On principal ideal multiplication modules
Let $R$ be a commutative ring with identity and let $M$ be a unitary $R$-module. A submodule $N$ of $M$ is said to be a multiple of $M$ if $N = rM$ for some $r \in R$. If every submodule of $M$ is a multiple of $M$, then $M$ is said to be a principal ideal multiplication module. We characterize pri...
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| author | Azizi, A. Jayaram, C. Азізі, А. Джаярам, К. |
| author_facet | Azizi, A. Jayaram, C. Азізі, А. Джаярам, К. |
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| description | Let $R$ be a commutative ring with identity and let $M$ be a unitary $R$-module. A submodule $N$ of $M$ is said to be a multiple of $M$ if $N = rM$ for some $r \in R$. If every submodule of $M$ is a multiple of $M$, then $M$ is said to be a principal ideal multiplication module. We characterize principal ideal multiplication modules and generalize some results from [Azizi A. Principal ideal multiplication modules // Algebra Colloq. – 2008. – 15. – P. 637 – 648]. |
| first_indexed | 2026-03-24T02:10:49Z |
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UDC 512.5
A. Azizi (College Sci., Shiraz Univ., Iran),
C. Jayaram (Univ. West Indies, Barbados)
ON PRINCIPAL IDEAL MULTIPLICATION MODULES
ПРО ГОЛОВНI IДЕАЛЬНI МУЛЬТИПЛIКАТИВНI МОДУЛI
Let R be a commutative ring with identity and let M be a unitary R-module. A submodule N of M is said to be a
multiple of M if N = rM for some r \in R. If every submodule of M is a multiple of M, then M is said to be a
principal ideal multiplication module. We characterize principal ideal multiplication modules and generalize some results
from [Azizi A. Principal ideal multiplication modules // Algebra Colloq. – 2008. – 15. – P. 637 – 648].
Нехай R — комутативне кiльце з одиницею, а M — унiтарний R-модуль. Субмодуль N модуля M називається
кратним M, якщо N = rM для деякого r \in R. Якщо кожний субмодуль модуля M є кратним для M, то
M називається головним iдеальним мультиплiкативним модулем. У роботi наведено характеристики головних
iдеальних мультиплiкативних модулiв та узагальнено деякi результати з роботи [Azizi A. Principal ideal multiplication
modules // Algebra Colloq. – 2008. – 15. – P. 637 – 648].
1. Introduction. Throughout this paper R denotes a commutative ring with identity and M denotes
a unitary R-module. Also L(R) (resp. L(M)) denotes the lattice of all ideals (resp. submodules) of
R (resp. M ).
A submodule N of M is proper if N \not = M. For any two submodules N and K of M, the
ideal \{ a \in R | aK \subseteq N\} will be denoted by (N : K). Thus (O : M) is the annihilator of M.
A module M is said to be faithful if (O : M) is the zero ideal of R. We say that a module M is a
multiplication module [7] if every submodule of M is of the form IM, for some ideal I of R.
According to [1] a submodule N of M is said to be a multiple of M if N = rM for some
r \in R. Also if every submodule of M is a multiple of M, then M is said to be a principal ideal
multiplication module or a PI -multiplication module, for abbreviation. In [1] some properties and
examples of PI -multiplication modules are given.
In this paper, we show that PI -multiplication modules are very close to cyclic modules over
principal ideal rings. We will give many conditions under each of which a PI -multiplication module
is a cyclic module over a principal ideal ring. Indeed our attempt for finding a noncyclic, nonzero
PI -multiplication module was unsuccessful.
A proper submodule N of M is a prime submodule, if for any r \in R and m \in M, rm \in N
implies either m \in N or r \in (N : M).
By a minimal prime submodule over a submodule N of M (or a prime submodule minimal
over N ), we mean a prime submodule which is minimal in the collection of all prime submodules
containing N. Minimal prime submodules over the zero submodule are simply called the minimal
prime submodules. It is well known that maximal submodules and prime submodules exist in
multiplication modules (for details, see [7]). It is well known that if M is a multiplication R-module
and P is a prime ideal of R containing (O : M) such that M \not = PM, then PM is a prime submodule
of M and every prime submodule of M is of the form PM for some prime ideal P of R containing
c\bigcirc A. AZIZI, C. JAYARAM, 2017
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 3 291
292 A. AZIZI, C. JAYARAM
(O : M) (see [7], Corollary 2.11). Also if M is a finitely generated multiplication R-module and
P is a prime ideal of R containing (O : M), then PM is a proper prime submodule of M (see
Lemma 1.1(iii), in the following). Further PM is minimal over a submodule N of M if and only if
P is minimal over the ideal (N : M) of R.
According to [14], a submodule N of M is called quasicyclic if (B \cap (K : N))N = BN \cap K
and (K +BN : N) = (K : N) +B for all ideals B of R and for all submodules K of M.
Note that N is quasicyclic, if and only if N is finitely generated and locally cyclic, if and only
if N is a finitely generated multiplication submodule (by [14], Theorem 6 and Lemma 1.1(v)(c), in
the following).
Recall that an R-module M is called a cyclic submodule module (CSM), if every submodule of
M is cyclic. A module M (resp. ring R) is said to be Laskerian [10], if every proper submodule
(resp. ideal) is a finite intersection of primary submodules (resp. ideals).
A module M is said to be finitely annihilated if there exists a finite subset T of M with (O :
T ) = (O : M). The finitely annihilated concept is due to G. Gabriel [8]. Evidently every finitely
generated module is finitely annihilated. Also according to [1] (Proposition 4.7(i)), every semi-
nontorsion module is finitely annihilated. Also note that the \BbbZ -module \BbbQ is finitely annihilated, but
not finitely generated.
Let M be an R-module and P a prime ideal of R. We define TP (M) = \{ x \in M | (1 - p)x =
= 0, for some p \in P\} . It is said that M is P -torsion if TP (M) = M. We say that M is P -cyclic
if there exist p0 \in P and m \in M such that (1 - p0)M \subseteq Rm.
For any a \in R, the principal ideal generated by a is denoted by (a). Recall that an ideal I of R
is called a multiplication ideal if I is a multiplication R-module.
Every multiplication ideal is locally principal (see Lemma 1.1(v)(b), in the following).
An ideal I of R is called a quasiprincipal ideal [15, p. 147] (Exercise 10) (or a principal element
of L(R) [17]) if I is quasicyclic as an element of L(R), that is it satisfies the identities (i) (A\cap (B :
I))I = AI \cap B and (ii) (A+BI : I) = (A : I) +B, for all A, B \in L(R).
Obviously, every quasiprincipal ideal is a multiplication ideal. It should be mentioned that every
quasiprincipal ideal is finitely generated and also a finite product of quasiprincipal ideals of R is again
a quasiprincipal ideal [15, p. 147] (Exercise 10). In fact, an ideal I of R is quasiprincipal if and only
if it is finitely generated and locally principal (see [4], Theorem 4), or [17], Theorem 2). It is well
known that R is a general ZPI-ring if and only if every ideal is quasiprincipal [11] (Theorem 2.2).
A ring R is a \pi -ring if every principal ideal is a finite product of prime ideals of R.
If \{ P\alpha \} is the collection of all minimal prime ideals of an ideal I of R, then by an isolated
P\alpha -primary component of I we mean the intersection Q\alpha of all P\alpha -primary ideals which contain I.
The kernel of I is the intersection of all Q\alpha
\prime s. It is well known that if R is an almost multiplication
ring, then every ideal is equal to its kernel [6] (Theorem 2.9).
For the convenience of reader, some results from our references, which are used frequently in
this paper, have been gathered in the following lemma.
Lemma 1.1. Let M be a nonzero R-module.
(i) ([1], Proposition 3.5(ii)). If M is finitely generated and every prime submodule of M is
finitely generated, then M is a Noetherian module.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 3
ON PRINCIPAL IDEAL MULTIPLICATION MODULES 293
(ii) ([7], Corollary 4.8). If M is a Noetherian multiplication module, then there exist ideals
A \subseteq B of R such that M \sim = B/A.
(iii) ([7], Theorem 3.1). Let M be a multiplication module. Then M is finitely generated, if and
only if M \not = PM, for each maximal ideal P of R containing (O : M), if and only if for any ideals
A,B of R containing (O : M), the inclusion AM \subseteq BM implies that A \subseteq B.
(iv) ([1], Theorem 4.9(i)). If R/(O : M) is a principal ideal ring and M is multiplication, then
M is PI-multiplication.
(v) (a) ([7], Theorem 3.7 and [13], Lemma 1). If M is multiplication and M (resp. R) has only
finitely many minimal prime submodules (resp. ideals), then M is finitely generated.
(b) ([7], Theorem 2.8 and [3], Proposition 4). If M is multiplication and M (resp. R) has only
finitely many maximal submodules (resp. ideals), then M is cyclic.
(c) ([3], Proposition 5). Let M be a finitely generated module. Then M is multiplication if and
only if it is locally cyclic.
(vi) ([13], Lemmas 4 and 3). If M is a faithful multiplication module, then M is a Laskerian
module if and only if R is a Laskerian ring. Furthermore, in this case M is finitely generated.
(vii) ([7], Corollary 3.9). If M is multiplication and R/(O : M) has ACC on semiprime ideals,
then M is finitely generated.
(viii) ([13], Lemma 6). If M is cyclic, then a submodule N of M is cyclic if and only if N is a
multiple of M.
(ix) ([5], Theorem 2). If M is a multiplication module and every minimal prime submodule is
finitely generated, then M has only finitely many minimal prime submodules.
(x) ([13], Lemma 7). If every cyclic submodule of M has primary decomposition, then any
locally cyclic submodule of M is finitely generated.
For general background and terminology, the reader is referred to [15] and [19].
2. PI-multiplication modules.
Proposition 2.1. Let M be a PI-multiplication R-module.
(i) If M is finitely generated, then R/(O : M) is a principal ideal ring.
(ii) R/(O : M) is a locally principal ideal ring.
Proof. Clearly, M is a multiplication module. Let \=I be an ideal of R/(O : M). Then \=I = J/(O :
M) for some J \in L(R) containing (O : M). By hypothesis, JM = rM for some r \in R. Then
JM = (Rr + (O : M))M, and according to Lemma 1.1(iii), J = Rr + (O : M). So \=I = J/(O :
M) = (Rr+(O : M))/(O : M) = (R/(O : M))(r+(O : M)). Therefore R/(O : M) is a principal
ideal ring.
(ii) Evidently M is a PI -multiplication \=R-module, where \=R = R/(O : M). Let P be a prime
ideal of R containing (O : M). Put \=P = P/(O : M). According to [1] (Lemma 4.4(ii)), M \=P is a PI -
multiplication \=R \=P -module. Now since \=R \=P is a local ring, by (1.1)(v)(b), M \=P is a PI -multiplication
cyclic \=R \=P -module.
Therefore by (i), \=R \=P is a principal ideal ring.
Example 2.1. Any Dedekind domain is a locally principal ring, but it is not necessarily a principal
ideal ring. For example, the integral closure of \BbbZ in \BbbQ
\bigl[ \surd
10
\bigr]
is a Dedekind domain, but it is not a
PID.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 3
294 A. AZIZI, C. JAYARAM
The set of all prime submodules of M is denoted by \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}M. Let N \in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}M. The height of
N, which is denoted by ht K, is n, if there exists a chain Nn \subset . . . \subset N2 \subset N1 \subset N0 = N in
\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}M and there does not exist such a chain of greater length (see [2]).
The following theorem generalizes most of the results in [1]. This theorem introduces many
conditions under which a PI -multiplication module is a cyclic module over a principal ideal ring.
Theorem 2.1 (main theorem). Suppose M is a nonzero R-module. Then the following are
equivalent:
(i) M is a cyclic module and R/(O : M) is a principal ideal ring.
(ii) M is finitely generated and every prime submodule of M is a multiple of M.
(iii) M is a finitely generated PI -multiplication module.
(iv) M is a finitely annihilated PI -multiplication module.
(v) M is a multiplication module and R/(O : M) is a principal ideal ring.
(vi) M is a PI -multiplication module and (O : M) contains an ideal which has a primary
decomposition.
(vii) M is a PI -multiplication module and a finitely generated submodule of M has a primary
decomposition.
(viii) M is a PI -multiplication module and R/(O : M) has ACC on semiprime ideals.
(ix) M is cyclic and every prime submodule of M is cyclic.
(x) M is a CSM (i.e., every submodule of M is a cyclic submodule).
(xi) M is a PI -multiplication module and every minimal prime submodule is finitely generated.
(xii) M is a PI -multiplication module and MP \not = O, for each maximal (prime) ideal P
containing (O : M).
(xiii) M is a PI -multiplication module and M is P -cyclic for each maximal ideal P containing
(O : M).
(xiv) M is a PI -multiplication module and there exists a nonnegative integer n such that
\{ N \in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}M | ht N = n\} is nonempty and finite.
Proof. (ii) =\Rightarrow (iii). By hypothesis and by Lemma 1.1(i), M is a Noetherian module. We claim
that every submodule of M is a multiple of M. Suppose not. Then by Zorn’s lemma, there exists a
submodule N such that N is a maximal element in the set of all nonmultiples of M. By hypothesis,
N is not a prime submodule. So there exist elements a \in R and x \in M such that ax \in N, x \not \in N
and aM \not \subseteq N. By the maximality of N, we have N +Rx = cM for some c \in R.
Put L = (N : Mc). We show that N = cL. Evidently cL \subseteq N. Now if y \in N, then since
N \subseteq N +Rx = cM, there exists z \in M such that y = cz. Then z \in L and so y = cz \in cL.
Obviously N \subseteq L. Now we show that N \not = L. Note that aM \not \subseteq N, then let am \not \in N, where
m \in M. We have cm \in cM = N + Rx, then cm = n + rx, for some n \in N and r \in R. Thus
c(am) = an+ rax \in N, which implies that am \in L \setminus N.
Now again by the maximality of N, we have L = dM, for some d \in R. Therefore, N = cL =
= (cd)M, which is a contradiction.
(iii) =\Rightarrow (i). By Proposition 2.1(i), R/(O : M) is a principal ideal ring.
Now since M is a nonzero Noetherian multiplication R/(O : M)-module, according to
Lemma 1.1(ii), M \sim = B/A, where A,B are ideals of R/(O : M) with A \subseteq B. Note that R/(O :
M) is a principal ideal ring, then the ideal B is principal, which implies that M is a cyclic R/(O :
M)-module and consequently a cyclic R-module.
(i) =\Rightarrow (iv). The proof is clear.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 3
ON PRINCIPAL IDEAL MULTIPLICATION MODULES 295
(iv) =\Rightarrow (ii). By Lemma 1.1(iii), it is enough to show that M \not = PM, for any maximal ideal of R
containing (O : M). On the contrary let M = mM, where m is a maximal ideal of R containing (O :
M). Note that Mm is a multiplication Rm-module and Rm is a local ring, then by Lemma 1.1(v)(b),
Mm is a cyclic module. Now since Mm = mmMm, by Nakayama’s lemma there exist r \in R and
s \in R \setminus m such that (r/s)Mm = O and 1 - (r/s) \in mm.
Suppose that T = \{ t1, t2, t3, . . . , tn\} is a finite subset of M with (O : T ) = (O : M). Then
(r/s)(ti/1) = 0, for each 1 \leq i \leq n. So there exists si \in R \setminus m with sirti = 0, for each 1 \leq i \leq n.
Put \=s = s1s2s3 . . . sn. Then \=srti = 0, for each 1 \leq i \leq n, which implies that \=sr \in (O : T ) = (O :
M) \subseteq m. Hence r \in m, and so r/s \in mm, which is a contradiction, since 1 - (r/s) \in mm.
(i) =\Rightarrow (v). The proof is clear.
(v) =\Rightarrow (vi). Note that every principal ideal ring is a Noetherian ring and any Noetherian
ring is Laskerian. Then (O : M) has a primary decomposition. Now the proof is completed by
Lemma 1.1(iv).
(vi) =\Rightarrow (ii). We show that M has only finitely many minimal prime submodules, and therefore
by Lemma 1.1(v)(c), M is finitely generated.
Let N be a minimal prime submodule of M and suppose that (O : M) contains an ideal I such that
I has a primary decomposition. Let I = \cap n
i=1Qi, where Qi is a primary ideal, for each 1 \leq i \leq n.
Then
\bigcap n
i=1
Qi = I \subseteq (O : M) \subseteq (N : M), and so
\bigcap n
i=1
\sqrt{}
Qi \subseteq (N : M). Now since (N :
M) is a prime ideal,
\sqrt{}
Qj \subseteq (N : M), for some 1 \leq j \leq n, and then
\sqrt{}
QjM \subseteq N. Now as\sqrt{}
QjM is a prime submodule of M and N is a minimal prime submodule,
\sqrt{}
QjM = N, which
completes the proof.
(vi) =\Rightarrow (vii). Evidently M is a faithful multiplication R/(O : M)-module and since parts (vi)
and (v) are equivalent, R/(O : M) is a Noetherian ring, and so it is a Laskerian ring. Then by
Lemma 1.1(vi), M is a Laskerian R/(O : M)-module and clearly a Laskerian R-module. Thus the
zero submodule has a primary decomposition.
(vii) =\Rightarrow (iii). Let N be a finitely generated submodule of M that has a primary decomposition.
We have two cases:
Case 1. N = 0.
Case 2. N \not = 0.
First suppose that Case 1 holds. Assume that 0 =
\bigcap n
i=1
Qi, where Qi is a primary submodule of
M, for each 1 \leq i \leq n. Hence (O : M) =
\bigcap n
i=1
(Qi : M), which is a primary decomposition of (0 :
M). Now the proof is given by (vi) =\Rightarrow (ii) =\Rightarrow (iii).
Now let Case 2 is satisfied. It is easy to see that M/N is a PI -multiplication module and since
it’s zero submodule has a primary decomposition, by Case 1, M/N is a finitely generated module
and since N is a finitely generated module, M is finitely generated.
(v) =\Rightarrow (viii). The proof follows from Lemma 1.1(iv).
(viii) =\Rightarrow (ii). By Lemma 1.1(vii), M is finitely generated.
(ii) =\Rightarrow (ix). Suppose (ii) holds. By (ii) and (i), M is a cyclic module. Now by Lemma 1.1(viii),
every prime submodule is a cyclic submodule.
(ix) =\Rightarrow (x). By Lemma 1.1(viii), every prime submodule is a multiple of M and by the proof
of (ii) =\Rightarrow (iii), every submodule is a multiple of M and since M is cyclic, every submodule is a
cyclic submodule of M.
(x) =\Rightarrow (ii). The proof is given by Lemma 1.1(viii).
(x) =\Rightarrow (xi). Note that (x) and (iii) are equivalent. So M is a PI -multiplication module.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 3
296 A. AZIZI, C. JAYARAM
(xi) =\Rightarrow (iii). By Lemma 1.1 parts (ix) and (v)(i), M finitely generated.
(iii) =\Rightarrow (xii). Suppose MP = O, where P is a prime ideal of R containing (O : M). Let
M be generated by x1, x2, . . . , xn. For each 1 \leq i \leq n we have xi/1 = 0 in MP , hence there
exists si \in R \setminus P such that sixi = 0. Consequently s1s2 . . . snxi = 0, for each 1 \leq i \leq n, that is
s1s2 . . . sn \in (O : M) \subseteq P. Hence sj \in P, for some 1 \leq j \leq n, which is a contradiction.
(xii) =\Rightarrow (iii). According to Lemma 1.1(iii), it is enough to show that M \not = PM, for any maximal
ideal P of R containing (O : M). If M = PM, where P is a maximal ideal of R containing (O :
M). Then PPMP = MP and since by Lemma 1.1(v)(b), MP is a cyclic RP -module, Nakayama’s
lemma implies that there exists r \in RP such that rMP = O and r - 1 \in PP . Now if r \not \in PP , then r
is a unit in RP , which implies that MP = O. Also if r \in PP , then 1 \in PP , which is a contradiction.
(xii) =\Rightarrow (xiii). Let P be a maximal ideal of R containing (O : M). By [7] (Theorem 1.2), M is
P -torsion or P -cyclic. If TP (M) = M, then for each x \in M there exists p \in P with (1 - p)x = 0.
So x/1 = 0 in MP , that is MP = O, which is a contradiction. Therefore M is P -cyclic.
(xiii) =\Rightarrow (xii). Let P be a maximal ideal of R containing (O : M). By our assumption
there exist p \in P and m \in M such that (1 - p)M \subseteq Rm. If MP = O, then m/1 = 0, and
so there exists s \in R \setminus P such that sm = 0. Hence s(1 - p)M = O, i.e., s(1 - p) \in (O :
M) \subseteq P, which is impossible.
(xiv) =\Rightarrow (i). Let N be an arbitrary prime submodule of M and let \=P = (N : M)/(0 :
M). Evidentely \=P is a prime ideal of \=R = R/(0 : M). Then ht \=P = \mathrm{d}\mathrm{i}\mathrm{m} \=R \=P and since by
Proposition 2.1, \=R \=P is a principal ideal ring, we have ht P = \mathrm{d}\mathrm{i}\mathrm{m} \=R \=P \leq 1.
Suppose that N1 \subset N2 \subset N3 \subset . . . \subset Nn = N is a chain of prime submodules of M. Evidently
(O : M) \subset (N1 : M) and if (Ni : M) = (Ni+1 : M), for some 1 \leq i \leq n - 1, and so Ni = (Ni :
M)M = (Ni+1 : M)M = Ni+1. Thus (N1 : M)/(O : M) \subset (N2 : M)/(O : M) \subset (N3 : M)/(O :
M) \subset . . . \subset (Nn : M)/(O : M) = \=P is a chain of prime ideals of \=R. Hence ht N \leq ht \=P \leq 1,
for any prime submodule N of M. Therefore for the integer n introduced in our assumption, either
n = 0 or n = 1.
If n = 0, that is M has only finitely many minimal prime submodules, then Lemma 1.1(v)(a)
implies that M is finitely generated. So the conditions of (iii) are satisfied and as (iii) implies (i), M
is a cyclic module.
Now suppose that n = 1. As ht K \leq 1 for any prime submodule K of M, the elements of
\{ N \in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}M | ht N = 1\} are exactly the maximal submodules of M. Thus there are only finitely
many maximal submodules. So in this case according to Lemma 1.1(v)(b), M is cyclic.
(i) =\Rightarrow (xiv). It is easy to see that if N is a minimal prime submodule of M, then (N : M)/(O :
M) is a minimal prime ideal of R/(O : M) and since R/(O : M) is a Noetherian ring, it has
only finitely many minimal prime ideals. Hence \{ N \in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}M | ht N = 0\} is finite. Also note
that according to [7] (Theorem 2.5), every multiplication module has a maximal submodule, and so
\{ N \in \mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}M | ht N = 0\} is nonempty.
As a consequence, we have the following result, which is due to I. M. Isaacs.
Corollary 2.1. A ring R is a principal ideal ring if and only if every prime ideal of R is a
principal ideal.
To illustrate the Theorem 2.1, we provide the following example.
Example 2.2. Let K be a field and R = K[x] and consider M = R/Rxn, where n \in \BbbN . Then
M is an R-module, and (0 : M) = Rxn = (0 : T ), where T = \{ 1 + Rxn\} , and so M is finitely
annihilated. Also R/(0 : M) is a principal ideal ring, since R is a PID. As
\sqrt{}
(0 : M) =
\surd
Rxn = Rx
is a maximal ideal of R, the ideal (0 : M) is a primary ideal of R, particularly (0 : M) has a primary
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 3
ON PRINCIPAL IDEAL MULTIPLICATION MODULES 297
decomposition. Every submodule N of M is of the form N = I/Rxn, where I is an ideal of R, and
since R is a PID, I is a principal ideal of R, let I = Rr, where r \in R. Thus N = rM, and so M
is a finitely generated (cyclic) PI-multiplication R-module. One can easily see that if P = p/Rxn
is a prime submodule of M, then p is a prime ideal of R and since xn \in p, we get x \in p, and so
Rx \subseteq p. Note that Rx is a maximal ideal of R, thus Rx = p. This shows that M has just one prime
submodule, which is Rx/Rxn.
For the proof of the following lemma, see [16] (Lemma 1.4).
Lemma 2.1. Suppose M is a faithful quasicyclic R-module and N a submodule of M. Then
the following statements are equivalent:
(i) N is quasicyclic.
(ii) (N : M) is quasiprincipal.
(iii) N = IM for some quasiprincipal ideal I of R.
Definition 2.1. An R-module M is said to be a general quasicyclic module if every submodule
of M is quasicyclic.
Note that if M is a CSM, then M is a general quasicyclic module. But the converse is not
true since in a ring R, quasiprincipal ideals need not be principal ideals. General ZPI -rings are
examples of general quasicyclics.
Example 2.3. Consider R = M = Z[
\surd
- 5]. Then M is a general quasicyclic R-module, but it
is not a CSM, as R is not a principal ideal ring.
The following theorem, establishes several characterizations for general quasicyclic modules.
Theorem 2.2. Suppose M is a faithful R-module. Then the following statements are equivalent:
(i) R is a general ZPI-ring and M is a multiplication module.
(ii) M is a multiplication module, locally PI -multiplication module and every cyclic submodule
has only finitely many prime submodules minimal over it.
(iii) M is a multiplication module, locally PI -multiplication module and R is a \pi -ring.
(iv) M is a locally PI -multiplication module in which every cyclic submodule has a primary
decomposition.
(v) M is a general quasicyclic module (i.e., every submodule of M is quasicyclic).
(vi) M is quasicyclic and every prime submodule is quasicyclic.
(vii) M is a multiplication module, locally PI -multiplication module and every minimal prime
submodule over any cyclic submodule is finitely generated.
Proof. (i) =\Rightarrow (ii). As R is a Laskerian ring, by Lemma 1.1(vi), M is a finitely generated
Laskerian module. Since R is a general ZPI -ring, it follows that R is an almost multiplication ring
and so R is a locally principal ideal ring.
As M is a finitely generated faithful multiplication module, it follows that MP
\sim = RP , for each
prime ideal P of R, and R is a locally principal ideal ring. So M is a locally PI -multiplication
module and (ii) holds.
(ii) =\Rightarrow (iii). Note that by Lemma 1.1(v)(a), M is finitely generated. Again since for each
maximal ideal P of R, MP
\sim = RP , it follows that RP is a principal ideal ring. Therefore, R is an
almost multiplication ring.
We show that every quasiprincipal ideal has only finitely many minimal prime ideals. Let I
be a quasiprincipal ideal. Then IM is finitely generated. Suppose IM =
\sum n
i=1
Rxi for some
x1, . . . , xn \in M. Then IM =
\sum n
i=1
Rxi =
\sum n
i=1
(Rxi : M)M =
\Bigl( \sum n
i=1
(Rxi : M)
\Bigr)
M, so by
Lemma 1.1(iii), I =
\sum n
i=1
(Rxi : M). Again since each Rxi has only finitely many minimal prime
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 3
298 A. AZIZI, C. JAYARAM
submodules and M is a faithful finitely generated multiplication module, it follows that each ideal
(Rxi : M) has only finitely many minimal primes. Let P be a prime ideal minimal over I. Then by
[4] (Theorem 1), IP is completely join irreducible in RP , so IP = (Rxi : M)P for some i and so
P is minimal over (Rxi : M). Therefore, I has only finitely many minimal prime ideals. As R is
an almost multiplication ring, by [6] (Theorems 2.7 and 2.9), every ideal is equal to its kernel and so
I has a primary decomposition. Again by [12] (Theorem 6), R is a \pi -ring.
(iii) =\Rightarrow (iv). As R is a \pi -ring, then R has finitely many minimal prime ideals, so by
Lemma 1.1(v)(a), M is finitely generated. Let Rx be a cyclic submodule. Then by Lemma 2.1, (Rx :
M) is quasiprincipal, so by [12] (Theorem 6), (Rx : M) =
\bigcap n
i=1
Qi for some primary ideals
Q1, . . . , Qn of R. So by [7] (Theorem 1.6), Rx = (Rx : M)M =
\Bigl( \bigcap n
i=1
Qi
\Bigr)
M =
\bigcap n
i=1
(QiM).
Again by Lemma 1.1(iii) and [20] (Corollary 1), each QiM is a primary submodule. Therefore,
every cyclic submodule has a primary decomposition. Thus (iv) holds.
(iv) =\Rightarrow (v). Note that M is locally cyclic, by Lemma 1.1(v)(b). So by Lemma 1.1(viii), every
submodule is locally cyclic. Now every submodule of M is finitely generated and locally cyclic, by
Lemma 1.1(x) and thus by [14] (Theorem 6), every submodule is quasicyclic.
(v) =\Rightarrow (vi). The proof is obvious.
(vi) =\Rightarrow (i). As M is quasicyclic, it follows that M is a finitely generated multiplication module.
Let P be a prime ideal of R. Then PM is a prime submodule of R, and so by our assumption PM
is quasicyclic. Now by Lemma 2.1, PM = IM, for some quasiprincipal ideal I of R, and according
to Lemma 1.1(iii), P = I. Hence P is quasiprincipal. As every prime ideal is quasiprincipal, it
follows that every ideal is quasiprincipal and hence R is a general ZPI -ring.
(ii) =\Rightarrow (vii). According to the proof of (iv) =\Rightarrow (v), every submodule of M is finitely generated.
(vii) =\Rightarrow (ii). As every minimal prime submodule over any cyclic submodule is finitely generated,
by Lemma 1.1(ix), every cyclic submodule has only finitely many prime submodules minimal over it.
Corollary 2.2. Suppose M is an R-module. Then the following statements are equivalent:
(i) R/(O : M) is a general ZPI-ring and M is a multiplication module.
(ii) M is a locally PI-multiplication R/(O : M)-module in which every cyclic submodule has a
primary decomposition.
Proof. The proof is clear by Theorem 2.2.
Corollary 2.3. If M is a multiplication R-module in which every prime submodule is cyclic,
then R/(O : M) is a general ZPI-ring and M is a Noetherian locally PI-multiplication R/(O :
M)-module.
Proof. Put \=R = R/(O : M). Then M is a faithful multiplication \=R-module in which every
prime submodule is cyclic. By Lemma 1.1(v)(b), for each prime ideal \=P of \=R, the \=R \=P -module M \=P
is cyclic. Now let N be an arbitrary prime submodule of M \=P . Then N c = \{ x \in M | x/1 \in N\}
is a prime submodule of M and since by our assumption N c is cyclic, N = (N c) \=P is cyclic.
Hence by Theorem 2.1(ix), M \=P is a PI -multiplication \=R \=P -module. Now we have the conditions of
Theorem 2.2(vii), which completes the proof.
If M is a noncyclic PI -multiplication R-module, then according to different parts of Theorem 2.1,
we have:
(i) None of the ideals of R contained in (0 : M) has a primary decomposition, particularly R is
neither a Noetherian ring nor an integral domain or a primary ring. Furthermore R does not have
ACC on semiprime ideals.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 3
ON PRINCIPAL IDEAL MULTIPLICATION MODULES 299
(ii) M is not a finitely annihilated module, particularly it is not a finitely generated module.
Also, none of the finitely generated submodules of M has a primary decomposition. Furthermore,
there exists a maximal ideal P of R with MP = 0, and PM = M, besides M has a nonfinitely
generated minimal prime submodule.
Therefore, according to the above discussion, if there is any nonzero noncyclic PI -multiplication
module, then it will be a very special example. This is a motivation for the following open problem.
Every nonzero PI -multiplication module is a cyclic module over a principal ideal ring.
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Received 09.03.14,
after revision — 11.12.16
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 3
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| id | umjimathkievua-article-1696 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:10:49Z |
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| publisher | Institute of Mathematics, NAS of Ukraine |
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| resource_txt_mv | umjimathkievua/d0/afd23086979a4e5296138ef877b66bd0.pdf |
| spelling | umjimathkievua-article-16962019-12-05T09:24:16Z On principal ideal multiplication modules Про головнi iдеальнi мультиплiкативнi модулi Azizi, A. Jayaram, C. Азізі, А. Джаярам, К. Let $R$ be a commutative ring with identity and let $M$ be a unitary $R$-module. A submodule $N$ of $M$ is said to be a multiple of $M$ if $N = rM$ for some $r \in R$. If every submodule of $M$ is a multiple of $M$, then $M$ is said to be a principal ideal multiplication module. We characterize principal ideal multiplication modules and generalize some results from [Azizi A. Principal ideal multiplication modules // Algebra Colloq. – 2008. – 15. – P. 637 – 648]. Нехай $R$ — комутативне кiльце з одиницею, а $M$ — унiтарний $R$-модуль. Субмодуль $N$ модуля $M$ називається кратним $M$, якщо $N = rM$ для деякого $r \in R$. Якщо кожний субмодуль модуля $M$ є кратним для $M$, то $M$ називається головним iдеальним мультиплiкативним модулем. У роботi наведено характеристики головних iдеальних мультиплiкативних модулiв та узагальнено деякi результати з роботи [Azizi A. Principal ideal multiplication modules // Algebra Colloq. – 2008. – 15. – P. 637 – 648]. Institute of Mathematics, NAS of Ukraine 2017-03-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1696 Ukrains’kyi Matematychnyi Zhurnal; Vol. 69 No. 3 (2017); 291-299 Український математичний журнал; Том 69 № 3 (2017); 291-299 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1696/678 Copyright (c) 2017 Azizi A.; Jayaram C. |
| spellingShingle | Azizi, A. Jayaram, C. Азізі, А. Джаярам, К. On principal ideal multiplication modules |
| title | On principal ideal multiplication modules |
| title_alt | Про головнi iдеальнi мультиплiкативнi модулi |
| title_full | On principal ideal multiplication modules |
| title_fullStr | On principal ideal multiplication modules |
| title_full_unstemmed | On principal ideal multiplication modules |
| title_short | On principal ideal multiplication modules |
| title_sort | on principal ideal multiplication modules |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1696 |
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