EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 12, No. 2, 2019, 544-552
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Weakly Prime and Weakly primary ideals in gamma
seminearrings

Waheed Ahmad Khan1, Abdelghani Taouti2,∗, Seema Karkain2,
Azar Salami2, Waqar Arif1

1 Department of Mathematics, University of Education, Attock Campus, Pakistan
2 ETS-Maths and NS Engineering division HCT, University City, P. O. Box: 7947
United Arab Emirates

Abstract. We introduce and discuss the weakly prime and weakly primary ideals of a gamma
seminearrings with illustrative examples. We also present few of characterizations of these ideals.

2010 Mathematics Subject Classifications: 16Y30, 16Y60

Key Words and Phrases: Gamma seminearrrings, prime ideals, primary ideals.

1. Introduction and Preliminaries

The concept of seminearring was introduced by W. G. van Hoorn et al. in [1]. Sem-
inearfields have been introduced in [5]. As a generalization of seminearrings that is Γ-
seminear-rings were introduced in [2]. Subsequently, prime and semiprime ideals in gamma
seminearrings have been explored in [3]. In a sequel, we introduce the notion of weakly
prime and weakly primary ideals Γ-seminearring and few of their characterizations. We
recall some useful concepts for the sake of completeness. A nonempty set R with two
binary operations ” + ”(addition) and ”.” (multiplication) is called a seminearring if it
satisfies (i) (R, +) and (R, .) are semigroups; (ii) (x+ y).z = x.z+ y.z for all x, y, z ∈ R.
In 2005, Krishna & Chatterjee [4], introduced the condition of minimality of generalized
linear sequential machines using the theory of near-semirings. Near-semirings have proven
to be useful in studying automata and formal languages. Following [3], Γ-seminearring is a
triple (R, +, Γ) where, (i) Γ is a non-empty set of binary operators on R such that for each
α ∈ Γ, (R, +, .) is a seminearring, (ii) xα(yβz) = (xαy)βz for all x, y, z ∈ R and α, β ∈
Γ. Similarly, let R be a Γ-seminearring, a subsemigroup A of (R,+) is called a left (resp.,
right) ideal of R if RΓA ⊆ A (resp., AΓR ⊆ A). A left and right ideal is called an ideal. Let

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v12i2.3397

Email addresses: sirwak2003@ yahoo.com (W. A. Khan), ganitaouti@yahoo.com.au (A. Taouti),
skarkain@hct.ac.ae (S. Karkain), asalami@hct.ac.ae (A. Salami),
waqarvicky6699@gmail.com (W. Arif)

http://www.ejpam.com 544 c© 2019 EJPAM All rights reserved.



Abdelghani Taouti et al. / Eur. J. Pure Appl. Math, 12 (2) (2019), 544-552 545

R be a Γ-seminearring and I, J ⊆ R. We denote it by IΓJ = {aαb | a, b ∈ R and α ∈ Γ}.
A mapping f : R → R′ between two gamma seminearrings is called a Γ-seminearring
homomorphism (Γ-homomorphism), if f(x + y) = f(x) + f(y) and f(xγy) = f(x)γf(y)
for all x, y ∈ R and γ ∈ Γ. Let R and R′ be a Γ-seminearrings and f : R → R′ be a
Γ-seminearring homomorphism. Then, (i) f(I1ΓI2) = f(I1)Γf(I2) for all I1, I2 ∈ R, (ii)
f−1(J1)Γf−1(J2) = f−1(J1ΓJ2) for all J1, J2 ∈ R′.

2. Weakly prime and weakly primary ideals in Γ-seminearrings

In this section we introduce the notion of weakly prime and weakly primary ideals in
Γ-seminearrings. By an ideal we mean two-sided ideal unless otherwise stated. We begin
with the following definition.

Definition 1. Let R be a Γ-seminearring. A proper ideal P of R is called weakly prime
if for ideals I and J , 0 6= IΓJ ⊆ P implies I ⊆ P or J ⊆ P.

Proposition 1. Let P be a proper ideal of a Γ-seminearring R. The following statements
are equivalent.

(i) P is weakly prime.
(ii) For ideals I and J of R, 0 6= (IΓJ) ⊆ P implies I ⊆ P or J ⊆ P .
(iii) For elements i and j in R, i /∈ P and j /∈ P implies 0 6= (i)Γ(j) * P.

Proof. Following definition1, clearly (i) and (ii) are equivalent. Now, (i) =⇒ (iii) Let
P be a weakly prime, i /∈ P and j /∈ P . Assume 0 6= (i) Γ(j) ⊆ P ⇒ (i) ⊆ P or (j) ⊆ P .
Hence, i ∈ P or j ∈ P , a contradiction. Thus, 0 6= (i)Γ(j) * P . (iii) =⇒ (i) Assume that
I * P and J * P . Then there exists i ∈ I\P and j ∈ J\P . Hence, 0 6= (i)Γ(j) ⊆ 0 6= IΓJ
but 0 6= (i)Γ(j) * P by (iii). Thus, 0 6= IΓJ * P.

Example 1. Let R = {0, 1, e, a, b, c} be a Γ-seminearring with Γ = {α, 1}.

+ 0 1 e a b c

0 0 1 e a a c

1 1 1 1 1 1 1

e e 1 e 1 1 e

a a 1 1 a a a

b a 1 1 a a a

c c 1 e a a c



Abdelghani Taouti et al. / Eur. J. Pure Appl. Math, 12 (2) (2019), 544-552 546

α 0 1 e a b c

0 0 0 0 0 0 0

1 0 1 e a b c

e 0 e e 0 c c

a 0 a 0 a a 0

b 0 b 0 a a 0

c 0 c 0 0 0 0

P = {0, a, b} of a seminearring R is a weakly prime ideal but not a prime ideal, since
cαc = 0 and c /∈ P . On the other hand, consider a prime ideal Q = {0, e, c} of R. It is
easy to show that Q is a weakly prime ideal. For this, let I and J are ideals of R, where
I = {0, e, c} and J = {0, a, b}. Then, 0 6= IΓJ ⊆ Q =⇒ I ⊆ Q =⇒ Q is a weakly prime
ideal. Hence every prime ideal of a gamma seminearring is a weakly prime ideal.

Example 2. Let R = {0, 1, e, a, b, c} be a Γ-seminearring with Γ = {α, 1} as defined
in example1. Let S = {0, 1, e, a, c} ⊆ R be a Γ-sub-seminearring of R with Γ = {1,
α}.Clearly, in S the ideals I = {0, a} and J = {0, c} are weakly prime ideals but not
prime.

Proposition 2. Let P be a proper ideal of a Γ-seminearring R and {0 6= aαrβb : r ∈ R,
α, β ∈ Γ} ⊆ P if and only if a ∈ P or b ∈ P, then P is a weakly prime ideal.

Proof. Let I and J are ideals of R with 0 6= IΓJ ⊆ P . Let I * P , and for a ∈ I\P ,
b ∈ J , we have {0 6= aαrβb : r ∈ R, α, β ∈ Γ} ⊆ IΓJ 6= 0 ⊆ P. Since a /∈ P and b ∈ P ⇒
J ⊆ P . Hence, P is a weakly prime ideal.

Proposition 3. Intersection of finite numbers of weakly prime ideals of a Γ-seminearring
R which are totally ordered by inclusion is a weakly prime ideal.

Proof. Let {Pα}α∈Λ be the family of weakly prime ideals which are totally ordered by

inclusion. Suppose I and J be ideals of R. If 0 6= IΓJ ⊆ ∩α∈ΛPα, then 0 6= IΓJ ⊆ Pα, for
all α ∈ Λ. Suppose that there exists α ∈ Λ such that I * Pα . Then, J ⊆ Pα and hence
J ⊆ Pβ for all β ≥ α. We assume that there exist γ < α such that J ⊆ Pγ .Then, I ⊆ Pγ
and hence I ⊆ Pα, which is impossible. Hence, J ⊆ Pβ for any β ∈ Λ. Thus, ∩α∈ΛPα is a
weakly prime ideal of a Γ-seminearring R.

Below we provide an illustrative example.

Example 3. Let S = {0, 1, e, a, c} with Γ = {1, α} be a Γ-seminearring defined in the
tables given below.

+ 0 1 e a c

0 0 1 e a c

1 1 1 1 1 1

e e 1 e 1 e

a a 1 1 a a

c c 1 e a c

α 0 1 e a c

0 0 0 0 0 0

1 0 1 e a c

e 0 e e 0 c

a 0 a 0 a 0

c 0 c 0 0 0



Abdelghani Taouti et al. / Eur. J. Pure Appl. Math, 12 (2) (2019), 544-552 547

Consider P1 = {0, c} and P2 = {0, e, c} are the weakly prime ideals of R and are
totally ordered by inclusion as well. Since, P1 ∩ P2 = P1, which is a weakly prime ideal of
R. Hence, ∩α∈APα is a weakly prime ideal.

Proposition 4. Let I be an ideal of a Γ-seminearring R with R + I ⊆ I and I +R ⊆ I.
Let P be a proper ideal of R containing I and ψ : R→ R/I be the canonical epimorphism.
Then, P is a weakly prime ideal if and only if ψ(P ) is a weakly prime.

Proof. Let P be a weakly prime ideal of R. Suppose J1 and J2 are ideals in R/I
such that 0 6= J1ΓJ2 ⊆ ψ(P ). Assume that ψ−1(J1) = I1 and ψ−1(J2) = I2. Then,
0 6= I1ΓI2 = 0 6= ψ−1(J1)Γψ−1(J2) ⊆ 0 6= ψ−1(J1ΓJ2) ⊆ 0 6= ψ−1(π(P )) = P . Since P is a
weakly prime ideal, it implies I1 ⊆ P or I2 ⊆ P . Hence, J1 = ψ(ψ−1(J1)) = ψ(I1) ⊆ ψ(P )
or J2 = ψ(ψ−1(J2)) = ψ(J2) ⊆ ψ(P ). Hence, ψ(P ) is a weakly prime. Conversely, suppose
ψ(P ) be a weakly prime ideal and let I1, I2 are ideals of R such that 0 6= I1ΓI2 ⊆ P .
Then, 0 6= ψ(I1)Γψ(I2) = 0 6= ψ(I1ΓI2) ⊆ ψ(P ).Since ψ(P ) is a weakly prime ideal, it
implies that ψ(I1) ⊆ ψ(P ) or ψ(I2) ⊆ ψ(P ). Thus, I1 ⊆ P or I2 ⊆ P , and hence P is a
weakly prime ideal of a Γ-seminearring R.

Definition 2. Let R be a Γ-seminearring and M be a non-empty subset of R. We call
M an m-system if for a, b ∈ M , there exist a1 ∈ (a), b1 ∈ (b) and α ∈ Γ such that
0 6= a1αb1 ∈M.

Proposition 5. Let P be a proper ideal of a Γ-seminearring R. Then, P is a weakly
prime ideal if and only if R\P is m-system.

Proof. Let P be a weakly prime ideal of a Γ-seminearring R. Consider a, b ∈ R\P
and 0 6= (a)Γ(b) * P . Let a1 ∈ (a), b1 ∈ (b) and α ∈ Γ such that 0 6= a1αb1 /∈ P , i.e.,
a1αb1 ∈ R\P . Thus, R\P is an m-system. Conversely, suppose R\P is an m-system and
let a, b ∈ R\P . Then, there exist a1 ∈ (a), b1 ∈ (b) and α ∈ Γ such that a1αb1 ∈ R\P .
Thus, 0 6= (a)Γ(b) * P and hence P is a weakly prime ideal of a Γ-seminearring R.

Definition 3. A subset A of a Γ-seminearring R is a subtractive, if a ∈ A and a+ b ∈ A
implies b ∈ A.

Proposition 6. Let R be a Γ-seminearring whose all ideals are subtractive, and let P be
a proper ideal of R. Then, P is a weakly prime if and only if for any ideals I, J of R,
P ⊂ I and P ⊂ J implies 0 6= IΓJ * P .

Proof. Suppose for any ideals I, J of R, P ⊂ I and P ⊂ J implies 0 6= IΓJ * P . Let
us suppose that I * P and J * P . Then there exist i ∈ I\P and j ∈ J\P and hence
P ⊂ P + (i). By hypothesis, 0 6= (P + (i))Γ(P + (j)) * P and so there exist i′ ∈ (i),
j′ ∈ (j), p, p′ ∈ P and α ∈ Γ such that 0 6= (p+i′)α(p′+j′) /∈ P . Since, 0 6= pα(p′+j′) ∈ P,
0 6= i′α(p′ + j′) /∈ P and P is an ideal, then i′ /∈ P and p′ + j′ /∈ P . Thus, i′ /∈ P and
j′ /∈ P because P is subtractive. It implies 0 6= (i′)Γ(j′) * P . But 0 6= (i′)Γ(j′) ⊆ 0 6=
IΓJ ⇒ 0 6= IΓJ * P . Hence, P is a weakly prime ideal. The converse is obvious by the
definition of a weakly prime ideal of a Γ-seminearring.



Abdelghani Taouti et al. / Eur. J. Pure Appl. Math, 12 (2) (2019), 544-552 548

Theorem 1. Let M be an m-system of a Γ-seminearring R whose each ideal is a subtrac-
tive. Let I be an ideal with I ∩M = ∅. Then, there exists a weakly prime ideal P such
that I ⊆ P and P ∩M = ∅.

Proof. Let = = {J : J is an ideal of R, I ⊆ J and J ∩M 6= ∅} . Then, = 6= ∅ and
let {Jα}α∈A be a chain in I which is ordered under set inclusion. Then, I ⊆ ∩α∈ΛJα and
(∪α∈ΛJα) ∩M = ∪α∈Λ(Jα ∩M) 6= ∅. Thus, ∪α∈ΛJα ∈ I. By Zorn’s Lemma, = has a
maximal element say P . We also claim that P is a weakly prime ideal. If P ⊂ K1 and
P ⊂ K2, then there exist k1 ∈ K1∩M , k2 ∈ K2∩M and α ∈ Γ such that 0 6= (k1)α(k2) ⊆
0 6= K1ΓK2 and there exist k′1 ∈ (k1) and k′2 ∈ (k2) such that 0 6= k′1αk

′
2 ∈M . Thus, 0 6=

k′1αk
′
2 ∈ 0 6= K1ΓK2 ∩M . Since P ∩M = ∅, (K1ΓK2) * P . Hence, P is a weakly prime

ideal.

Now we present few results about such a Γ-seminearring R in which each ideal is weakly
prime.

Proposition 7. Every ideal of a Γ-seminearring R is a weakly prime if and only if for
any ideals I, J , K of R, IΓJ = I, IΓJ = J, IΓJ = K where K is the ideal contained in
both I and J , or IΓJ = 0.

Proof. Suppose that every ideal of R is a weakly prime. Let I, J are ideals of a
Γ-seminearring R. If IΓJ 6= R, then IΓJ is a weakly prime. If 0 6= IΓJ ⊆ IΓJ , then we
have I ⊆ IΓJ or J ⊆ IΓJ i.e., I = IΓJ or J = IΓJ . If IΓJ = K then clearly K = I ∩J is
a weakly prime ideal then by proposition3, K ⊂ I and K ⊂ J . Finally, if IΓJ = R, then
we have I = J = R and hence RΓR = R.

Conversely, let L be any proper ideal of R and suppose that 0 6= IΓJ ⊆ L for ideals I
and J of R. Then, we have either I = IΓJ ⊆ L or J = IΓJ ⊆ L. And if K = IΓJ ⊆ L,
where K ⊂ I ∩ J and hence K ∩ I ⊆ I and K ∩ J ⊆ L.

Example 4. Refer to the Γ-seminearring S defined by tables in example3. Clearly, S has
four ideals, I = {0, a}, J = {0, c}, K = {0, e, c} and L = {0, a, c}. Now, IΓJ = {0},
IΓK = {0}, IΓL = I, JΓI = {0}, JΓK = {0}, JΓL = {0}, KΓI = {0}, KΓJ = J,
KΓL = J where J ⊂ K and J ⊂ L, LΓI = I, LΓJ = {0}, LΓK = {0}. Hence, we can
check easily that every ideal of S is a weakly prime ideal.

Corollary 1. Let R be a Γ-seminearring in which every ideal of R is a weakly prime.
Then for any ideal I of R, either IΓI = I2 = I or IΓI = I2 = 0.

Example 5. Refer to example4, since I = {0, a} be the weakly prime ideal of S and hence
IΓI = I2 = I. Also, for another weakly prime ideal J = {0, c} of S we have JΓJ = J2 = 0.

In the above example the ideal K = {0, e, c} we have KΓK = K2 = {0, e} which
is a subset of Γ-seminearring but not an ideal. And for the ideal L = {0, a, c} we have
LΓL = L2 = {0, a}which is a weakly prime ideal of S.)

Proposition 8. Suppose that every ideal of a Γ-seminearring R is a weakly prime. If M1

and M2 are two maximal ideals of R then M1ΓM2 = 0 or M1ΓM2 = N = M1 ∩M2.



Abdelghani Taouti et al. / Eur. J. Pure Appl. Math, 12 (2) (2019), 544-552 549

Proof. Suppose every ideal of a Γ-seminearring R is a weakly prime ideal. Let M1

and M2 be the two distinct maximal ideals. Since, M1 ∩M2 is a weakly prime and hence
M1ΓM2 ⊆M1∩M2, we must have M1ΓM2 = 0 and similarlyM2ΓM1 = 0, or M2ΓM1 = N ,
being every ideal a weakly prime ideal of R, the result follows from proposition3.

Example 6. Refer to the Γ-seminearring S defined in tables of an example3. Let I = {0,
a, c} and J = {0, e, c} be the two maximal ideals of S. Clearly IΓJ = 0 and JΓI = {0,
c} = I ∩ J = {0, c}.

Corollary 2. Let every ideal of a Γ-seminear-ring R is a weakly prime. Then, every
nonzero ideal of R/N(R) is prime.

Corollary 3. Suppose that every ideal of a Γ-seminear-ring R is a weakly prime. Then
(N(R))Γ(N(R)) = 0 and every prime ideal P (R) contains N(R). There are three possi-
bilities.

(a) N(R) = R.
(b) N(R) = P (R) is the smallest prime ideal and all other prime ideals are idempotent

and are linearly ordered. If N(R) 6= 0, then it is the only non-idempotent prime ideal.
(c) N(R) = P (R) is not a prime ideal. And in such case there exist two nonzero minimal
prime ideals J1 and J2 with N(R) = J1 ∩ J2 and J1ΓJ2 = {0} or (d), J2ΓJ1 = {0} or (c).
All other ideals containing N(R) also contain J1 + J2 and they are linearly ordered.

We elaborate the above proposition in the below example.

Example 7. Refer to the Γ-seminearring S defined in tables of an example3. In S, N(S) =
{0, c} and we have (N(S))2 = (N(R)) Γ(N(R)) = {0, c}2 = {0}. As, N(S) = P (S) is
not a prime ideal and possibility (c) of the above corollary3 is valid for this i.e., there
exist two nonzero minimal prime ideals J1 and J2 with N(R) = J1 ∩ J2 and J1ΓJ2 = {0}
and J2ΓJ1 = (c) = {0, c}. All other ideals containing N(R) also contain J1 + J2 and are
linearly ordered. Let J1 = {0, a, c} and J2 = {0, e, c} be the minimal prime ideals of S.
We have N(S) = {0, c} = J1 ∩ J2 and J1J2 = J2J1 = 0. Beside these two ideals another
ideal of S is S itself and clearly it contains N(S) and also J1 + J2 where J1 + J2 = S.

Example 8. Let T = {0, a, b} be a right seminearring under the operations defined in
given below tables.

+ 0 a b

0 0 a b

a a a a

b b b b

. 0 a b

0 0 0 0

a 0 a a

b 0 a b

Here N(T ) = P (T ) = {0} and it is the smallest prime ideal. Possibility (b) of above
corollary3 is valid for this seminearring.

Definition 4. Let R be a Γ-seminearring under the mapping from R × Γ × R into R,
say f , and D be the set of all destributive elements of R, i.e., D = {d ∈ R | dα(a + b) =



Abdelghani Taouti et al. / Eur. J. Pure Appl. Math, 12 (2) (2019), 544-552 550

dαa + dαb for all a, b ∈ R and α ∈ Γ}. Then R is called distributively generated (in
short, d.g.) if the set D is non empty subset of R which fD×Γ×D : D × Γ×D → D and

(< D,+ >) = (R,+) where < D >= {
m∑
i=1

αidi | m,αi ∈ N and di ∈ D for all i }. In fact,

< D >= {
∑n

i=1 di | n ∈ N and di ∈ D} where all d’is in
∑
di may not be distinct. In

addition, (< D,+ >) = (R,+) means that every element in R can be written as a finite
sum of desrtributive elements.

Example 9. Refer to the Γ-seminearring S defined in tables of an example3. Let D = {0,
1}, where all elements of D are distributive elements of R i.e., D = {d ∈ R | dα(a+ b) =
dαa + dαb for all a, b ∈ R and α ∈ Γ}. S is called distributively generated because the
set D = {0, 1} is a nonempty subset of R which satisfies fD×Γ×D : D × Γ×D → D and
(< D >,+) = (R, +).

Theorem 2. Let R be a distributively generated Γ-seminearring.
(1) If A is weakly prime ideal of R and B is a nonempty subset of R. Then, AΓB is

a weakly prime ideal of R.
(2) If A and B are weakly prime ideals of R, then AΓB is an ideal of R.

Example 10. Refer to the Γ-seminearring S defined in tables of an example3. Let A = {0,
a} be a weakly prime ideal of R and B = {1, e} be a nonempty subset of R. Clearly
AΓB = {0, a} is a weakly prime ideal of S. Let C = {0, c} be another weakly prime ideal.
Also AΓC = {0} and it is a minimal prime ideal of S.

Weakly primary ideals

Definition 5. Let R be a Γ-seminearring. A proper ideal P of R is said to be a weakly
primary ideal if 0 6= pγq ∈ P implies p ∈ P or qn ∈ P .

+ 0 1 e a b c

0 0 1 e a a c

1 1 1 1 1 1 1

e e 1 e 1 1 e

a a 1 1 a a a

b a 1 1 a a a

c c 1 e a a c

α 0 1 e a b c

0 0 0 0 0 0 0

1 0 1 e a b c

e 0 e e 0 c c

a 0 a 0 a a 0

b 0 b 0 a a 0

c 0 c 0 0 0 0



Abdelghani Taouti et al. / Eur. J. Pure Appl. Math, 12 (2) (2019), 544-552 551

Example 11. Let R = {0, 1, e, a, b, c} be a Γ-seminearring with Γ = {α, 1} defined
in example1. Here I = {0, a}, J = {0, a, c} are weakly primary ideals but not a weakly
prime. Clearly, bαb = a ∈ I but b2 = a ∈ I. Similarly, J is also a weakly primary but
not a weakly prime ideal. neither prime because in J , as e.b = c ∈ J. Clearly, e, b /∈ J but
b2 = a ∈ J.

Proposition 9. Every weakly prime ideal is a weakly primary ideal but converse is not
true.

Example 12. Let R = {0, 1, e, a, b, c} be a Γ-seminearring with Γ = {α, 1}. In R the
ideal I = {0, a, b} is weakly prime and also by above proposition it is weakly primary but
it is not prime b/c cαc = 0 and c /∈ I. Another ideal J = {0, a} is weakly primary but not
weakly prime neither prime b/c bαb = a ∈ I. Clearly, b /∈ I but b2 = a ∈ I.

Proposition 10. Intersection of finite numbers of weakly primary ideals of a Γ-seminearring
R which are totally ordered by inclusion is a weakly primary ideal.

Proof. Let {Pα}α∈Λ be the family of weakly primary ideals which are totally ordered
by inclusion. Suppose I and J be ideals of R. If 0 6= IΓJ ⊆ ∩α∈ΛPα, then 0 6= IΓJ ⊆ Pα,
for all α ∈ Λ. Suppose that there exists α ∈ Λ such that I * Pα . Then, Jn ⊆ Pα and
hence Jn ⊆ Pβ for all β ≥ α. We assume that there exist γ < α such that Jn ⊆ Pγ .Then,
I ⊆ Pγ and hence I ⊆ Pα, which is impossible. Hence, Jn ⊆ Pβ for any β ∈ Λ. Thus,
∩α∈ΛPα is a weakly primary ideal of a Γ-seminearring R.

Example 13. Let R = {0, 1, e, a, b, c} be a Γ-seminearring with Γ = {α, 1}. Here
I = {0, a}, J = {0, a, c} are weakly primary ideals but not weakly prime and K = {0, a, b, c}
is prime and hence weakly primary because every prime ideal is weakly primary. Clearly,
these ideals are totally ordered by inclusion i.e. I ⊆ J ⊆ K. Since, I ∩J ∩K = I = {0, a},
which is also a primary ideal b/c b.b = a ∈ I. Clearly, b /∈ I but b2 = a ∈ I.

Proposition 11. Every ideal of a Γ-seminearring R is a weakly primary if and only if
for any ideals I, J , K of R, IΓJ = I, IΓJ = J, IΓJ = K where K is the ideal contained
in both I and J or either in I or in J i.e.K ⊆ I, J or K ⊆ I or K ⊆ J, or IΓJ = 0.

Proof. Suppose that every ideal of R is a weakly prime. Let I, J are ideals of a
Γ-seminearring R. If IΓJ 6= R, then IΓJ is a weakly prime. If 0 6= IΓJ ⊆ IΓJ , then we
have I ⊆ IΓJ or Jn ⊆ IΓJ i.e., I = IΓJ or Jn = IΓJ . If IΓJ = K then clearly K = I ∩J
is a weakly primary ideal then, K ⊂ I and K ⊂ Jn. Finally, if IΓJ = R, then we have
I = J = R and hence RΓR = R.

Conversely, let L be any proper ideal of R and suppose that 0 6= IΓJ ⊆ L for ideals I
and J of R. Then, we have either I = IΓJ ⊆ L or Jn = IΓJ ⊆ L. And if K = IΓJ ⊆ L,
where K ⊂ I ∩ J and hence K ∩ I ⊆ I and K ∩ J ⊆ L.

Example 14. Let R = {0, 1, e, a, b, c} be a Γ-seminearring with Γ = {1, α}. As R has six
different ideals i.e. I = {0, a}, J = {0, c}, K = {0, e, c}, L = {0, a, c}, M = {0, a, b}, and



REFERENCES 552

N = {0, a, b, c}. Now, IΓI = I, IΓJ = {0}, IΓK = {0}, IΓL = I, IΓM = I, IΓN = I,
JΓI = {0}, JΓJ = {0}, JΓK = {0}, JΓL = {0}, JΓM = {0}, JΓN = {0}, KΓI = {0},
KΓJ = J, KΓK = K, KΓL = J, KΓM = J where J ⊆ K , KΓN = J, where J ⊆ K and
J ⊆ N, LΓI = I, LΓJ = {0}, LΓK = {0}, LΓL = I, where I ⊆ L, LΓM = I where I ⊆ L
and I ⊆ M, LΓN = I, where I ⊆ L and I ⊆ N, MΓI = I, MΓJ = {0}, MΓK = {0},
MΓL = I where I ⊆ M and I ⊆ L, MΓM = I, where I ⊆ M, MΓN = I, where I ⊆ M
and I ⊆ N. Hence, we can easily check that every ideal of R is weakly primary.

Proposition 12. Suppose that every ideal of a Γ-seminearring R is a weakly primary.
If M1 and M2 are two maximal ideals of R then either M1ΓM2 = 0 or M1ΓM2 = N =
M1 ∩M2.

Example 15. Let R = {0, 1, e, a, b, c} be a Γ- seminearring with Γ = {1, α}. Let I =
{0, a, b, c} and J = {0, e, c} be the two maximal ideals of R. Clearly IΓJ = {0} and
JΓI = {0, c} = I ∩ J.

References

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