On a class of dual Rickart modules
UDC 512.5 Let $R$ be a ring and let $\Omega_R$ be the set of maximal right ideals of $R$. An $R$-module $M$ is called an sd-Rickart module if for every nonzero endomorphism $ f$ of $M$, $\Im f$ is a fully invariant direct summand of $M$. We obtain a characterization for an arbitrary direct sum of sd...
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Institute of Mathematics, NAS of Ukraine
2020
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860512222155898880 |
|---|---|
| author | Tribak, R. Tribak, R. |
| author_facet | Tribak, R. Tribak, R. |
| author_sort | Tribak, R. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2022-03-26T11:01:56Z |
| description | UDC 512.5
Let $R$ be a ring and let $\Omega_R$ be the set of maximal right ideals of $R$. An $R$-module $M$ is called an sd-Rickart module if for every nonzero endomorphism $ f$ of $M$, $\Im f$ is a fully invariant direct summand of $M$. We obtain a characterization for an arbitrary direct sum of sd-Rickart modules to be sd-Rickart. We also obtain a decomposition of an sd-Rickart $R$-module $M$, provided $R$ is a commutative noetherian ring and $Ass(M) \cap \Omega_R$ is a finite set. In addition, we introduce and study ageneralization of sd-Rickart modules. |
| doi_str_mv | 10.37863/umzh.v72i7.6021 |
| first_indexed | 2026-03-24T03:25:21Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v72i7. 6021
UDC 512.5
R. Tribak (Centre Régional des Métiers de l’Education et de la Formation (CRMEF-TTH), Tanger, Morocco)
ON A CLASS OF DUAL RICKART MODULES
ПРО КЛАС ДУАЛЬНИХ МОДУЛIВ РIКАРТА
Let R be a ring and let \Omega R be the set of maximal right ideals of R. An R-module M is called an sd-Rickart module if for
every nonzero endomorphism f of M , Imf is a fully invariant direct summand of M . We obtain a characterization for an
arbitrary direct sum of sd-Rickart modules to be sd-Rickart. We also obtain a decomposition of an sd-Rickart R-module
M , provided R is a commutative noetherian ring and \mathrm{A}\mathrm{s}\mathrm{s}(M) \cap \Omega R is a finite set. In addition, we introduce and study a
generalization of sd-Rickart modules.
Нехай R — кiльце, а \Omega R — множина максимальних правих iдеалiв R. R-модуль M називається sd-модулем Рiкарта,
якщо для будь-якого ненульового ендоморфiзму f на M множина Imf є повнiстю iнварiантним прямим доданком
M . Отримано характеризацiю умов, за яких довiльна пряма сума sd-модулiв Рiкарта є sd-модулем Рiкарта. Також
отримано розклад R-модуля M, що є sd-модулем Рiкарта, за умови, що R — комутативне нетерове кiльце та
\mathrm{A}\mathrm{s}\mathrm{s}(M) \cap \Omega R є скiнченною множиною. На додаток визначаються та вивчаються узагальнення sd-модулiв Рiкарта.
1. Introduction. Throughout this article, all rings are associative and have an identity. All modules
are unital right modules. R is a ring, \Omega R is the set of its maximal right ideals and M is an R-module.
We use J(R), \mathrm{E}\mathrm{n}\mathrm{d}R(M), and E(M) to denote the Jacobson radical of R, the endomorphism ring
of M, and the injective hull of M, respectively. If R is a commutative Noetherian ring, then \mathrm{A}\mathrm{s}\mathrm{s}(M)
will denote the set of all prime ideals associated to M. By \BbbQ and \BbbZ we denote the ring of rational
and integer numbers, respectively. If p is a prime number, then \BbbZ p\infty will denote the Prüfer p-group.
Recently, the notion of dual Baer modules was introduced and studied in [8]. A module M is said to
be dual Baer if for every submodule N of M, there exists an idempotent e \in S = \mathrm{E}\mathrm{n}\mathrm{d}R(M) such
that D(N) = eS, where D(N) = \{ \varphi \in S | \mathrm{I}\mathrm{m}\varphi \subseteq N\} . In [9], Lee, Rizvi and Roman provided
some motivations to study the concept of a dual Rickart module which is a related concept to that of
a dual Baer module. A module M is called dual Rickart (or d-Rickart) if \mathrm{I}\mathrm{m} f is a direct summand
of M for every f \in \mathrm{E}\mathrm{n}\mathrm{d}R(M). A submodule N of M is called fully invariant if f(N) is contained
in N for every R-endomorphism f of M. In [3], Călugăreanu and Schultz introduced and studied
the notion of stable modules. A module M is said to be stable if all endomorphic images are fully
invariant. Clearly, if M is an R-module such that \mathrm{E}\mathrm{n}\mathrm{d}R(M) is a commutative ring, then M is
a stable module. Abelian groups whose endomorphism ring is commutative was characterized in
[13, 14]. It is of interest to investigate the intersection \scrC of the class of dual Rickart modules and
the class of stable modules. Any element in \scrC will be called a strongly dual Rickart module. So, an
R-module M is strongly dual Rickart if and only if the image of any single element of \mathrm{E}\mathrm{n}\mathrm{d}R(M)
is a fully invariant direct summand of M.
A module M is said to have the (D2) condition if for every submodule N of M for which M/N
is isomorphic to a direct summand of M, N is a direct summand of M. In Section 2, we study the
notion of sd-Rickart modules. We show that this notion is distinct from that of d-Rickart modules.
It is shown that the class of sd-Rickart modules is precisely the class of d-Rickart modules M for
which every direct summand of M is fully invariant in M. Also, we characterize an sd-Rickart
module having the (D2) condition as the one which has a strongly regular ring as its endomorphism
ring. In particular, the R-module RR is sd-Rickart if and only if R is a strongly regular ring. Then we
c\bigcirc R. TRIBAK, 2020
960 ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 7
ON A CLASS OF DUAL RICKART MODULES 961
study the question of when is the direct sum of sd-Rickart modules, sd-Rickart? We first show that the
class of sd-Rickart modules is not closed under finite direct sums. Then we obtain a characterization
for an arbitrary direct sum of sd-Rickart modules to be sd-Rickart. We also prove that if R is a
commutative Noetherian ring and M is an R-module with \mathrm{A}\mathrm{s}\mathrm{s}(M)\cap \Omega R is a finite set, then M is an
sd-Rickart module if and only if M \sim = N\oplus
\bigl(
\oplus k
i=1R/Pi
\bigr)
for some distinct maximal ideals P1, . . . , Pk
of R and an sd-Rickart module N such that \mathrm{R}\mathrm{a}\mathrm{d}(N) = N and \mathrm{H}\mathrm{o}\mathrm{m}R(\oplus k
i=1R/Pi, N) = 0.
In Section 3, we introduce the concept of a generalized strongly dual Rickart module. A module
M is called a generalized strongly dual Rickart (gsd-Rickart for short) module if for every nonzero
endomorphism f of M, \mathrm{I}\mathrm{m} f contains a nonzero fully invariant direct summand of M. A result
(Proposition 3.1) and an example (Example 3.1) are provided to show that the concept of a gsd-
Rickart module is distinct from that of an sd-Rickart module. Then we show that there are many
similarities between the two concepts. For example, we show that if a module M is a direct
sum of indecomposable submodules, then M is an sd-Rickart module if and only if M is a gsd-
Rickart module. Also, we determine the structure of injective gsd-Rickart modules over commutative
Noetherian hereditary rings. We conclude the paper by describing the structure of sd-Rickart modules
and gsd-Rickart modules over discrete valuation rings.
2. Strongly dual Rickart modules.
Definition 2.1. A module M is called a strongly dual Rickart (or an sd-Rickart) module if for
every nonzero endomorphism f of M, \mathrm{I}\mathrm{m} f is a fully invariant direct summand of M.
A ring R is said to be abelian (or normal) if all its idempotents are central elements (see [4]
(15.31) and [15, p. 37]). Recall that a ring R is called strongly regular if for each x \in R, there exists
y \in R such that x2y = x. This is equivalent to the condition that R is an abelian von Neumann
regular ring (see [7], Theorem 3.5).
The next result will be of interest.
Proposition 2.1. Every module which has a strongly regular endomorphism ring, is sd-Rickart.
Proof. Let M be an R-module such that S = \mathrm{E}\mathrm{n}\mathrm{d}R(M) is a strongly regular ring. Let \varphi \in S.
By [15] (Proposition 7.9), there exists 0 \not = e2 = e \in S such that \mathrm{I}\mathrm{m}\varphi = e(M). Since the idempotent
e is a central element in S, e(M) is a fully invariant submodule of M.
Example 2.1. (i) It follows from Proposition 2.1 that every indecomposable nonsingular injective
module is sd-Rickart (see [1], Lemma 25.4, and [7], Corollary 1.23). In particular, for any nonsingular
uniform module M, E(M) is an sd-Rickart module. So, the \BbbZ -module \BbbQ is an sd-Rickart \BbbZ -module.
(ii) It easy to see that every module, for which all nonzero endomorphisms are epimorphisms,
is sd-Rickart. Thus every indecomposable d-Rickart module is sd-Rickart (see [8], Corollary 2.2,
and [9], Corollary 4.3). In particular, for any prime integer p, the \BbbZ -modules \BbbZ p\infty and \BbbZ /p\BbbZ are
sd-Rickart.
Let e = e2 \in R. Then e is called a left semicentral idempotent if xe = exe for all x \in R.
Equivalently, eR is a two-sided ideal of R (see [2], Lemma 2.1). Recall that a module M is called
weak duo if every direct summand of M is fully invariant (see [11]). The next result presents other
ways of stating an “sd-Rickart module”.
Proposition 2.2. The following statements are equivalent for a module M :
(i) M is an sd-Rickart module;
(ii) M is a d-Rickart weak duo module;
(iii) M is a d-Rickart module and every idempotent of \mathrm{E}\mathrm{n}\mathrm{d}R(M) is left semicentral.
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962 R. TRIBAK
Proof. (i) \Rightarrow (ii) Clearly, M is a d-Rickart module. Let K be a direct summand of M and let \pi :
M \rightarrow K be the projection map. By hypothesis, \mathrm{I}\mathrm{m}\pi = K is fully invariant in M. Therefore M is
a weak duo module.
(ii) \Rightarrow (i) This is immediate.
(ii) \leftrightarrow (iii) This follows from [1] (Corollary 5.8) and [2] (Lemma 1.9(ii)).
Corollary 2.1. Every direct summand of an sd-Rickart module M is sd-Rickart.
Proof. This follows easily from Proposition 2.2, Propositions 2.8 [9] and 1.8 [11].
Consider the following condition:
(D2) If N is a submodule of M such that M/N is isomorphic to a direct summand of M, then
N is a direct summand of M.
It is well known that every projective module has the condition (D2). The following example
shows that, in general, an sd-Rickart module need not have the condition (D2).
Example 2.2. Consider the \BbbZ -module M = \BbbZ p\infty , where p is a prime number. Let L be any
nonzero proper submodule of M. Then M/L \sim =M, but L is not a direct summand of M. Therefore,
M does not satisfy (D2). On the other hand, M is an sd-Rickart module.
The next proposition characterizes sd-Rickart modules satisfying (D2).
Proposition 2.3. The following statements are equivalent for an R-module M :
(i) M is an sd-Rickart module having the (D2) condition;
(ii) \mathrm{E}\mathrm{n}\mathrm{d}R(M) is a strongly regular ring.
Proof. (i) \Rightarrow (ii) By [9] (Theorem 3.8), S = \mathrm{E}\mathrm{n}\mathrm{d}R(M) is a von Neumann regular ring. Let e
be an idempotent of S. By Proposition 2.2, e is left semicentral. Therefore eS is a two-sided ideal
of S by [2] (Lemma 2.1). So e is central by [7] (Lemma 3.1). Therefore, S is a strongly regular
ring.
(ii) \Rightarrow (i) This follows from Proposition 2.1 and Theorem 3.8 [9].
The next result characterizes the class of rings R for which the R-module RR is sd-Rickart.
Proposition 2.4. The following conditions are equivalent for a ring R:
(i) RR is an sd-Rickart R-module;
(ii) for every two-sided ideal I of R, IR is an sd-Rickart R-module;
(iii) R has a free sd-Rickart R-module F ;
(iv) R is a strongly regular ring.
Proof. (i) \Rightarrow (iv) This follows easily from Proposition 2.3 since the R-module RR satisfies (D2).
(iv) \Rightarrow (ii) Let I be a two-sided ideal of R. By [7] (Theorem 3.7), \mathrm{E}\mathrm{n}\mathrm{d}R(IR) is a strongly
regular ring. Therefore, IR is an sd-Rickart R-module by Proposition 2.3.
(ii) \Rightarrow (iii) This is obvious.
(iii) \Rightarrow (i) This follows from Corollary 2.1.
It is clear that every sd-Rickart module is d-Rickart. Next, we exhibit an example of a module
for which the converse is not true.
Example 2.3. Let R be a von Neumann regular ring which is not strongly regular. Then the
R-module RR is d-Rickart by [9] (Remark 2.2). On the other hand, the R-module RR is not sd-
Rickart by Proposition 2.4. As an example of a ring satisfying these conditions, we can consider the
endomorphism ring of a nonzero vector space V over a division ring such that the dimension of V
is different from 1.
The next result is taken from [11] (Lemma 1.9).
Lemma 2.1. Let a module M = M1 \oplus M2 be a direct sum of submodules M1 and M2. Then
M1 is a fully invariant submodule of M if and only if \mathrm{H}\mathrm{o}\mathrm{m}R(M1,M2) = 0.
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ON A CLASS OF DUAL RICKART MODULES 963
Next, we will be concerned with the study of direct sums of sd-Rickart modules. We begin with
a result which shows that a direct sum of sd-Rickart modules may not be sd-Rickart.
Proposition 2.5. For any nonzero module M, the module M \oplus M is not sd-Rickart.
Proof. This is a direct consequence of Proposition 2.2 and Lemma 2.1.
Remark 2.1. Let (P ) be a property of modules which is closed under finite direct sums (e.g.,
being injective or projective). Proposition 2.5 shows that it is not possible to characterize the class
of rings R for which every R-module satisfying (P ) is an sd-Rickart R-module.
The next result is a characterization for an arbitrary direct sum of sd-Rickart modules to be
sd-Rickart.
Theorem 2.1. Let a module M = \oplus i\in IMi be a direct sum of submodules Mi, i \in I. Then the
following statements are equivalent:
(i) M is an sd-Rickart module;
(ii) Mi is an sd-Rickart submodule of M for every i \in I and \mathrm{H}\mathrm{o}\mathrm{m}R(Mi,Mj) = 0 for all
distinct i, j \in I.
Proof. (i) \Rightarrow (ii) This follows from Proposition 2.2, Corollary 2.1 and Lemma 2.1.
(ii) \Rightarrow (i) By Lemma 2.1, Mi is a fully invariant submodule of M for every i \in I. Therefore
M is a d-Rickart module by [9] (Proposition 5.14). Let N be a direct summand of M. Then
there exists an epimorphism \psi \in \mathrm{E}\mathrm{n}\mathrm{d}R(M) such that N = \psi (M). Therefore, N =
\sum
i\in I
\psi (Mi).
Since \psi (Mi) \subseteq Mi for all i \in I, we have \psi (Mi) = N \cap Mi for all i \in I. This implies that
N = \oplus i\in I(N \cap Mi). Note that the modules Mi, i \in I, are weak duo modules by Proposition 2.2.
Applying [11] (Theorem 2.7), we see that M is a weak duo module. Consequently, M is an sd-
Rickart module (see Proposition 2.2).
It is well-known that every semisimple module is a d-Rickart module. The next result that is a
direct consequence of Theorem 2.1 and Lemma 2.1 provides more examples of d-Rickart modules
which are not sd-Rickart.
Corollary 2.2. The following statements are equivalent for a nonzero semisimple module M :
(i) M is an sd-Rickart module;
(ii) M = \oplus i\in ISi is a direct sum of simple submodules Si, i \in I, such that Si is not isomorphic
to Sj for all i \not = j in I.
Lemma 2.2. Let a module M =M1\oplus M2 be a direct sum of submodules M1 and M2 such that
\mathrm{R}\mathrm{a}\mathrm{d}(M1) = M1 and \mathrm{R}\mathrm{a}\mathrm{d}(M2) = 0. If \mathrm{H}\mathrm{o}\mathrm{m}R(M2,M1) = 0, then M1 and M2 are fully invariant
submodules of M.
Proof. It is clear that M1 = \mathrm{R}\mathrm{a}\mathrm{d}(M). Hence M1 is fully invariant in M by [1] (Proposi-
tion 9.14). Also, since \mathrm{H}\mathrm{o}\mathrm{m}R(M2,M1) = 0, M2 is fully invariant in M by Lemma 2.1.
Proposition 2.6. Let a module M =M1\oplus M2 be a direct sum of submodules M1 and M2 such
that \mathrm{R}\mathrm{a}\mathrm{d}(M1) =M1 and \mathrm{R}\mathrm{a}\mathrm{d}(M2) = 0. Then the following statements are equivalent:
(i) M is an sd-Rickart module;
(ii) M1 and M2 are sd-Rickart modules with \mathrm{H}\mathrm{o}\mathrm{m}R(M2,M1) = 0.
Proof. (i) \Rightarrow (ii) This follows from Theorem 2.1.
(ii) \Rightarrow (i) Let \varphi : M1 \rightarrow M2 be a homomorphism. Since \mathrm{I}\mathrm{m}\varphi \sim = M1/Ker\varphi , \mathrm{R}\mathrm{a}\mathrm{d}(\mathrm{I}\mathrm{m}\varphi ) =
= \mathrm{I}\mathrm{m}\varphi . But \mathrm{R}\mathrm{a}\mathrm{d}(M2) = 0. Then \varphi = 0. Therefore, \mathrm{H}\mathrm{o}\mathrm{m}R(M1,M2) = 0. The result follows from
Theorem 2.1 and Lemma 2.2.
Proposition 2.7. Let a module M =M1\oplus M2 be a direct sum of submodules M1 and M2 such
that \mathrm{R}\mathrm{a}\mathrm{d}(M1) = M1 and M2 is semisimple. Then M is a d-Rickart module if and only if M1 is a
d-Rickart module and \mathrm{H}\mathrm{o}\mathrm{m}R(M2,M1) = 0.
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964 R. TRIBAK
Proof. The sufficiency follows from Lemma 2.2 and [9] (Proposition 5.14). Conversely, suppose
that M is a d-Rickart module. By [9] (Proposition 2.8), M1 is a d-Rickart module. Now let
f : M2 \rightarrow M1 be a homomorphism. Let \pi 2 : M \rightarrow M2 denote the projection map and let \mu 1 :
M1 \rightarrow M denote the inclusion map. Then \mu 1f\pi 2 is an endomorphism of M. So \mathrm{I}\mathrm{m}(\mu 1f\pi 2) = \mathrm{I}\mathrm{m} f
is a direct summand of M1. This gives \mathrm{R}\mathrm{a}\mathrm{d}(\mathrm{I}\mathrm{m} f) = \mathrm{I}\mathrm{m} f. But \mathrm{I}\mathrm{m} f is a semisimple submodule of
M1. Then f = 0. It follows that \mathrm{H}\mathrm{o}\mathrm{m}R(M2,M1) = 0.
Remark 2.2. It is obvious from [9] (Theorem 2.29) that every indecomposable injective module
over a right hereditary ring, is an sd-Rickart module. In particular, for every prime ideal P of a
Dedekind domain R, E(R/P ) is an sd-Rickart R-module.
Proposition 2.8. Let R be a Dedekind domain. The following statements are equivalent for a
nonzero torsion R-module M :
(i) M is an sd-Rickart R-module;
(ii) there exist distinct maximal ideals Pi, i \in I, of R and submodules Mi, i \in I, of M such
that M = \oplus i\in IMi and for each i \in I, either Mi
\sim = E(R/Pi) or Mi
\sim = R/Pi.
Proof. (i) \Rightarrow (ii) By Proposition 2.2 and [11] (Theorem 3.10), there exist distinct maximal ideals
Pi, i \in I, of R and submodules Mi, i \in I, of M such that M = \oplus i\in IMi, and for each i \in I,
either Mi
\sim = E(R/Pi) or Mi
\sim = R/Pni
i for some positive integer ni. Since each Mi is an sd-Rickart
module (see Corollary 2.1), we have Mi
\sim = E(R/Pi) or Mi
\sim = R/Pi by [9] (Proposition 4.13).
(ii) \Rightarrow (i) Clearly, each Mi is the Pi-primary component of M. So each Mi is a fully invariant
submodule of M. In addition, we have \mathrm{H}\mathrm{o}\mathrm{m}R(Mi,Mj) = 0 for all distinct i, j in I. Also, note
that each Mi is an sd-Rickart module (see Remark 2.2). Therefore, M is an sd-Rickart module by
Theorem 2.1.
Lemma 2.3. Let I be a finitely generated ideal of a commutative ring R. If M is a d-Rickart
R-module, then MI is a direct summand of M.
Proof. By assumption, we have I = \alpha 1R+ . . .+\alpha kR for some positive integer k and elements
\alpha i, 1 \leq i \leq k, of R. Fix 1 \leq i \leq k and consider the homomorphism fi : M \rightarrow M defined
by fi(x) = x\alpha i for all x \in M. It is clear that \mathrm{I}\mathrm{m} fi = M\alpha i. Since M is a d-Rickart module,
M\alpha i is a direct summand of M. In addition, M has the SSP by [9] (Proposition 2.11). Then
MI =
\sum k
i=1
M\alpha i is a direct summand of M.
A module M is said to be radical if \mathrm{R}\mathrm{a}\mathrm{d}(M) = M. The sum of all radical submodules of a
module M will be denoted by P (M). A module M is said to be reduced if P (M) = 0.
Theorem 2.2. Let R be a commutative Noetherian ring and let \Omega R be the set of maximal ideals
of R. Let M be an R-module such that \mathrm{A}\mathrm{s}\mathrm{s}(M) \cap \Omega R is a finite set. Then the following conditions
are equivalent:
(i) M is a d-Rickart R-module;
(ii) M = M1 \oplus M2 is a direct sum of a d-Rickart submodule M1 and a semisimple submodule
M2 such that M1 = \mathrm{R}\mathrm{a}\mathrm{d}(M1) = \mathrm{R}\mathrm{a}\mathrm{d}(M) = P (M) and \mathrm{H}\mathrm{o}\mathrm{m}R(M2,M1) = 0.
Proof. (i) \Rightarrow (ii) Without loss of generality we can assume that \mathrm{R}\mathrm{a}\mathrm{d}(M) \not = M. Note that
\mathrm{R}\mathrm{a}\mathrm{d}(M) =
\bigcap
m\in \Omega R
Mm by [5] (Lemma 3). Then there exists a maximal ideal m1 of R such
that Mm1 \not = M. Since R is Noetherian, the ideal m1 is finitely generated. Therefore, Mm1 is a
direct summand of M by Lemma 2.3. Let N1 be a submodule of M such that M = Mm1 \oplus N1.
Clearly, N1m1 = 0. So N1 is semisimple. As N1 \not = 0, N1
\sim = (R/m1)
\Lambda 1 for some index set
\Lambda 1. Assume that \mathrm{R}\mathrm{a}\mathrm{d}(Mm1) \not = Mm1. Then there exists a maximal ideal m2 of R such that
Mm1m2 \not = Mm1 (see [5], Lemma 3). Note that m1 \not = m2 since Mm2
1 = Mm1. Moreover,
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ON A CLASS OF DUAL RICKART MODULES 965
Mm1 is a d-Rickart module by [9] (Proposition 2.8). By the same method as before, we conclude
that Mm1
\sim = Mm1m2 \oplus (R/m2)
\Lambda 2 for some index set \Lambda 2. Thus M \sim = Mm1m2 \oplus (R/m2)
\Lambda 2 \oplus
\oplus (R/m1)
\Lambda 1 . We continue in this way. Since \mathrm{A}\mathrm{s}\mathrm{s}(M) \cap \Omega R is a finite set, this process must
terminate. So there exist submodules M1 and M2 of M such that \mathrm{R}\mathrm{a}\mathrm{d}(M1) = M1 and M2 is
semisimple. By Proposition 2.7, M1 is a d-Rickart module and \mathrm{H}\mathrm{o}\mathrm{m}R(M2,M1) = 0. It is easily
seen that M1 = \mathrm{R}\mathrm{a}\mathrm{d}(M) = P (M).
(ii) \Rightarrow (i) This is immediate by Proposition 2.7.
Remark 2.3. From the proof of Theorem 2.2, it follows that if M is a nonzero d-Rickart module
over a commutative Noetherian ring with \mathrm{R}\mathrm{a}\mathrm{d}(M) \not =M, then M has a simple direct summand.
Combining Corollary 2.2, Proposition 2.6 and Theorem 2.2, we obtain the following result.
Corollary 2.3. Let R be a commutative Noetherian ring and \Omega R the set of maximal ideals of R.
Let M be a nonzero R-module such that \mathrm{A}\mathrm{s}\mathrm{s}(M)\cap \Omega R is a finite set. Then the following conditions
are equivalent:
(i) M is an sd-Rickart R-module;
(ii) M \sim = N \oplus (\oplus k
i=1R/Pi) for some distinct maximal ideals P1, . . . , Pk of R and an sd-Rickart
module N such that \mathrm{R}\mathrm{a}\mathrm{d}(N) = N and \mathrm{H}\mathrm{o}\mathrm{m}R(\oplus k
i=1R/Pi, N) = 0.
The next example shows that Theorem 2.2 and Corollary 2.3 are not true if the condition
”\mathrm{A}\mathrm{s}\mathrm{s}(M) \cap \Omega R is a finite set” is deleted from their hypotheses.
Example 2.4. Consider the reduced \BbbZ -module M =
\prod
p prime
\BbbZ /p\BbbZ . As
\mathrm{E}\mathrm{n}\mathrm{d}\BbbZ (M) \sim =
\prod
p prime
\BbbZ /p\BbbZ
is a strongly regular ring, M is an sd-Rickart module (see Proposition 2.1). On the other hand, it is
clear that the module M is not semisimple.
Corollary 2.4. Assume that R is a Dedekind domain with quotient field K. Let M be a nonzero
R-module such that \mathrm{A}\mathrm{s}\mathrm{s}(M) is a finite set. Then the following statements are equivalent:
(i) M is an sd-Rickart R-module;
(ii) M \sim = K \oplus (\oplus k
i=1R/Pi) for some distinct maximal ideals P1, . . . , Pk of R,
or M \sim =
\bigl(
\oplus n
i=1 E(R/Pi)
\bigr)
\oplus (\oplus m
j=n+1R/Pj), where Pi, 1 \leq i \leq m, are distinct maximal ide-
als of R.
Proof. This follows from Theorem 2.1, Proposition 2.8 and Corollary 2.3 and the fact that an
R-module M is injective if and only if \mathrm{R}\mathrm{a}\mathrm{d}(M) =M.
Next, we show that torsion-free sd-Rickart modules over a commutative domain are injective.
Proposition 2.9. Let M be a torsion-free module over a commutative domain R. If M is an
sd-Rickart R-module, then M is injective.
Proof. Assume that M is a d-Rickart R-module. We only need to show that M is a divisible
R-module (see [12], Proposition 2.7). Suppose that there exists a nonzero element r \in R such that
Mr \not =M. Consider the homomorphism f : M \rightarrow M defined by f(x) = xr for all x \in M. Since M
is d-Rickart, it follows that \mathrm{I}\mathrm{m} f = Mr is a direct summand of M. Let N be a nonzero submodule
of M such that Mr \oplus N =M. Then clearly, we have Nr = 0, a contradiction.
Example 2.5. (i) From Corollary 2.4, it follows that the \BbbZ -module \BbbQ is the only torsion-free
sd-Rickart \BbbZ -module. So the \BbbZ -module M = \BbbQ \oplus \BbbQ is a torsion-free injective d-Rickart \BbbZ -module
which is not sd-Rickart (see [9], Theorem 2.29).
The torsion submodule of an sd-Rickart R-module M over a commutative domain R is not, in
general, a direct summand of M as shown below.
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966 R. TRIBAK
Remark 2.4. Consider again the sd-Rickart \BbbZ -module M =
\prod
p prime
\BbbZ /p\BbbZ (see Example 2.4).
It is easy to see that the torsion submodule of M is T (M) = \oplus p prime\BbbZ /p\BbbZ and T (M) is not a
direct summand of M.
3. Strongly dual Rickart modules versus gsd-Rickart modules. In this section, we introduce
a generalization of sd-Rickart modules and then we compare the two notions.
Definition 3.1. A module M is called a generalized strongly dual Rickart (or a gsd-Rickart)
module if for every nonzero endomorphism f of M, \mathrm{I}\mathrm{m} f contains a nonzero fully invariant direct
summand of M.
Recall that a ring R is called semipotent (or an I0-ring) if every right ideal that is not contained
in J(R) contains a nonzero idempotent (see [10, p. 257; 15, p. 131, 132]). A ring without nonzero
nilpotent elements is called a reduced ring.
Proposition 3.1. The following conditions are equivalent for a ring R:
(i) RR is a gsd-Rickart R-module;
(ii) R is a reduced semipotent ring with J(R) = 0;
(iii) R is an abelian semipotent ring with J(R) = 0.
Proof. (i) \Rightarrow (ii) For every r \in R, let \varphi r : R \rightarrow R denote the R-endomorphism of the right
R-module RR defined by \varphi r(x) = rx for all x \in R. By (i), it follows that for every 0 \not = r \in R,
rR = \mathrm{I}\mathrm{m}\varphi r contains a nonzero direct summand of RR. Therefore, R is a semipotent ring with
J(R) = 0 (see also [15], Remark 15.3(3)). Now let 0 \not = a \in R. Then \mathrm{I}\mathrm{m}\varphi a = aR. By assumption,
there is b \in R such that 0 \not = ab = (ab)2 and (ab)R is a fully invariant submodule of RR (see [1],
Proposition 7.1). It is easily seen that (ab)R is a two-sided ideal of R. By [7] (Lemma 3.1), it
follows that ab is a central idempotent of R. Therefore, R is a reduced semipotent ring by [15]
(Remark 15.3(4)).
(ii) \Rightarrow (i) This follows from [15] (Remark 15.3(4)) and the fact that every endomorphism of RR
is a left multiplication by an element of R.
(ii) \leftrightarrow (iii) This follows from [15] (Remark 15.3(4)).
It is clear that every sd-Rickart module is gsd-Rickart. The next example shows that the class of
gsd-Rickart modules contains strictly the class of sd-Rickart modules.
Example 3.1. (i) From Propositions 2.4 and 3.1, it follows that if R is a commutative semipotent
ring such that J(R) = 0 and R is not von Neumann regular, then RR is a gsd-Rickart R-module
which is not sd-Rickart.
(ii) Consider the ring D =
\prod \infty
i=1
Ai, where each Ai = \BbbQ the field of rational numbers. Let R
be the subring of D generated by the ideal \oplus \infty
i=1Ai and by the subring \{ (x, x, x, . . .) | x \in \BbbZ \} . By
[15] (Example 15.7(9)), the ring R is a commutative semipotent ring such that J(R) = 0 and R is
not von Neumann regular. From (i), we conclude that RR is a gsd-Rickart R-module, but is not an
sd-Rickart R-module.
Note that if R is the ring given in Example 3.1(ii), then the R-module RR is gsd-Rickart, while
RR is not a d-Rickart R-module by [9] (Remark 2.2). The next example exhibits a d-Rickart module
which is not gsd-Rickart.
Example 3.2. Let R be a von Neumann regular ring which is not strongly regular (see Example
2.3). By [7] (Theorem 3.5), R is not an Abelian ring. So RR is not a gsd-Rickart R-module by
Proposition 3.1. However, it is not hard to see that RR is a d-Rickart R-module (see [9], Remark
2.2).
Proposition 3.2. If a module M is gsd-Rickart, then so are its direct summands.
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ON A CLASS OF DUAL RICKART MODULES 967
Proof. Let M = N \oplus L and let \varphi be an endomorphism of N. Consider the endomorphism
\psi = \mu \varphi \pi of M, where \pi : M \rightarrow N is the projection map and \mu : N \rightarrow M is the inclusion map. As
M is gsd-Rickart, there exists a fully invariant direct summand K of M such that K \subseteq \mathrm{I}\mathrm{m}\psi . Note
that \mathrm{I}\mathrm{m}\psi = \mathrm{I}\mathrm{m}\varphi . It follows that K is a fully invariant direct summand of N with K \subseteq \mathrm{I}\mathrm{m}\varphi .
Proposition 3.3. The following statements are equivalent for an indecomposable module M :
(i) M is an sd-Rickart module;
(ii) M is a d-Rickart module;
(iii) M is a gsd-Rickart module;
(iv) every endomorphism of M is an epimorphism.
Proof. The proof is immediate.
Proposition 3.4. Let M be a gsd-Rickart module. Then:
(i) Every indecomposable direct summand N of M is fully invariant in M.
(ii) If M = N \oplus L is a direct sum of an indecomposable submodule N and a submodule L,
then \mathrm{H}\mathrm{o}\mathrm{m}R(N,L) = 0.
(iii) If N and K are indecomposable direct summands of M such that N \cap K = 0 and N \oplus K
is a direct summand of M, then \mathrm{H}\mathrm{o}\mathrm{m}R(N,K) = \mathrm{H}\mathrm{o}\mathrm{m}R(K,N) = 0.
Proof. (i) Consider the projection map \pi : M \rightarrow N \leq M. Since M is gsd-Rickart, N = \mathrm{I}\mathrm{m}\pi
contains a nonzero fully invariant direct summand A of M. Clearly, A = N.
(ii) This follows from (i) and Lemma 2.1.
(iii) This follows from (ii).
The next corollary is a direct consequence of Proposition 3.4. This result and Example 3.3 show
that the gsd-Rickart property does not go to direct sums of gsd-Rickart modules.
Corollary 3.1. For any nonzero indecomposable module M, the module M \oplus M is not gsd-
Rickart.
Example 3.3. Let p be a prime integer. It is clear that the \BbbZ -module \BbbZ p\infty is gsd-Rickart, while
the \BbbZ -module \BbbZ p\infty \oplus \BbbZ p\infty is not gsd-Rickart by Corollary 3.1.
Remark 3.1. (i) Every ring R has an R-module which is not gsd-Rickart. In fact, if S is any
simple R-module, then the R-module S \oplus S is not gsd-Rickart (see Corollary 3.1).
(ii) Let (P ) be a property of modules which is closed under finite direct sums. Assume that there
is at least one indecomposable module which satisfies the property (P ). Corollary 3.1 shows that it
is not possible to characterize the class of rings R for which every R-module satisfying (P ) is a
gsd-Rickart module.
Proposition 3.5. Let a module M = \oplus i\in IMi be a direct sum of indecomposable submodules
Mi, i \in I. Then the following conditions are equivalent:
(i) M is an sd-Rickart module;
(ii) M is a gsd-Rickart module;
(iii) Mi is an sd-Rickart submodule of M for all i \in I, and \mathrm{H}\mathrm{o}\mathrm{m}R(Mi,Mj) = 0 for all distinct
i, j \in I.
Proof. (i) \Rightarrow (ii) is obvious.
(ii) \Rightarrow (iii) follows from Propositions 3.2, 3.3 and 3.4.
(iii) \Rightarrow (i) by Theorem 2.1.
Combining Proposition 3.5 with Corollary 2.2, we obtain the following result.
Corollary 3.2. The following conditions are equivalent for a nonzero semisimple module M :
(i) M is sd-Rickart;
(ii) M is gsd-Rickart;
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968 R. TRIBAK
(iii) M = \oplus i\in ISi is a direct sum of simple submodules Si, i \in I, such that Si is not isomorphic
to Sj whenever i \not = j.
Corollary 3.3. Let M be a nonzero module that has either the ascending or the descending chain
condition on direct summands. Then M is an sd-Rickart module if and only if M is gsd-Rickart.
Proof. This follows from Proposition 3.5 and [1] (Proposition 10.14).
Recall that a module M is said to have finite uniform dimension if M does not contain an infinite
independent set of submodules (see [4] (5.1) or [6, p. 79]).
A module M is said to have finite hollow dimension if for some n \in \BbbN , there exists an epimor-
phism from M to a direct sum of n nonzero modules but no epimorphism from M to a direct sum
of more than n nonzero modules (see [4] (5.2)).
Corollary 3.4. Let M be a nonzero module. If M satisfies one of the following conditions:
(i) M is Noetherian;
(ii) M is Artinian;
(iii) M has finite hollow dimension;
(iv) M has finite uniform dimension.
Then M is an sd-Rickart module if and only if M is a gsd-Rickart module.
Proof. This follows from Corollary 3.3, [4] (5.3) and [6] (Theorem 3.14).
Corollary 3.5. Let M be a nonzero Noetherian module over a commutative ring R. Then the
following statements are equivalent:
(i) M is an sd-Rickart R-module;
(ii) M is a gsd-Rickart R-module;
(iii) M \sim = \oplus k
i=1R/Pi for some positive integer k and distinct maximal ideals Pi, 1 \leq i \leq k,
of R.
Proof. (i) \leftrightarrow (ii) follows from Corollary 3.4.
(i) \leftrightarrow (iii) This is a consequence of [9] (Proposition 4.13) and Corollary 3.2.
Applying Corollary 3.2 to semisimple modules over local rings, we get the following example.
Example 3.4. (i) Let R be a local ring with maximal right ideal m. If M is a semisimple R-
module, then M is sd-Rickart if and only if M is gsd-Rickart if and only if M = 0 or M \sim = RR/m.
(ii) If R is a division ring, then an R-module M is sd-Rickart if and only if M is gsd-Rickart if
and only if M = 0 or M \sim = RR.
Proposition 3.6. Assume that R is a commutative Noetherian hereditary ring (e.g., R is a
Dedekind domain). Then the following conditions are equivalent for a nonzero injective R-module M :
(i) M is an sd-Rickart R-module;
(ii) M is a gsd-Rickart R-module;
(iii) there exist distinct prime ideals Pi, i \in I, of R and submodules Mi, i \in I, of M such that:
(a) M = \oplus i\in IMi,
(b) for each i \in I, Mi
\sim = E(R/Pi),
and
(c) Pi \nsubseteq Pj whenever i \not = j.
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ON A CLASS OF DUAL RICKART MODULES 969
Proof. (i) \leftrightarrow (ii) follows from [1] (Theorem 25.6) and Proposition 3.5.
(i) \Rightarrow (iii) follows from [1] (Theorem 25.6), [12] (Corollary on page 53 and Proposition 4.21)
and Proposition 3.5.
(iii) \Rightarrow (i) Note that each Mi, i \in I, is an sd-Rickart R-module by [12, p. 53] (Corollary) and
Remark 2.2. Moreover, applying [12] (Proposition 4.21), we conclude that \mathrm{H}\mathrm{o}\mathrm{m}R(Mi,Mj) = 0 for
all distinct i, j \in I. Using Theorem 2.1, we see that M is an sd-Rickart R-module.
The next consequence of Proposition 3.6 describes the structure of injective sd-Rickart \BbbZ -
modules.
Corollary 3.6. The following statements are equivalent for a nonzero injective \BbbZ -module M :
(i) M is an sd-Rickart \BbbZ -module;
(ii) M is a gsd-Rickart \BbbZ -module;
(iii) M \sim = \BbbQ or M = \oplus i\in I\BbbZ p\infty i
, where pi, i \in I, are distinct prime integers.
Proposition 3.7. Assume that R is a commutative Noetherian local ring with maximal ideal m.
Then every nonzero gsd-Rickart reduced R-module is a simple R-module.
Proof. Let M be a nonzero gsd-Rickart reduced R-module. Then Mm \not = M as \mathrm{R}\mathrm{a}\mathrm{d}(M) =
= Mm. Assume that Mm \not = 0. So there exists a nonzero element a \in m such that Ma \not = 0.
Consider the endomorphism \varphi of M defined \varphi (x) = xa for all x \in M. Since M is gsd-Rickart,
there exists a nonzero direct summand A of M such that A \subseteq Ma \subseteq Mm. Let B be a submodule of
M such that A\oplus B =M. Then Am\oplus Bm =Mm and A = A\cap Mm = Am\oplus (Bm\cap A) = Am.
Hence \mathrm{R}\mathrm{a}\mathrm{d}(A) = Am = A, a contradiction. It follows that Mm = 0 and M is semisimple.
Therefore, M \sim = R/m (see Example 3.4).
We conclude this paper by describing the structure of sd-Rickart modules and gsd-Rickart mod-
ules over discrete valuation rings.
Proposition 3.8. Let R be a discrete valuation ring with quotient field Q and maximal ideal m.
Then the following statements are equivalent for an R-module M :
(i) M is an sd-Rickart module;
(ii) M is a gsd-Rickart module;
(iii) M = 0 or M \sim = R/m or M \sim = Q or M \sim = Q/R or M \sim = Q\oplus R/m.
Proof. (i) \Rightarrow (ii) This is clear.
(ii) \Rightarrow (iii) Let M1 be the largest divisible submodule of M. Then there exists a reduced sub-
module M2 of M such that M = M1 \oplus M2. By Propositions 3.2 and 3.6, M1 = 0 or M1
\sim = Q
or M1
\sim = Q/R. Now applying Propositions 3.2 and 3.7, we conclude that M2 = 0 or M2
\sim = R/m.
Therefore, M = 0 or M \sim = R/m or M \sim = Q or M \sim = Q/R or M \sim = Q\oplus R/m by Proposition 3.5.
(iii) \Rightarrow (i) This follows from Proposition 2.6 and Remark 2.2.
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Received 26.05.17
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| id | umjimathkievua-article-6021 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T03:25:21Z |
| publishDate | 2020 |
| publisher | Institute of Mathematics, NAS of Ukraine |
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| resource_txt_mv | umjimathkievua/cd/561ec4e2d9add984304c4d53a56dc3cd.pdf |
| spelling | umjimathkievua-article-60212022-03-26T11:01:56Z On a class of dual Rickart modules On a class of dual Rickart modules Tribak, R. Tribak, R. UDC 512.5 Let $R$ be a ring and let $\Omega_R$ be the set of maximal right ideals of $R$. An $R$-module $M$ is called an sd-Rickart module if for every nonzero endomorphism $ f$ of $M$, $\Im f$ is a fully invariant direct summand of $M$. We obtain a characterization for an arbitrary direct sum of sd-Rickart modules to be sd-Rickart. We also obtain a decomposition of an sd-Rickart $R$-module $M$, provided $R$ is a commutative noetherian ring and $Ass(M) \cap \Omega_R$ is a finite set. In addition, we introduce and study ageneralization of sd-Rickart modules. УДК 512.5 Про клас дуальних модулiв Рiкарта Нехай $R$ є кiльце, а $\Omega_R$ є множиною максимальних правих iдеалiв $R$. $R$-модуль $M$ називається sd-модулем Рiкарта, якщо для будь-якого ненульового ендоморфiзму $f$ на $M$ множина $\Im f$ є повнiстю iнварiантним прямим доданком $M$. Отримано характеризацiю умов, за яких довiльна пряма сума sd-модулiв Рiкарта є sd-модулем Рiкарта. Також отримано розклад $R$-модуля $M$, що є sd-модулем Рiкарта, за умови, що $R$ є комутативним ньотеровим кiльцем та$Ass(M) \cap \Omega_R$ є скiнченною множиною. На додаток визначаються та вивчаються узагальнення sd-модулiв Рiкарта. Institute of Mathematics, NAS of Ukraine 2020-07-15 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6021 10.37863/umzh.v72i7.6021 Ukrains’kyi Matematychnyi Zhurnal; Vol. 72 No. 7 (2020); 960-970 Український математичний журнал; Том 72 № 7 (2020); 960-970 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6021/8732 |
| spellingShingle | Tribak, R. Tribak, R. On a class of dual Rickart modules |
| title | On a class of dual Rickart modules |
| title_alt | On a class of dual Rickart modules |
| title_full | On a class of dual Rickart modules |
| title_fullStr | On a class of dual Rickart modules |
| title_full_unstemmed | On a class of dual Rickart modules |
| title_short | On a class of dual Rickart modules |
| title_sort | on a class of dual rickart modules |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6021 |
| work_keys_str_mv | AT tribakr onaclassofdualrickartmodules AT tribakr onaclassofdualrickartmodules |