Semiprime submodules of a module and related concepts

Abstract

In this paper, we consider the notion of semiprime submodules, which is a natural generation of semiprime ideals. The main purpose of this paper is to give local-global properties of semiprimes and semiprime radicals as well as results on finitely generated multiplication modules. Also, we prove that for every submodule N of a module M, the semiprime radical of N in M is the direct limit of the ascending chain EM(N) ⊆ EM(EM(N)) ⊆⋯ of submodules of M, and we prove that N satisfies the semiprime radical formula if and only if its envelope EM(N) in M is semiprime.

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Journal of Algebra and Its Applications
(2019) 1950147 (11 pages)
c
World Scientific Publishing Company
DOI: 10.1142/S0219498819501470
Semiprime submodules of a module and related concepts
Sang Cheol Lee
Department of Mathematics Education
Institute of Pure and Applied Mathematics
Chonbuk National University, Jeonju
Jeonbuk 54896, South Korea
scl@jbnu.ac.kr
Rezvan Varmazyar∗
Department of Mathematics, Khoy Branch
Islamic Azad University, Khoy 58168-44799, Iran
varmazyar@iaukhoy.ac.ir
Received 6 February 2018
Accepted 16 July 2018
Published 29 August 2018
Communicated by E. Gorla
In this paper, we consider the notion of semiprime submodules, which is a natural
generation of semiprime ideals. The main purpose of this paper is to give local-global
properties of semiprimes and semiprime radicals as well as results on finitely generated
multiplication modules. Also, we prove that for every submodule Nof a module M,
the semiprime radical of Nin Mis the direct limit of the ascending chain EM(N)⊆
EM(EM(N)) ⊆···of submodules of M, and we prove that Nsatisfies the semiprime
radical formula if and only if its envelope EM(N)inMis semiprime.
Keywords: Semiprime submodules; weakly semiprime submodules; semiprime radical
formula.
Mathematics Subject Classification: 13C05, 18E40, 13B30, 16D60, 13B25
1. Introduction and Preliminaries
Throughout this paper, every ring is a commutative ring with identity and every
module is a unitary module over a ring.
For a submodule Nof an R-module M, let’s consider the four conditions as
follows:
(1) For a∈R, m ∈Mwith am ∈N,eithera∈(N:RM)orm∈N.
(2) For a, b ∈Rand m∈Mwith (ab)m∈N,eitheram ∈Nor bm ∈N.
∗Corresponding author.
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(3) For each a∈R,m∈Mwith a2m∈N,am ∈N.
(4) For each a∈Rwith a2M⊆N,aM ⊆N.
A proper submodule Nof an R-module Mis called prime if it satisfies the
condition (1). It is shown in [11, Sec. 2] that there is a nonzero module Msuch that
Spec(M)=∅. However, for any finitely generated R-module M, Spec(M)=∅.It
can be easily seen that (1) ⇒(2). However, the converse is not true in general. An
example of this is given as follows: For a prime number pof the ring Zof integers,
let R=Zp3.LetM=R[x], the polynomial ring over Rwith an indeterminate
x,andletN=px, the submodule of Mgenerated by px.ThenNsatisfies
the condition (2). However, Nis not prime. For, px ∈N, but (N:RM)=0,
and p=0inR,sothatp/∈(N:RM). Moreover, x/∈N. This shows that the
condition (2) is weaker than the condition (1). Hence, a proper submodule Nof an
R-module Mis called a weakly prime submodule if Nsatisfies the condition (2).
Adding ab ∈(N:RM) to the conclusion statement of the condition (2), we can
get the definition of a 2-absorbing submodule [19], which is a natural generalization
of a 2-absorbing ideal [4]. Further, the notion of an n-absorbing ideal and then an
n-absorbing submodule was introduced in [1, 5].
It can be easily proven that (2) ⇒(3). However, the converse is not true in
general. For example, the proper submodule 6Zof the Z-module Zsatisfies the
condition (3). However, (2 ·3) ·1∈6Z, but 2 ·1/∈6Zand 3 ·1/∈6Z. This shows
that the condition (3) is weaker than the condition (2). Hence, a proper submodule
Nof an R-module Mis weakly prime and is said to be semiprime if it satisfies the
condition (3) (see [15]).
It can be easily proven that (3) ⇒(4). However, the converse is not true in
general. For example, we may consider R=Z,M=Qand N=Zand observe
that (N:M) = 0 (and hence Nsatisfies the condition (4)) but 2 ·1
4=1
2/∈N
whereas 4 ·1
4=1∈N. This shows that the condition (4) is weaker than the
condition (3). Hence, a proper submodule of an R-module Mis said to be weakly
semiprime if it satisfies the condition (4).
We summarize the results as follows:
prime ⇒
weakly prime ⇒
semiprime ⇒
weakly semiprime
A number of authors have studied the subject for a couple of decades and many
results have been given to explore the nature of semiprime submodules and related
concepts.
For any submodule Nof an R-module M,GM(N):=∪r∈R(rM ∩(N:Mr)).
For an e l e me nt rof Rand for a submodule Nof an R-module M,wecansee
that N⊆(N:Mr). Since N=M∩N=1M∩(N:M1), it follows that N⊆
GM(N)⊆GM(N). Here, GM(N)denotes the submodule of Mgenerated by the
set GM(N). It can be easily seen that GM(N)⊆Nif and only if GM(N)⊆N.
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Semiprime submodules of a module and related concepts
Also, the envelope of Nin Mis defined to be EM(N):=∪r∈R(rM ∩(N:Mr))=
GM(N).
In Sec. 2, several properties of semiprimes, their radicals, and their localizations
will be given. The semiprime radical of a submodule Nin Mis defined to be the
intersection of all semiprime submodules of Mcontaining N, and is denoted by
sradM(N). The semiprime radical of Mis defined to be sradM(0), and is denoted
by srad(M). We describe Sem(MS) for a multiplicatively closed subset of a ring R
and for an R-module Mand then give a new proof of [11, Proposition 1, p. 3742]. We
then show that for every multiplicatively closed subset Sof a ring Rand for every
subset Aof an R-module M,(srad
M(A))S⊆sradMS(AS)(Theorem2.6).IfMis a
finitely generated multiplication R-module, then we show that for every submodule
Nof M,srad
M(N)=
(N:M)M(Theorem 2.7). Moreover, if Mis generated
by m1,m
2,...,m
n, then we use this to show that srad(M)=n
i=1annR(mi)mi
(Theorem 2.10). If Mis a cyclic R-module, or Mis a projective R-module, or Ris
a Dedekind domain, then srad(M)=rad(M)=W(M)(Theorem2.12).
In Sec. 3, we say that a submodule Nof Msatisfies the semiprime radical
formula of Mif sradM(N)=EM(N). Moreover, an R-module Mis said to satisfy
the semiprime radical formula if every submodule of Msatisfies the semiprime
radical formula of M. We characterize modules satisfying the semiprime radical
formula (Theorem 3.4). Finally, if Mis a finitely generated multiplication R-module
and if Sis a multiplicatively closed subset of R, then we show, in Theorem 3.7,
that Msatisfies globally (and hence locally) the semiprime radical formula and
for every submodule Nof M,srad
MS(NS)=(srad
MN)S.Itisproventhatfor
every multiplicatively closed subset Sof a ring Rand for every subset Aof an
R-module M,(srad
M(A))S⊆sradMS(AS). We characterize modules satisfying the
semiprime radical formula. Finally, it will be considered under what conditions M
satisfies globally (and hence locally) the SRF and for every submodule Nof M,
sradMS(NS)=(srad
MN)S. It will be shown that if Nis a submodule of a module
M, then the semiprime radical of Nin Mis the direct limit of the ascending chain
EM(N)⊆EM(EM(N)) ⊆ ··· (Theorem 3.8), and a submodule of a module M
satisfies the semiprime radical formula if and only if its envelope in Mis semiprime
(Corollary 3.9).
Notations: Let Spec(M), and Sem(M) be the spectrum of prime submodules, and
the class of semiprime submodules, respectively. For a submodule Nof an R-module
Mand an ideal aof a ring R,wewrite
V(N)={P∈Spec(M)|P⊇N},
SVM(N)={P∈Sem(M)|P⊇N}.
2. Semiprimes, Radicals, and Localizations
In this section, several properties of semiprimes, their radicals, and their localiza-
tions are given.
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An R-module Mis called simple if M= 0 and the only submodules of Mare 0
and Mitself. We won’t use the term “irreducible” like other papers do instead of
the term “simple” in order to avoid confusion in further discussion.
Let abe any proper ideal of a ring R. Then by [17, Corollary 4.35] ahas a
minimal primary decomposition, say q1∩···∩qn. By [2, Exercise 1.13(iii)]√a=
√q1∩···∩√qn.Foreachi∈{1,...,n}let pi=√qi.Then
√a=p1∩···∩pn.By
[2, Proposition 4.1] each piis prime. Each piare distinct, so that√ais semiprime.
This shows that for every proper ideal aof a ring R,√ais semiprime. It is well
knownthatanArtinianringRis semiprime if and only if the zero ideal of R
is a product of distinct maximal ideal of Rif and only if Ris semisimple (see
[6, Theorem 2.14], the Chinese Remainder Theorem, and [18, Proposition 3.3]).
For an R-module M,letZ(M)={x∈M|(0 :Rx) is essential in R}.Thesec-
ond singular submodule of an R-module Mis the submodule Z2(M)ofMcontaining
Z(M) such that Z2(M)/Z (M)=Z(M/Z(M)). Notice that
M/Z2(M)∼
=M/Z(M)
Z2(M)/Z (M)=M/Z(M)
Z(M/Z(M)) .
Then if Mis a module over a nonsingular ring R, we can see that the factor
R-module M/Z2(M) is nonsingular.
Lemma 2.1. An R-module Mis semisimple if and only if there exists a submodule
Kof Msuch that M=Z2(M)⊕K, and Z2(M)is nonsingular semisimple and K
is semisimple.
The semiprime radical of Nin Mis defined to be the intersection of all the
semiprime submodules of Mcontaining N, and is denoted by sradM(N). That is
sradM(N)=∩P∈SVM(N)P. In particular, the semiprime radical of the zero sub-
module in Mis called the semiprime radical of M, and is denoted by srad(M).
That is srad(M):=srad
M(0). It can be proven that for every submodule Nof M,
sradM(N)issemiprime,sothatsrad
M(sradM(N)) = sradM(N); hence
N⊆sradM(N)=srad
M(sradM(N)) = ···⊂M.
However, we have a filtration
M⊃srad(M)⊃srad(srad(M)) ⊃···.
For every submodule Nof M
(sradM(N))/N =srad
M/N (N/N)=srad
M/N (0) = srad(M/N)
in the residue class R-module M/N. More generally, for every submodule Kof
Mcontaining N,srad
M/N (K/N)=(srad
M(K))/N in the residue class R-module
M/N.
Lemma 2.2. Let Mand ¯
Mbe R-modules and let ϕ:M→¯
Mbe an epimorphism
of R-modules with its kernel K.LetNbe a submodule of Msuch that K⊆N.
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Then the following statements are true:
(1) (a) If Pis a semiprime [respectively weakly semiprime]submodule of Mcon-
taining N, then ϕ(P)is a semiprime [respectively weakly semiprime]sub-
module of ¯
Mcontaining ϕ(N).
(b) If ¯
Pis a semiprime [respectively weakly semiprime]submodule of ¯
Mcon-
taining ϕ(N),then ϕ−1(¯
P)is a semiprime [respectively weakly semiprime]
submodule of Mcontaining N.
(2) ϕ(sradM(N)) = srad ¯
M(ϕ(N)).
(3) If sradM(N)=EM(N),then srad ¯
M(ϕ(N)) = E¯
M(ϕ(N)).
Lemma 2.3. Let Sbe a multiplicatively closed subset of a ring R.LetMbe an
R-module. Then there is a surjective map from the set of all the semiprime sub-
modules Pof Msuch that (P:Rx)∩S=∅for some x∈Mto the set of all the
semiprime submodules of the RS-module MS.
Corollary 2.4. Let Sbe a multiplicatively closed subset of a ring R.LetMbe an
R-module. Then
Sem(MS)={PS|P∈Sem(M)such that (P:Rx)∩S=∅for some x∈M}.
Let Sbe a multiplicatively closed subset of a ring R.LetNbe a submodule of
an R-module M. Then the S-component of Nin Mis defined to be a submodule
of the R-module Mcontaining N
NS:= {m∈M|sm ∈Nfor some s∈S}=∪s∈S(N:Ms).
Here, notice that the union of submodules is not a submodule in general, but NS
is a submodule.
Lemma 2.5. Let Sbe a multiplicatively closed subset of a ring R.LetMbe an
R-module. Then the following are true:
(1) If Nis a submodule of M, then NS=(NS)Sin the RS-module MS.
(2) Let A, B be submodules of M.IfAS=BSin the RS-module MS,then AS=BS
in the R-module M.
(3) If Pis a prime submodule of Msuch that (P:RM)∩S=∅,then P=PS.
A new proof of [11, Proposition 1, p. 3742]. There is a map Φ from the set
of all prime submodules Pof Msuch that (P:RM)∩S=∅onto Spec(MS). Now,
use Lemma 2.5(2) and (3) to show that Φ is 1-1. Hence, Φ is a 1-1 correspondence.
Theorem 2.6. Let Sbe a multiplicatively closed subset of a ring Rand let M
be an R-module. If Ais a submodule of M, then (sradM(A))S⊆sradMS(AS)and
there are 3submodules of the RS-module MSwhose lattice diagram is given as
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follows :
(3) (∩P∈X PS)S
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(7) (∩P∈X (PS)S)S
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(8) ((∩P∈X PS)S)S
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(6) ((∩P∈X P)S)S
Here,X={P∈SVM(A)|(P:Rx)∩S=∅for some x∈M}.
Proof. Using Lemma 2.3, we can see that there is a surjective map X→SVMS×
(AS). So, SVMS(AS)={PS|P∈X}⊆{PS|P∈SVM(A)}. Hence,
(sradM(A))S⊆∩
P∈SVM(A)PS⊆sradMS(AS)=∩P∈X PS.
Clearly, the inclusion relations of the diagram above hold. The equalities of the
diagram above, except the equality ‘(4) = (5)’, come from Lemma 2.5(1). Also,
direct computation gives the equality ‘(1) = (5)’, or the equality ‘(4) = (5)’.
Theorem 2.7. Let Mbe a finitely generated,faithful multiplication R-module and
Nbe a submodule of M.ThensradM(N)=(N:RM)M.
Proof. Let Pbe any semiprime submodule of Mcontaining N.Then(P:RM)is
a radical ideal of R,so
(N:RM)M⊆(P:RM)M=(P:RM)M=P. Hence,
(N:RM)M⊆sradM(N).
Conversely, let pbe any prime ideal of Rcontaining (N:RM). Then pM
is a prime submodule of M,sothatpMis semiprime. Hence, sradM(N)⊆pM.
This shows that sradM(N)⊆∩
p∈V(N:RM)(pM). Moreover, by [7, Theorem 1.6] and
[17, 3.48],

p∈V(N:RM)
(pM)=

p∈V(N:RM)
p
M=(N:RM)M.
Hence, sradM(N)⊆(N:RM)M.
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Corollary 2.8. If Mis a cyclic,faithful R-module generated by an element mof
M, then srad(M)=
annR(m)m.
Proof. It is well known that every cyclic R-module is a multiplication R-module.
Clearly, annR(M)=ann
R(m). By Theorem 2.7, srad(M)=
annR(M)M=
annR(m)Rm =annR(m)m.
For an R-module M,W(M):=m∈MannR(m)m. (see [9] for more details.)
Lemma 2.9. Let Mbe an R-module. Then W(M)⊆srad(M).
Proof. Let m∈W(M). Then there exist r1,r
2,...,r
n∈R,m1,m
2,...,m
n∈M
such that m=r1m1+r2m2+···+rnmnand there exist k1,k
2,...,k
n∈Z+
such that rk1
1m1=0,r
k2
2m2=0,...,r
kn
nmn=0.(Ofcourse,wemaytakek=
max{k1,k
2,...,k
n}and proceed to show that m∈srad(M).)
Now, let Nbe an arbitrary semiprime submodule of M.Thenrk1
1m1=0∈
N,rk2
2m2=0∈N,...,r
kn
nmn=0∈N.SinceNis semiprime, we can see that
r1m1∈N,r2m2∈N,...,r
nmn∈N. Hence, m=r1m1+r2m2+···+rnmn∈N.
This shows that m∈srad(M). Therefore, W(M)⊆srad(M).
Theorem 2.10. Let Mbe a finitely generated,faithful multiplication module over
aringR.Ifm1,m
2,...,m
nare generators of M, then
srad(M)=
annR(M)M=annR(m1)m1+···+annR(mn)mn=W(M).
Proof. By Theorem 2.7, and Lemma 2.9, we have
srad(M)=
annR(M)M⊆annR(m1)m1+···+annR(mn)mn
⊆
m∈MannR(m)m=W(M)⊆srad(M).
Lemma 2.11. Consider the fol lowing three conditions :(1)Mis a cyclic R-module;
(2) Mis a projective R-module;(3)Ris a Dedekind domain. If one of these con-
ditions is true,then rad(M)=W(M).
Proof. Case (1) follows from [9, p. 3596], Case (2) from [9, Corollary 8], and Case
(3) from [9, Theorem 9].
Theorem 2.12. Let Mbe an R-module. If one of (1),(2),and (3) in Lemma 2.11
is true,then srad(M)=rad(M)=W(M).
Proof. By Lemmas 2.9, and 2.11, we have W(M)⊆srad(M)⊆rad(M)=W(M).
Hence, the proof is completed.
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3. Modules Satisfying the Semiprime Radical Formula
In this section, we will deal with modules satisfying the semiprime radical formula
and get some results.
Let Mbe an R-module and Nbe a submodule of M.Letx∈EM(N). Then
since EM(N) is generated by ∪r∈R(rM ∩(N:Mr)), there exist r1,...,r
n∈Rand
x1,...,x
n∈Msuch that x=r1x1+···+rnxnand each rixi∈(N:Mri). Now, let
Pbe any member of SVM(N). Then for each i∈{1,...,n},r2
ixi∈N⊆P.Since
Pis semiprime, we have rixi∈P. This shows that x∈P. Hence EM(N)⊆P.
Thus, EM(N)⊆∩
P∈SVM(N)P=srad
M(N). Therefore EM(N)⊆sradM(N).
Definition 3.1. A submodule Nof an R-module Mis said to satisfy the semiprime
radical formula if sradM(N)=EM(N). (See, for example, [3, 9, 10, 12, 13] and [16].)
Definition 3.2. An R-module Mis called a module satisfying the semiprime
radical formula if every submodule of Msatisfies the semiprime radical formula.
Every vector space is a module satisfying the semiprime radical formula. In
fact, every proper subspace of a vector space is prime and hence semiprime. Also,
every finitely generated multiplication module over a ring is a module satisfying
the semiprime radical formula.
Proposition 3.3. Let Mbe an R-module. If every submodule of the factor
R-module M/EM(0) is semiprime,then Msatisfies the semiprime radical formula.
Proof. Let Nbe a submodule of M.SinceEM(0) ⊆EM(N), we have EM(N)/
EM(0) is a submodule of M/EM(0). By the hypothesis, we can see that EM(N)/
EM(0)issemiprime.Now,wehave
EM(N)/EM(0) = sradM/EM(0)(EM(N)/EM(0)) = (sradM(EM(N)))/EM(0),
so that EM(N)=srad
M(EM(N)). Moreover, since N⊆EM(N), it follows that
sradM(N)⊆sradM(EM(N)). Hence by the statement just prior to Definition 3.1,
EM(N)=srad
M(N). Therefore, Msatisfies the semiprime radical formula.
Theorem 3.4. The following are equivalent for every R-module M:
(a) Msatisfies the semiprime radical formula.
(b) Every homomorphic image of Msatisfies the semiprime radical formula.
(c) For every submodule Nof M, EM(EM(N)) = EM(N).
Proof. (a) ⇔(b) Assume (a). It suffices to prove that every residue class R-module
M/K satisfies the semiprime radical formula. Consider the canonical map π:M→
M/K.Thenπis an epimorphism with kernel K.Let ¯
Nbe any submodule of M/K.
Then there exists a submodule Nof Mcontaining Ksuch that ¯
N=π(N). By the
assumption, sradM(N)=EM(N). Hence, it follows from Lemma 2.2(3) that ¯
N
satisfies the semiprime radical formula.
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Conversely, Mis a homomorphic image of Mitself, so that Msatisfies the
semiprime radical formula.
(a) ⇔(c). Assume (a). Then since for every submodule Nof M,EM(N)is
submodule of M,andsrad
M(N) is semiprime, it follows that
EM(EM(N)) = sradM(sradM(N)) = sradM(N)=EM(N).
Conversely, assume (c). Now, let Nbe any submodule of M. Hence, EM(N)is
a semiprime submodule of Mcontaining N. Hence sradM(N)⊆EM(N). By the
statement just prior to Definition 3.1, sradM(N)=EM(N).
Now, we consider localizations of modules satisfying the semiprime radical
formula.
Lemma 3.5. Let Sbe a multiplicatively closed subset of a ring R.LetMbe an
R-module. Then for every submodule Nof M, (EM(N))S=EMS(NS).
Corollary 3.6. An R-module Msatisfies (globally)the semiprime radical formula
if and only if Msatisfies locally the semiprime radical formula.
Proof. This follows from Theorem 3.4, Lemma 3.5, and [8, Lemma 1.3.15, p. 20].
Theorem 3.7. Let Mbe a finitely generated multiplication R-module and Sbe a
multiplicatively closed subset of R. Then the following are true:
(1) Msatisfies globally (and hence local ly)the semiprime radical formula.
(2) For every submodule Nof M, sradMS(NS)=(srad
M(N))S.
Proof. (1) Let Mbe a finitely generated multiplication module over Rand Nbe
a submodule of M. By [14, Theorem 4.4],
EM(N)=(EM(N):
RM)M=(rad
M(N):
RM)M=rad
M(N)⊇sradM(N).
It follows that sradM(N)=EM(N). This shows that Msatisfies globally the
semiprime radical formula. By Corollary 3.6, Msatisfies locally the semiprime
radical formula.
(2) Let Nbe an arbitrary submodule of M. Then by (1) and Lemma 3.5, we
get sradMS(NS)=EMS(NS)=(EM(N))S=(srad
M(N))S.
The previous result shows that the equality in Theorem 2.6 holds for every
finitely generated multiplication module.
For a submodule Nof a module M, we define E(n)
M(N) inductively to be
E(1)
M(N)=EM(N),E
(2)
M(N)=EM(EM(N)),E
(3)
M=EM(EM(EM(N))),....
Theorem 3.8. If Nis a submodule of a module M, then
sradM(N) = lim
−→ E(n)
M(N).
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S. C. Lee & R. Varmazyar
Proof. Since
E(1)
M(N)⊆E(2)
M(N)⊆E(3)
M(N)⊆···,
we have
lim
−→ E(n)
M(N)=∪∞
n=1E(n)
M(N).
Now, it suffices to proof that ∪∞
n=1E(n)
M(N)=srad
M(N). Clearly, ∪∞
n=1E(n)
M(N)⊆
sradM(N). Conversely, suppose that the inclusion is proper. Then, we can take
f∈sradM(N)\∪
∞
n=1 E(n)
M(N). Set
Σ={N⊆M|N⊆N,f/∈∪
∞
n=1E(n)
M(N)}.
N∈Σ=∅. We can show that every chain in Σ has an upper bound in Σ. By
the Zorn Lemma, Σ has a maximal member, say P.Pis semiprime. For otherwise,
there exist r∈R,x∈Msuch that r2x∈Pbut rx /∈P. This implies
PrM ∩(P:Mr)⊆EM(P).
By the maximality of P,wemusthave
f∈∪
∞
n=1E(n)
M(EM(P)) = ∪∞
n=1E(n+1)
M(P)=∪∞
r=1E(r)
M(P),
a contradiction. Moreover, f/∈∪
∞
n=1E(n)
M(P), and so, f/∈P. Hence f/∈sradM(N).
This contradiction shows that sradM(N)⊆∪
∞
n=1E(n)
M(N). Therefore, the proof is
completed.
Corollary 3.9. A submodule of a module Msatisfies the semiprime radical formula
if and only if its envelope in Mis semiprime.
If Ris a simple ring, then every submodule of an R-module satisfies the
semiprime radical formula. If every ideal of a ring Ris radical, then every module
over the ring Rsatisfies the semiprime radical formula. If every cyclic ideal of a ring
Ris idempotent, then every module over the ring Ralso satisfies the semiprime
radical formula.
A ring is semisimple if and only if it is Artinian and its Jacobson radical is zero.
A ring is semisimple if and only if every ideal is generated by an idempotent. If R
is a semisimple ring, then every submodule of an R-module satisfies the semiprime
radical formula.
Acknowledgments
The draft of this paper was written during the first author’s stay at Department
of Mathematics of the University of Colorado at Boulder. Professor Dr. Eric Stade
hosted me in the Department and provided me with office space, staff assistance,
copy machine use, library privileges, and so on. He proofread the draft as well. At
the University of Colorado, I was participating in Kempner Colloquium series and
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2nd Reading
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Semiprime submodules of a module and related concepts
in the Algebraic Lie Theory seminar, both of which was organized by Dr. Richard
M. Green, and was also participating in the Logic Seminar, which was organized
by Professor Dr. Keith Kearnes. Furthermore, I was talking with Professor Satya
Mandal in the Department of Mathematics of the University of Kansas at Lawrence.
I would like to express sincere gratitude to all of them.
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