EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 3, 2018, 730-739
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

On α-prime and weakly α-prime submodules

Thawatchai Khumprapussorn

Department of Mathematics, Faculty of Science
King Mongkut’s Institute of Technology Ladkrabang, Bangkok 10520, Thailand

Abstract. We have introduced the notion of α-prime and weakly α-prime submodules as a gener-
alization of prime submodules. Some basic properties of α-prime and weakly α-prime submodules
are the extension of prime submodules. Finally, after introducing the notion of α-prime submod-
ules, we also define and study the concept of α-prime ideals in a ring.

2010 Mathematics Subject Classifications: 13C99

Key Words and Phrases: α-prime submodules, weakly α-prime submodules, α-prime ideals,
weakly α-prime ideals

1. Introduction

All rings are assumed to be commutative with nonzero identity and all modules are left
unital. Let (G,+) be a group. For a subset H of G, denote α(H) = {h ∈ G | h+ h ∈ H}
and β(H) = {h+ h | h ∈ H}. It is clear that β(H) ⊆ H ⊆ α(H). If I is an ideal of a ring
R, then α(I) and β(I) are ideals of R. If N is a submodule of a module M , then α(N)
and β(N) are submodules of M . We recall the definition of prime submodules from [1]. A
proper submodule P of a left R-module M is called prime if rm ∈ P for some r ∈ R and
m ∈M , then r ∈ (P : M) or m ∈ P where (N : M) = {r ∈ R | rM ⊆ N}.

Let M be a left R-module, m ∈ M and N be a submodule of M . For convenience,
we denote (0 : m) = {r ∈ R | rm = 0} and (N : m) = {r ∈ R | rm ∈ N}. With these
notations, we have both of (N : m) and (0 : m) are ideals of R.

It is well known that there are several authors have extended the notion of prime
submodules. All of those definitions focus on multiplication between element of rings
and of modules. This motivates us to study α-prime submodules by taking care on all
operations of a left module structure. Our extension obtains a generalization of prime
submodules which call α-prime submodules. Its definition and results appear in section 1.

In section 2, we introduce α-prime submodules and also give some examples of an
α-prime submodule which is not a prime submodule. Characterization of α-prime sub-
modules of Z-module Z is completely given.

DOI: https://doi.org/10.29020/nybg.ejpam.v11i3.3275

Email address: thawatchai.kh@kmitl.ac.th (Thawatchai Khumprapussorn)

http://www.ejpam.com 730 c© 2018 EJPAM All rights reserved.



T. Khumprapussorn / Eur. J. Pure Appl. Math, 11 (3) (2018), 730-739 731

In section 3, we extend the notion of α-prime submodules to weakly α-prime sub-
modules. We study properties the product of submodules in the Cartesian product of
modules.

In section 4, we move the investigation of α-prime submodules to α-prime ideals.

2. α-prime submodules

First, we present fundamental definitions of α-prime submodules which will be studied
in this paper.

Definition 1. Let P be a proper submodule of M . We call P is α-prime if for any
element r ∈ R and m ∈ M such that r(m + m) ∈ P , we have r + r ∈ (P : M) or
m+m ∈ P .

By this definition, every prime submodule is an α-prime submodule, but the converse
is not true in general.

Example 1. Let Z be an Z-module and p ∈ Z. Then pZ is an α-prime submodule of Z if
and only if p = 0 or p is a prime number or p = 2q where q is a prime number.

Proof. (→) Assume that pZ is an α-prime submodule of Z. Suppose that p 6= 0 and p
is not prime number. Then p = ab for some integers a and b with 1 < a, b < p. We see
that p | a(b + b). This implies that p | a + a or p | b + b. Now, we assume that p | a + a.
This means p ≤ 2a. Hence ab ≤ 2a. Therefore b ≤ 2. That is b = 2. Next, suppose that a
is not a prime number. Then a = cd for some integers c and d with 1 < c, d < a. We have
p = 2a = 2cd = c(d + d). Since pZ is an α-prime submodule of Z, p | c + c or p | d + d.
Hence a | c or a | d. This implies that a ≤ c or a ≤ d which is a contradiction. This prove
that p = 2q for some prime numbers q.

(←) It is clear that pZ is an α-prime submodule of Z where p = 0 or p is a prime
number or p = 2q for some prime numbers q.

Example 1 obtains that 4Z is α-prime but is not prime submodule of Z. The following
first result gives the characterization of α-prime submodules.

Theorem 1. Let P be a proper submodule of an R-module M . The following statements
are equivalent.

(i) P is an α-prime submodule of M .

(ii) For all ideals I of R and for all submodules N of M ,

if Iβ(N) ⊆ P , then I ⊆ α((P : M)) or N ⊆ α(P ).

(iii) For all a ∈ R and for all submodules N of M ,

if aβ(N) ⊆ P , then a ∈ α((P : M)) or N ⊆ α(P ).



T. Khumprapussorn / Eur. J. Pure Appl. Math, 11 (3) (2018), 730-739 732

(iv) For all ideals I of R and for all m ∈M ,

if I(m+m) ⊆ P , then I ⊆ α((P : M)) or m ∈ α(P ).

(v) For all a ∈ R and for all m ∈M ,

if aR(m+m) ⊆ P , then a ∈ α((P : M)) or m ∈ α(P ).

(vi) For all m ∈M , if m+m /∈ P , then α((P : M)) = α((P : m)).

Proof. (i) → (ii) Assume that P is an α-prime submodule of M . Let I be an ideal
of R and N be a submodule of M such that Iβ(N) ⊆ P and N * α(P ). To show that
I ⊆ α((P : M)), let r ∈ I and n ∈ N be such that n /∈ α(P ). Then n + n /∈ P and
n + n ∈ β(N). This implies that r(n + n) ∈ P . Since P is an α-prime submodule of M
and n+ n /∈ P , r + r ∈ (P : M). Hence I ⊆ α((P : M)).

(ii)→ (iii) Assume that (ii) holds. Let a ∈ R and N be a submodule of M such that
aβ(N) ⊆ P . Then (Ra)β(N) = R(aβ(N)) ⊆ RP ⊆ P . By (ii), we have Ra ⊆ α((P : M))
or N ⊆ α(P ). Therefore a ∈ α((P : M)) or N ⊆ α(P ).

(iii) → (iv) Assume that (iii) holds. To prove that (iv) holds, let I be an ideal of R
and m ∈ M such that I(m+m) ⊆ P and m /∈ α(P ). Let a ∈ I. Then aβ(Rm) ⊆ P . By
(iii) and m /∈ α(P ), a ∈ α((P : M)). Hence I ⊆ α((P : M)).

(iv)→ (v), (v)→ (i) and (vi)→ (i) are obvious.
(i) → (vi) Assume that P is an α-prime submodule of M . Let m ∈ M be such that

m + m /∈ P . It is clear that α((P : M)) ⊆ α((P : m)). Let r ∈ α((P : m)). Then
r + r ∈ (P : m). Hence r(m + m) = (r + r)m ∈ P . Since P is α-prime and m + m /∈ P ,
r + r ∈ (P : M). That is r ∈ α((P : M)). Therefore α((P : M)) = α((P : m)).

Lemma 1. Let φ : M1 → M2 be an R-module homomorphism, P be a submodule of M1

and K be a submodule of M2. Then

(i) If φ is an epimorphism and r + r ∈ (P : M1), then r + r ∈ (φ(P ) : M2).

(ii) If r + r ∈ (K : M2), then r + r ∈ (φ−1(K) : M1).

Proof. (i) Assume that φ is an epimorphism and (r + r)M1 ⊆ P . Let m2 ∈M2. Then
φ(m1) = m2 for some m1 ∈ M1. Thus (r + r)m1 ∈ P . This implies that (r + r)m2 =
(r + r)φ(m1) ∈ φ(P ). That is r + r ∈ (φ(P ) : M2).

(ii) Assume that (r+r)M2 ⊆ K. Let m1 ∈M1. Then φ((r+r)m1) = (r+r)φ(m1) ∈ K.
Hence (r + r)m1 ∈ φ−1(K). Therefore r + r ∈ (φ−1(K) : M1).

Proposition 1. Let φ : M1 →M2 be an R-module homomorphism. Then

(i) If φ is an epimorphism and P is an α-prime submodule of M1 containing kerφ, then
φ(P ) is an α-prime submodule of M2.

(ii) If K is an α-prime submodule of M2, then φ−1(K) is an α-prime submodule of M1.



T. Khumprapussorn / Eur. J. Pure Appl. Math, 11 (3) (2018), 730-739 733

Proof. (i) Assume that φ is an epimorphism and P is an α-prime submodule of
M1 containing kerφ. Let r ∈ R and m ∈ M2 be such that r(m + m) ∈ φ(P ). There
exist elements n ∈ M1 and p ∈ P such that r(m + m) = φ(p) and φ(n) = m. Then
φ(p) = r(m + m) = r(φ(n) + φ(n)) = r(φ(n + n)) = φ(r(n + n)). This implies that
r(n+ n)− p ∈ kerφ. Since kerφ ⊆ P , r(n+ n) ∈ P . Since P is an α-prime submodule of
M1, r+ r ∈ (P : M1) or n+n ∈ P . Since φ is onto, r+ r ∈ (φ(P ) : M2) or m+m ∈ φ(P ).
Hence φ(P ) is an α-prime submodule of M2.

(ii) Assume that K is an α-prime submodule of M2. Let r ∈ R and m ∈ M be such
that r(m+m) ∈ φ−1(K). Then r(φ(m) + φ(m)) ∈ K. Since K is an α-prime submodule
of M2, r + r ∈ (K : M2) or φ(m) + φ(m) ∈ K. This implies that r + r ∈ (φ−1(K) : M1)
or m+m ∈ φ−1(K). Hence φ−1(K) is an α-prime submodule of M1.

Corollary 1. Let N be a submodule of M . Then

(i) If P is an α-prime submodule of M and K is a submodule of M contained in P ,
then P/K is an α-prime submodule of M/K .

(ii) If K
′

is an α-prime submodule of M/N , then K
′

= K/N . for some α-prime submodule
K of M .

Proof. (i) Assume that P is an α-prime submodule of M and K is a submodule of M
contained in P . Define a homomorphism ϕ : M → M/K by ϕ(m) = m+K for all m ∈M .
Then ϕ is an epimorphism and kerϕ = K. By Proposition 1 (i), ϕ(P ) = P/K is an α-prime
submodule of M/K .

(ii) Assume that K
′

is an α-prime submodule of M/N . Then the set K = {x ∈
M | x+N ∈ K ′} is an α-prime submodule of M . Clearly, K

′
= K/N .

For subgroups A and B of a group (G,+), we have A ⊆ α(B) if and only if β(A) ⊆ B.

Definition 2. Let R be a ring and M be an R-module. A nonempty set S ⊆ M\{0} is
called an α-multiplicative system if for all ideal I of R and for all submodules K and

N of M , if

(
K+β(I)M

)
∩S 6= ∅ and

(
K+β(N)

)
∩S 6= ∅, then

(
K+Iβ(N)

)
∩S 6= ∅.

Proposition 2. Let P be a submodule of an R-module M . Then P is an α-prime sub-
module of M if and only if M\P is an α-multiplicative system.

Proof. (→) Assume that P is an α-prime submodule of M . Let I be an ideal of

R and let K and N be submodules of M such that

(
K + Iβ(N)

)
∩M\P = ∅. Then

K + Iβ(N) ⊆ P . It follows that K ⊆ P and Iβ(N) ⊆ P . Since P is an α-prime
submodule of M , I ⊆ α((P : M)) or N ⊆ α(P ). This implies that β(I) ⊆ (P : M) or

β(N) ⊆ P . Hence K + β(I)M ⊆ P or K + β(N) ⊆ P . Hence

(
K + β(I)M

)
∩M\P = ∅

or

(
K + β(N)

)
∩M\P = ∅. This shows that M\P is an α-multiplicative system.



T. Khumprapussorn / Eur. J. Pure Appl. Math, 11 (3) (2018), 730-739 734

(←) Assume that M\P is an α-multiplicative system. Let I be an ideals of R and N

be a submodule of M such that Iβ(N) ⊆ P . Hence

(
Iβ(N)

)
∩M\P = ∅. Since M\P

is an α-multiplicative system,

(
β(I)M

)
∩M\P = ∅ or

(
β(N)

)
∩M\P = ∅. That is,

β(I)M ⊆ P or β(N) ⊆ P . We already show that β(I) ⊆ (P : M) or β(N) ⊆ P . This
means I ⊆ α((P : M)) or N ⊆ α(P ). Therefore P is an α-prime submodule of M .

Proposition 3. Let M be an R-module and X be an α-multiplicative system. If P is
a submodule of M maximal with respect to the property that P ∩ X = ∅, then P is an
α-prime submodule of M .

Proof. Assume that P is a submodule of M maximal with respect to the property
that P ∩ X = ∅. Let I be an ideal of R and N be a submodule of M . Now, assume

that I * α((P : M)) and N * α(P ). Hence β(I)M * P and β(N) * P . Then

(
P +

β(I)M

)
∩ X 6= ∅ and

(
P + β(N)

)
∩ X 6= ∅. Since X is an α-multiplicative system,(

P + Iβ(N)

)
∩X 6= ∅. Since P ∩X = ∅, Iβ(N) * P . This implies that P is an α-prime

submodule of M .

Definition 3. Let M be an R-module and N be a submodule of M . If there is an α-prime
submodule of M containing N , then we define

α
√
N = {x ∈M | every α-multiplicative system containing x meets N}.

If there is no a α-prime submodule of M containing N , then we define α
√
N = M .

Theorem 2. Let M be an R-module and N be a submodule of M . Then either α
√
N = M

or α
√
N is the intersection of all α-prime submodule of M containing N .

Proof. Assume that α
√
N 6= M . Let x ∈ α

√
N and P be an α-prime submodule of M

containing N . By Proposition 2, M\P is an α-multiplicative system and N ∩ (M\P ) = ∅.
Hence x ∈ P . Conversely, let x ∈M be such that x /∈ β

√
N . Let S be an α-multiplicative

system such that x ∈ S and S ∩ N = ∅. By Zorn’s Lemma on the set of submodule J
of M containing N and S ∩ J = ∅, there exists a maximal submodule K of M such that
S ∩K = ∅. By Proposition 3, K is a α-prime submodule of M . Hence x /∈ K.

3. Weakly α-prime submodules

In this section we begin with the definition of weakly α-prime submodules which is
a generalization of α-prime submodules. In [2], S.E. Atani and F. Farzalipour gave the
notion of weakly prime submodules stated that a proper submodule P of a left R-module
M is called weakly prime if 0 6= rm ∈ P for some r ∈ R and m ∈M , then r ∈ (P : M) or
m ∈ P where (N : M) = {r ∈ R | rM ⊆ N}.



T. Khumprapussorn / Eur. J. Pure Appl. Math, 11 (3) (2018), 730-739 735

Definition 4. Let P be a proper submodule of M . We call P is weakly α-prime if for
any elements r ∈ R and m ∈ M such that r(m + m) ∈ P\{0}, we have r + r ∈ (P : M)
or m+m ∈ P .

Every α-prime submodule is weakly α-prime submodule. But the converse need not
be true. For example, {0̄} is weakly α-prime but is not α-prime submodule of Z-module
Z8 because 2 · (2̄ + 2̄) = 2 · 4̄ = 8̄ = 0̄ and (2 + 2)Z8 * {0̄} and 2̄ + 2̄ 6= 0̄.

Next we give several characterizations of weakly α-prime submodules.

Theorem 3. Let M be an R-module and P be a submodule of M . The following statements
are equivalent.

(i) P is a weakly α-prime submodule of M .

(ii) For any m ∈M , if m+m /∈ P , then (P : m+m) = α((P : M)) ∪ α((0 : m)).

(iii) For any m ∈ M , if m+m /∈ P , then (P : m+m) = α((P : M)) or (P : m+m) =
α((0 : m)).

Proof. (i)→ (ii) Assume that P is a weakly α-prime submodule of M . Let m ∈M be
such that m+m /∈ P . Let r ∈ (P : m+m). Then r(m+m) ∈ P . If r(m+m) = 0, then
r ∈ α((0 : m)). Suppose that r(m+m) 6= 0. Since P is weakly α-prime and m+m /∈ P ,
r + r ∈ (P : M). That is r ∈ α((P : M)). Conversely, let r ∈ α((P : M)) ∪ α((0 : m)).
Then r + r ∈ (P : M) or rm+ rm = 0. These implie that r ∈ (P : m+m).

(ii)→ (iii) Obvious.
(iii) → (i) Assume that (iii) holds. Let r ∈ R and m ∈ M be such that r(m + m) ∈

P\{0} and m+m /∈ P . Then r ∈ (P : m+m). Since r(m+m) 6= 0, r /∈ α((0 : m)). By
(iii), (P : m+m) = α((P : M)). Hence r ∈ α((P : M)). Therefore r+ r ∈ (P : M). This
proves that P is a weakly α-prime submodule of M .

Let M1 and M2 be R-modules. Then M1 ×M2 is an R-module under the operation
(a, b) + (c, d) = (a+ c, b+ d) and r(a, b) = (ra, rb) for all a, c ∈M1, b, d ∈M2 and r ∈ R.
We denote this module by M1 ⊕M2.

Proposition 4. Let N1 be a submodule of M1 and N2 be a submodule of M2. If N1 ×N2

is a weakly α-prime submodule of M1 ⊕M2, then N1 is a weakly α-prime submodule of
M1 and N2 is a weakly α-prime submodule of M2.

Proof. It is straightforward.

Let R1 and R2 be commutative rings with identity, Mi be a unital Ri-module where
i = 1, 2. Then M1 ×M2 is an (R1 × R2)-module under the operation (r1, r2)(m1,m1) =
(r1m1, r2m2) for all (r1, r2) ∈ R1×R2 and (m1,m2) ∈M1×M2. We set up these notation
for the next two results.

Proposition 5. Let R = R1 ×R2 and M = M1 ×M2 and let N1 be an R1-submodule of
M1. Consider the following statements.



T. Khumprapussorn / Eur. J. Pure Appl. Math, 11 (3) (2018), 730-739 736

(i) N1 is an α-prime submodule of M1.

(ii) N1 ×M2 is an α-prime submodule of M1 ×M2.

(iii) N1 ×M2 is a weakly α-prime submodule of M1 ×M2.

Then (i)→ (ii)→ (iii). Moreover, if β(M2) 6= {0}, then (i), (ii) and (iii) are equivalent.

Proof. (i) → (ii) Assume that N1 is an α-prime submodule of M1. Let (a, b) ∈
R1 × R2 and (x, y) ∈ M1 × M2 be such that (a, b)[(x, y) + (x, y)] ∈ N1 × M2. Then
[a(x+ x), b(y+ y)] ∈ N1×M2. Thus a(x+ x) ∈ N1. Since N1 is an α-prime submodule of
M1, a+a ∈ (N1 : M1) or x+x ∈ N1. This leads to (a+a, b+ b) ∈ (N1×M2 : M1×M2) or
(x, y) + (x, y) ∈ N1×M2. Therefore N1×M2 is an α-prime prime submodule of M1×M2.
(ii)→ (iii) It is obvious.

Next, let w ∈ M2 be such that w + w 6= 0 and assume that N1 ×M2 is a weakly
α-prime submodule of M1 ×M2. Let r ∈ R1 and m ∈ M1 such that r(m + m) ∈ N1.
Then (r, 1)[(m,w) + (m,w)] = (r(m+m), w+w) ∈ N1×M2\{(0, 0)}. Since N1×M2 is a
weakly α-prime submodule of M1 ×M2, we have (r+ r, 1 + 1) ∈ (N1 ×M1 : M1 ×M2) or
(m,w) + (m,w) ∈ N1 ×M2. This implies that r + r ∈ (N1 : M1) or m+m ∈ N1. Hence
N1 is an α-prime submodule of M1.

The following example shows that, in general, the condition β(M2) 6= {0} in Proposi-
tion 5 can not be omitted.

Example 2. Let M1 = Z8, M2 = {0}, R1 = R2 = Z. It is clear that {0̄}×{0} is a weakly
α-prime submodule of M1 ×M2. However, {0̄} is not an α-prime submodule of Z-module
Z8.

Proposition 6. Let M1,M2 be R1, R2-modules respectively and N1 ×N2 be a submodule
of M1 ×M2. Then β(N1 ×N2) = {(0, 0)} if and only if β(N1) = {0} and β(N2) = {0}.

Proof. It is evident.

Proposition 7. Let M1,M2 be R1, R2-modules respectively. Then

(i) If N1 ×N2 is a weakly α-prime submodule of M1 ×M2, then either β(N1) = {0} or
α(N1) = M1 or α(N2) = M2.

(ii) If N1 ×N2 is a weakly α-prime submodule of M1 ×M2, then either β(N2) = {0} or
α(N1) = M1 or α(N2) = M2.

(iii) If N1×N2 is a weakly α-prime submodule of M1×M2, then β(N1) = {0} or α(N2) =
M2 or N1 ×N2 is an α-prime submodule of M1 ×M2.

(iv) If N1×N2 is a weakly α-prime submodule of M1×M2, then β(N2) = {0} or α(N1) =
M1 or N1 ×N2 is an α-prime submodule of M1 ×M2.



T. Khumprapussorn / Eur. J. Pure Appl. Math, 11 (3) (2018), 730-739 737

Proof. (i) Assume that N1 × N2 is a weakly α-prime submodule of M1 × M2 and
β(N1) 6= {0} and α(N1) 6= M1. Let a ∈ N1 be such that a+ a 6= 0. Let r ∈ (N2 : M2) and
y ∈M2. Then (0, 0) 6= (a+ a, r(y + y)) = (1, r)[(a, y) + (a, y)] ∈ N1 ×N2. Since N1 ×N2

is a weakly α-prime submodule of M1 ×M2, we have (1 + 1, r+ r)
(
M1 ×M2

)
⊆ N1 ×N2

or (a, y) + (a, y) ∈ N1 × N2. This implies that (1 + 1)M1 ⊆ N1 or y + y ∈ N2. Since
α(N1) 6= M1, there is m ∈ M1 such that m + m /∈ N1. This means (1 + 1)M1 * N1.
Therefore y ∈ α(N2).

(ii) The proof is similar to (i).
(iii) Assume that N1×N2 is a weakly α-prime submodule of M1×M2 and β(N1) 6= {0}

and α(N2) 6= M2. By (i), α(N1) = M1. Let (r1, r2) ∈ R1 × R2 and (m1,m2) ∈ M1 ×M2

be such that (r1, r2)[(m1,m2) + (m1,m2)] ∈ N1 × N2. Then r1(m1 + m1) ∈ N1 and
r2(m2+m2) ∈ N2. Let a ∈ N1 be such that a+a 6= 0. Then (0, 0) 6= (a+a, r2(m2+m2)) =
(1, r2)[(a,m2) + (a,m2)] ∈ N1 × N2. Since N1 × N2 is a weakly α-prime submodule of

M1×M2, we have (1+1, r2+r2)
(
M1×M2

)
⊆ N1×N2 or (a,m2)+(a,m2) ∈ N1×N2. Since

N1×N2 is a submodule of M1×M2 and α(N1) = M1, (r1+r1, r2+r2)
(
M1×M2

)
⊆ N1×N2

or (m1,m2) + (m1,m2) ∈ N1 ×N2. This implies that N1 ×N2 is an α-prime submodule
of M1 ×M2.

(iv) The proof is similar to (iii).

The following example obtains that the assumption α(N2) 6= M2 in the proof of Propo-
sition 7 (iii) is necessary.

Example 3. Consider a submodule 4Z× 3Z of a Z×Z-module Z× 3Z, by Proposition 5,
4Z× 3Z is a weakly α-prime submodule of Z× 3Z. However, 4Z× 3Z is not an α-prime

submodule of Z× 3Z because (1, 3)[(2, 1) + (2, 1)] = (4, 6) ∈ 4Z× 3Z and (2, 6)
(
Z× 3Z

)
*

4Z× 3Z and (4, 2) /∈ 4Z× 3Z. In particular, α(3Z) = 3Z.

4. The traveling of α-prime from modules to rings

In this section we apply the notion of (weakly) α-prime submodules to (weakly) α-
prime ideals.

Definition 5. A proper ideal P of a ring R is called an α-prime ideal of R if P is an
α-prime submodule of an R-modules R.

Similarly, a proper ideal P of a ring R is called a weakly α-prime ideal of R if P is
an weakly α-prime submodule of an R-modules R.

It is easy to show that for an ideal P of R, P is an α-prime ideal of R if and only if for
all a, b ∈ R, if a(b+ b) ∈ P , then a+ a ∈ P or b+ b ∈ P . Similarly, P is a weakly α-prime
ideal of R if and only if for all a, b ∈ R, if a(b+ b) ∈ P\{0}, then a+ a ∈ P or b+ b ∈ P .

Proposition 8. If P is an α-prime submodule of an R-module M , then (P : M) is an
α-prime ideal of R.



T. Khumprapussorn / Eur. J. Pure Appl. Math, 11 (3) (2018), 730-739 738

Proof. Assume that P is an α-prime submodule of an R-module M . Let a, b ∈ R
be such that a(b + b) ∈ (P : M) and b + b /∈ (P : M). Then there exists an element
m ∈M such that (b+ b)m /∈ P and a(b+ b)m ∈ P . Since P is α-prime and (b+ b)m /∈ P ,
a+ a ∈ (P : M). Therefore (P : M) is an α-prime ideal of R.

Let R be a ring. The Cartesian product R×R is a ring under componentwise addition
and the multiplication (a, b) ∗ (c, d) = (ac, ad + bc). We use the notation R(+)R for this
ring.

Proposition 9. If I is an α-prime ideal of a ring R, then I × R is an α-prime ideal of
R(+)R.

Proof. It is straightforward.

Example 4. We know that 4Z and 6Z are α-prime ideal of Z. In Z(+)Z, we have
(2, 1)[(1, 1) + (1, 1)] = (2, 1)(2, 2) = (4, 6) ∈ 4Z × 6Z. However, (4, 2) /∈ 4Z × 6Z and
(2, 2) /∈ 4Z × 6Z. This is an example shows that I × J may be not an α-prime ideal of
R(+)R even if I and J are α-prime ideals of R.

Proposition 10. If P is a weakly α-prime submodule of M and (P : M)β(P ) 6= 0, then
P is an α-prime submodule of M

Proof. Assume that P is a weakly α-prime submodule of M and (P : M)β(P ) 6= 0.
Let r ∈ R and m ∈ M be such that r(m+m) ∈ P . If r(m+m) 6= 0, r + r ∈ (P : M) or
m+m ∈ P . Assume that r(m+m) = 0. We consider the following two cases.
Case 1. rβ(P ) 6= 0.
Then r(n0 +n0) 6= 0 for some n0 ∈ P . Hence r(m+m+n0 +n0) = r(n0 +n0) ∈ P . Since
P is a weakly α-prime submodule of M , r + r ∈ (P : M) or m+m+ n0 + n0 ∈ P . Since
no ∈ P , r + r ∈ (P : M) or m+m ∈ P . Hence P is an α-prime submodule of M .
Case 2. rβ(P ) = 0.

Subcase 2.1. (P : M)(m+m) 6= 0.
Let k ∈ (P : M) be such that k(m+m) 6= 0. Then (r + k)(m+m) = k(m+m) ∈ P .

Since P is a weakly α-prime submodule of M , r + k + r + k ∈ (P : M) or m + m ∈ P .
Since k ∈ (P : M), r + r ∈ (P : M) or m+m ∈ P . Hence P is an α-prime submodule of
M .

Subcase 2.2. (P : M)(m+m) = 0.
Since (P : M)β(P ) 6= 0, we have k(n + n) 6= 0 for some k ∈ (P : M) and n ∈ P . Then
(r+k)(m+m+n+n) = r(m+m)+r(n+n)+k(m+m)+k(n+n) = k(n+n) ∈ P . Since
P is a weakly α-prime submodule of M , r + k + r + k ∈ (P : M) or m+m+ n+ n ∈ P .
Since k ∈ (P : M) and n ∈ P , r + r ∈ (P : M) or m + m ∈ P . Hence P is an α-prime
submodule of M .

The following result directly implies from Proposition 8 and 10.

Corollary 2. If P is a weakly α-prime submodule of M and (P : M)β(P ) 6= 0, then
(P : M) is a weakly α-prime ideal of R.



REFERENCES 739

We prove in Proposition 8 that if P is an α-prime submodule of an R-module M , then
(P : M) is an α-prime ideal of R. However, this situation is false for weakly α-prime
submodules.

Example 5. In Z/8Z as a Z-module, we have {0̄} is a weakly α-prime submodule of Z/8Z.
However, ({0̄} : Z/8Z) = 8Z is not a α-prime ideal of Z.

References

[1] R. Ameri, On the prime submodules of multiplication modules, International Journal
of Mathematics and Mathematical Sciences, 27: 1715-1724, (2003).

[2] S.E. Atani and F. Farzalipour, On Weakly Prime Submodules, Tamkang Journal of
Mathematics, 38(3): 247-252, (2007).