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EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 9, No. 3, 2016, 244-249

ISSN 1307-5543 – www.ejpam.com

Comultiplication Modules

Jafar A’zami, M. Khajepour∗

Faculty of mathematical sciences,Department of Mathematics, University of Mohaghegh Ardabili,

Ardabil, Iran

Abstract. Let R be a commutative ring. An R-module M is comutiplication if for every submodule N

of M there exists an ideal I of R such that N = (0 :M I). This paper is devoted to study some properties

of comultiplication rings and modules.

2010 Mathematics Subject Classifications: 13C13

Key Words and Phrases: Multiplication modules, Comultiplication modules

1. Introduction

Throughout this paper, R will denote a commutative ring with identity. We recall that

R-module M is comutiplication if for every submodule N of M there exists an ideal I of R

such that N = (0 :M I). It was shown that M is comultiplication if and only if for each

submodule N of M , N = (0 :M AnnR(N)) [4]. Also a Noetherian local ring R is a gorenstein

ring if in jdimR<∞ [6]. In this article, among other results, we will show that if R is a local

Artinian ring, then R is comultiplication if and only if R is gorenstein. An R-module M is called

generalized hopfian, if every surjective endomorphism of M has a small kernel. It is proved

that every comultiplication module is generalized hopfian. At last but not at least, we consider

the direct sum of comultiplication modules, it is shown that M =
⊕

i∈I Mi , is comultiplication

if and only if for each i ∈ I , Mi is a comultiplication module and for each submodule N of

M , N =
⊕

i∈I(N
⋂

Mi).

2. Auxiliary Results

In this section we will provide the definitions and results which are necessary in the next

section.

Definition 1.

∗Corresponding author.

Email addresses: jafar.azami@gmail.com, azami@uma.ac.ir (J. A’zami),

maryamkhajepour@uma.ac.ir (M. Khajepour)

http://www.ejpam.com 244 c© 2016 EJPAM All rights reserved.



J. A’zami, M. Khajepour / Eur. J. Pure Appl. Math, 9 (2016), 244-249 245

(1) Let M be an R− module. A submodule N of M is said to be large (resp. small) if for every

non-zero submodule K of M, we have N
⋂

K 6= 0 (resp. N + K 6= M).

(2) An R-module M is called generalized hopfian, if every subjective endomorphism of M has a

small kernel.

(3) An R-module M is called weakly co-hopfian, if every injective endomorphism of M has a

large image.

(4) Let I be an ideal of R. We say that I is a second ideal of R, if for each r ∈ R, r I = 0 or

r I = I .

(5) An R-module M is called uniform, if every submodule of M is large.

(6) An ideal I of R is a pure ideal if for each ideal J of R, I J = I ∩ J.

(7) A submodule N of M is a copure submodule if for each ideal I of R, (N :M I) = N+(0 :M I).

(8) An R-module M is called weak comultiplication if for every prime submodule N of M, there

exists an ideal I of R such that N = (0 :M I)

Theorem 1. Let R be a discrete valuation ring with the unique maximal ideal m. If R-module M

is comultiplication, then M ∼= E(R/m) or M ∼= R/mn, for some n ∈ N.

Proof. See [1] and [2].

Theorem 2 ([4]). Let R be a Noetherian ring, and M be a comultiplication R-module so M is

Artinian.

Theorem 3 ([3]). Let R be a Noetherian ring, and M be an injective multiplication R-module, so

M is comultiplication.

Lemma 1 ([5]). If M is a comultiplication R -module, then for each endomorphism f of M,

Imf = AnnR(ker f )M.

3. Main Results

Lemma 2. Let R be a Noetherian ring. Then the following statements are equivalent:

(1) R is a comultiplication ring;

(2) For all P ∈ Spec(R), RP is a comultiplication ring;

(3) For all P ∈ Max(R), RP is a comultiplication ring.

Proof. (1→2) Let J be an ideal of RP , then there exists an ideal I of R such that J = IP .

Now I = AnnAnnI and so J = IP = AnnAnnIP = AnnAnnJ .

(2→3) It is clear.

(3→1) Let I be an ideal of R. For all P ∈ Max(R), we have IP = AnnAnnIP = (AnnAnnI)P
and so I = AnnAnnI .



J. A’zami, M. Khajepour / Eur. J. Pure Appl. Math, 9 (2016), 244-249 246

Theorem 4. Let M be a comultiplication R-module then

(1) If I is a second ideal of R, then N = (0 :M I) is a prime submodule of M.

(2) If N is a second submodule of M, then AnnRN is a prime ideal of R.

(3) If M is faithful, and N a submodule of M such that AnnRN is a large ideal of R, then N is

a small submodule of M.

(4) If N be a submodule of M such that AnnN is a pure ideal of R, then N is a copure submodule

of M.

Proof.

(1) Let r ∈ R and m ∈ M be elements such that rm ∈ N and r 6∈ (N :R M). Therefore

rM 6⊆ N and so rM I 6= 0. This shows that r I 6= 0, and by hypothesis r I = I . Since

rm ∈ N = (0 :M I), it follows that rmI = 0 and so mI = 0, that implies m ∈ (0 :M I) = N .

(2) Let N be a second submodule of M . Set I := AnnRN and so N = (0 :M I). Suppose that

x , y be two elements of R such that x y ∈ I but x 6∈ I and y 6∈ I . Now x y ∈ I , implies that

x yN = 0 and hence (x y)nN 6= N for each n ∈ N . Since x , y 6∈ I , it follows that there

exists n ∈ N such that xnN = N and ynN = N . Consequently (x y)nN = xn ynN = N ,

which is a contradiction.

(3) Let there exists a submodule K of M such that M = N + K . So

M = N + K = (0 :M AnnN) + (0 :M AnnK) = (0 :M AnnN
⋂

AnnK).

Therefore AnnN
⋂

AnnK ⊆ AnnM = 0, and consequently AnnK = 0 which implies that

K = (0 :M AnnK) = M .

(4) We show that for each ideal I of R, (N :M I) = N +(0 :M I). Note that for each ideal I of

R there exists a submodule K of M such that (0 :M I) = (0 :M AnnK), so

(N :M I) =((0 :M AnnN) :M I) = ((0 :M I) :M AnnN) = ((0 :M AnnK) :M AnnN)

=(0 :M AnnKAnnN) = (0 :M AnnK
⋂

AnnN)

=(0 :M AnnN) + (0 :M AnnK) = N + (0 :M I).

Theorem 5. Every comultiplication module is a generalized hopfian and weakly co-hopfian mod-

ule.

Proof. Let M be a comultiplication module and f be a surjective endomorphism of M .

Suppose that there exists a submodule N of M such that M = ker f + N . In this case

f (M) = f (ker f + N) = f (N). Therefore M = f (N) = (0 :M (0 :R f (N))).



J. A’zami, M. Khajepour / Eur. J. Pure Appl. Math, 9 (2016), 244-249 247

Since M is a comultiplication R-module, it follows that f (N) ⊆ N and so we have:

M = (0 :M (0 :R f (N))) ⊆ (0 :M (0 :R N)) = N ,

and hence ker f is a small submodule of M . Now let f be an injective endomorphism of M and

N be a submodule of M such that Imf
⋂

N = 0. By the previous lemma, Imf = AnnR(ker f )M ,

and so AnnR(ker f )M
⋂

N = 0. But ker f = 0 and hence N = M
⋂

N = 0.

Theorem 6. let (R, m) be a local Artinian ring, then the following statements are equivalent:

(1) R is a comultiplication ring;

(2) R is a gorenstein ring;

(3) soc(R)≈ R/m;

(4) E(R/m) is a multiplication R module.

Proof. (1→ 2) It is enough to show that r(R) = 1. Suppose on the contrary that r(R) 6= 1,

so r(R) = 0 or r(R)> 1. If r(R) = 0, then r(R) = dimkHomR(R/m,R) = 0 and so

(0 :R m)≈ HomR(R/m,R) = 0. Which is a contradiction, because the annihilator of any proper

ideal of an Artinian local ring is non-zero. Now suppose thatr(R)> 1, so there exist two ideals

I and J of R such that (0 :R m) = I
⊕

J = (0 :R AnnI)
⊕
(0 :R AnnJ) = (0 :R AnnI

⋂
AnnJ), this

means that (0 :R AnnI
⋂

AnnJ) 6= 0. On the other hand

I
⋂

J = AnnAnnI
⋂

AnnAnnJ = Ann(AnnI + AnnJ) 6= 0,

Which is a contradiction.

(2→3) Since R is gorenstein, it follows from [6], for all non-zero ideals I and J of R,

I
⋂

J 6= 0. Now let S1 and S2 be two simple submodules of R, then S1

⋂
S2 6= 0 and conse-

quently S1 = S2.

(3→4) Since soc(R) = (0 :R m) ≈ R/m, it follows that r(R) = 1 and so R is a gorenstein

ring. On the other hand dimR = in jdimR = 0. Thus R is an injective R module and so

R≈ E(R/m), by [6].

(4→1) E(R/m) is multiplication and Artinian, it follows that E(R/m) is cyclic and so

E(R/m) ≈ R. Now R is an injective and multiplication R module, then by [3] R is comultipli-

cation.

Theorem 7. Let (R, m) be a local Artinian ring, and M be a faithful comultiplication R-module.

Then M is uniform.

Proof. Let (R, m) be a local Artinian ring, and M be a faithful comultiplication R-module

and N be a submodule of M such that N ∩K = 0, for some submodule K of M . Then we have

N ∩ K = (0 :M AnnR(N))∩ (0 :M AnnR(K)) = (0 :M AnnR(N) + AnnR(K)).



J. A’zami, M. Khajepour / Eur. J. Pure Appl. Math, 9 (2016), 244-249 248

Now if AnnR(N) + AnnR(K) = 0, then

N ∩ K = (0 :M 0) = M 6= 0

So suppose that AnnR(N) + AnnR(K) 6= 0, then AnnR(N) + AnnR(K) ⊆ m. Therefore

0= N ∩ K = (0 :M AnnR(N) + AnnR(K)). (1)

Hence

Ann(AnnR(N) + AnnR(K)) ⊆ Ann(M) = 0 (2)

which is a contradiction, because R is Artinian.

Lemma 3. Let (R, m) be an Artinian local ring. Then R is comultiplication if and only if

(0 :R mn+1)/(0 :R mn)≃ mn/mn+1 for all n≥ 0.

Proof. Let R be a comultiplication ring, then by Theorem 6, soc(R) ≈ R/m and so we have

(0 :R m)≈ R/m. Now let n≥ 1, consider the following exact sequence:

0→ mn/mn+1→ R/mn+1→ R/mn→ 0

Since R is gorenstein, it follows that 0 = dimR ≤ in jdimR = depthR ≤ dimR = 0, and so

R is injective R-module. Therefore we have the following exact sequence.

0→ (0 :R mn)→ (0 :R mn+1)→ Hom(mn/mn+1,R)→ 0

On the other hand r(R) = 1 and we have

Hom(mn/mn+1,R)≈ Hom(

t⊕

i=1

R/m,R)≈
t⊕

i=1

Hom(R/m,R)≈
t⊕

i=1

R/m= mn/mn+1.

Therefore by the last exact sequence (0 :R mn)/(0 :R mn+1)≈ mn/mn+1.

Lemma 4. Let M1, M2 be two submodules of a comultiplication R-module M such that

M = M1

⊕
M2. Then HomR(M1, M2) = HomR(M2, M1) = 0.

Proof. Let f : M1→ M2 be a homomorphism. Since M is comultiplication, f (M1) ⊆ M1, by

[2]. On the other hand f (M1) ⊆ M2 and so f (M1) ⊆ M1

⋂
M2 = 0. This shows that f = 0.

Theorem 8. Let R be a Dedekind domain, and M be a comultiplication module, then there exist

distinct maximal ideals Pi i∈I of R and submodules Mi , i ∈ I of M, such that M =
⊕

i∈I Mi and

for each i ∈ I , Mi
∼= E(R/Pi) or Mi

∼= R/P
ni

i
, for some ni ∈ N.



REFERENCES 249

Proof. Let R be a Dedekind domain and M be comultiplication, so R is Noetherian and M is

Artinian by [4]. Set M(P) := {m ∈ M | ∃n ∈ N , Pnm= 0} for P ∈ Spec(R). There exist distinct

maximal ideal {Pi}i∈I such that M =
⊕

i∈I M(Pi). Let Mi = M(Pi), since M is comultiplication,

it follows that each Mi is also comultiplication. On the other hand each Mi is an RPi
-module.

So by Theorem 8 for each i ∈ I , Mi
∼= E(RPi

/PiRPi
) or Mi

∼= RPi/PiR
ni

Pi
, since Pi

⋂
(R \ Pi) = ;,

it follows that E(RPi
/PiRPi

)∼= E(R/Pi), also RPi
/PiR

ni

Pi

∼= R/P
ni

i
.

Theorem 9. Let R ⊆ R be an integral extension and R be weak comultiplication, then R is weak

comultiplication.

Proof. Let P ∈ Spec(R) so there exists a prime ideal q of R such that p = qc so

p = qc = (0 :R AnnR(q))
c ⊇ (0 :R AnnR(q

c)) = (0 :R AnnR(p)). (3)

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