soc(RR)M⊆Soc(M). Then M=Soc(M). Since semisimple modules are
zero radical, we get M=Rad(M) = 0. This means that Ris a left max
ring.
Now we characterize the left δss-perfect rings via a different kind of pro-
jective δ-covers. Let Mbe a module and f:P−→ Mbe an epimorphism.
We call the module Paδss-cover of Mif ker(f) is semisimple and δ-small in
P, and call a δss-cover Paprojective δss -cover of Min case Pis projective.
Theorem 5.6. Let Mbe a projective module. Then the following conditions
are equivalent.
(1) Mis δss-supplemented,
(2) every submodule of Mhas a δss-supplement that is a direct summand of
M,
(3) for any submodule Nof M,Mhas the decomposition M=N0⊕Ksuch
that N0⊆Nand N∩K⊆Socδ(M),
(4) every factor module of Mhas a projective δss-cover.
Proof. (1) =⇒(4) Let Ube a submodule of M. It follows that Uhas a δss-
supplement, say V, in M. Since M=U+V, the homomorphism g:V−→ M
U
via g(v) = v+Uis an epimorphism. Let π:M−→ M
Ube the canonical
projection. Since Mis projective, there exists a homomorphism f:M−→ V
such that gf =π. Then it can be seen that M=U+f(M). Applying the
modular law, we get V=U∩V+f(M). Therefore we can write V=S⊕f(M)
for some projective semisimple submodule Sof Vbecause U∩VδV. Since
U∩f(M)⊆U∩VδV, then U∩f(M)δVby Lemma 2.1 (2). It
follows from [15, Lemma 1.3 (3)] that U∩f(M)δf(M) since f(M) is a
direct summand of V. This means that f(M) is a δss-supplement of Uin M.
Since Mis projective and δss -supplemented, by Proposition 4.2, it is amply
δss-supplemented and so f(M) has a δss-supplement U0⊆Uin M. Therefore
f(M) and U0are mutual δss-supplements in M. Using [7, Lemma 2.15], we
obtain that f(M) is projective.

δss-supplemented modules and rings 213
Now we consider the epimorphism ϕ:f(M)−→ M
Uvia ϕ(x) = x+Ufor
all x∈f(M). Since M=U+f(M), we obtain that ker(ϕ) = U∩f(M) is
semisimple and δ-small in f(M). Hence f(M) is a projective δss-cover of M
U
as desired.
(4) =⇒(3) It follows from [15, Lemma 2.4].
(3) =⇒(2) and (2) =⇒(1) Clear.
The next result is crucial.
Corollary 5.7. The following conditions are equivalent for a ring R.
(1) Ris a left δss-perfect ring,
(2) every left R-module has a projective δss-cover,
(3) every semisimple left R-module has a projective δss-cover,
(4) every simple left R-module has a projective δss-cover.
Proof. (1) =⇒(2) Let Mbe a left R-module. Then there exist a projective
module Pand an epimorphism Ψ : P−→ M. By the assumption and Theorem
5.3, we get that Pis δss-supplemented. It follows from Theorem 5.6 that M
has a projective δss-cover as a factor module of P.
(2) =⇒(3) and (3) =⇒(4) are clear.
(4) =⇒(1) It follows from [15, Lemma 2.4] and Theorem 5.3.
Proposition 5.8. A commutative δss-perfect domain is field.
Proof. Let Rbe a commutative δss-perfect domain and a∈R. It follows
that Ris a local ring. If a∈R\Rad(R), we have that Ra =Rand so ais
an invertible element of R. Suppose that a∈Rad(R). By Proposition 5.4,
a2∈(Rad(R))2= 0. Therefore a= 0 since Ris a domain. Thus Ris field.
Let Rbe a ring. Next we will give a necessary and sufficient condition for
the δss-perfect ring Rto be ss-supplemented as a left R-module. Recall from
Lomp [8] that a module Mis said to be semilocal if M
Rad(M)is semisimple, and
a ring Ris said to be semilocal if it is semilocal as a left (right) module over
itself. It is shown in [8, Teorem 3.5] that a ring Ris semilocal if and only if
every left R-module is semilocal.
It is shown in [4, Proposition 4.2] that a projective semilocal, δ-supplemented
module Mwith small radical is supplemented. From this fact we see that the
condition ”small radical” is necessary for Mto be a supplemented. How-
ever, we show by the following proposition that a projective semilocal, δss -
supplemented module is ss-supplemented without necessity of this condition.

δss-supplemented modules and rings 214
Proposition 5.9. Let Mbe a projective module. If Mis semilocal and δss-
supplemented, then it is ss-supplemented.
Proof. Let Mbe a semilocal and δss -supplemented module. Then Soc(M) =
X⊕Socs(M), where X⊆S oc(M). Since Mis semilocal, we can write M=
X+Yand X∩Y⊆Rad(M) for some submodule Yof M. Now X∩Y⊆
X∩Rad(M) = [X∩Soc(M)] ∩Rad(M) = X∩[S oc(M)∩Rad(M)] = X∩
Socs(M) = 0. Therefore M=X⊕Yand Soc(Y)⊆Rad(Y) = Rad(M). Then
Yis projective as a direct summand of the projective module M. By the proof
of [4, Proposition 4.2], we have Rad(Y) = δ(Y). Since Mis δss-supplemented,
it follows from Proposition 4.14 that Yis δss-supplemented.
Let Ube a submodule of Y. By Theorem 5.6, there exists a direct summand
Vof Ysuch that Y=U+Vand U∩V⊆Socδ(V). Then U∩V⊆δ(V)⊆
δ(Y) = Rad(Y) and hence U∩Vis small in Y. It follows from [13, 19.3 (5)]
that U∩VV. This means that Yis ss-supplemented. Hence M=X⊕Y
is ss-supplemented by [6, Corollary 3.13].
For a ring R, let X(R) = Soc(RR)
Socs(RR)as in [4].
Corollary 5.10. Let Rbe a ring. Then the following statements are equiva-
lent:
(1) RRis ss-supplemented,
(2) Ris left δss-perfect and semilocal,
(3) Ris left δss-perfect and X(R)is finitely generated.
Proof. (1) ⇐⇒ (2) By Proposition 5.9.
(2) ⇐⇒ (3) It follows from [4, Lemma 4.1].
Observe from Corollary 5.10 that if a left δss-perfect ring is left noetherian,
then it is a left artinian ring.
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δss-supplemented modules and rings 216
Burcu Ni¸sancı T¨urkmen,
Amasya University,
Faculty of Art and Science,
Department of Mathematics,
Ipekkoy, 05100, Amasya, Turkey.
Email: burcunisancie@hotmail.com
Erg¨ul T¨urkmen,
Amasya University,
Faculty of Art and Science,
Department of Mathematics,
Ipekkoy, 05100, Amasya, Turkey.
Email: ergulturkmen@hotmail.com