Properties of a certain product of submodules
Let R be a commutative ring with identity, M an R-module and K1, . . . , Kn submodules of M. In this article, we construct an algebraic object, called product of K1, . . . , Kn. We equipped this structure with appropriate operations to get an R(M)-module. It is shown that R(M)-module Mⁿ = M . . ....
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irk-123456789-1660262020-02-19T01:26:06Z Properties of a certain product of submodules Heidari, S. Nikandish, R. Nikmehr, M.J. Статті Let R be a commutative ring with identity, M an R-module and K1, . . . , Kn submodules of M. In this article, we construct an algebraic object, called product of K1, . . . , Kn. We equipped this structure with appropriate operations to get an R(M)-module. It is shown that R(M)-module Mⁿ = M . . . M and R-module M inherit some of the most important properties of each other. For example, we show that M is a projective (flat) R-module if and only if Mⁿ is a projective (flat) R(M)-module Припустимо, що R — комутативне кiльце з одиницею, M — R-модуль i K1, . . . , Kn — пiдмодулi M. Побудовано алгебраїчний об’єкт, що називається добутком пiдмодулiв K1, . . . , Kn. Цю структуру оснащено вiдповiдними операцiями для отримання R(M)-модуля. Показано, що R(M)-модуль Mⁿ = M . . . M та R-модуль M успадковують деякi з найбiльш важливих властивостей один одного. Наприклад, показано, що M є проективним (плоским) R-модулем тодi i тiльки тодi, коли Mⁿ — проективний (плоский) R(M)-модуль. 2011 Article Properties of a certain product of submodules / S. Heidari, R. Nikandish, M.J. Nikmehr // Український математичний журнал. — 2011. — Т. 63, № 4. — С. 502–512. — Бібліогр.: 13 назв. — англ. 1027-3190 http://dspace.nbuv.gov.ua/handle/123456789/166026 512.5 en Український математичний журнал Інститут математики НАН України |
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Статті Статті |
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Статті Статті Heidari, S. Nikandish, R. Nikmehr, M.J. Properties of a certain product of submodules Український математичний журнал |
| description |
Let R be a commutative ring with identity, M an R-module and K1, . . . , Kn submodules of M. In this article,
we construct an algebraic object, called product of K1, . . . , Kn. We equipped this structure with appropriate
operations to get an R(M)-module. It is shown that R(M)-module Mⁿ = M . . . M and R-module M
inherit some of the most important properties of each other. For example, we show that M is a projective (flat)
R-module if and only if Mⁿ is a projective (flat) R(M)-module |
| format |
Article |
| author |
Heidari, S. Nikandish, R. Nikmehr, M.J. |
| author_facet |
Heidari, S. Nikandish, R. Nikmehr, M.J. |
| author_sort |
Heidari, S. |
| title |
Properties of a certain product of submodules |
| title_short |
Properties of a certain product of submodules |
| title_full |
Properties of a certain product of submodules |
| title_fullStr |
Properties of a certain product of submodules |
| title_full_unstemmed |
Properties of a certain product of submodules |
| title_sort |
properties of a certain product of submodules |
| publisher |
Інститут математики НАН України |
| publishDate |
2011 |
| topic_facet |
Статті |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/166026 |
| citation_txt |
Properties of a certain product of submodules / S. Heidari, R. Nikandish, M.J. Nikmehr // Український математичний журнал. — 2011. — Т. 63, № 4. — С. 502–512. — Бібліогр.: 13 назв. — англ. |
| series |
Український математичний журнал |
| work_keys_str_mv |
AT heidaris propertiesofacertainproductofsubmodules AT nikandishr propertiesofacertainproductofsubmodules AT nikmehrmj propertiesofacertainproductofsubmodules |
| first_indexed |
2025-07-14T20:30:32Z |
| last_indexed |
2025-07-14T20:30:32Z |
| _version_ |
1837655689044951040 |
| fulltext |
UDC 512.5
M. J. Nikmehr, R. Nikandish, S. Heidari (K. N. Toosi Univ. Technology, Iran)
PROPERTIES OF A CERTAIN PRODUCT OF SUBMODULES
ВЛАСТИВОСТI ПЕВНОГО ДОБУТКУ ПIДМОДУЛIВ
Let R be a commutative ring with identity, M an R-module and K1, . . . ,Kn submodules of M. In this article,
we construct an algebraic object, called product of K1, . . . ,Kn. We equipped this structure with appropriate
operations to get an R(M)-module. It is shown that R(M)-module Mn = M . . .M and R-module M
inherit some of the most important properties of each other. For example, we show that M is a projective (flat)
R-module if and only if Mn is a projective (flat) R(M)-module.
Припустимо, що R — комутативне кiльце з одиницею, M — R-модуль i K1, . . . ,Kn — пiдмодулi
M. Побудовано алгебраїчний об’єкт, що називається добутком пiдмодулiв K1, . . . ,Kn. Цю струк-
туру оснащено вiдповiдними операцiями для отримання R(M)-модуля. Показано, що R(M)-модуль
Mn = M . . .M та R-модуль M успадковують деякi з найбiльш важливих властивостей один одного.
Наприклад, показано, що M є проективним (плоским) R-модулем тодi i тiльки тодi, коли Mn —
проективний (плоский) R(M)-модуль.
1. Introduction and preliminaries. In this paper, all rings are commutative with identity
and all modules are unitary. Let M be an R-module; there are some attempts to define
a product between submodules of M, see for example [5, p. 386]. Based on this idea,
in this article, we introduce and investigate a kind of product of submodules of M and
especially we study R(M)-module Mn = M . . .M, in which R(M) is idealization
of M. It is worthy to mention that Nagata introduced the notion of idealization and
the idea to use idealization is due to him. Idealization is useful for extending results
about ideals to submodules and constructing examples of commutative rings with zero-
divisors. The theme throughout is how properties of R-module M are related to those of
R(M)-module Mn and this is the main goal of this article. For example, in Section 2,
we show that M is a projective (flat) R-module if and only if Mn is a projective (flat)
R(M)-module and in Section 3, we find primary and secondary representation for Mn
by means of those of M and conversely. Now, we define the concepts that we will need.
Recall that R(M) = R(+)M with coordinate-wise addition and multiplication
(r1,m1)(r2,m2) = (r1r2, r1m2 + r2m1),
is a commutative ring with identity, called the idealization of M. Note that R naturally
embeds into R(M) via r −→ r(+)0, if N is a submodule of M, then 0(+)N is an ideal
of R(M), 0(+)M is a nilpotent ideal of R(M) of index 2, every ideal that contains
0(+)M has the form I(+)M for some ideal I of R, and every ideal that is contained in
0(+)N has the form 0(+)K for some submodule K of N. The purpose of idealization
is to put M inside a commutative ring A so that the structure of M as an R-module
is essentially the same as that of M as an A-module, that is, an ideal of A. Since
R ∼= R(M)/0(+)M, I −→ I(+)M gives a one-to-one correspondence between ideals
of R and ideals of R(M) that contains 0(+)M. Thus the prime (maximal) ideals of
R(M) have the form P (+)M where P is a prime (maximal) ideal of R. Some basic
results on idealization can be found in [10].
An R-module M is said to be multiplication if every submodule N of M has the
form IM for some ideal I of R. Equivalently, N = [N : M ]M. A submodule K of M
is multiplication if and only if N ∩K = [N : K]K for all submodules N of M. See for
example [5], for more details.
c© M. J. NIKMEHR, R. NIKANDISH, S. HEIDARI, 2011
502 ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 4
PROPERTIES OF A CERTAIN PRODUCT OF SUBMODULES 503
Generalizing the case for ideals, an R-module M is called a cancellation (weak
cancellation) module if IM = JM for ideals I and J of R implies I = J (I +
+ annM = J + annM), see [3]. Examples of cancellation modules include invertible
ideals, free modules, and finitely generated faithful multiplication modules [4] (Corollary
to Theorem 9). The trace ideal of an R-moduleM is Tr(M) =
∑
f∈Hom(M,R)
f(M). If
M is projective, thenM = Tr(M)M and ann(M) = ann Tr(M) [9] (Proposition 3.30).
The set of all prime ideals of ring R is denoted by Spec(R). Moreover, we denote by
ZdvR(M) the set of all zero-divisors of module M over ring R. Also, for unexplained
definitions and terminologies we refer to [6, 9].
2. The multiplication property and product. We begin with the following definition
which plays an important role in this article.
Definition. LetK1,K2, . . . ,Kn be submodules of an R-moduleM. Define product
of K1, K2,. . . ,Kn as follows:
K1K2 . . .Kn =
{
(1R, k1, k2, . . . , kn) | ki ∈ Ki, for all 1 6 i 6 n}.
One can easily check that K1K2 . . .Kn forms an R(M)-module under the below
operations:
(1R, k1, k2, . . . , kn) + (1R, k
′
1, k
′
2, . . . , k
′
n) =
= (1R, (k1 + k′1), (k2 + k′2), . . . , (kn + k′n)),
(r,m)(1R, k1, k2, . . . , kn) = (1R, rk1, rk2, . . . , rkn).
For convenience, 1R and M . . .M (n-times) will be denoted by 1 and Mn, respecti-
vely.
In this section, we shall investigate the multiplication, quasimultiplication, projective,
flat, faithfully flat, cancellation and weak cancellation properties under this new product
submodules.
Let P be a maximal ideal of R and let TP (M) = {m ∈M | (1− p)m = 0, for some
p ∈ P}. Then TP (M) is a submodule of M. An R-module M is called P -torsion if
TP (M) = M. On the other hand, M is called P -cyclic provided there exist x ∈M and
q ∈ P such that (1 − q)M ⊆ Rx. El-Bast and Smith [8] (Theorem 1.2), showed that
M is multiplication if and only if M is P -torsion or P -cyclic for each maximal ideal P
of R.
Next we prove that ifK1 . . .Kn is a multiplication module, then eachKi, 1 ≤ i ≤ n,
is a multiplication module. But first we need the following lemma.
Lemma 1. Let K1, . . . ,Kn be submodules of an R-module M and P a maximal
ideal of R. Then:
(i) TP (+)M (K1 . . .Kn) = {(1,m1,m2, . . . ,mn) | mi ∈ TP (Ki)}. In particular,
K1 . . .Kn is P (+)M -torsion if and only if each Ki, 1 ≤ i ≤ n, is P -torsion.
(ii) K1 . . .Kn is P (+)M -cyclic if and only if each Ki (1 ≤ i ≤ n) is P -cyclic.
Proof. (i) Let (1,m1,m2, . . . ,mn) ∈ TP (+)M (K1 . . .Kn). Then there exists
(p,m) ∈ P (+)M such that ((1, 0)− (p,m))(1,m1,m2, . . . ,mn) = (1, 0, 0, . . . , 0). So
(1, (1−p)m1, (1−p)m2, . . . , (1−p)mn) = (1, 0, 0, . . . , 0). It follows thatmi ∈ TP (Ki),
for all 1 ≤ i ≤ n. Now, suppose that mi ∈ TP (Ki), 1 ≤ i ≤ n. Then there exists pi ∈ P
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 4
504 M. J. NIKMEHR, R. NIKANDISH, S. HEIDARI
such that (1− pi)mi = 0. If we put q = 1−
∏n
i=1
(1− pi) ∈ P, then (1− q)mi = 0,
for each 1 ≤ i ≤ n and hence ((1, 0) − (q, 0))(1,m1,m2, . . . ,mn) = (1, 0, 0, . . . , 0).
Thus (1,m1,m2, . . . ,mn) ∈ TP (+)M (K1 . . .Kn).
(ii) To see why (ii) is true, let Ki, 1 ≤ i ≤ n, be P -cyclic. Then there exist mi ∈ Ki
and pi ∈ P such that (1 − pi)Ki ⊆ Rmi. If we put q = 1 −
∏n
i=1
(1 − pi) ∈ P, then
((1, 0)− (q, 0))K1 . . .Kn ⊆ R(M)(1,m1,m2, . . . ,mn). Hence K1 . . .Kn is P (+)M -
cyclic.
Conversely, let K1 . . .Kn be P (+)M -cyclic. Then there exist (1, k1, . . . , kn) ∈
∈ K1 . . .Kn and (p,m) ∈ P (+)M such that ((1, 0) − (p,m))K1 . . .Kn ⊆
⊆ R(M)(1, k1, . . . , kn). Thus (1 − p)Ki ⊆ Rki, for all 1 ≤ i ≤ n and so Ki is
P -cyclic, as required.
Lemma 1 is proved.
Theorem 1. Let K1, . . . ,Kn be submodules of an R-module M. If K1 . . .Kn is
a multiplication R(M)-module, then each Ki is a multiplication R-module. Moreover,
if M is a multiplication R-module, then Mn is a multiplication R(M)-module.
Proof. Let K1 . . .Kn be a multiplication R(M)-module and P a maximal ideal of
R. If each Ki, 1 ≤ i ≤ n, is P -torsion, there is nothing to prove. Now, assume that there
exists i, 1 ≤ i ≤ n, such that Ki is not P -torsion. By Lemma 1(i), K1 . . .Kn is not
P (+)M -torsion and so there exist (p,m) ∈ P (+)M and (1, k1, . . . , kn) ∈ K1 . . .Kn
such that
((1, 0)− (p,m))K1 . . .Ki . . .Kn ⊆ R(+)M(1, k1, . . . , ki, . . . , kn).
Hence (1− p)Ki ⊆ Rki, i.e., Ki is P -cyclic. It follows that each Ki is a multiplication
submodule of M.
Now, let M be a multiplication R-module and P (+)M be a maximal ideal of
R(M). Suppose that Mn is not P (+)M -torsion. By Lemma 1(i), M is not P -torsion
and hence there exist p ∈ P and m ∈ M such that (1 − p)M ⊆ Rm. It follows
that ((1, 0)− (p, 0))Mn ⊆ R(+)M(1,m, . . . ,m). Thus Mn is P (+)M -cyclic and this
completes the proof.
For an R-module M, following [7], we set
M(P ) = {x ∈M | sx ∈ PM, for some s ∈ R \ P},
in which P is a prime ideal of R. In [7], it is shown that M(P ) = M or M(P ) is a
submodule of M, for every P ∈ Spec(R). As usual, we will denote the Support of M
by
SuppRM =
{
P ∈ Spec(R) | there exists 0 6= x ∈M such that ann(x) ⊆ P
}
.
Recall that an R-module M is called quasimultiplication if M(P ) = PM, for all
P ∈ SuppRM. For a reference on quasimultiplication module see [7].
The next result will be used in the Theorem 2.
Lemma 2. Let M be an R-module. Then:
(i) SuppR(M)M
n = {P (+)M | P ∈ SuppRM}.
(ii) Mn(P (+)M) = {(1,m1, . . . ,mn) ∈Mn | mi ∈M(P ), for all 1 ≤ i ≤ n}.
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 4
PROPERTIES OF A CERTAIN PRODUCT OF SUBMODULES 505
Proof. (i) If P ∈ SuppRM, then there exists 0 6= x ∈ M such that ann(x) ⊆ P.
Clearly, ann(1, x, x, . . . , x) = {(r,m) | r ∈ ann(x)} ⊆ P (+)M. Hence P (+)M ∈
∈ SuppR(+)M Mn and so {P (+)M | P ∈ SuppRM} ⊆ SuppR(M)M
n. Now, let
P (+)M ∈ SuppR(+)M Mn. Then there exists 0 6= (1, x1, . . . , xn) ∈ Mn such that
ann(1, x1, . . . , xn) ⊆ P (+)M, and hence annx1 ∩ annx2 ∩ . . . ∩ annxn ⊆ P. Since
P is a prime ideal, there exists 1 ≤ i ≤ n such that annxi ⊆ P, i.e., P ∈ SuppRM
and so SuppR(M)M
n ⊆ {P (+)M | P ∈ SuppRM}. By the above argument the proof
is finished.
(ii) Let T = {(1,m1, . . . ,mn) ∈ Mn | mi ∈ M(P ) for all 1 ≤ i ≤ n}, and
(1,m1,m2, . . . ,mn) ∈ Mn(P (+)M). Then there exists (s′,m′) ∈ R(+)M \ P (+)M
such that (s′,m′)(1,m1,m2, . . . ,mn) ∈ (P (+)M)Mn. Hence s′mi ∈ PM, for all
1 ≤ i ≤ n and so mi ∈M(P ) for all 1 ≤ i ≤ n. It follows that Mn(P (+)M) ⊆ T.
Conversely, let (1,m1, . . . ,mn) ∈ T. For every 1 ≤ i ≤ n, there exists si ∈
∈ R \ P such that simi ∈ PM. Put s = s1 . . . sn. Clearly, s /∈ P and smi ∈ PM,
for each 1 ≤ i ≤ n. So (s, 0)(1,m1,m2, . . . ,mn) ∈ (P (+)M)Mn and therefore,
(1,m1, . . . ,mn) ∈ Mn(P (+)M). Thus T ⊆ Mn(P (+)M). By the above argument it
follows that
Mn(P (+)M) =
{
(1,m1, . . . ,mn) ∈Mn | mi ∈M(P ), for all 1 ≤ i ≤ n
}
.
Lemma 2 is proved.
The next result shows how quasimultiplication property of an R-module M can be
transferred to an R(M)-module Mn and conversely.
Theorem 2. An R-module M is quasimultiplication if and only if Mn is a quasi-
multiplication R(M)-module.
Proof. First, note that M(P ) = M if and only if Mn(P (+)M) = Mn. Suppose
that M is a quasimultiplication R-module and P (+)M ∈ SuppR(+)M Mn. Then
Mn(P (+)M) = {(1,m1, . . . ,mn) ∈ Mn | mi ∈ M(P ) = PM, for all 1 ≤ i ≤ n}.
It follows that for all 1 ≤ i ≤ n there exist pij ∈ P and m′ij ∈ M such that
mi =
∑t
j=1
pijm
′
ij . Hence
(1,m1, . . . ,mn) =
1,
t∑
j=1
p1jm
′
1j , . . . ,
t∑
j=1
pnjm
′
nj
=
t∑
j=1
(1, p1jm
′
1j , . . . , pnjm
′
nj).
Therefore, (1,m1, . . . ,mn) ∈ (P (+)M)Mn. This yields that
Mn(P (+)M) ⊆ (P (+)M)Mn.
Now, let
∑m
i=1
(1, pimi1, . . . , pimin) ∈ (P (+)M)Mn. Since M(P ) = PM, pimij ∈
∈M(P ), for each 1 ≤ i ≤ m and 1 ≤ j ≤ n. It follows that(
1,
m∑
i=1
pimi1, . . . ,
m∑
i=1
pimin
)
⊆ (1,M(P ), . . . ,M(P )),
and by Lemma 2,
∑m
i=1
(1, pimi1, . . . , pimin) ∈Mn(P (+)M). Therefore,
(P (+)M)Mn ⊆Mn(P (+)M),
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 4
506 M. J. NIKMEHR, R. NIKANDISH, S. HEIDARI
i.e., Mn is a quasimultiplication R(M)-module.
Conversely, let Mn be a quasimultiplication R(M)-module and P ∈ SuppRM.
Then P (+)M ∈ SuppR(M)M
n, by Lemma 2. Since Mn is quasimultiplication,
Mn(P (+)M) = (P (+)M)Mn, and hence M(P ) = PM. Thus M is a quasimultipli-
cation R-module.
Theorem 2 is proved.
The following question is araised: Does Mn as an R(M)-module have all properties
of R-module M? It is easily checked that Q is a faithful Z-module, but ann(Qn) =
= ann(Q)(+)Q = 0(+)Q. In fact, it shows that every property of M can not be
transferred to Mn.
Before we state and prove our next corollary, we need the following proposition.
Proposition 1. For an R-module M
Tr(Mn) = Tr(M)(+)
∑
g∈Hom(M,M)
g(M) = Tr(M)(+)M.
Proof. Let f ∈ Hom(Mn, R(+)M). It is clear that there exist g1 ∈ Hom(M,R)
and g2 ∈ Hom(M,M) such that f = g1(+)g2. Hence
Tr(Mn) =
∑
f∈Hom(Mn,R(+)M)
f(Mn) =
=
∑
g1∈Hom(M,R),g2∈Hom(M,M)
g1(M)(+)g2(M) =
=
∑
g1∈Hom(M,R)
g1(M) +
∑
g2∈Hom(M,M)
g2(M) ⊆ Tr(M)(+)M.
Conversely, let g ∈ Hom(M,R). Define f : Mn −→ R(+)M as follows: for each
(1,m1,m2, . . . ,mn) ∈Mn, f(1,m1,m2, . . . ,mn) = g(m1+m2+. . .+mn)(+)id(m1+
+m2+. . .+mn). It is clear that f is well-defined andR(M)-homomorphism. Therefore,
Tr(M)(+)M =
∑
g∈Hom(M,R)
g(M)(+)M ⊆
∑
f(Mn) ⊆ Tr(Mn).
It follows that Tr(Mn) = Tr(M)(+)M.
Proposition 1 is proved.
Lemma 3. Let M be a projective R-module. Then Tr(M) is a finitely generated
ideal of R if and only if Tr(Mn) is a finitely generated ideal of R(M).
Proof. By Proposition 1 and [9] (Proposition 3.3), Tr(Mn) = Tr(M)(+)M =
= Tr(M)(+) Tr(M)M. Hence, Tr(Mn) is a finitely generated ideal of R(M) if and
only if Tr(M) is a finitely generated ideal of R [1] (Theorem 7(1)).
Lemma 3 is proved.
It is shown in [9] (Lemma 3.23) that an R-module M is projective if and only
if there exist a family of elements {mi}i∈I in M and family {fi}i∈I of elements in
M∗ = HomR(M,R) such that every m ∈ M is a finite sum m = Σmifi(m), where
fi(m) = 0 almost for every i ∈ I. In the next theorem we prove that M is a projective
R-module if and only if Mn is a projective R(M)-module.
ISSN 1027-3190. Укр. мат. журн., 2011, т. 63, № 4
PROPERTIES OF A CERTAIN PRODUCT OF SUBMODULES 507
Theorem 3. An R-module M is projective if and only if Mn is a projective
R(M)-module.
Proof. Let M be a projective R-module and (1,m1,m2, . . . ,mn) ∈ Mn. There
exist a family {mi,j}(i,j)∈I in M and family {f(i,j)}(i,j)∈I of elements in M∗ =
= HomR(M,R) such that mj =
∑
i
m(i,j)f(i,j)(mj). Hence,
(1,m1,m2, . . . ,mn) = (1,Σm(i,1)f(i,1)(m1), . . . ,Σm(i,n)f(i,n)(mn)) =
= Σ(1,m(i,1)f(i,1)(m1), . . . ,m(i,n)f(i,n)(mn)) =
= Σ(f(i,1)(m1), 0)(1,m(i,1), 0, . . . , 0) + . . .+ (f(i,n)(mn), 0)(1, 0, . . . ,m(i,n)).
For each (i, j) ∈ I, define g(i,j) : Mn → R(M) as follows:
for each (1,m1,m2, . . . ,mn) ∈ Mn, g(i,j)(1,m1,m2, . . . ,mn) = (f(i,j)(mj), 0).
It is clear that g(i,j) is well-defined and R(M)-homomorphism. Let mi,j =
(
1, 0, . . . ,
. . . ,
j th︷ ︸︸ ︷
m(i,j), . . . , 0
)
. So (1,m1,m2, . . . ,mn) = Σg(i,j)(1,m1,m2, . . . ,mn)mi,j and
hence Mn is a projective R(M)-module.
Conversely, let Mn be a projective R(M)-module and m ∈ M. Since Mn is
projective, there exist a family of elements fi ∈ Hom(Mn, R(+)M) and family
(1,m1,i, . . . ,mn,i) of elements Mn such that
(1,m, . . . , 0) = Σfi(1,m, . . . , 0)(1,m1,i, . . . ,mn,i).
Because fi ∈ Hom(Mn, R(+)M), there exist gi ∈ Hom(M,R) and g′i ∈ Hom(M,M)
such that fi = gi(+)g′i. Thus m = Σgi(m)m1,i, i.e., M is a projective R-module.
Theorem 3 is proved.
It is well-known that a projective module is weak cancellation if and only if its trace
is a finitely generated ideal [14] (Theorem 4.1). The following question raises: If M is a
weak cancellation module, can we deduce that Mn is a weak cancellation module? The
following corollary gives an affirmative answer in case the projective modules.
Corollary 1. Let M be a projective R-module. Then M is a weak cancellation
R-module if and only if Mn is a weak cancellation R(M)-module.
Proof. Let M be a projective weak cancellation R-module. Then Tr(M) is fi-
nitely generated by [13] (Theorem 4.1). By the above theorem, Lemma 3 and [13]
(Theorem 4.1), Mn is a weak cancellation module. The proof of the converse is similar.
Our the following result is taken from [13] (Theorem 4.2).
Corollary 2. Let M be a projective R-module. Then M is a cancellation R-module
if and only if Mn is a cancellation R(M)-module.
Proof. Let M be a projective cancellation R-module. By [13] (Theorem 4.2),
Tr(M) = R. By Proposition 1, Tr(Mn) = Tr(M)(+)M = R(M). Hence, by [13]
(Theorem 4.2), Mn is a cancellation module. The proof of the converse is similar.
It is shown in [11] (Theorem 7.6), that M is flat if and only if for every pair
of finite subsets {x1, . . . , xn} and {a1, . . . , an} of M and R, respectively, such that∑n
i=1
aixi = 0 there exist elements z1, . . . , zk ∈ M and bij ∈ R, i = 1, . . . , n, j =
= 1, . . . , k such that
∑n
i=1
bijai = 0 (j = 1, . . . , k) and xi =
∑k
j=1
bijzj . Our main
concern in this part is to show that M is flat if and only if Mn is flat.
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508 M. J. NIKMEHR, R. NIKANDISH, S. HEIDARI
Theorem 4. An R-module M is flat if and only if Mn is a flat R(M)-module.
Proof. Let M be a flat R-module and
∑m
i=1
(ai, 0)(1, x1,i, x2,i, . . . , xn,i) = 0.
Then
∑m
i=1
aixj,i = 0, for every 1 ≤ j ≤ n. Since M is flat, there exist elements
zj,1, . . . , zj,s ∈M and bj,ik ∈ R, i = 1, . . . ,m, k = 1, . . . , s and j = 1, . . . , n such that∑m
i=1
bj,ikai = 0, k = 1, . . . , s and xj,i =
∑s
k=1
bj,ikzj,k, for j = 1, . . . , n. Hence∑m
i=1
(bj,ik, 0)(ai, 0) = 0 and
(1, x1,i, . . . , xn,i) =
=
s∑
k=1
(b1,ik, 0)(1, z1,k, 0, . . . , 0) + . . .+ (bn,ik, 0)(1, 0, . . . , zn,k), 1 ≤ i ≤ m.
Therefore, Mn is a flat R(M)-module.
Conversely, let Mn be flat. Suppose that {x1, . . . , xm} and {a1, . . . , am} are two
finite subsets of M and R, respectively, such that
∑m
i=1
aixi = 0. This implies that
m∑
i=1
(ai, 0)(1, xi, 0, . . . , 0) = 0.
Since Mn is flat, there exist elements (1, z1, . . . , 0), . . . , (1, zs, . . . , 0) ∈ Mn and
(bi,k, 0) ∈ R(M) (i = 1, . . . ,m, k = 1, . . . , s), such that
m∑
i=1
(bi,k, 0)(ai, 0) = 0, k = 1, . . . , s,
and
(1, xi, 0, . . . , 0) =
s∑
k=1
(bi,k, 0)(1, zk, 0, . . . , 0).
Hence
m∑
i=1
bi,kai = 0, xi =
s∑
k=1
bi,kzk.
By the above argument, M is a flat R-module.
Theorem 4 is proved.
It is known that an R-module M is called faithfully flat if M is flat and N ⊗M 6= 0
for any non-zero R-module N. Equivalently, M is flat and PM 6= M for every maximal
ideal P of R [11] (Theorem 7.2). We next show that if M is faithfully flat, then Mn is
faithfully flat.
Corollary 3. An R-module M is faithfully flat if and only if Mn, n ≥ 2, is a
faithfully flat R(M)-module.
Proof. It is easy to check that for every maximal ideal P of R, PM 6= M if and
only if (P (+)M)Mn 6= Mn. The result now follows by Theorem 4.
3. Product submodules and decomposition. Recall that a proper submodule N of a
module M is said to be primary submodule if the condition ra ∈ N, r ∈ R and a ∈M,
implies that a ∈ N or rnM ⊆ N, for some positive integer n. Let T = K1 . . .Kn be a
primary submodule of Mn and riki ∈ Ki where ri ∈ R, ki ∈M. Then
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PROPERTIES OF A CERTAIN PRODUCT OF SUBMODULES 509
(ri, 0)
(
1, 0, . . . ,
i th︷︸︸︷
ki , . . . , 0
)
∈ K1 . . .Ki . . .Kn.
Since K1 . . .Ki . . .Kn is primary, either (1, 0, . . . , ki, 0, . . . , 0) ∈ K1 . . .Ki . . .Kn or
(ri, 0)mMn ⊆ K1 . . .Ki . . .Kn, for some positive integer m and this means that ki ∈
∈ Ki or rmi M ⊆ Ki. By the above argument it follows that Ki is a primary submodule
for all 1 ≤ i ≤ n.
Conversely, let N be a primary submodule and (r,m)(1,m1, . . . ,mn) ∈ Nn, for
some (r,m) ∈ R(M), and (1,m1, . . . ,mn) ∈Mn. Hence rmi ∈ N for each 1 ≤ i ≤ n.
Since N is primary, mi ∈ N or rtM ⊆ N, for some positive integer t. It follows that
Nn is a primary submodule of Mn.
By the above argument we have the following theorem.
Theorem 5. Let T = K1 . . .Kn be a primary submodule of Mn. Then Ki is a
primary submodule of M for all 1 ≤ i ≤ n. Furthermore, let N be a primary submodule
of M. Then Nn, n ≥ 2, is a primary submodule of Mn.
Before we state and prove our next theorem, the following lemma is needed.
Lemma 4. Let Q1, . . . , Qm be submodules of an R-module M. Then:
(i) (Q1 +Q2 + . . .+Qm)n = Qn
1 +Qn
2 + . . .+Qn
m,
(ii) (Q1 ∩Q2 ∩ . . . ∩Qm)n = Qn
1 ∩Qn
2 ∩ . . . ∩Qn
m.
Proof. (i) First suppose that m = 2. We have only to prove that (Q1+Q2)n ⊆ Qn
1 +
+Qn
2 . Let (1,m1, . . . ,mn) ∈ (Q1 +Q2)n. Then there exist qi ∈ Q1 and q′i ∈ Q2 such
that (1,m1, . . . ,mn) = (1, q1 + q′1, . . . , qn + q′n) = (1, q1, . . . , qn) + (1, q′1, . . . , q
′
n) ∈
∈ Qn
1 +Qn
2 and hence (Q1+Q2)n ⊆ Qn
1 +Qn
2 . Ifm > 2, then the assertion follows by the
case m = 2 and induction on m. Hence (Q1+Q2+ . . .+Qm)n = Qn
1 +Qn
2 + . . .+Qn
m.
(ii) It is easy to check that (Q1 ∩ Q2)n = Qn
1 ∩ Qn
2 . Induction on m shows that
(Q1 ∩Q2 ∩ . . . ∩Qm)n = Qn
1 ∩Qn
2 ∩ . . . ∩Qn
m, as desired.
Lemma 4 is proved.
We record the following theorem.
Theorem 6. Let N be a submodule of an R-module M. If N has a primary
decomposition, then Nn, n ≥ 2, has a primary decomposition.
Proof. Suppose thatN has a primary decomposition. ThenN = Q1∩Q2∩. . .∩Qm,
where each Qi is a Pi-primary submodule of M. Hence Nn = Qn
1 ∩Qn
2 ∩ . . .∩Qn
m, by
Lemma 5. To see why this is a primary decomposition of Nn, note first that
Mn
Qn
i
6= 0,
because
M
Qi
6= 0. Next, if (r,m) ∈ ZdvR(M)
(
Mn
Qn
i
)
, then there exists a positive integer
n1 such that rn1
(
M
Qi
)
= 0. Since rn1M ⊆ Qi, (r,m)n1Mn = (rn1 , n1r
n1−1m)Mn ⊆
⊆ Qn
i and hence (r,m)n1
Mn
Qn
i
= 0. It remains to be shown that Qn
i is Pi(+)M -primary(
where Pi = rad
(
annR
M
Qi
))
. Let (t,m) ∈ rad
(
annR(M)
Mn
Qn
i
)
= Pi(+)M. Then
there exists a positive integer n1 such that (tn1 , n1t
n1−1m)Mn ⊆ Qn
i , so tn1M ⊆ Qi.
It follows that t ∈ rad
(
annR
M
Qi
)
= Pi. Therefore, rad
(
annR(M)
Mn
Qn
i
)
⊆ Pi(+)M.
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510 M. J. NIKMEHR, R. NIKANDISH, S. HEIDARI
Now, let (pi,m) ∈ Pi(+)M = rad
(
annR
M
Qi
)
(+)M. There exists a positive integer
n1 such that pn1
i M ⊆ Qi and so (pi,m)n1Mn ⊆ Qn
i . Since (pi,m)n1
Mn
Qn
i
= 0,
(pi,m) ∈ rad
(
annR(M)
Mn
Qn
i
)
. So Pi(+)M⊆ rad
(
annR(M)
Mn
Qn
i
)
, as required.
Theorem 6 is proved.
It is naturally to ask when the converse of Theorem 6 is true. See the next theorem.
Theorem 7. Let Nn be a submodule of Mn, n ≥ 2. If Nn has a primary
decomposition of the form Nn = Kn
1 ∩ Kn
2 ∩ . . . ∩ Kn
m, where Kn
i , 1 ≤ i ≤ m, is
Pi(+)M -primary, then N has a primary decomposition of the form N = K1 ∩K2 ∩ . . .
. . . ∩Km.
Proof. We show that N = K1 ∩ K2 ∩ . . . ∩ Km is a primary decomposition
for N. First, it is clear that N = K1 ∩ K2 ∩ . . . ∩ Km. Next, we show that each
Ki, for 1 ≤ i ≤ m, is Pi-primary. Suppose that a ∈ ZdvR
(
M
Ki
)
. Then there
exists m ∈ M \ Ki such that am ∈ Ki. Since m /∈ Ki, (1,m, . . . ,m, . . . ,m) /∈
/∈ Kn
i . But (a, 0)(1,m, . . . ,m, . . . ,m) = (1, am, . . . , am, . . . , am) ∈ Kn
i . So (a, 0) ∈
∈ Zdv
(
Mn
Kn
i
)
. Because Kn
i is Pi(+)M -primary, there exists a positive integer n1
such that (a, 0)n1
(
Mn
Kn
i
)
= 0, and hence an1
(
M
Ki
)
= 0. It remains to be shown
rad
(
annR
M
Ki
)
= Pi. Let r ∈ rad
(
annR
M
Ki
)
. Then rn1M ⊆ Ki, for some positive
integer n1. Thus (r, 0)n1Mn ⊆ Kn
i and so (r, 0) ∈ rad
(
annR(M)
Mn
Kn
i
)
= Pi(+)M,
i.e., r ∈ Pi. Therefore, rad
(
annR
M
Ki
)
⊆ Pi.
Conversely, let r ∈ Pi. Then (r, 0) ∈ Pi(+)M = rad
(
annR(M)
Mn
Kn
i
)
and hence
rn1M ⊆ Ki, for some integer n1. It follows that Pi ⊆ rad
(
annR
M
Ki
)
.
Theorem 7 is proved.
I. G. Macdonald has developed the theory of attached prime ideals and secondary
representations of a module, which is, in a certain sense, dual to the theory of associated
prime ideals and primary decompositions. Let us recall from [11], the definition of
secondary module. An R-module M is said to be secondary if M 6= 0 and, for each
a ∈ R the endomorphism ϕa : M →M defined by ϕa(m) = am (for m ∈M) is either
surjective or nilpotent.
If M is secondary, then P =
√
annM is a prime ideal, and M is said to be P -
secondary. A secondary representation of an R-module M is an expression of M as a
finite sum of secondary submodules:
M = N1 +N2 + . . .+Nn.
Let N be a submodule of an R-module M, in the following theorem we investigate
secondary representation of Nn.
Theorem 8. Let N be a submodule of an R-module M. If N has a secondary
representation, then Nn, n ≥ 2, has a secondary representation. Conversely, let Nn
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PROPERTIES OF A CERTAIN PRODUCT OF SUBMODULES 511
has a secondary representation of the form Nn = Qn
1 + Qn
2 + . . . + Qn
m in which
every Qn
i is Pi(+)M -secondary. Then N has a secondary representation of the form
N = Q1 +Q2 + . . .+Qm.
Proof. Let N = Q1 + Q2 + . . . + Qm be a secondary representation of N with
rad
(
annQi
)
= Pi. Then Nn = Qn
1 + Qn
2 + . . . + Qn
m, by Lemma 1. To see why
this is a secondary representation of Nn note first that for each (r,m) ∈ R(M),
the endomorphism φ(r,m) : Qn
i → Qn
i defined by φ(r,m)((1, q1,i, q2,i, . . . , qn,i)) =
= (r,m)(1, q1,i, q2,i, . . . , qn,i) = (1, rq1,i, rq2,i, . . . , rqn,i) induces endomorphism ϕr :
Qi → Qi defined by ϕr(q) = rq. Since Qi is secondary, ϕr is either surjective or
nilpotent. If ϕr is surjective, then it is clear that φ(r,m) is surjective. If ϕr is nilpotent,
then there exists a positive integer m1 such that (ϕr)m1 = 0 and therefore, rm1q = 0,
for all q ∈ Q. It follows that (r,m)m1(1, q1,i, q2,i, . . . , qn,i) = (1, rm1q1,i, r
m1q2,i, . . .
. . . , rm1qn,i) = (1, 0, 0, . . . , 0) for all (1, q1,i, q2,i, . . . , qn,i) ∈ Qn
i and hence φm1
(r,m) =
= 0, i.e., φ(r,m) is nilpotent. This yields that Qn
i is a secondary submodule of Mn. It
remains to be shown that rad
(
annQn
i
)
= Pi(+)M, for all 1 ≤ i ≤ n. Suppose that
(t,m) ∈ rad
(
annQn
i
)
. Then there exists a positive integer m1 such that tm1Qi = 0.
So t ∈ rad
(
annQi
)
= Pi. It turns out that rad
(
annQn
i
)
⊆ Pi(+)M. One can check
that Pi(+)M ⊆ rad
(
annQn
i
)
. Thus Qn
i is Pi(+)M -secondary.
Conversely, let Nn has a secondary representation of the form Nn = Qn
1 +Qn
2 + . . .
. . . + Qn
m, in which Qn
i is Pi(+)M -secondary, for all 1 ≤ i ≤ n. Clearly, N =
= Q1 +Q2 + . . .+Qm. We show that each Qi is a Pi-secondary submodule of M. Let
ϕr : Qi → Qi be an endomorphism defined by ϕr(q) = rq. We show that ϕr is either
surjective or nilpotent. The endomorphism ϕr induces endomorphism φ(r,0) : Qn
i → Qn
i
defined by φ(r,0)(1, q1, q2, . . . , qn) = (1, rq1, rq2, . . . , rqn). Since Qn
i is a secondary
submodule, φ(r,0) is either surjective or nilpotent, and so ϕr is either surjective or
nilpotent. It is easy to check that rad
(
annQi
)
= Pi. Hence Qi is Pi-primary.
Example 1. Let R be an integral domain and K be quotient field R. Then Kn is
a 0(+)M secondary R(+)K-module. In particular, Q2 is a 0(+)Q secondary Z(+)Q-
module.
Example 2. If P is a maximal ideal of R, then
Rn
(Pm)n
is a P (R)-secondary
R(R)-module, for every positive integer m.
Example 3. Let R be a local ring with maximal ideal P. If every element of P is
nilpotent, then Rn is a P (R)-secondary R(R)-module.
Acknowledgements. The authors would like to express their deep aptitude to the
referee for the careful reading of the paper and her/his fruitful comments.
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Received 08.06.10,
after revision — 06.02.11
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