Generalizations of $\oplus$-supplemented modules
We introduce $\oplus$-radical supplemented modules and strongly $\oplus$-radical supplemented modules (briefly, $srs^{\oplus}$-modules) as proper generalizations of $\oplus$-supplemented modules. We prove that (1) a semilocal ring $R$ is left perfect if and only if every left $R$-module is an $\opl...
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| author | Pancar, A. Türkmen, B. N. Пансар, А. Тюркмен, Б. Н. |
| author_facet | Pancar, A. Türkmen, B. N. Пансар, А. Тюркмен, Б. Н. |
| author_sort | Pancar, A. |
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| description | We introduce $\oplus$-radical supplemented modules and strongly $\oplus$-radical supplemented modules (briefly, $srs^{\oplus}$-modules) as proper generalizations of $\oplus$-supplemented modules.
We prove that (1) a semilocal ring $R$ is left perfect if and only if every left $R$-module is an $\oplus$-radical supplemented module;
(2) a commutative ring $R$ is an Artinian principal ideal ring if and only if every left $R$-module is a $srs^{\oplus}$-module;
(3) over a local Dedekind domain, every $\oplus$-radical supplemented module is a $srs^{\oplus}$-module. Moreover, we completely determine the structure of these modules over local Dedekind domains. |
| first_indexed | 2026-03-24T02:23:27Z |
| format | Article |
| fulltext |
UDC 512.5
B. N. Türkmen, A. Pancar (Ondokuz Mayıs Univ., Samsun, Turkey)
GENERALIZATIONS OF ⊕-SUPPLEMENTED MODULES
УЗАГАЛЬНЕННЯ ⊕-ДОПОВНЮВАНИХ МОДУЛIВ
We introduce ⊕-radical supplemented modules and strongly ⊕-radical supplemented modules (briefly, srs⊕-modules) as
proper generalizations of ⊕-supplemented modules. We prove that (1) a semilocal ring R is left perfect if and only if every
left R-module is an ⊕-radical supplemented module; (2) a commutative ring R is an Artinian principal ideal ring if and
only if every left R-module is a srs⊕-module; (3) over a local Dedekind domain, every ⊕-radical supplemented module
is a srs⊕-module. Moreover, we completely determine the structure of these modules over local Dedekind domains.
Введено поняття ⊕-радикальних доповнюваних модулiв та сильно ⊕-радикальних доповнюваних модулiв (скорочено
srs⊕-модулiв) як вiдповiдних узагальнень ⊕-доповнюваних модулiв. Доведено, що: (1) напiвлокальне кiльце R є
досконалим злiва тодi i тiльки тодi, коли кожен лiвий R-модуль є ⊕-радикальним доповнюваним модулем; (2) кому-
тативне кiльце R є артiновим кiльцем головних iдеалiв тодi i тiльки тодi, коли кожен лiвий R-модуль є srs⊕-модулем;
(3) над локальною дедекiндовою областю кожен ⊕-радикальний доповнюваний модуль є srs⊕-модулем. Повнiстю
визначено структуру цих модулiв над локальними дедекiндовими областями.
1. Introduction. Throughout the whole text, all rings are to be associative, unit and all modules are
left unitary. Let M be such a module. We shall write N ≤M (N �M ) if N is a submodule of M
(small in M ). By Rad(M) we denote the radical of M. Let U, V ≤M. V is called a supplement of
U in M if it is minimal with respect to M = U + V. V is a supplement of U in M if and only if
M = U +V and U ∩V � V (see [12]). A module M is called supplemented (weakly supplemented
in [10]) if every submodule of M has a supplement in M, and it is called ⊕-supplemented if every
submodule of M has a supplement that is a direct summand of M. Clearly ⊕-supplemented modules
are supplemented.
In [13], Zöschinger introduced a notion of modules whose radical has supplements called radical
supplemented. The author determined in the same paper and in [15] the structure of radical sup-
plemented modules. Motivated by this, Büyükaşık and Türkmen call a module M strongly radical
suplemented (or briefly a srs-module) if every submodule containing radical has a supplement [2]. So
it is natural to introduce another notion that we called ⊕-radical supplemented. A module M is called
⊕-radical supplemented if Rad(M) has a supplement that is a direct summand of M. We call also
a module M strongly ⊕-radical supplemented (or briefly srs⊕-module) provided every submodule
containing radical has a supplement that is a direct summand of M.
In this paper, we obtain various properties of ⊕-radical supplemented and srs⊕-modules as a
proper generalization of ⊕-supplemented modules. We show that the class of srs⊕-modules and
⊕-radical supplemented modules are closed under finite direct sums. A semilocal ring R is left
perfect if and only if every left R-module is ⊕-radical supplemented, and a commutative ring R is
an Artinian principal ideal ring if and only if every left R-module is a srs⊕-module. We prove also
that a non-zero projective module M with cofinite radical is ⊕-supplemented if and only if it is a
srs⊕-module if and only if it is ⊕-cofinitely supplemented. Over a local Dedekind domain every
⊕-radical supplemented module is a srs⊕-module, and over a local Dedekind domain the structure
of these modules is completely determined.
c© B. N. TÜRKMEN, A. PANCAR, 2013
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 4 555
556 B. N. TÜRKMEN, A. PANCAR
2. Modules over any rings. Recall that a module M is called radical if M has no maximal
submodules, that is, Rad(M) = M. For a module M, P (M) will indicate the sum of all radical
submodules of M. if P (M) = 0, M is called reduced. Note that P (M) is the largest radical
submodule of M.
Now we have the following simple fact, which plays a key role in our working.
Lemma 2.1. P (M) is a srs⊕-module for every R-module M.
Proof. Let M be any R-module. We know that Rad(P (M)) = P (M). So P (M) has trivial
supplement 0 in P (M). Consequently, P (M) is a srs⊕-module.
We begin by giving some examples of module to seperate ⊕-supplemented, srs⊕-module, ⊕-
radical supplemented and radical supplemented.
Example 2.1. Let R be a non-local Dedekind domain with quotient field K. Consider the
R-module M = K(N). Since P (M) = M, M is a srs⊕-module by Lemma 2.1. If K(N) is ⊕-
supplemented, K is supplemented as a factor module of M and so, by [14], R is a local ring. This
contradicts the assumption. Hence M is not ⊕-supplemented.
Note that every ⊕-supplemented with zero radical is semisimple.
Example 2.2. (1) Consider the non-Noetherian ring R which is the direct product
∏∞
i≥1
Fi,
where Fi = F is any field. Clearly Rad(R) = 0 and so the left R-module R is ⊕-radical supple-
mented. On the other hand, the left R-module R is not a srs⊕-module since it is not semisimple.
(2) Let M =Z Z, where Z is the ring of integers. It is well known that M is not semisimple and
Rad(M) = 0. Hence M is ⊕-radical supplemented, but it is not a srs⊕-module.
Example 2.3. Let R = Z and I be a collection of distinct maximal ideal of Z. Consider the
left Z-module M =
∏
p∈I
(
Z
p2
)
. Then M is radical supplemented. However, it is not ⊕-radical
supplemented (see [13]).
Now we shall show that in general srs-modules need not be a srs⊕-module. To see this, we need
to the following lemma.
Lemma 2.2. Let M be a module. Suppose that Rad(M) is small in M. Then M is a srs⊕-
module if and only if it is ⊕-supplemented.
Proof. (=⇒) Let N be any submodule of M. Then Rad(M) ⊆ Rad(M) +N ⊆ M. Since M
is a srs⊕-module, we have M = Rad(M) +N +L, (Rad(M) +N)∩L� L and M = L⊕L′ for
two submodules L,L′ ≤ M. Since Rad(M) � M, we get M = N + L and N ∩ L � L. So L is
a supplement of N in M such that L is a direct summand of M. Therefore M is a ⊕-supplemented
module.
(⇐=) Clear.
Example 2.4 (see [9], Corollary 2.4). Let F be any field and R = F [[X,Y ]], the ring of formal
power series over F indeterminates X,Y. Then R is a local commutative Noetherian domain. Now
suppose that M =R Rad(R). So M = RX + RY. Since R is local, by [12] (42.6), M is supple-
mented and so it is a srs-module. It follows from [9] (Corollary 2.4) that M is not ⊕-supplemented.
Therefore, by Lemma 2.2, M is not a srs⊕-module.
Recall from [3] that a ring R is a left Bass ring if every non-zero left R-module has a maximal
submodule. It is known that the ring R is left Bass if and only if Rad(M) is small in M for every
non-zero left R-module M. By using Lemma 2.2, we obtain the following important corollary.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 4
GENERALIZATIONS OF ⊕-SUPPLEMENTED MODULES 557
Corollary 2.1. Every srs⊕-module over a left Bass ring is ⊕-supplemented.
A module M is called coatomic if every proper submodule of M is contained in a maximal
submodule of M. Note that coatomic modules have a small radical and so every coatomic module is
⊕-radical supplemented.
Corollary 2.2. Let M be a coatomic module. Then M is a srs⊕-module if and only if it is
⊕-supplemented.
Proof. It follows from Lemma 2.2.
Now we shall prove that the class of srs⊕-modules and ⊕-radical supplemented modules are
closed under finite direct sums.
Theorem 2.1. Let Mi, i = 1, 2, . . . , n, be any finitely collection of modules and M = M1 ⊕
⊕M2 ⊕ . . .⊕Mn. Then:
(1) M is ⊕-radical supplemented if Mi is ⊕-radical supplemented for each 1 ≤ i ≤ n;
(2) M is a srs⊕-module if Mi is a srs⊕-module for each 1 ≤ i ≤ n.
Proof. (1) The proof can be made similar to (2).
(2) Let Mi be a srs⊕-module for each 1 ≤ i ≤ n. To prove that M is a srs⊕-module, it is
sufficient by induction on n to prove this is the case when n = 2. Hence suppose n = 2. Let U
be any submodule of M with Rad(M) ⊆ U. Then M = M1 +M2 + U so that M1 +M2 + U
has a supplement 0 in M. Since M = M1 ⊕ M2, then Rad(M2) ⊆ U + M1. It follows that
Rad(M2) ⊆ M2 ∩ (U +M1) has a supplement H in M2 such that H is a direct summand of M2.
By [5] (Lemma 1.3), H is a supplement of M1 +U in M. Moreover Rad(M1) ⊆ U +H. Since M1
is a srs⊕-module, M1 ∩ (U + H) has a supplement K in M1 such that K is a direct summand of
M1. Again applying [5] (Lemma 1.3), we have that H +K is a supplement of U in M. It is clear
that H +K is a direct summand of M. Therefore M is a srs⊕-module.
Now we shall give another example of a non-radical module which is a srs⊕-module but not
⊕-supplemented.
Example 2.5. Consider the left Z-module M = Q ⊕ Zp, where p is a prime integer. Note that
M has a unique maximal submodule, which means that Rad(M) 6= M. According to Lemma 2.1,
the left Z-module Q is a srs⊕-module. By Theorem 2.1 (2), M is a srs⊕-module as a direct sum of
two srs⊕-modules. On the other hand, M is not ⊕-supplemented because it is not torsion.
Proposition 2.1. Let M be a non-radical module. If M is a ⊕-radical supplemented, then M
contains a radical direct summand. In particular, if P (M) = 0, then Rad(M)�M.
Proof. Suppose that Rad(M) 6=M. By the hypothesis, there exist submodules V, V ′ of M such
that M = Rad(M) + V, Rad(V ) = V ∩ Rad(M) � V and M = V ⊕ V ′. It follows from [12]
(21.6 (5)) that Rad(M) = Rad(V )⊕ Rad(V ′). So M = Rad(M) + V = Rad(V ′)⊕ V. Therefore
by modularity, V ′ = Rad(V ′)⊕ (V ∩ V ′) = Rad(V ′), that is, V ′ is radical.
Suppose that P (M) = 0. Then V ′ = 0, which shows that V =M. Hence Rad(M)�M.
Recall that a subset X of a ring R is called right t-nilpotent if, for every sequence x1, x2, . . . of
elements in X, there exists a k ∈ N with x1x2 . . . xk = 0. A ring R is called left perfect if R is
semilocal and Rad(R) is right t-nilpotent [12] (43.9).
Theorem 2.2. Let R be any ring. Then Rad(R) is right t-nilpotent if and only if every projec-
tive left R-module is ⊕-radical supplemented.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 4
558 B. N. TÜRKMEN, A. PANCAR
Proof. (=⇒) Let M be any projective left R-module. By [8] (9.2.1), Rad(M) = Rad(R)M
and so, by [12] (43.5), Rad(M)�M as required.
(⇐=) Let M = R(N). Again applying [8] (9.2.1), we have Rad(M) = Rad(R)M. Since M is
⊕-radical supplemented, there exist submodules V, V ′ of M such that M = Rad(M)+V, Rad(V ) =
= V ∩Rad(M)� V and V ⊕V ′ =M. So V ′ is radical. It follows from [12] (22.3 (2)) that V ′ = 0,
which means that V = M. Hence Rad(M) is small in M and, by [12] (43.5), Rad(R) is right
t-nilpotent.
Corollary 2.3. A semilocal ring R is left perfect if and only if every left R-module is ⊕-radical
supplemented.
Proof. It follows from Theorem 2.2 and [12] (49.9).
Note that the condition “semilocal” in the above corollary is necessary. We see, for example, the
left Bass rings which are not left perfect.
Proposition 2.2. A non-zero projective srs⊕-module is ⊕-supplemented.
Proof. Let M be any non-zero projective srs⊕-module. Therefore, it is ⊕-radical supplemented.
Then there exist submodules V, V ′ of M such that M = Rad(M) + V, Rad(V ) � V and M =
= V ⊕ V ′. So V ′ is radical. By [12] (22.3(2)), V ′ = 0 . It follows that Rad(M)�M. Hence M is
⊕-supplemented by Lemma 2.2.
It is well known that a ring R is semiperfect if and only if every finitely generated free R-
module is ⊕-supplemented. By Lemma 2.2, we know that every finitely generated srs⊕-module is
⊕-supplemented. Using these facts we obtain the following corollary.
Corollary 2.4. For any ring R with identity element, R is semiperfect if and only if every finitely
generated free R-module is a srs⊕-module.
Proof. Let F = R(I) be any free R-module for some finite set I. Since R is semiperfect, by [9]
(Theorem 2.1), the left R-module R is ⊕-supplemented and so the module is a srs⊕-module. Hence
F is a srs⊕-module by Theorem 2.1 (2). Conversely, suppose that every finitely generated free
R-module is a srs⊕-module. Then the left R-module R is a srs⊕-module. By Lemma 2.2, R is
semiperfect.
Let R be any ring. R is called FGC ring if every finitely generated R-module decomposes into a
direct sum of cyclic submodules. If R is a local FGC ring, then R is an almost maximal valuation
ring [1] (Theorem 4.4). It is proved [6] (Proposition 1.3) that a commutative local ring R is an almost
maximal valuation ring if and only if every finitely generated R-module is ⊕-supplemented. Now we
have the following corollary.
Corollary 2.5. For a commutative ring R, R is a finitely product of almost maximal valuation
rings if and only if every finitely generated R-module is a srs⊕-module.
Lemma 2.3. Let M be an indecomposable module. If M is a srs⊕-module, then M is radical
or M is local.
Proof. Suppose that Rad(M) 6=M. Then M contains a maximal submodule K. By the hypoth-
esis, there exists a direct summand V of M such that M = K + V and K ∩ V � V. It follows
from [12] (41.1(3)) that V is local. Since M is an indecomposable module and K is a maximal
submodule of M, we get V =M. Thus M is local.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 4
GENERALIZATIONS OF ⊕-SUPPLEMENTED MODULES 559
Theorem 2.3. Let R be a local commutative ring and M be a uniform R-module. Then every
submodule of M is a srs⊕-module if and only if M is uniserial.
Proof. (=⇒) By [11] (Lemma 6.2), it sufficies to show that every finitely generated submodule of
M is local. Let N be any finitely generated submodule of M. By assumption, N is indecomposable.
So, by Lemma 2.3, N is local.
(⇐=) Since M is uniserial, every submodule of M is hollow by [3] (2.17). Therefore every
submodule of M is a srs⊕-module.
Corollary 2.6. Let R be a local commutative ring. Suppose that every submodule of
E
(
R
Rad(R)
)
is a srs⊕-module, where E
(
R
Rad(R)
)
is the injective hull of the simple module
R
Rad(R)
. Then R is a uniserial ring.
Proof. Since E
(
R
Rad(R)
)
is uniform, the hypothesis implies that E
(
R
Rad(R)
)
is uniserial
by Theorem 2.3. It follows from [11] (Lemma 6.2) that R is a uniserial ring.
It is shown [6] (Theorem 1.1) that a commutative ring R is an artinian principal ring if and only
if every left R-module is ⊕-supplemented. Now we generalize this fact.
Theorem 2.4. A commutative ring R is an artinian principal ideal ring if and only if every left
R-module is a srs⊕-module.
Proof. Suppose that every left R-module is a srs⊕-module. Then, by Lemma 2.2, the left R-
module R is ⊕-supplemented and so R is semiperfect. By [12] (42.6), R is semilocal. It follows
from Corollary 2.3 that R is left perfect. Since R is semiperfect, we can write, [12] (42.6), R =
= Re1 ⊕ Re2 ⊕ . . . ⊕ Ren such that ei is local orthogonal idempotent for 1 ≤ i ≤ n with n ∈ N.
For all 1 ≤ i ≤ n, Rei is commutative and it is not difficult to see that every Rei-module is a srs⊕-
module by assumption. Now Corollary 2.6 implies that Rei is an uniserial ring for every 1 ≤ i ≤ n.
By [11] (Lemma 6.3), Rei is a principal ideal ring, which shows that R is an artinian principal ideal
ring.
Proposition 2.3. Let R be a ring and M be a ⊕-radical supplemented R-module with
Rad(M) 6=M. If its ring of endomorphism is quasi local, then M is local.
Proof. By the hypothesis, there exist submodules U,U ′ of M such that M = Rad(M) + U,
Rad(M) ∩ U � U and M = U ⊕ U ′. By [11] (Proposition 3.11), M is an indecomposable module.
So U ′ = 0, that is, U =M. Thus Rad(M)�M. By Lemma 2.2, M is ⊕-supplemented. Let N be
any proper submodule of M. It follows that M = N + T, N ∩ T � T and M = T ⊕ T ′ for some
submodules T, T ′ ⊆ M. Since M is an indecomposable module, M = T. Then N � M. Therefore
M is hollow. By [12] (41.4), M is local.
Example 2.6 (see [7], Example 2.3). Let R be a commutative local ring which is not a valuation
ring. Let x and y be elements of R, neither of them divides the other. By taking a suitable quotient
ring, we may assume that (x) ∩ (y) = 0 and xP = yP = 0, where P is the unique maximal ideal
of R. Let F be a free module with generators a1, a2, a3. Let N be the submodule generated by
xa1 − ya2 and let M =
F
N
. By Theorem 2.1 (2), F is a srs⊕-module. Suppose that M is a srs⊕-
module. It is clear that M is finitely generated and it follows that Rad(M) � M. By Lemma 2.2,
M is ⊕-supplemented. This is a contradiction.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 4
560 B. N. TÜRKMEN, A. PANCAR
Now we give some properties of factor modules of srs⊕-modules. Recall from [12] that a sub-
module U of an R-module M is called fully invariant if f(U) is contained in U for every R-
endomorphism f of M. Let M be an R-module and τ be a preradical for the category of R-modules.
Then τ(M) is a fully invariant submodule of M. We prove the following proposition which is a
modified form of [7] (Proposition 2.5).
Proposition 2.4. If M is a srs⊕-module, then
M
U
is a srs⊕-module for every fully invariant
submodule U of M.
Proof. Let U be any fully invariant submodule of M and let
V
U
be any submodule of
M
U
with
Rad
(
M
U
)
⊆ V
U
. Since
Rad(M) + U
U
⊆ Rad
(
M
U
)
, we have Rad(M) ⊆ V. By the hypothesis,
we have M = V +T, V ∩T � T and M = T ⊕T ′ for some submodules T, T ′ of M. Then by [14]
(Lemma 1.2(d)),
(T + U)
U
is a supplement of
V
U
in
M
U
. Since U is a fully invariant submodule of
M, we have U = (T ∩ U) + (T ′ ∩ U) by [7] (Lemma 2.4). Note that
M
U
=
(T + U)
U
+
(T ′ + U)
U
and
(T + U)
U
∩ (T ′ + U)
U
= 0,
i.e.,
(T + U)
U
is a direct summand of
M
U
. Hence
M
U
is a srs⊕-module.
Proposition 2.5. Let M be a ⊕-radical supplemented module. Then
M
P (M)
has a small radi-
cal.
Proof. Since P (M) is a fully invariant submodule of M, by Proposition 2.4, the factor module
M
P (M)
is ⊕-radical supplemented. Note that
M
P (M)
is reduced. It follows from Proposition 2.1 that
M
P (M)
has a small radical.
Proposition 2.6. LetM be a srs⊕-module. Suppose that
M
Rad(M)
is projective. Then Rad(M)
is ⊕-supplemented if and only if M is ⊕-supplemented.
Proof. (=⇒) Let Rad(M) be a ⊕-supplemented module. By the hypothesis, we have M =
= Rad(M) ⊕ N for some submodule N of M. Since M is a srs⊕-module, by Proposition 2.4,
M
Rad(M)
is semisimple and so N is semisimple. Therefore N is ⊕-supplemented. By [5] (Theo-
rem 1.4), M is ⊕-supplemented.
(⇐=) Since Rad(M) is a fully invariant submodule of M and M is ⊕-supplemented, Rad(M)
is ⊕-supplemented by [7] (Proposition 2.5).
A submodule N of M is said to be cofinite if
M
N
is finitely generated.
Proposition 2.7. Let M be a srs⊕-module. Suppose that a cofinite fully invariant submodule
K of M is a direct summand of M. Then K is a srs⊕-module.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 4
GENERALIZATIONS OF ⊕-SUPPLEMENTED MODULES 561
Proof. Let U be any submodule of K with Rad(K) ⊆ N. By the hypothesis, we have M =
= K ⊕ L for some finitely generated submodule L of M. Then Rad(L) � L. Clearly Rad(M) ⊆
⊆ U + Rad(L). And so there exist submodules V, V ′ of M such that M = U + Rad(L) + V,
(U +Rad(L))∩V � V and M = V ⊕V ′. Since Rad(L)� L, we have M = U +V, U ∩V � V
and M = V ⊕ V ′. It follows that K = U + (K ∩ V ) and U ∩ (K ∩ V ) � M. Since K is a fully
invariant submodule of M, then K = (K ∩ V ) ⊕ (K ∩ V ′). Note that U ∩ (K ∩ V ) � K ∩ V.
Therefore K is a srs⊕-module.
Corollary 2.7. Let M be a srs⊕-module and let τ(M) be a cofinite direct summand of M, then
τ(M) is a srs⊕-module.
Lemma 2.4. Let M be an R-module and Rad(M) ⊆ N. If N is a direct summand of M, then
Rad(M) = Rad(N). In particular, if Rad(M) is a direct summand of M, Rad(M) = P (M).
Proof. By the hypothesis, we have M = N⊕N ′ for some submodule N ′ of M. Then Rad(M) =
= Rad(N)⊕Rad(N ′) by [8] (9.1.5). Since Rad(M) ⊆ N, Rad(M) = Rad(N)⊕ (N ∩Rad(N ′)).
Note that N ∩ Rad(N ′) ⊆ N ∩N ′ = 0. Hence Rad(M) = Rad(N). Now we take N = Rad(M)
under the similar condition. So M = Rad(M) ⊕ X for some submodule X of M. It follows that
Rad(M) = Rad(Rad(M)) ⊕ Rad(X). Since Rad(M) ∩ X = 0, we have Rad(X) = 0 and so
Rad(M) = Rad(Rad(M)), i.e., Rad(M) is radical. Consequently, Rad(M) = P (M).
Let R be a ring and let M be an R-module. We consider the following condition.
(D3) If M1 and M2 are direct summands of M with M = M1 +M2, then M1 ∩M2 is also a
direct summand of M.
Proposition 2.8. LetM be a srs⊕-module with (D3) and letN be a submodule with Rad(M) ⊆
⊆ N. If N is a direct summand of M, N is a srs⊕-module.
Proof. Let U be a submodule of N such that Rad(N) ⊆ U. By Lemma 2.4, Rad(M) =
= Rad(N). Since M is a srs⊕-module, there exist submodules V, V ′ of M such that M = U + V,
U ∩ V � V and M = V ⊕ V ′. Then N = U + (N ∩ V ). Since M satisfies (D3), N ∩ V is a direct
summand of M. Then there exists a submodule X of M such that M = (N ∩ V ) ⊕ X. It follows
that U ∩ (N ∩ V )� N ∩ V and N = (N ∩ V )⊕ (N ∩X). Therefore N is a srs⊕-module.
Corollary 2.8. Let M be a UC-extending module. If M is a srs⊕-module, then every direct
summand of M containing Rad(M) is a srs⊕-module.
Recall that an R-module M has summand sum property (SSP ) if the sum of two direct sum-
mands of M is again a direct summand of M. In [4], a module M is called ⊕-cofinitely supple-
mented if every cofinite submodule of M has a supplement that is a direct summand of M. It is well
known [4] (Theorem 2.3) that a module M with (SSP ) is ⊕-cofinitely supplemented if and only if
every maximal submodule of M has a supplement that is a direct summand of M. We don’t know
whether srs⊕-modules are ⊕-cofinitely supplemented, but we have the following fact.
Theorem 2.5. Let M be a srs⊕-module with (SSP ). Then M is ⊕-cofinitely supplemented.
Proof. Let U be any maximal submodule of M. Then Rad(M) ⊆ U. By the hypothesis, U has a
supplement that is a direct summand of M. By [4] (Theorem 2.3), M is ⊕-cofinitely supplemented.
The following example shows that a ⊕-cofinitely supplemented module is not a srs⊕-module.
Example 2.7. Consider that the ring Zp consisting all rational numbers of the form
a
b
, where
p - b. Then Zp is a local ring, which is not left perfect. So, by [4] (Theorem 2.9), every left free
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 4
562 B. N. TÜRKMEN, A. PANCAR
Zp-module is ⊕-cofinitely supplemented. Since Zp is not left perfect, there exists an infinite index
set I such that Z(I)
p is not ⊕-supplemented. By Proposition 2.2, Z(I)
p is not a srs⊕-module.
Proposition 2.9. Let M be a module and Rad(M) be cofinite. If M is ⊕-cofinitely supple-
mented, then M is a srs⊕-module.
Proof. Let N be any submodule of M with Rad(M) ⊆ N. Note that(
M
Rad(M)
)
(
N
Rad(M)
) ∼= M
N
.
Since
M
Rad(M)
is finitely generated, N is a cofinite submodule of M. By the hypothesis, N has a
supplement that is a direct summand of M. Therefore M is a srs⊕-module.
Theorem 2.6. Let M be a non-zero projective module with cofinite radical. Then the following
statements are equivalent:
(1) M is a ⊕-supplemented module;
(2) M is a ⊕-cofinitely supplemented module;
(3) M is a srs⊕-module.
Proof. (1) ⇒ (2) Obvious.
(2) ⇒ (3) This implication follows from Proposition 2.9.
(3) ⇒ (1) By Proposition 2.2.
3. Modules over Dedekind domains. Throughout this section R will denote a Dedekind domain
unless otherwise specified.
Proposition 3.1. Let M be an R-module. Then M is ⊕-radical supplemented if and only if
M
P (M)
has a small radical.
Proof. (=⇒) By Proposition 2.5.
(⇐=) Since R is Dedekind domain, P (M) is injective and so there exists a submodule N of M
such that M = P (M) ⊕N. By the hypothesis, N is ⊕-radical supplemented. Thus, by Lemma 2.1
and Theorem 2.1 (1), M is ⊕-radical supplemented.
Note that from [14] (Lemma 2.1), over a local Dedekind domain module with small radical is
coatomic. By using this fact and Proposition 3.1, we obtain the following corollary.
Corollary 3.1. Let R be a local Dedekind domain and M be a module over such a ring R.
Then M is ⊕-radical supplemented if and only if
M
P (M)
is coatomic.
Proposition 3.2. Let M be an R-module. Then M is srs⊕ if and only if
M
P (M)
is a srs⊕-
module.
Proof. We know that P (M) is a fully invariant submodule of M. So, by Proposition 2.4,
M
P (M)
is a srs⊕-module. Conversely, suppose that
M
P (M)
is a srs⊕-module. Since R is a Dedekind domain,
we have M = P (M) ⊕ N for some submodule N of M. By the hypothesis, N is a srs⊕-module.
Hence M is a srs⊕-module by Theorem 2.1 (2) and Lemma 2.1.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 4
GENERALIZATIONS OF ⊕-SUPPLEMENTED MODULES 563
Corollary 3.2. Let R be a local Dedekind domain and M be an R-module. Then M is ⊕-radical
supplemented if and only if it is a srs⊕-module.
Proof. Suppose that M is ⊕-radical supplemented. By Corollary 3.1,
M
P (M)
is coatomic and
so, by [14] (Lemma 2.1)
M
P (M)
is ⊕-supplemented, which shows that
M
P (M)
is a srs⊕-module. By
Proposition 3.2, M is a srs⊕-module.
Theorem 3.1. Let R be a local Dedekind domain and M be an R-module. Then the following
statements are equivalent:
(1) M is ⊕-radical supplemented;
(2) M is a srs⊕-module;
(3) M ∼= K(I) ⊕
(
K
R
)(J)
⊕R(n) ⊕N, where K is the quotient field of R, I and J denote any
index sets, n is a non-negative integer and N is a bounded R-module.
Proof. (1)⇐⇒ (2) It is clear from Corollary 3.2.
(3) =⇒ (2) The module K(I) ⊕
(
K
R
)(J)
is radical and so, by Lemma 2.1, K(I) ⊕
(
K
R
)(J)
is
a srs⊕-module. By [14] (Lemma 2.1), R(n) ⊕ N is ⊕-supplemented. Hence the direct sum K(I) ⊕
⊕
(
K
R
)(J)
⊕R(n) ⊕N is a srs⊕-module by Theorem 2.1 (2).
(2) =⇒ (3) By Corollary 3.1,
M
P (M)
is coatomic. Then by [14] (Lemma 2.1), we have
M
P (M)
∼=
∼= R(n) ⊕ N, where n is non-negative integer and N is bounded. Since P (M) is radical, P (M) ∼=
∼= K(I) ⊕
(
K
R
)(J)
for some index sets I and J. Thus M ∼= K(I) ⊕
(
K
R
)(J)
⊕R(n) ⊕N.
We know that every ⊕-radical supplemented module is radical supplemented. In Example 2.3,
we showed that a radical supplemented module need not be ⊕-radical supplemented. Now we shall
prove that the converse of this fact is true for torsion modules over local Dedekind domains.
Proposition 3.3. Let R be a local Dedekind domain and M be a torsion R-module. Then M
is radical supplemented if and only if it is ⊕-radical supplemented.
Proof. Suppose that M is radical supplemented. By [13] (Proposition 3.1),
M
P (M)
is bounded
since M is torsion. Hence M is ⊕-radical supplemented by Theorem 2.1 (1).
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Received 31.07.11
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 4
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| id | umjimathkievua-article-2439 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:23:27Z |
| publishDate | 2013 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/7d/4ac94aa8f294bb4037f0a116c4d7817d.pdf |
| spelling | umjimathkievua-article-24392020-03-18T19:15:36Z Generalizations of $\oplus$-supplemented modules Узагальнення $\oplus$-доповнюваних модулiв Pancar, A. Türkmen, B. N. Пансар, А. Тюркмен, Б. Н. We introduce $\oplus$-radical supplemented modules and strongly $\oplus$-radical supplemented modules (briefly, $srs^{\oplus}$-modules) as proper generalizations of $\oplus$-supplemented modules. We prove that (1) a semilocal ring $R$ is left perfect if and only if every left $R$-module is an $\oplus$-radical supplemented module; (2) a commutative ring $R$ is an Artinian principal ideal ring if and only if every left $R$-module is a $srs^{\oplus}$-module; (3) over a local Dedekind domain, every $\oplus$-radical supplemented module is a $srs^{\oplus}$-module. Moreover, we completely determine the structure of these modules over local Dedekind domains. Введено поняття $\oplus$-радикальних доповнюваних модулiв та сильно $\oplus$-радикальних доповнюваних модулiв (скорочено $srs^{\oplus}$-модулiв) як вiдповiдних узагальнень $\oplus$-доповнюваних модулiв. Доведено, що: (1) напiвлокальне кiльце $R$ є досконалим злiва тодi i тiльки тодi, коли кожен лiвий $R$-модуль є $\oplus$-радикальним доповнюваним модулем; (2) комутативне кiльце $R$ є артiновим кiльцем головних iдеалiв тодi i тiльки тодi, коли кожен лiвий $R$-модуль є $srs^{\oplus}$-модулем; (3) над локальною дедекiндовою областю кожен $\oplus$-радикальний доповнюваний модуль є $srs^{\oplus}$-модулем. Повнiстю визначено структуру цих модулiв над локальними дедекiндовими областями. Institute of Mathematics, NAS of Ukraine 2013-04-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2439 Ukrains’kyi Matematychnyi Zhurnal; Vol. 65 No. 4 (2013); 555-564 Український математичний журнал; Том 65 № 4 (2013); 555-564 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2439/1644 https://umj.imath.kiev.ua/index.php/umj/article/view/2439/1645 Copyright (c) 2013 Pancar A.; Türkmen B. N. |
| spellingShingle | Pancar, A. Türkmen, B. N. Пансар, А. Тюркмен, Б. Н. Generalizations of $\oplus$-supplemented modules |
| title | Generalizations of $\oplus$-supplemented modules |
| title_alt | Узагальнення $\oplus$-доповнюваних модулiв |
| title_full | Generalizations of $\oplus$-supplemented modules |
| title_fullStr | Generalizations of $\oplus$-supplemented modules |
| title_full_unstemmed | Generalizations of $\oplus$-supplemented modules |
| title_short | Generalizations of $\oplus$-supplemented modules |
| title_sort | generalizations of $\oplus$-supplemented modules |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2439 |
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