A study on dual square free modules
Let M be an H-supplemented coatomic module with FIEP. Then we prove that M is dual square free if and only if every maximal submodule ofM is fully invariant. Let M = ⊕ i∈I Mi be a direct sum, such that M is coatomic. Then we prove that M is dual square free if and only if each Mi is dual square free...
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irk-123456789-1887532023-03-15T01:27:28Z A study on dual square free modules Medina-Bárcenas, M. Keskin Tütüncü, D. Kuratomi, Y. Let M be an H-supplemented coatomic module with FIEP. Then we prove that M is dual square free if and only if every maximal submodule ofM is fully invariant. Let M = ⊕ i∈I Mi be a direct sum, such that M is coatomic. Then we prove that M is dual square free if and only if each Mi is dual square free for all i ∈ I and, Mi and ⊕ j̸≠i Mj are dual orthogonal. Finally we study the endomorphism rings of dual square free modules. Let M be a quasi-projective module. If EndR(M) is right dual square free, then M is dual square free. In addition, if M is finitely generated, then EndR(M) is right dual square free whenever M is dual square free. We give several examples illustrating our hypotheses. 2021 Article A study on dual square free modules / M. Medina-Bárcenas, D. Keskin Tütüncü, Y. Kuratomi // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 2. — С. 267-279. — Бібліогр.: 17 назв. — англ. 1726-3255 DOI:10.12958/adm1512 2020 MSC: 16D40, 16D70 http://dspace.nbuv.gov.ua/handle/123456789/188753 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Let M be an H-supplemented coatomic module with FIEP. Then we prove that M is dual square free if and only if every maximal submodule ofM is fully invariant. Let M = ⊕ i∈I Mi be a direct sum, such that M is coatomic. Then we prove that M is dual square free if and only if each Mi is dual square free for all i ∈ I and, Mi and ⊕ j̸≠i Mj are dual orthogonal. Finally we study the endomorphism rings of dual square free modules. Let M be a quasi-projective module. If EndR(M) is right dual square free, then M is dual square free. In addition, if M is finitely generated, then EndR(M) is right dual square free whenever M is dual square free. We give several examples illustrating our hypotheses. |
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Medina-Bárcenas, M. Keskin Tütüncü, D. Kuratomi, Y. |
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Medina-Bárcenas, M. Keskin Tütüncü, D. Kuratomi, Y. A study on dual square free modules Algebra and Discrete Mathematics |
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Medina-Bárcenas, M. Keskin Tütüncü, D. Kuratomi, Y. |
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Medina-Bárcenas, M. |
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A study on dual square free modules |
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A study on dual square free modules |
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A study on dual square free modules |
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A study on dual square free modules |
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A study on dual square free modules |
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study on dual square free modules |
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Інститут прикладної математики і механіки НАН України |
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A study on dual square free modules / M. Medina-Bárcenas, D. Keskin Tütüncü, Y. Kuratomi // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 2. — С. 267-279. — Бібліогр.: 17 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT medinabarcenasm astudyondualsquarefreemodules AT keskintutuncud astudyondualsquarefreemodules AT kuratomiy astudyondualsquarefreemodules AT medinabarcenasm studyondualsquarefreemodules AT keskintutuncud studyondualsquarefreemodules AT kuratomiy studyondualsquarefreemodules |
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2025-07-16T10:57:27Z |
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© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 32 (2021). Number 2, pp. 267ś279
DOI:10.12958/adm1512
A study on dual square free modules
M. Medina-Bárcenas, D. Keskin Tütüncü,
and Y. Kuratomi
Communicated by R. Wisbauer
Abstract. Let M be an H-supplemented coatomic module
with FIEP. Then we prove that M is dual square free if and only if
every maximal submodule ofM is fully invariant. LetM =
⊕
i∈I Mi
be a direct sum, such that M is coatomic. Then we prove that M
is dual square free if and only if each Mi is dual square free for all
i ∈ I and, Mi and
⊕
j ̸=iMj are dual orthogonal. Finally we study
the endomorphism rings of dual square free modules. Let M be
a quasi-projective module. If EndR(M) is right dual square free,
then M is dual square free. In addition, if M is őnitely generated,
then EndR(M) is right dual square free whenever M is dual square
free. We give several examples illustrating our hypotheses.
Introduction
We consider associative rings R with identity and all modules consid-
ered are unitary right R-modules. The notations Rad(M) and EndR(M)
denote the radical and the endomorphism ring of any module M , respec-
tively.
A module M is said to be dual square free or brieŕy DSF if whenever
its factor module is isomorphic to N2 = N ⊕N for some module N , then
N = 0. Note that any factor module of a DSF module is also DSF. A ring
R is said to be right (resp. left) dual square free if it is dual square free as
a right (resp. left) R-module. This concept was introduced őrst in [6]. We
2020 MSC: 16D40, 16D70.
Key words and phrases: dual square free module, endoregular module, (őnite)
internal exchange property.
https://doi.org/10.12958/adm1512
268 Dual Square Free Modules
note that R is right DSF if and only if every cyclic right R-module is DSF.
We also know that a module M is DSF if and only if M has no proper
submodules A and B with M = A+B and M/A ∼=M/B (see [12]).
Recall that a module M is coatomic if every proper submodule of M
is contained in a maximal submodule. It is not difficult to see that M is
coatomic if and only if every nonzero factor module of M has a maximal
submodule. Let M and N be two right R-modules. M and N are called
dual orthogonal if, no nonzero factor module of M is isomorphic to a factor
module of N (it is called factor-orthogonal in [10]).
Let {Mi | i ∈ I} be a family of modules. Recall that the direct sum
decomposition M =
⊕
I Mi is said to be exchangeable if, for any direct
summandX ofM , there existM ′
i ⊆Mi (i ∈ I) such thatM = X⊕(⊕IM
′
i).
A module M is said to have the (őnite) internal exchange property (or
brieŕy, (F)IEP) if, any (őnite) direct sum decomposition M = ⊕IMi is
exchangeable.
The organization of our paper is as follows:
In the őrst section, we investigate some properties of DSF modules.
We also prove that for an H-supplemented coatomic module M with FIEP,
M is DSF if and only if every maximal submodule of M is fully invariant.
We illustrate our hypotheses in this section, as well.
In the second section, we work on direct sums of DSF modules. Let
M =
⊕
i∈I Mi be a direct sum, such that M is coatomic. Then M is
DSF if and only if each Mi is DSF for all i ∈ I and, Mi and
⊕
j ̸=iMj are
dual orthogonal. As a corollary we obtain that if M = A ⊕ B where A
is a őnitely generated DSF module and B =
⊕
i∈I Si is a direct sum of
non-isomorphic simple modules, then M is a DSF module if and only if
A and B are dual orthogonal.
In the last section, we investigate the endomorphism rings of DSF
modules. In [10, Example 2.5], they prove that a strongly regular ring is
a DSF ring. In [15], it is presented a module-theoretic version of strongly
regular rings called abelian endoregular modules. As a generalization of
[10, Example 2.5] we prove that if M is an endoregular quasi-projective
module, then M is abelian if and only if M is a DSF. It sounds interesting
to know when the endomorphism ring of a DSF module is a DSF ring and
the converse. In this vein we prove that for any quasi-projective module
M , if EndR(M) is right DSF, then M is DSF. In addition, if M is őnitely
generated, then EndR(M) is right DSF whenever M is DSF. Again we
give examples illustrating our hypotheses in this section.
For undeőned notions we refer to [12] and [13].
Medina-Bárcenas, Keskİn Tütüncü, Kuratomi 269
1. Dual square free modules
We start with the following result which can be established using
the same arguments in [10, Proposition 2.13 and Proposition 2.15]. We
just point out that, in Theorem 1, the implication (1) ⇒ (2) is the proof
of (1) ⇒ (2) in [12, Lemma 2.6] and for the implication (2) ⇒ (3), if
Rad(M) =M , then trivially M satisőes (3).
Theorem 1. (compare with [10, Theorem 2.16]) Consider the following
conditions for a module M :
1) M is DSF,
2) For any simple module S and every nonzero homomorphisms f, g
from M to S, Kerf = Kerg,
3) Every maximal submodule of M is fully invariant.
Then (1) ⇒ (2) ⇒ (3). If M is coatomic, then (2) ⇒ (1). In addition, if
M is quasi-projective, then (3) ⇒ (1).
The following examples illustrate that coatomic and quasi-projective
hypotheses on Theorem 1 are not superŕuous.
Example 1. Let A = B = Z(p∞) and C = Zq, where p and q are primes.
Put GZ = A⊕B⊕C. Note that G is not coatomic and not quasi-projective.
Also G is not DSF since it has the part A⊕B.
Since Rad(A⊕B) = A⊕B, A⊕B does not have a maximal submodule.
On the other hand, A⊕ B is the unique maximal submodule of G. Say
X = A⊕B.
Now let f : G → G be any endomorphism of G. If f(X) ⊈ X, then
G = X + f(X). Hence C ∼= G/X ∼= f(X)/(X ∩ f(X)). By considering
the epimorphism X → f(X)/(X ∩ f(X)) we have the epimorphism
α : X → C. Then Kerα is a maximal submodule of X, a contradiction.
Therefore f(X) ⊆ X.
Example 2. Consider the above example. The Z-moduleG is not coatomic
and not DSF. Since it has a unique maximal submodule, condition (2) in
Theorem 1 is satisőed.
Example 3. We take the next example from [3, Example 1.12]. Let
R = Z2 ⋊ (Z2 ⊕ Z2) be the trivial extension of Z2 by Z2 ⊕ Z2. This ring
can be described as
R ∼=
{(
a 0 0
x a 0
y 0 a
)
| a, x, y ∈ Z2
}
270 Dual Square Free Modules
with the usual operations of matrices. This ring is a őnite local ring, hence
there is only one simple rightR-module up to isomorphism, say S. Consider
an injective hull E(S) of S. The right R-module E(S) can be seen as the
abelian group M1×3(Z2) with action the product of vectors by matrices.
In [4, Section 3, Example 4] it is proved that every submodule of E(S)
is fully invariant. Thus, E(S) satisőes (3) in Theorem 1 and is coatomic.
Consider the lattice of submodules {0, S,K,L,N,E(S)} of E(S), which
is drawn in [3, Example 1.12]. The module E(S) is not DSF because
E(S)/S ∼= S ⊕ S. On the other hand, assume E(S) is quasi-projective.
Consider the following diagram
E(S)
f
��
α
{{
E(S) π
// S ⊕ S
where π is the composition E(S) ↠ E(S)/S ∼= S ⊕ S and f is the
composition E(S) ↠ E(S)/N ∼= S →֒ S ⊕ S. Suppose there exists
α : E(S) → E(S) such that f = πα. Note that α cannot be the zero
homomorphism neither an isomorphism. We have that every proper factor
module of E(S) is semisimple, hence α(E(S)) ⊆ S. This implies that
0 = πα = f , which is a contradiction. Thus E(S) is not quasi-projective.
Example 4. Let Zp̂ be the ring of p-adic integers and Qp̂ its quotient őeld.
It is known that Zp̂ is a Dedekind domain which is a complete discrete
valuation ring. Note that Qp̂ is a nonsingular injective Zp̂-module. By
[16, Lemma 5.1], Qp̂ is quasi-projective. Let M = Qp̂ ⊕ Qp̂ be a right
Zp̂-module. It follows from the fact that M is an injective module over
a PID that M has no maximal submodules. Hence, we have that M is
quasi-projective and every maximal submodule of M is fully invariant.
Note that M is not coatomic neither DSF.
Recall that a module A is said to be weakly generalized (epi-)B-
projective if, for any homomorphism (epimorphism) f : A → X and
any epimorphism g : B → X, there exist a small epimorphism ρ : X → Y
for some module Y , decompositions A = A1⊕A2,B = B1⊕B2, a homomor-
phism (an epimorphism) h1 : A1 → B1 and an epimorphism h2 : B2 → A2
such that ρ(f |A1
) = ρgh1 and ρ(g|B2
) = ρfh2.
Proposition 1. Let M be an H-supplemented coatomic module with
FIEP. Then M is DSF if and only if every maximal submodule of M is
fully invariant.
Medina-Bárcenas, Keskİn Tütüncü, Kuratomi 271
Proof. łOnly ifž part: By Theorem 1.
łIfž part: Suppose that M is not DSF. Then, by the assumption,
there exists an epimorphism ρ : M → S2 with S a simple module. Let
pi : S2 = S1 ⊕ S2 → Si (i = 1, 2) be the projections. Since M is H-
supplemented, there exist a decomposition M =M1⊕M2 and a submodule
K1 of M such that K1/M1 is small in M/M1 and K1/Kerp1ρ is small in
M/Kerp1ρ. ThenM = Kerp1ρ+M2. As Kerp1ρ is a maximal submodule of
M , Kerp1ρ = K1. Clearly, we have M = Kerp1ρ+Kerp2ρ =M1+Kerp2ρ.
Let p :M =M1⊕M2 →M2 be the projection. Now M2 = p(Kerp2ρ) and
since every maximal submodule of M is fully invariant, M2 ⊆ Kerp2ρ.
Let πi :M →M/Kerpiρ (i = 1, 2) be the natural epimorphisms. Then
πi|Mj
: Mj → M/Kerpiρ (i ≠ j) is onto. By [13, Proposition 2.4], M1 is
weakly generalized epi-M2-projective. Hence there exist decompositions
Mi = M ′
i ⊕M ′′
i and epimorphisms hi : M
′
i → M ′′
j (i, j = 1, 2, i ≠ j)
such that (π1|M2
)h1 = α(π2|M ′
1
) and α(π2|M1
)h2 = π1|M ′
2
, where α :
M/Kerp2ρ→ S1 → S2 →M/Kerp1ρ is the natural isomorphism.
In the case thatM ′
1 is not contained in Kerp2ρ, then M =M ′
1+Kerp2ρ.
Deőne φ : M = M ′
1 ⊕M ′′
1 ⊕M2 → M ′′
2 by φ(m′
1 +m′′
1 +m2) = h1(m
′
1),
where m′
1 ∈ M ′
1, m
′′
1 ∈ M ′′
1 and m2 ∈ M2. Since φ(Kerp1ρ) ⊆ Kerp1ρ
(every maximal submodule is fully invariant), for any m′
1 ∈M ′
1 −Kerp2ρ,
0 ̸= α(π2|M1
)(m′
1) = (π1|M2
)h1(m
′
1) = (π1|M2
)φ(m′
1) ∈ π1(Kerp1ρ) = 0,
a contradiction.
In the case of M ′
1 ⊆ Kerp2ρ, then 0 = απ2(M
′
1) = (π1|M2
)h1(M
′
1)
and so M ′′
2 = h1(M
′
1) ⊆ Kerp1ρ. Hence M ′
2 is not contained in Kerp1ρ.
Deőne ψ : M = M1 ⊕M ′
2 ⊕M ′′
2 → M ′′
1 by ψ(m1 +m′
2 +m′′
2) = h2(m
′
2),
where m1 ∈M1, m
′
2 ∈M ′
2 and m′′
2 ∈M ′′
2 . Since ψ(Kerp2ρ) ⊆ Kerp2ρ, for
any m′
2 ∈ M ′
2 − Kerp1ρ, 0 ̸= (π1|M2
)(m′
2) = απ2h2(m
′
2) = απ2ψ(m
′
2) ∈
απ2(Kerp2ρ) = 0, a contradiction.
Therefore M is DSF.
Example 5. Following the notation in Example 3, set M = E(S). Then
every maximal submodule of M is fully invariant, M is coatomic and
satisőes FIEP because it is uniform. Consider K ⩽ M in Example 3.
Note that the unique submodule X∗ of M satisfying X∗/K ≪ M/K is
K = X∗, and there is no direct summand A of M such that K/A≪M/A.
Thus, M is not H-supplemented neither DSF.
Example 6. Let A = B = Z(p∞) and put M = A⊕B. Let f : A→ X
be a nonzero homomorphism and g : B → X be an epimorphism for some
module X. If f is onto, then Kerf ⊆ Kerg or Kerg ⊆ Kerf since Z(p∞)
272 Dual Square Free Modules
is a uniserial module. Hence there exists an epimorphism h : A→ B such
that gh = f or an epimorphism h : B → A such that fh = g by Kerf and
Kerg is small in Z(p∞). If f is not onto, then f(A) is small in X since X
is hollow. Let ρ : X → X/f(A) be the natural epimorphism and let h′ = 0.
Then ρ is a small epimorphism and ρf = 0 = ρgh′. Thus A is weakly
generalized B-projective. By [13, Theorem 2.7], M is H-supplemented.
In addition, since M has no maximal submodules, it satisőes that every
maximal submodule of M is fully invariant. Moreover, M is injective and
so it satisőes the exchange property. Thus M is H-supplemented with the
exchange property. But it is not coatomic, not DSF.
Let UR be a module. U is called quasi-small if given a family of
modules {Uα : α ∈ Γ} such that U is isomorphic to a direct summand
of ⊕α∈ΓUα, there exists a őnite subset F ⊆ Γ such that U is isomorphic
to a direct summand of ⊕α∈FUα. Suppose that U is a uniserial module.
Then the endomorphism ring E = EndR(U) has two (two sided) ideals
L = {f ∈ E : f is not injective} and K = {f ∈ E : f is not surjective}
such that every proper right ideal of E is contained either in L or in K
(see [11, Proposition 3.7] and [7, Theorem 1.2]).
Now we give the following example which is important in terms of
existing of an H-supplemented module that does not satisfy FIEP.
Example 7. Let U be a uniserial right R-module which is not quasi-
small with E = EndR(U). Let K = {f ∈ E : f is not surjective} and
L = {f ∈ E : f is not injective} be two two-sided ideals of E. Consider
the subset {fn : n ∈ N∗} of E such that fn+1fn = fn for all n ∈ N∗
and K =
∑∞
n=1
fnE. Fix m ∈ N∗ such that fm ̸∈ KJ(E). By [12,
Example 3.3], the right E-module E/fmE is radical projective hollow
and EndE(E/fmE) is not local. By [13, Theorem 2.7] and [1, Proposition
12.10], E/fmE⊕E/fmE is an H-supplemented E-module which does not
satisfy FIEP.
In [9, Lemma 2.2] it is proved that a module M is distributive if and
only if every submodule of M is DSF. The next proposition adds new
equivalent conditions to that lemma and makes a connection with the
general distributivity presented in [2]. For, we introduce some terminology
from [2]. Given a cardinal ω, we write ω+ for the smallest cardinal larger
that ω. By crs(R) we denote the cardinality of all (non-isomorphic) simple
left R-modules and if M is a semisimple module, then dimM denotes the
cardinal number of simple summands of M . Let ω be a cardinal. It is
said that a module M is ω-thick provided dimS < ω for any semisimple
subfactor S of M .
Medina-Bárcenas, Keskİn Tütüncü, Kuratomi 273
Proposition 2. Let M be a module. Then the following are equivalent:
1) M is distributive.
2) X/Rad(X) is DSF for any submodule X of M .
3) M is crs(R)+-thick. (In particular M is crs(R)+-distributive)
Proof. (1)⇒(2) By Lemma 2.2 in [9], M is distributive if and only if any
factor module of M is square free if and only if any submodule of M is
DSF. Hence it is clear that, if M is distributive, then X/Rad(X) is DSF
for any submodule X of M .
(2)⇒(3) We claim that Soc(M/X) is square free for any submodule
X of M . Suppose that Soc(M/X) is not square free. Then there exists
a submodule N of Soc(M/X) such that N = S1 ⊕ S2, where S1 ∼= S2
is simple. Let π : M → M/X be the natural epimorphism and put
T = π−1(N), f = π |T . Then Kerpif is maximal in T , where pi : N =
S1 ⊕ S2 → Si is the projection. By Rad(T ) ⊆ Kerp1f ∩Kerp2f = Kerf ,
there exists an epimorphism from T/Rad(T ) to T/Kerf ∼= N = S1 ⊕ S2.
Since T/Rad(T ) is DSF,N = 0, a contradiction. Thus Soc(M/X) is square
free for any submodule X ofM . Now, let S be a semisimple subfactor ofM ,
that is, S = N/X for submodules N and X of M . Since S is semisimple,
S ⩽ Soc(M/X). By the claim above, the homogeneous components of S
have size one. This implies that dimS ⩽ crs(R) < crs(R)+. Thus, M is
crs(R)+-thick.
(3)⇒(1) Let N be a submodule of M . Suppose that there is an epi-
morphism ρ : N → L ⊕ L for some module L. If L is nonzero, there is
a semisimple subfactor S ⊕ S of N for some simple module S. It follows
from [2, Corollary 4.5(c)] that S ⊕ 0 is fully invariant in S ⊕ S which
cannot be. Thus, L = 0. This implies that every submodule of M is DSF.
Hence, M is distributive by Lemma 2.2 in [9].
2. Direct sums of dual square modules
Proposition 3. Let M =
⊕
i∈I Mi be a direct sum, such that M is
coatomic. Then M is DSF if and only if each Mi is DSF for all i ∈ I and,
Mi and
⊕
j ̸=iMj are dual orthogonal.
Proof. (⇒) It is clear.
(⇐) Let f : M → S1 ⊕ S2 be an epimorphism, with 0 ̸= S1 ∼= S2.
Since any nonzero factor module of M has a maximal submodule, we can
assume S1 is simple. Put N = S1 ⊕ S2.
Since f ≠ 0, there is i ∈ I such that f(Mi) ̸= 0. If f(Mi) ∩ Sj ̸= 0
for each j = 1, 2, then f(Mi) = N = S1 ⊕ S2, a contradiction. Then,
274 Dual Square Free Modules
suppose f(Mi) ∩ S2 = 0. Then f(Mi) ⊕ S2 = f(Mi) + f
(
⊕
j ̸=iMj
)
. If
this sum is not direct, then f(Mi) is a direct summand of f
(
⊕
j ̸=iMj
)
or f
(
⊕
j ̸=iMj
)
is a direct summand of f(Mi), which is a contradic-
tion because Mi and
⊕
j ̸=iMj are dual orthogonal. Therefore f(Mi) ⊕
S2 = f(Mi) ⊕ f
(
⊕
j ̸=iMj
)
. This implies that f(Mi) ∼= S1 ∼= S2 ∼=
f
(
⊕
j ̸=iMj
)
, a contradiction. Thus S1 = 0 = S2.
The proof of next lemma can be found in [8, Corollary 5], we write
the statement here for the convenience of the reader.
Lemma 1. Let M and N be two coatomic modules. Then M ⊕ N is
coatomic.
Corollary 1. Let M1, ...,Mn be coatomic modules and M =
⊕n
i=1
Mi.
Then M is DSF if and only if each Mi is DSF for all 1 ⩽ i ⩽ n and, Mi
and
⊕
j ̸=iMj are dual orthogonal.
The following corollary should be compared with Proposition 2.8
in [10].
Corollary 2. Let M = A⊕B where A is a őnitely generated DSF module
and B =
⊕
i∈I Si is a direct sum of non-isomorphic simple modules. Then
M is a DSF module if and only if A and B are dual orthogonal.
3. Endomorphism rings of dual square free modules
Next proposition is a generalization of [10, Example 2.5].
Proposition 4. If M is an endoregular quasi-projective module, then M
is abelian if and only if M is DSF.
Proof. (⇒) LetA andB proper submodules ofM such thatM = A+B and
M/A ∼= M/B. Since M is quasi-projective, EndR(M) = HomR(M,A) +
HomR(M,B). This implies that M = HomR(M,A)M +HomR(M,B)M .
Set A′ = HomR(M,A)M and B′ = HomR(M,B)M . Then M = A′ +B′
and there are epimorphisms ρ1 :M/A′ →M/A and ρ2 :M/B′ →M/B.
By [15, Proposition 2.9], A′ and B′ are fully invariant submodules of M .
Hence A′∩B′ is fully invariant. It follows thatM/A′∩B′ is quasi-projective
and by [15, Proposition 2.8], it is an abelian endoregular module. Moreover,
Medina-Bárcenas, Keskİn Tütüncü, Kuratomi 275
M/A′ ∩ B′ = (A′/A′ ∩ B′)⊕ (B′/A′ ∩ B′) with A′/A′ ∩ B′ ∼= M/B′ and
B′/A′ ∩B′ ∼=M/A′. By [15, Proposition 2.11(d)],
0 = HomR
(
A′/A′ ∩B′, B′/A′ ∩B′
)
= HomR(M/B′,M/A′).
SinceM/A′∩B′ is quasi-projective,M/B′ isM/A′-projective. This implies
that the following diagram can be completed commutatively only with
the zero homomorphism
M/B′ 0 //
ρ2
��
M/A′
ρ1
��
M/B ∼=
//M/A.
Thus ρ2 = 0, that is, M = B. Analogously M = A. Therefore, M is DSF.
(⇐) LetA andB beM -generated submodules ofM such thatA∩B = 0
and A ∼= B. There exists a nonzero homomorphism α : M → A. Then
there exists B′ ⩽ B such that α(M) ∼= B′ and α(M) ∩ B′ = 0. Hence,
without loss of generality we can take A = α(M) and B = B′. Since M is
endoregular, A and B are direct summands of M . The module M satisőes
SSP (sum of any two direct summands of M is again a direct summand
of M) by [14, Theorem 2.4]. Then A ⊕ B is a direct summand of M ,
that is, there exists, D ⩽M such that M = A⊕B ⊕D. It follows that
M/(B ⊕D) ∼= A ∼= B ∼=M/(A⊕D). Since M is DSF, A⊕D = B ⊕D.
In particular, A ⊆ B ⊕D. Therefore, 0 = A ∩ (B ⊕D) = A. Analogously
B = 0. Hence M is abelian endoregular by [15, Proposition 2.11(c)].
Corollary 3. ([10, Example 2.5]) Let R be a ring. Then, R is strongly
regular if and only if R is a regular and left (right) DSF module.
Remark 1. Any DSF module M with SSP and SIP (intersection of any
two direct summands of M is again a direct summand of M) satisőes the
internal cancellation property. For, let M be a DSF module with SSP and
SIP. Let M = A ⊕ C = B ⊕ D and let f : A → B be an isomorphism.
Since M satisőes SIP, there exist decompositions A = (A ∩D)⊕A′ and
D = (A ∩D)⊕D′. Then B = f(A ∩D)⊕ f(A′). Hence M = B ⊕D =
f(A∩D)⊕(A∩D)⊕f(A′)⊕D′. Since M is DSF with SSP, A∩D = 0 and
A⊕D is a direct summand of M . Put M = (A⊕D)⊕K. By M = B⊕D,
there exists a decomposition B = B′ ⊕ B′′ such that K ∼= B′. Hence
M = A⊕D⊕K = f−1(B′)⊕ f−1(B′′)⊕D⊕K and f−1(B′) ∼= B′ ∼= K.
276 Dual Square Free Modules
Since M is DSF, B′ = 0 and hence K = 0. Thus we see M = A⊕D and
so C ∼= D. This means that M satisőes the internal cancellation property.
Now by [14, Theorem 2.4], any endoregular module satisőes the SSP
and SIP properties. Hence any DSF endoregular module satisőes the
internal cancellation property.
Theorem 2. Let M be a quasi-projective module. Consider the following
conditions:
1) S = EndR(M) is a right DSF ring.
2) M is a DSF module.
Then (1) ⇒ (2). In addition, if M is őnitely generated, then (2) ⇒ (1).
Proof. (1) ⇒ (2): Let A and B submodules of M such that M = A+B
with M/A ∼= M/B. Since M is quasi-projective, S = HomR(M,A) +
HomR(M,B). Let f : M/A → M/B be an isomorphism, and let πA :
M →M/A and πB :M →M/B be the natural epimorphisms.
M
g
πA //M/A
f
M
j
HH
πB
//M/B
f−1
II
Since M is quasi-projective, there exists g ∈ S such that πBg =
fπA. It follows that g(A) ⊆ B. We claim that θ : S/HomR(M,A) →
S/HomR(M,B) given by θ(1 + HomR(M,A)) = g +HomR(M,B) is an
isomorphism of right S-modules. Let h ∈ HomR(M,A). Then gh(M) ⊆
g(A) ⊆ B. Thus gh ∈ HomR(M,B) and so θ is well deőned. It is clear that
θ is an S-homomorphism of right S-modules. Now, if h + HomR(M,A)
is such that θ(h + HomR(M,A)) = 0, that is, gh ∈ HomR(M,B), then
0 = πBgh = fπAh. Since f is an isomorphism, πAh = 0. This implies that
h(M) ⊆ A and hence h ∈ HomR(M,A). Thus, θ is a monomorphism. If we
take now f−1 :M/B →M/A, there exists j ∈ S such that πAj = f−1πB
because M is quasi-projective. Note that πB = ff−1πB = fπAj = πBgj.
Let l +HomR(M,B) ∈ S/HomR(M,B). Then,
πB(gjl − l) = πBgjl − πBl = πBl − πBl = 0.
It follows that gjl − l ∈ HomR(M,B). Therefore, θ(jl +HomR(M,A)) =
gjl+HomR(M,B) = l+HomR(M,B). Hence θ is an epimorphism and so
an isomorphism. Since S is right DSF, HomR(M,A) = S = HomR(M,B).
Hence M = SM = HomR(M,A)M ⊆ A. Analogously, B =M . Thus, M
is DSF.
Medina-Bárcenas, Keskİn Tütüncü, Kuratomi 277
Now, assume that M is a őnitely generated quasi-projective module.
(2) ⇒ (1): Let I and J be right ideals of S such that S = I + J and
S/I ∼= S/J . It follows that M = IM + JM . Let θ : S/I → S/J be
an isomorphism of right S-modules. Write h + J = θ(1 + I). Let f :
M/IM → M/JM be given by f(m + IM) = h(m) + JM . Suppose
m − n ∈ IM . Therefore m − n =
∑ℓ
i=1
gi(ki) with gi ∈ I and ki ∈ M .
Then, 0 = θ(gi + I) = θ(1 + I)gi = (h+ J)gi = hgi + J , for all 1 ⩽ i ⩽ ℓ.
Thus, hgi ∈ J for all 1 ⩽ i ⩽ ℓ. Hence,
h(m− n) = h(
∑
gi(ki)) =
∑
hgi(ki) ∈ JM.
It is clear that f is an R-homomorphism. Analogously, if θ−1(1+J) = h′+I
then we have an R-homomorphism g :M/JM →M/IM given by g(m+
JM) = h′(m)+IM . Note that 1+I = θ−1θ(1+I) = θ−1(h+J) = h′h+I,
hence 1− h′h ∈ I. Analogously, 1− hh′ ∈ J . Now, let m+ IM ∈M/IM .
Then
gf(m+ IM) = g(h(m) + JM) = h′h(m) + IM = m+ IM,
and
fg(m+ JM) = f(h′(m) + IM) = hh′(m) + JM = m+ JM.
It follows thatM/IM ∼=M/JM . Since M is DSF, IM =M = JM and so
S = HomR(M, IM) = HomR(M,JM). We have that I = HomR(M, IM)
and J = HomR(M,JM) because M is őnitely generated and quasi-
projective [17, 18.4]. Thus, I = S = J , that is, S is right DSF.
Example 8. Let K be a őeld and A a hereditary K-algebra. Let P be an
indecomposable projective A-module. By [15, Example 2.2(v)] we know
that, EndA(P ) ∼= K. Therefore EndA(P ) is a right (and left) DSF ring.
Then from the above Theorem, P is a DSF A-module.
Example 9. (see [5, Example 2.3]) Let K be a őeld and let A be the
hereditary K-algebra given by the quiver •
1
•
3
•
2
. Let M
be the indecomposable injective left A-module I(3) =
1 2
3
. Here the
indecomposable projective non-isomorphic left A-modules are P (1) =
1
3
,
278 Dual Square Free Modules
P (2) =
2
3
and P (3) = 3. Then the lattice of submodules of M is
M
P (1) P (2)
P (3)
0
Note that the projective left A-module P = P (1)⊕ P (2) is the projective
cover of M . Since M is őnitely generated, P is őnitely generated. On the
other hand, EndA(P ) ∼= K⊕K. Therefore EndA(P ) is not a left (and right)
DSF ring. Hence P is not DSF left A-module by Theorem 2. But from
Example 8, each P (1) and P (2) is DSF. Clearly, M = P (1) + P (2) and
M/P (1) ≇M/P (2), where P (1) and P (2) are the only proper submodules
of M with sum M . Therefore M is clearly a DSF left A-module. Also
EndA(M) is a left (right) DSF ring since EndA(M) ∼= K.
As we see in the following example, in Theorem 2 (1) ⇒ (2), łquasi-
projectivež hypothesis is not superŕuous.
Example 10. Consider the module E(S) in Example 3. E(S) is not
quasi-projective and it is not DSF. On the other hand, since E(S) is
indecomposable injective, EndR(E(S)) is local, which is a right and left
DSF ring.
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Contact information
Mauricio
Medina-Bárcenas
Facultad de Ciencias Físico-Matemáticas,
Benemérita Universidad Autónoma de Puebla,
Av. San Claudioy 18 Sur, Col.San Manuel,
Ciudad Universitaria, 72570, Puebla, México
E-Mail(s): mmedina@fcfm.buap.mx
Derya Keskin
Tütüncü
Department of Mathematics, Hacettepe
University, 06800, Beytepe, Ankara, Turkey
E-Mail(s): keskin@hacettepe.edu.tr
Yosuke Kuratomi Department of Mathematics, Faculty of Science,
Yamaguchi University, Yamaguchi, Japan
E-Mail(s): kuratomi@yamaguchi-u.ac.jp
Received by the editors: 11.12.2019
and in őnal form 04.02.2021.
mailto:mmedina@fcfm.buap.mx
mailto:keskin@hacettepe.edu.tr
mailto:kuratomi@yamaguchi-u.ac.jp
Medina-Bárcenas, Keskİn Tütüncü, Kuratomi
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