# Generalized⊕-Radical Supplemented Modules

Authors: Burcu Nişancı Türkmen; Ali Pancar
Journal: ISRN Algebra
 (2014)

---

## Abstract

Çalışıcı and Türkmen called a moduleM generalized ⊕-supplemented if every
submodule has a generalized supplement that is a direct summand of M. Motivated
by this, it is natural to introduce another notion that we called generalized ⊕-radical
supplemented modules as a proper generalization of generalized ⊕-supplemented modules. In this paper, we obtain various properties of generalized ⊕-radical supplemented modules.
We show that the class of generalized ⊕-radical supplemented modules is closed under
finite direct sums. We attain that over a Dedekind domain a module M is generalized
⊕-radical supplemented if and only if M
/
P
(
M
) is generalized ⊕-radical supplemented. We
completely determine the structure of these modules over left V-rings. Moreover, we
characterize semiperfect rings via generalized ⊕-radical supplemented modules.


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## Body

## 1. Introduction

Throughout the whole text, all rings are to be associative; unit and all modules are left unitary. We specially mention [1–4] among books concerning the structures of modules and rings. 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 [4]. A module M is called supplemented if every submodule of M has a supplement, and it is called ⊕
-supplemented if every submodule of M has a supplement that is a direct summand of M [5]. Clearly every ⊕-supplemented module is supplemented. In [6], Zöschinger introduced a notion of modules whose radical has supplements called radical supplemented. Xue defined generalized supplemented modules as another generalization of supplemented modules [7]. Let U be any submodule of M. If there exists a submodule V of M such that M
=
U
+
V and U
∩
V
⊆
Rad
(
V
), V is called a generalized supplement of U in M. M is called generalized supplemented if every submodule of M has a generalized supplement in M. Also Çalışıcı and Türkmen called a module M  generalized
⊕
-supplemented if every submodule has a generalized supplement that is a direct summand of M as a generalization of ⊕-supplemented modules [8]. So it is natural to introduce another notion that we called generalized ⊕-radical supplemented modules. A module M is called generalized
⊕
-radical supplemented if every submodule containing radical has a generalized supplement that is a direct summand of M.In this paper we obtain various properties of generalized⊕-radical supplemented modules as a proper generalization of generalized ⊕-supplemented modules. We prove the following indications.(i)
Every generalized⊕-radical supplemented module has a radical direct summand.
(ii)
P
(
M
) is a generalized ⊕-radical supplemented module for every R-module M.
(iii)
The class of generalized⊕-radical supplemented modules is closed under finite direct sums.
(iv)
IfM is a generalized ⊕-radical supplemented module, then M
/
U is a generalized ⊕-radical supplemented module for every fully invariant submodule U of M.
(v)
Let{
M
i
}
i
∈
I be any infinite collection of generalized ⊕-radical supplemented modules, and let M
=
⊕
i
∈
I
M
i. If M is a duo module, then M is generalized ⊕-radical supplemented.
(vi)
Over a Dedekind domain a moduleM is generalized ⊕-radical supplemented if and only if M
/
P
(
M
) is generalized ⊕-radical supplemented.
(vii)
Every generalized⊕-radical supplemented R-module is injective if and only if R is left Noetherian V-ring.
(viii)
A ringR is semiperfect if and only if every finitely generated free R-module is generalized ⊕-radical supplemented.


## 2. Generalized⊕-Radical Supplemented Modules

Definition 1.
A moduleM is called generalized
⊕-radical supplemented if every submodule containing radical has a generalized supplement that is a direct summand of M.Recall that a moduleM 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 [4].Now we have the following simple fact, which lays a key role in our study.Lemma 2.
LetM be an R-module. If Rad
(
M
)
=
M, then M is a generalized ⊕-radical supplemented module.Proof.
SinceRad
(
M
)
=
M, M has the trivial generalized supplement 0 in M. Consequently, M is a generalized ⊕-radical supplemented module.Corollary 3.
P
(
M
) is a generalized ⊕-radical supplemented module for every R-module M.Proposition 4.
Every generalized⊕-radical supplemented module has a radical direct summand.Proof.
LetM be a generalized ⊕-radical supplemented module. Then there exist submodules V and V
′ of M such that M
=
Rad
(
M
)
+
V, Rad
(
M
)
∩
V
⊆
Rad
(
V
), and M
=
V
⊕
V
′. Since Rad
(
M
)
=
Rad
(
V
)
⊕
Rad
(
V
′
) by [4, Theorem 21.6], then M
=
Rad
(
V
′
)
+
V. Applying the modular law, we obtain that Rad
(
V
′
)
=
V
′.Recall from [4] that a submodule U of an R-module M is called fully invariant if f
(
U
) is contained in U for every R-endomorphism f of M. A module M is called duo, if every submodule of M is fully invariant [9].Theorem 5.
The following statements hold over a ringR.  (1)
Let{
M
i
}
i
∈
{
1,2
,
…
,
n
} be any finite collection of modules and M
=
M
1
⊕
M
2
⊕
⋯
⊕
M
n. Then M is generalized ⊕-radical supplemented if, for each 1
≤
i
≤
n, M
i is generalized ⊕-radical supplemented.
(2)
Let{
M
i
}
i
∈
I be any infinite collection of modules and =
⊕
i
∈
I
M
i. If M is a duo module, then M is generalized ⊕-radical supplemented if, for each i
∈
I, M
i is generalized ⊕-radical supplemented.Proof.
(
1
) Let M
i be a generalized ⊕-radical supplemented module for each 1
≤
i
≤
n. To prove that M is a generalized ⊕-radical supplemented module it is sufficient by induction on n to prove this is the case when n
=
2. Hence, suppose that n
=
2. Let N be any submodule of M with Rad
(
M
)
⊆
N. Then M
=
M
1
+
M
2
+
N so that M
1
+
M
2
+
N has a generalized supplement 0 in M. Since M
=
M
1
⊕
M
2, then Rad
(
M
2
)
⊆
N
+
M
1. It follows that Rad
(
M
2
)
⊆
M
2
∩
(
N
+
M
1
) has a generalized supplement H in M
2 such that H is a direct summand of M
2. By [10, Proposition 2.4], H is a generalized supplement of N
+
M
1 in M. Moreover, Rad
(
M
1
)
⊆
N
+
H. Since M
1 is a generalized ⊕-radical supplemented module, M
1
⊆
(
N
+
H
) has a generalized supplement G in M
1 such that G is a direct summand of M
1. Again applying [10, Proposition 2.4], H
+
G is a generalized supplement of N in M. It is clear that H
+
G is a direct summand of M. Therefore, M is a generalized ⊕-radical supplemented module.
(
2
) Let U be any submodule such that Rad
(
M
)
⊆
U. Then Rad
(
M
i
)
⊆
U
∩
M
i for every i
∈
I. By this hypothesis, there exist submodules V
i and V
i
′ of M
i such that M
i
=
(
U
∩
M
i
)
+
V
i, (
U
∩
M
i
)
∩
V
i
⊆
Rad
(
V
i
), and M
i
=
V
i
⊕
V
i
′ for every i
∈
I. Since U is a fully invariant submodule of M, then U
=
⊕
i
∈
I
(
U
∩
M
i
). Let V
=
⊕
i
∈
I
V
i and V
′
=
⊕
i
∈
I
V
i
′. Then there exist submodules U and V of M such that M
=
⊕
i
∈
I
M
i
=
[
⊕
i
∈
I
(
U
∩
M
i
)
]
+
(
⊕
i
∈
I
V
i
)
=
U
+
V, U
∩
V
=
⊕
i
∈
I
(
U
∩
M
i
)
∩
(
⊕
i
∈
I
V
i
)
=
⊕
i
∈
I
[
(
U
∩
M
i
)
∩
V
i
]
⊆
⊕
i
∈
I
Rad
(
V
i
)
=
Rad
(
V
) by [4, Theorem 21.6] and V
⊕
V
′
=
(
⊕
i
∈
I
V
i
)
⊕
(
⊕
i
∈
I
V
i
′
)
=
⊕
i
∈
I
M
i
=
M. Thus, M is a generalized ⊕-radical supplemented module.Proposition 6.
IfM is a generalized ⊕-radical supplemented module, then M
/
U is a generalized ⊕-radical supplemented module for every fully invariant submodule U of M.Proof.
LetU be any fully invariant submodule of M and let K
/
U be any submodule of M
/
U with Rad
(
M
/
U
)
⊆
K
/
U. Since (
Rad
(
M
)
+
U
)
/
U
⊆
Rad
(
M
/
U
), we have Rad
(
M
)
⊆
K. By the hypothesis, we have M
=
K
+
V, K
∩
V
⊆
Rad
(
V
), and M
=
V
⊕
V
′ for some submodules V and V
′ of M. By [11, Lemma 2.3] (
V
+
U
)
/
U is generalized supplement of K
/
U in M
/
U. Since U is a fully invariant submodule of M, we have U
=
(
U
∩
V
)
+
(
U
∩
V
′
) and [
(
V
+
U
)
/
U
]
∩
[
(
V
′
+
U
)
/
U
]
=
0. Then M
/
U
=
[
(
V
+
U
)
/
U
]
⊕
[
(
V
′
+
U
)
/
U
]. Hence, M
/
U is a generalized ⊕-radical supplemented module.Corollary 7.
LetR be a Dedekind domain and let M be an R-module. M is a generalized ⊕-radical supplemented module if and only if M
/
P
(
M
) is a generalized ⊕-radical supplemented module.Proof.
We know thatP
(
M
) is a fully invariant submodule of M. So, by Proposition 6, M
/
P
(
M
) is a generalized ⊕-radical supplemented module. Conversely, suppose that M
/
P
(
M
) is a generalized ⊕-radical supplemented 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 generalized ⊕-radical supplemented module. Hence, M is a generalized ⊕-radical supplemented module by Theorem 5(1) and Corollary 3.Example 8.
Consider the leftℤ-module ℚ. According to Lemma 2, ℚ is a generalized ⊕-radical supplemented module. On the other hand, ℚ is not ⊕-supplemented because it is not torsion.Theorem 9.
LetM be a module. Suppose that Rad
(
M
) is small in M. Then M is a generalized ⊕-radical supplemented 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 generalized ⊕-radical supplemented module, we have M
=
Rad
(
M
)
+
N
+
L, (
Rad
(
M
)
+
N
)
∩
L
⊆
Rad
(
L
), and M
=
L
⊕
L
′ for submodules L
,
L
′
≤
M. Since Rad
(
M
)
≪
M, we have M
=
N
+
L, N
∩
L
≪
L. Therefore M is ⊕-supplemented.
(
⇐
) is trivial.Corollary 10.
Every generalized⊕-radical supplemented module over a left Bass ring is ⊕-supplemented.Proposition 11.
Every direct summand ofP
(
M
) is a generalized ⊕-radical supplemented module for every R-module M.Proof.
LetN be any direct summand of P
(
M
). Then there exists a submodule L of P
(
M
) such that P
(
M
)
=
N
⊕
L. Since P
(
M
)
=
Rad
(
P
(
M
)
)
=
Rad
(
N
)
⊕
Rad
(
L
) by [4, Theorem 21.6], then N
=
Rad
(
N
)
⊕
[
N
∩
Rad
(
L
)
]
=
Rad
(
N
). By Lemma 2, N is generalized ⊕-radical supplemented.Proposition 12.
LetM be a generalized ⊕-radical supplemented module. Suppose that a cofinite fully invariant submodule K of M is a direct summand of M. Then K is generalized ⊕-radical supplemented.Proof.
LetU be any submodule of K with Rad
(
K
)
⊆
U. 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
). So there exist submodules V
and
V
′ of M such that M
=
U
+
Rad
(
L
)
+
V, (
U
+
Rad
(
L
)
)
∩
V
⊆
Rad
(
V
), and M
=
V
⊕
V
′. Since Rad
(
L
)
≪
L, we have M
=
U
+
V, U
∩
V
⊆
Rad
(
V
) and M
=
V
⊕
V
′. It follows that K
=
U
+
(
K
∩
V
) and U
∩
(
K
∩
V
)
⊆
Rad
(
M
). Since K is a fully invariant submodule of M, then K
=
(
K
∩
V
)
⊕
(
K
∩
V
′
). Note that U
∩
(
K
∩
V
)
⊆
Rad
(
K
∩
V
). Therefore, K is generalized ⊕-radical supplemented.Lemma 13.
LetM be a module and N a submodule such that 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, then Rad
(
M
)
=
P
(
M
).Proof.
LetN be any direct summand of M such that Rad
(
M
)
⊆
N. Then there exists a submodule N
′ of M such that M
=
N
⊕
N
′. By [4, Theorem 21.6], Rad
(
M
)
=
Rad
(
N
)
⊕
Rad
(
N
′
). Since Rad
(
M
)
⊆
N, Rad
(
M
)
=
[
Rad
(
N
)
⊕
Rad
(
N
′
)
]
∩
N
=
Rad
(
N
)
⊕
[
Rad
(
N
′
)
∩
N
]
=
Rad
(
N
). Now we take N
=
Rad
(
M
) under the similar condition. Rad
(
M
)
=
Rad
(
Rad
(
M
)
); that is, Rad
(
M
) is radical. Consequently, Rad
(
M
)
=
P
(
M
).LetR be a ring and let M be an R-module. We consider the following condition.(
D
3
) If M
1 and M
2 are direct summands of M with M
=
M
1
+
M
2, then M
1
∩
M
2 is also a direct summand of M.Proposition 14.
LetM be a generalized ⊕-radical supplemented module with (
D
3
) and let N be a submodule with Rad
(
M
)
⊆
N. If N is a direct summand of M, N is generalized ⊕-radical supplemented.Proof.
LetU be a submodule of N such that Rad
(
N
)
⊆
U. By Lemma 13, Rad
(
M
)
=
Rad
(
N
). Since M is a generalized ⊕-radical supplemented module, there exist submodules V
and
V
′ of M such that M
=
U
+
V, U
∩
V
⊆
Rad
(
V
), and M
=
V
⊕
V
′. Then N
=
U
+
(
N
∩
V
). Since M satisfies (
D
3
), 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
)
⊆
Rad
(
N
∩
V
) and N
=
(
N
∩
V
)
⊕
(
N
∩
X
). Therefore N is a generalized ⊕-radical supplemented module.A moduleM is called local if M
/
Rad
(
M
) is simple and Rad
(
M
) is small in M [10]. The module M is called w-local if M
/
Rad
(
M
) is simple [12].The following fact is clear.Corollary 15.
Everyw-local module is generalized ⊕-radical supplemented.Lemma 16.
LetM be an indecomposable module. If M is a generalized ⊕-radical supplemented module, then M
=
Rad
(
M
) or M is w-local.Proof.
Suppose thatRad
(
M
)
≠
M. Then M contains at least one maximal submodule, say K. Since Rad
(
M
)
⊆
K, then there exist submodules V
and
V
′ of M such that M
=
K
+
V, K
∩
V
⊆
Rad
(
V
), and M
=
V
⊕
V
′. We have V
=
M because M is indecomposable. Since V is a generalized supplement of K, V is w-local by [12, Lemma 3.3]. So M is w-local.Theorem 17.
LetR be a local commutative ring and M be a uniform R-module. Every submodule of M is generalized ⊕-radical supplemented if and only if it is uniserial.Proof.
(
⇒
) By [13, Lemma 6.2], it suffices 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 16, N is w-local. Since N is finitely generated, N is local.
(
⇐
) Since M is uniserial, every submodule of M is hollow by [3, 2.17]. So it is easy to see that every submodule of M is a generalized ⊕-radical supplemented module.Corollary 18.
LetR be a local commutative ring. Suppose that every submodule of E
(
R
/
Rad
(
R
)
) is generalized ⊕-radical supplemented, where E
(
R
/
Rad
(
R
)
) is the injective hull of the simple module R
/
Rad
(
R
). Then R is a uniserial ring.Proof.
SinceE
(
R
/
Rad
(
R
)
) is uniform, the hypothesis implies that E
(
R
/
Rad
(
R
)
) is uniserial by Theorem 17. It follows from [13, Lemma 6.2] that R is a uniserial ring.Theorem 19.
The following statements are equivalent for a ringR.  (1)
R is semiperfect.
(2)
Every finitely generated freeR-module is generalized ⊕-radical supplemented.Proof.
By [14, Theorem 2.1] and Theorem 9, the proof is clear.Lemma 20.
LetM be a module. Suppose that Rad
(
M
)
=
0. Then M is a generalized ⊕-radical supplemented module if and only if it is semisimple.Proof.
This is clear by [15, Proposition 3.3].A ringR is called left V
-ring if every simple left R-module is injective. The ring R is called an SSI ring if semisimple left R-module is injective. Let R be a commutative ring. R is regular if and only if every simple left R-module is injective [13, Theorem 2.14].Proposition 21.
LetR be a left V-ring and M an R-module. Then M is generalized ⊕-radical supplemented if and only if M is semisimple.Corollary 22.
LetR be a commutative regular ring and M an R-module. Then M is generalized ⊕-radical supplemented if and only if M is semisimple.Proposition 23.
The following statements are equivalent for a ringR.  (1)
Every generalized
⊕
-radical supplemented
R
-module is injective.
(2)
R
is left Noetherian
V
-ring.Proof.
By [16, Proposition 5.3], the proof is clear.



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*Source: 603851-2014-01-14.xml*