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EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 8, No. 2, 2015, 232-238

ISSN 1307-5543 – www.ejpam.com

On a Proper Subclass of Primeful Modules Which Contains the

Class of Finitely Generated Modules Properly

Hosein Fazaeli Moghimi∗, Fatemeh Rashedi

1,2 Department of Mathematics, University of Birjand, P.O. Box 97175-615, Birjand, Iran

Abstract. Let R be a commutative ring with identity and M a unital R-module. Moreover, let PSpec(M)

denote the primary-like spectrum of M and Spec(R/Ann(M)) the prime spectrum of R/Ann(M). We

define an R-module M to be a φ-module, if φ : PSpec(M)→ Spec(R/Ann(M)) given by

φ(Q) =
p

(Q : M)/Ann(M) is a surjective map. The class of φ-modules is a proper subclass of primeful

modules, calledψ-modules here, and contains the class of finitely generated modules properly. Indeed,

φ and ψ are two sides of a commutative triangle of maps between spectrums. We show that if R is an

Artinian ring, then all R-modules are φ-modules and the converse is true when R is a Noetherian ring.

2010 Mathematics Subject Classifications: 13C13, 13C99

Key Words and Phrases: Primary-like submodule, φ-module, Prime submodule, ψ-module

1. Introduction

Throughout this paper, all rings are commutative with identity and all modules are unital.

For a submodule N of an R-module M , (N : M) denotes the ideal {r ∈ R | rM ⊆ N} and

annihilator of M , denoted by Ann(M), is the ideal (0 : M). A submodule P of an R-module

M is said to be prime (or p-prime) if P 6= M and for p = (P : M), whenever rm ∈ P (where

r ∈ R and m ∈ M) then m ∈ P or r ∈ p [5, 11, 12]. The collection of all prime (resp. p-

prime) submodules of M , denoted by Spec(M) (resp. Specp(M)), is called the prime (resp.

p-prime) spectrum of M . Also the intersection of all prime submodules of M containing a

submodule N is called the radical of N and is denoted by rad N . In the ideal case, we denote

the radical of an ideal I of R by
p

I . An R-module M is said to be a primeful module or a

ψ-module if either M = (0) or M 6= (0) and the map ψ : Spec(M) → Spec(R/Ann(M)),

defined by ψ(P) = (P : M)/Ann(M), is surjective [9]. If M/N is a ψ-module over R, then
p

(N : M) = (rad N : M) [9, Proposition 5.3]. A submodule Q of M is said to be primary-

like if Q 6= M and whenever rm ∈ Q (where r ∈ R and m ∈ M) implies r ∈ (Q : M) or

∗Corresponding author.

Email addresses: hfazaeli@birjand.ac.ir (HF. Moghimi), fatemehrashedi@birjand.ac.ir (F.
Rashedi)

http://www.ejpam.com 232 c© 2015 EJPAM All rights reserved.



H. Moghimi, F. Rashedi / Eur. J. Pure Appl. Math, 8 (2015), 232-238 233

m ∈ radQ [7]. The primary-like spectrum PSpec(M) is defined to be the set of all primary-like

submodules N of M such that M/N is a ψ-module over R. In [7, Lemma 2.1] it is shown that,

if Q ∈ PSpec(M), then (Q : M) is a primary ideal of R and so p =
p

(Q : M) is a prime ideal of

R. In this case, the primary-like submodule Q is also called a p-primary-like submodule of M .

Definition 1. We say that an R-module M is a φ-module if either M = (0) or M 6= (0) and the

map φ : PSpec(M)→ Spec(R/Ann(M)) defined by φ(Q) =
p

(Q : M)/Ann(M) is surjective.

The saturation of a submodule N of an R-module M with respect to a prime ideal p of R is

the contraction of Np in M and designated by Sp(N). It is known that [4, 10]

Sp(N) = {m ∈ M | rm ∈ N for some r ∈ R\p}.

If p ∈ Spec(R) and N is a submodule of an R-module M such that (N : M) ⊆ p and M/N

is a ψ-module over R, then Sp(N + pM) is a p-prime submodule of M [9, Proposition 4.4].

Therefore ρ : PSpec(M) → Spec(M) defined by ρ(Q) = Sp(Q + pM) is a well-defined map,

where p =
p

(Q : M). It is easy to see that φ = ψ ◦ ρ, ψ composed with ρ. Thus, if φ is a

surjective map, so is ψ. This means that every φ-module is a ψ-module. We give an example

of a ψ-module module which is not a φ-module (Example 1). An R-module M is said to be

multiplication module if every submodule N of M is of the form I M for some ideal I of R

[6]. We show that the multiplicationψ-modules, finitely generated modules, free modules (of

finite or infinite rank), faithful projective modules over domains and modules over Artinian

rings are φ-modules (Theorem 1, Corollary 1, Theorem 2, Theorem 3 and Theorem 4).

2. φ-Modules

We will use X , Xp and Xp to represent PSpec(M), Specp(M) and

{Q ∈ PSpec(M) | p(Q : M) = p} respectively. Also V (Ann(M)) will be the set of all prime

ideals containing Ann(M). We begin with a lemma which will be referred to in the rest of this

section.

Lemma 1 (cf. [9, Theorem 2.1]). Let M be a non-zero R-module. Then the following statements

are equivalent.

(1) M is a ψ-module;

(2) Xp 6= ; for every p ∈ V (Ann(M));

(3) pMp 6= Mp for every p ∈ V (Ann(M));

(4) Sp(pM) is a p-prime submodule for every p ∈ V (Ann(M)).

Theorem 1. Every φ-module M over a ring R is a ψ-module, and the converse is true in each of

the following cases.

(1) M is a multiplication R-module.



H. Moghimi, F. Rashedi / Eur. J. Pure Appl. Math, 8 (2015), 232-238 234

(2) M is a non-zero faithfully flat (or in particular a projective) R-module.

(3) M/Sp(pM) is a ψ-module over R for every p ∈ V (Ann(M)).

Proof. Since φ = ψ ◦ ρ, every φ-module is a ψ-module. (1) Let p ∈ V (Ann(M)). Then

there exists a prime submodule P such that (P : M) = p. Since M is a multiplication module

P = pM . Suppose q ∈ Spec(R) and p ⊆ q. By Lemma 1, there exists a prime submodule P ′

such that (P ′ : M) = q. It follows that P = pM ⊆ qM = P ′. Hence M/P is a ψ-module and

so P ∈ PSpec(M). Now from φ(P) = p/Ann(M), we conclude that φ is surjective, i.e., M is

a φ-module. (2) Let p ∈ V (Ann(M)) and (P : M) = p. If M is a projective module, then pM

is a prime submodule of M by [1, Corollary 2.3]. Also if M is a faithfully flat module, then

pM is a prime submodule by [3, Corollary 2.6]. On the other hand M/pM is a ψ-module and

(pM : M) = p by [9, Corollary 4.3 and Proposition 4.5]. Consequently pM ∈ Xp. Thus M is

a φ-module. (3) Since M is a ψ-module, Sp(pM) is a p-prime submodule of M by Lemma 1.

Hence Sp(pM) ∈ Xp. Thus M is a φ-module.

The following example shows that a ψ-module is not necessarily a φ-module.

Example 1 (cf. [9, Example 1]). Let Ω be the set of all prime integers, M =
∏

p∈Ω
Z
pZ and

M ′ =
⊕

p∈Ω
Z
pZ , where p runs through Ω. Hence M is a faithful ψ-module over Z and

Spec(M) = {M ′ = S0(0)} ∪ {pM : p ∈ Ω}. Now if φ is surjective, then there exists N ∈ X
such that φ(N) =

p

(N : M) = 0. It follows that (N : M) = 0. Since M/N is a ψ-module, we

have N ⊆ ∩p∈ΩpM = 0. But 0 is not prime and so is not primary-like because rad0 = 0. Hence

N /∈ X , a contradiction. Thus M is not a φ-module.

Corollary 1. Every finitely generated R-module M is a φ-module, hence so is the factor module

M/N of M by any submodule N of M.

Proof. Follows from Lemma 1 and Theorem 1.

Corollary 2. Let R be a ring of (Krull) dimension 0 and M be a non-zero R-module. Then the

following statements are equivalent.

(1) mM 6= M for every m ∈ V (Ann(M))∩Max(R);

(2) M is a ψ-module;

(3) M is a φ-module.

Proof. (1)⇔ (2) follows from [9, Result 3].

(2) ⇒ (3) Suppose M is a ψ-module. We show that M/Sp(pM) is a ψ-module for every

p ∈ V (Ann(M)). Assume (Sp(pM) : M) ⊆ q for a prime ideal q of R. Hence p ⊆ q. Since

dim(R) = 0, then p = q. Hence Sq(qM) is a q-prime submodule containing Sp(pM). Thus M

is a φ-module by Theorem 1.

(3)⇒ (2) follows from Theorem 1.



H. Moghimi, F. Rashedi / Eur. J. Pure Appl. Math, 8 (2015), 232-238 235

Corollary 3. Let R be a domain which is not a field. If a non-zero R-module M is either a divisible

module or a faithful torsion module, then M is not a φ-module.

Proof. Use Theorem 1 and [9, Proposition 2.6].

Theorem 2. Every free module is a φ-module.

Proof. Suppose F is a free R-module and p ∈ Spec(R/Ann(F)). It is easy to see that pF is

a prime, and hence a primary-like submodule, of F . Now we show that F/pF is a ψ-module.

Assume q is a prime ideal of R containing (pF : F). It follows from [13, Proposition 2.2] that

(qF : F) = q and hence qF 6= F . Thus qF is a q-prime submodule of F containing pF [11,

Theorem 3]. It implies that F/pF is a ψ-module.

Theorem 3. Let R be a domain and M be a faithful projective R-module. Then M is a φ-module.

Proof. Assume M 6= (0) and p ∈ Spec(R). We show that pM ∈ X . By [9, Corollary 3.4], M

is a ψ-module and hence pM 6= M by [9, Result 2]. It follows from [11, Theorem 3] that pM

is a p-prime, and hence a p-primary-like, submodule of M . It remains to show that M/pM is

a ψ-module. Suppose q is a prime ideal of R containing p = (pM : M). Therefore pM ⊆ qM

and qM ∈ Xq. Thus M/pM is a ψ-module and so M is a φ-module.

Proposition 1. Let M be a non-zero φ-module over a ring R. Then the following statements hold.

(1) Let I be a radical ideal of R. Then (I M : M) = I if and only if I ⊇ Ann(M).

(2) mM ∈ X for every m ∈ V (Ann(M))∩Max(R).

(3) If M is faithful, then M is flat if and only if M is faithfully flat.

Proof. (1) follows from [9, Proposition 3.1] and Theorem 1.

(2) By Theorem 1, M is aψ-module. Hence by [9, Result 2], mM 6= M . Thus mM is a m-prime,

and hence m-primary-like, submodule of M . It remains to show that M/mM is a ψ-module.

Assume p is a prime ideal of R containing (mM : M). Since m ∈ Max(R), then m = p and so

M/mM is a ψ-module. Thus mM ∈ X .

(3) The sufficiency is clear. Suppose that M is flat. Hence by part (2), we have mM 6= M for

every m ∈ Max(R). This implies that M is faithfully flat.

We give an elementary example of a module which is not a φ-module.

Example 2. The Z-module Q is flat and faithful, but not faithfully flat. So, Q is not a φ-module,

by Proposition 1.

Proposition 2. Let M be a non-zero φ-module over a ring R. Then Mp is a non-zero φ-module

over Rp for every p ∈ V (Ann(M)).



H. Moghimi, F. Rashedi / Eur. J. Pure Appl. Math, 8 (2015), 232-238 236

Proof. Suppose M is a non-zeroφ-module over R. Hence Mp 6= (0) for every p ∈ V (Ann(M)).

Assume q′ ∈ Spec(Rp/Ann(Mp)). We set q = (q′)c , the contraction of q′ in R. It is easy to check

that q is a prime ideal of R. We show that there exists a q′-primary-like submodule Qp of Mp

such that Mp/Qp is a ψ-module. Since Rp is a local ring, pp ⊇ q′ ⊇ Ann(Mp) ⊇ (Ann(M))p.

Taking the contraction of each term of this sequence of ideals in R, we have that

p ⊇ q ⊇ Ann(Mp)∩ R ⊇ Sp(Ann(M)) ⊇ Ann(M).

Hence q ∈ Spec(R/Ann(M)). Since M is a φ-module over R, there exists Q ∈ X such that
p

(Q : M) = q. Thus by [7, Theorem 3.8] Qp is a q′-primary-like submodule in Mp such that

Mp/Qp is a ψ-module and hence Mp is a φ-module over Rp for every p ∈ V (Ann(M)).

Theorem 4. Let R be a ring. Consider the following statements.

(1) R is an Artinian ring.

(2) Every R-module is a φ-module.

(3) Every R-module is a ψ-module.

(4) mM 6= M for every R-module M and m ∈ V (Ann(M))∩Max(R).

(5) dim(R) = 0

Then (1) ⇒ (2) ⇒ (3) ⇒ (4) ⇒ (5). Furthermore, if R is a Noetherian ring, then the above

statements are equivalent.

Proof. (1)⇒ (2) Let M be a non-zero R-module. Then Ann(M) 6= R. Since R is Artinian,

we have R= R1× · · · ×Rn, where n ∈ N and each Ri is an Artinian local ring. First we assume

that n = 1, i.e., R is an Artinian local ring with maximal ideal m. Since J(R), the Jacobson

radical of R, equals to m and J(R) is T-nilpotent, mM 6= M by [8, Theorem 23.16]. Thus every

R-module is aφ-module, by Corollary 2. Now assume n≥ 2 and let mi be the maximal ideal of

the local ring Ri for every 1≤ i ≤ n. Let m be a maximal ideal of R containing Ann(M). Clearly

m is the form R1×· · ·Ri−1×mi ×Ri+1×· · ·×Rn for some i. Without loss of generality we may

assume that i = 1, i.e., m = m1 × R2 × · · · × Rn. Again, by Corollary 2, it suffices to show that

mM = (m1×R2× · · · ×Rn)M 6= M . On the contrary, suppose that (m1×R2× · · · ×Rn)M = M .

Take M1 = (R1 × (0) × · · · × (0))M . It is easy to verify that R1
∼= R/(0) × R2 × · · · × Rn and

hence M1 can be expressed as an R1-module by defining r1 x1 = r1(1,0, · · · , 0)x1 for r1 ∈ R1

and x1 ∈ M1. We may assume that M1 6= 0, for otherwise we have

R1 × (0)× · · · × (0) ⊆ Ann(M) ⊆ m1 × R2 × · · · × Rn,

a contradiction. Thus m1M1 6= M1 by using case n= 1. On the other hand, for each

x ∈ M , (1,0, · · · , 0)x ∈ M = (m1 × R2 × · · · × Rn)M . Thus for each x ∈ M ,

(1,0, · · · , 0)x = ∑sj=1(p1 j , r2 j , · · · , rn j)x j for some s ∈ N, x j ∈ M , p1 j ∈ M1 and ri j ∈ R,



REFERENCES 237

where 2 ≤ i ≤ n and 1 ≤ j ≤ s. Multiplying the former equation by (1,0, · · · , 0), we get

(1,0, · · · , 0)x ∈ (m1 × (0)× · · · × (0))M for each x ∈ M . It follows that

(R1 × (0)× · · · × (0))M ⊆ (m1 × (0)× · · · × (0))M
and so m1M1 = M1, a contradiction.

(2)⇒ (3) follows from Theorem 1.

(3)⇒ (4) follows from [9, Result 2].

(4) ⇒ (5) Suppose p be a prime ideal of R and K the quotient field of R/p. We know that

K is a non-zero divisible R/p-module. Let 0 6= r + p ∈ R/p. Then (r + p)K = K implies

that Ann(K) + R/p(r + p) = R/p. Otherwise, if Ann(K) + R/p(r + p) 6= R/p, then there is a

maximal ideal m/p of R/p containing Ann(K) + R/p(r + p). Thus K = (r + p)K ⊆ (m/p)K
follows that (m/p)K = K , contradicting the assumption in (4). Now, let Ann(K) 6= (0). Take

r+p ∈ Ann(K) and hence by the above argument Ann(K) = R/p, i.e., K = (0), a contradiction.

Thus Ann(K) = (0). Hence R/p(r + p) = R/p for any 0 6= r + p ∈ R/p. Thus dim(R) = 0.

(4)⇒ (5) follows from [2, Theorem 8.5].

The following is now immediate.

Corollary 4. Let R be a domain. Then the following statements are equivalent.

(1) Every R-module is a φ-module;

(2) Every R-module is a ψ-module;

(3) R is a field.

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