Proof. Let x ∈ Vβα (eαf ) and let g = f βxαe. Then
(eαf )βgα(eαf ) = eα(f βf )βxα(eαe)αf
= eαf βxαeαf
= eαf
Now,
gα(eαf )βg = f βxα(eαe)α(f βf )βxαe
= f β(xαeαf βx)αe
= f βxαe
= g
Hence g ∈ Vβα (eαf ).
Moreover,
gαg = f β(xαeαf βx)αe
= f βxαe
= g
Consequently g ∈ E.
Finally it is clear that
gαe = f βxα(eαe)
= f βxαe
= g
106
and
f βg = (f βf )βxαe
= f βxαe
= g
Hence g ∈ Sαβ (e, f ).
Remark 4.2.1. The set Sαβ (e, f ) is called the (α, β) sandwich set of e and f. It has
an obvious alternative characterization Sαβ (e, f ) = {gαg = g ∈ S; gαe = g =
f βg, eαgαf = eαf }
Lemma 4.2.3. Let S be a regular Γ-semiring with set E of idempotents and let e, f ∈
E. Then Sαβ (e, f ) is a sub Γ-semiring of S and it is a rectangular Γ-band.
Proof. Let g, h ∈ Sαβ (e, f ).
Then
gαhαg = gα(eαhαf )βg
= (gαe)α(f βg)
= gαg
= g − − − (i)
It follows that (gαh)α(gαh) = gαh and so gαh is α-idempotent.
Moreover,
(gαh)αe = gα(hαe)
= gαh
f β(gαh) = (f βg)αh
= gαh
107
and
eα(gαh)αf = (eαgαf )βhαf
= eα(f βh)αf
= eαhαf
= eαf
Hence gαh ∈ Sαβ (e, f ). From (i), we deduce that Sαβ (e, f ) is a rectangular Γ-band.
Lemma 4.2.4. Let S be a regular Γ-semiring. Let a, b ∈ S and α, β, γ, δ ∈ Γ. Suppose
a0 ∈ Vαβ (a), b0 ∈ Vγδ (b). Then for each g ∈ Sαδ (a0 βa, bγb0 ),
b0 δgαa0 ∈ Vγβ (aαb).
Proof.
(aαb)γ(b0 δgαa0 )β(aαb) = aα(bγb0 δg)αa0 βaαb
= aα(gαa0 βa)αb
= aαgαb
= aα(a0 βaαgαbγb0 )δb
= (aαa0 βa)α(bγb0 δb)
= aαb
Moreover,
(b0 δgαa0 )β(aαb)γ(b0 δgαa0 ) = b0 δ(gαa0 βa)α(bγb0 δg)αa0
= b0 δ(gαg)αa0
= b0 δgαa0
Hence b0 δgαa0 ∈ Vγβ (aαb).
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Theorem 4.2.5. Let S be a regular Γ-semiring with set E of idempotents. Then the
following statements are equivalent:
(i) S is orthodox
(ii) If e = eαe and f = f βf are any two idempotents of S, then f βe ∈ Sαβ (e, f ) where
α, β ∈ Γ.
(iii) For all a, b ∈ S, there exist α, β, γ, δ ∈ Γ such that Vγδ (b)ΓVαβ (a) ⊆ Vγβ (aαb).
(iv) For every e ∈ E, there exist α, β ∈ Γ such that Vαβ (e) ⊆ E.
Proof.
(i) ⇒ (ii): Suppose that S is orthodox. Let e = eαe, f βf = f ∈ E and g = f βe.
Then
gαe = f β(eαe)
= f βe
= g
f βg = (f βf )βe
= f βe
= g
and eαgαf = eαf βeαf
= eαf
By remark 4.2.1, g = f βe ∈ Sαβ (e, f ).
(ii) ⇒ (iii): Let a, b ∈ S and α, β, γ, δ ∈ Γ. Suppose a0 ∈ Vαβ (a) and b0 ∈ Vγδ (b).
Then by lemma 4.2.4, b0 δgαa0 ∈ Vγβ (aαb) for all g in Sαδ (a0 βa, bγb0 ). From (ii), it
follows that (bγb0 )δ(a0 βa) ∈ Sαδ (a0 βa, bγb0 ). By lemma 4.2.4, b0 δ(bγb0 δa0 βa)αa0 ∈
Vγβ (aαb). Hence b0 δa0 ∈ Vγβ (aαb).
109
(iii) ⇒ (iv): Let e ∈ E and α, β ∈ Γ. Suppose x ∈ Vαβ (e). Then xβeαx = x and
eαxβe = e. Since xβe is α-idempotent and eαx is β-idempotent, xβe ∈ Vαα (xβe) and
eαx ∈ Vββ (eαx). By (iii), (eαx)β(xβe) ∈ Vβα (xβeαeαx) which implies (eαx)β(xβe) ∈
Vβα (x). Now,
x = xβ(eαxβxβe)αx
= (xβeαx)β(xβeαx)
= xβx
Hence x is idempotent.
(iv) ⇒ (i): Let eαe = e, f βf = f ∈ E and α, β ∈ Γ. By lemma 4.2.2, there exists
an idempotent g in Vαβ (f βe). Since (f βe) being an (β, α) inverse of the idempotent
g is itself an idempotent, S is orthodox.
Theorem 4.2.6. Let S be an orthodox Γ-semiring with set E of idempotents. Let
α, β, γ ∈ Γ. For all a in S, e in E and a0 ∈ Vαγ (a), the element aαeβa0 is γ-idempotent
and the element a0 βeαa is α-idempotent.
Proof.
(aαeβa0 )γ(aαeβa0 ) = aα(eβa0 γaαeβa0 γa)αa0
= aαeβ(a0 γaαa0 )
= aαeβa0
and (a0 βeαa)α(a0 βeαa) = a0 γ(aαa0 βeαaαa0 βe)αa
= (a0 γaαa0 )βeαe
= a0 βeαa
as required.
110
Theorem 4.2.7. Every inverse Γ-semiring S is an orthodox Γ-semiring
Proof. Let e be an α-idempotent and f be a β-idempotent of S. We first show that
eαf is a β-idempotent. Now eαf ∈ S. There exist γ, δ ∈ Γ and x ∈ S such that
x ∈ Vδγ (eαf ). Then (eαf )δxγ(eαf ) = eαf and xγ(eαf )δx = x. Let g = f δxγeαf
and h = f δxγe. It can be seen easily that gβg = g.
Also,
(f δxγeαf )β(f δxγe)α(f δxγeαf ) = f δ(xγeαf δx)γ(eαf δxγeαf )
= f δxγeαf
= g
This shows that gβhαg = g. Similarly hαgβh = h. Hence, g ∈ Vαβ (h). Also, eαf ∈
Vαβ (h). Since S is an inverse Γ-semiring, g = eαf . Hence eαf is a β-idempotent.
Proceeding as above, we can show that f αe is a β-idempotent and both eβf and f βe
are α-idempotents. Hence S is an orthodox Γ-semiring.
Theorem 4.2.8. A regular Γ-semiring S is an orthodox Γ-semiring if and only if
for any α-idempotent e ∈ S, where α ∈ Γ, if Vαβ (e) 6= φ and Vβα (e) 6= φ, then each
member of Vαβ (e) and Vβα (e) is a β-idempotent.
Proof. Suppose S is an orthodox Γ-semiring. Let e be an α-idempotent of S and let
x ∈ Vαβ (e). Then eαxβe = e and xβeαx = x. Now, eαx is a β-idempotent and xβe
is an α-idempotent. Then x = (xβe)α(eαx) is a β-idempotent. Next let y ∈ Vβα (e).
Then eβyαe = e and yαeβy = y. Now yαe is a β-idempotent and eβy is an αidempotent. Then y = (yαe)α(eβy) is a β-idempotent. Conversely suppose that S
satisfies the given conditions. Let e be an α-idempotent and f a β-idempotent of S.
111
Let us now consider eαf . Now eαf ∈ S and since S is regular, there exists x ∈ S
and γ, δ ∈ Γ such that (eαf )γxδ(eαf ) = eαf , xδ(eαf )γx = x. Let g = f γxδe. Then
gαg = f γxδeαf γxδe
= f γ(xδeαf γx)δe
= f γxδe
= g
Further
(eαf )β(f γxδe)α(eαf ) = eαf γxδeαf
= eαf
and (f γxδe)α(eαf )β(f γxδe) = (f γxδe)α(f γxδe)
= f γxδe
Hence eαf ∈ Vαβ (f γxδe). Then by the given condition, eαf is a β-idempotent.
Dually, we can prove that f αe is a β-idempotent.
Lemma 4.2.9. Let S be an inverse Γ-semiring. Let a, b ∈ S and a0 ∈ Vαα12 (a), b0 ∈
Vββ12 (b) where α1 , α2 , β1 , β2 ∈ Γ. Then b0 β2 a0 ∈ Vβα12 (aα1 b) and b0 α1 a0 ∈ Vβα12 (aβ2 b).
Proof. Since a0 ∈ Vαα12 (a) and b0 ∈ Vββ12 (b),
aα1 a0 α2 a = a, a0 α2 aα1 a0 = a0 and
bβ1 b0 β2 b = b, b0 β2 bβ1 b0 = b0 . Now a0 α2 a is an α1 -idempotent and bβ1 b0 is a β2 idempotent. From theorem 4.2.7, it follows that a0 α2 aα1 bβ1 b0 is a β2 -idempotent,
bβ1 b0 β2 a0 α2 a is an α1 -idempotent, a0 α2 aβ2 bβ1 b0 is an α1 -idempotent and bβ1 b0 α1 a0 α2 a
is a β2 -idempotent.
112
Now,
(aα1 b)β1 (b0 β2 a0 )α2 (aα1 b) = aα1 a0 α2 aα1 bβ1 b0 β2 a0 α2 aα1 bβ1 b0 β2 b
= aα1 (a0 α2 aα1 bβ1 b0 )β2 (a0 α2 aα1 bβ1 b0 )β2 b
= aα1 a0 α2 a)α1 (bβ1 b0 β2 b)
= aα1 b
(b0 β2 a0 )α2 (aα1 b)β1 (b0 β2 a0 ) = b0 β2 bβ1 b0 β2 a0 α2 aα1 bβ1 b0 β2 a0 α2 aα1 a0
= b0 β2 (bβ1 b0 β2 a0 α2 a)α1 (bβ1 b0 β2 a0 α2 a)α1 a0
= (b0 β2 bβ1 b0 )β2 (a0 α2 aα1 a0 )
= b 0 β 2 a0
Hence b0 β2 a0 ∈ Vβα12 (aα1 b). Similarly we can prove that b0 α1 a0 ∈ Vβα12 (aβ2 b)
Theorem 4.2.10. A regular Γ-semiring S is an orthodox Γ-semiring if and only if
for a, b ∈ S, α1 , α2 , β1 , β2 ∈ Γ, a0 ∈ Vαα12 (a), b0 ∈ Vββ2
(b) imply that b0 β2 a0 ∈ Vβα12 (aα1 b)
1
and b0 α1 a0 ∈ Vβα12 (aβ2 b).
Proof. Let us assume that S is an orthodox Γ-semiring. Let a0 ∈ Vαα12 (a) and b0 ∈
Vββ2
(b). Then aα1 a0 α2 a = a, a0 α2 aα1 a0 = a0 , bβ1 b0 β2 b = b, b0 β2 bβ1 b0 = b0 . Then by
1
lemma 4.2.9, b0 β2 a0 ∈ Vβα12 (aα1 b) and b0 α1 a0 ∈ Vβα12 (aβ2 b).
Conversely assume that the given conditions hold in S. Let e be an α-idempotent
and f be a β-idempotent of S. Now, f ∈ Vββ (f ) and e ∈ Vαα (e). Then by the given
conditions eαf ∈ Vαβ (f βe) − − − (i) and eβf ∈ Vαβ (f αe) − − − (ii). From (i), we get
(eαf )β(f βe)α(eαf ) = eαf . Hence (eαf )β(eαf ) = eαf . Thus eαf is a β-idempotent.
From (ii), we get (f αe)α(eβf )β(f αe) = f αe. Hence, (f αe)β(f αe) = f αe. So, f αe
is a β-idempotent. Hence S is an orthodox Γ-semiring.
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4.3
Homomorphisms on Orthodox Γ-semirings
In this section we deal with homomorphisms on orthodox Γ-semirings.
Lemma 4.3.1. Let S be a regular Γ-semiring and S 0 be a Γ0 -semiring. Let (f,g) be
a homomorphism from (S, Γ) onto (S 0 , Γ0 ). Let e0 be an α0 -idempotent of S 0 . Then
f −1 (e0 ) contains an idempotent of S.
Proof. Let a ∈ S be such that f (a) = e0 = e0 α0 e0 where α0 ∈ Γ0 . Let α ∈ Γ be such that
g(α) = α0 . Now let us consider the element aαa. As S is a regular Γ-semiring, there
exist β, γ ∈ Γ and c ∈ S such that (aαa)βcγ(aαa) = aαa and cγ(aαa)βc = c. Now,
aβcγa is an α-idempotent in S since (aβcγa)α(aβcγa) = aβ(cγaαaβc)γa = aβcγa.
Also,
f (aβcγa) = f (a)g(β)f (c)g(γ)f (a)
= e0 α0 e0 g(β)f (c)g(γ)e0 α0 e0
= f (aαa)g(β)f (c)g(γ)f (aαa)
= f (aαaβcγaαa)
= f (aαa) = e0 α0 e0 = e0
Hence f −1 (e0 ) contains an idempotent of S.
Lemma 4.3.2. Let S be a regular Γ-semiring and let S 0 be a Γ0 -semiring. Let (f,g)
be a homomorphism from (S, Γ) onto (S 0 , Γ0 ). Then S 0 is a regular Γ0 -semiring.
Proof. Let a0 ∈ S 0 . There exists a ∈ S such that f (a) = a0 . Since S is a regular
Γ-semiring, there exists b ∈ S and α, β ∈ Γ such that a = aαbβa.
114
Then
a0 = f (a)
= f (aαbβa)
= f (a)g(α)f (b)g(β)f (a)
= a0 g(α)f (b)g(β)a0
Thus a0 is a regular element of S 0 . Hence S 0 is a regular Γ0 -semiring.
Theorem 4.3.3. Let S be an orthodox Γ-semiring and S 0 be a Γ0 -semiring. Let (ϕ, ψ)
be a homomorphism from (S, Γ) onto (S 0 , Γ0 ). Then S 0 is an orthodox Γ0 -semiring.
Proof. Since S is regular, by lemma 4.3.2, it follows that S 0 is also regular Γ0 -semiring.
Let e0 = e0 α0 e0 and f 0 = f 0 β 0 f 0 where α0 , β 0 ∈ Γ0 be two arbitrary idempotents of S 0 .
By lemma 4.3.1, ϕ−1 (e0 ) and ϕ−1 (f 0 ) both contain idempotents of S. Let ES denote
the set of idempotents of S. Let e ∈ ϕ−1 (e0 ) ∩ ES and eαe = e, where ψ(α) = α0
and f ∈ ϕ−1 (f 0 ) ∩ ES and f βf = f where ψ(β) = β 0 . Now since S is an orthodox
Γ-semiring, eαf , f αe are β-idempotents and eβf , f βe are α-idempotents. Then
ϕ(eαf ) = e0 α0 f 0 ∈ ES 0 and e0 α0 f 0 is β 0 -idempotent. Similarly f 0 α0 e0 is a β 0 -idempotent.
Hence S 0 is an orthodox Γ0 -semiring.
115
Chapter 5
Pseudo Symmetric and
Semipseudo Symmetric Γ-semirings
In this chapter we introduce the notions of pseudo symmetric ideals in Γ-semirings
and pseudo symmetric Γ-semirings. We characterize pseudo symmetric ideals in
Γ-semirings and exhibit some examples and some classes of pseudo symmetric Γsemirings. We also introduce a class of Γ-semirings known as semipseudo symmetric
Γ-semiring and characterize Archimedian Γ-semirings of this class. We obtain necessary conditions for semipseudo symmetric Γ-semirings to possess semisimple elements. We also characterize maximal ideals in Archimedian semipseudo symmetric
Γ-semirings.
116
5.1
Basic Definitions and Results
In order to make this chapter reasonably self contained, we present here some
basic definitions and a result which we require for the development of this chapter.
Definition 5.1.1. Let S and Γ be two additive commutative semigroups. Then S is
called Γ-semiring if there exists a mapping S × Γ × S → S(image to be denoted by
aαb for a, b ∈ S, α ∈ Γ) satisfying the following conditions.
(i) aα(b + c) = aαb + aαc
(ii) (a + b)αc = aαc + bαc
(iii) a(α + β)b = aαb + aβb
(iv) aα(bβc) = (aαb)βc, for all a, b, c ∈ S and for all α, β ∈ Γ.
It is obvious that every semiring S is a Γ-semiring where Γ = S and xαy denotes
the product of the elements x, y, α ∈ S. Every Γ-ring is also a Γ-semiring.
A Γ-semiring S is said to be commutative if aαb = bαa, for all a, b ∈ S and for all
α ∈ Γ.
Let S be a Γ-semiring and let A be a non empty subset of S. A is called a sub
Γ-semiring of S if A is a sub semigroup of (S, +) and AΓA ⊆ A.
Definition 5.1.2. A Γ-semiring S is said to be left(right) pseudo commutative provided aΓbΓc = bΓaΓc(aΓbΓc = aΓcΓb) for all a, b, c ∈ S.
Definition 5.1.3. A Γ-semiring S is said to be quasi commutative provided for any
a, b ∈ S, there exists a natural number n such that aΓb = (bΓ)n−1 bΓa.
Definition 5.1.4. An element ‘a’ of a Γ-semiring S is said to be an r-element provided
aΓs = sΓa, for all s ∈ S and if x, y ∈ S, we have aΓxΓy = bΓyΓx, for some b ∈ S.
117
Definition 5.1.5. A Γ-semiring S is said to be a generalised commutative Γ-semiring
provided S contains 1 as an r-element.
Definition 5.1.6. A Γ-semiring S is said to be normal provided aΓS = SΓa, for all
a ∈ S.
Definition 5.1.7. An element ‘a’ of a Γ-semiring S is said to be an idempotent
provided aαa = a, for all α ∈ Γ.
A Γ-semiring S is said to be an idempotent
Γ-semiring provided every element is an idempotent.
Definition 5.1.8. An element ‘a’ of a Γ-semiring S is said to be a mid unit provided
xΓaΓy = xΓy for any x, y ∈ S.
Definition 5.1.9. An element ‘a’ of a Γ-semiring S is said to be intra regular provided
a = x(aα)1 aβy, for some x, y ∈ S and α, β ∈ Γ.
Definition 5.1.10. A Γ-semiring S is said to be an Archimedian Γ-semiring provided
for any a, b ∈ S, there exists a natural number n such that (aΓ)n−1 a ⊆ SΓbΓS.
Definition 5.1.11. A Γ-semiring S is said to be a globally idempotent Γ-semiring
provided (SΓ)1 S = S.
Definition 5.1.12. Let S be a Γ-semiring with a zero element 0 and let A be an ideal
of S. Then A is said to be nilpotent if (AΓ)n−1 A = 0, for some integer n > 0.
Definition 5.1.13. An ideal A of a Γ-semiring S is called a prime ideal provided
XΓY ⊆ A ; X,Y are ideals of S, then either X ⊆ A or Y ⊆ A or xΓSΓy ⊆ A implies
either x ∈ A or y ∈ A. An ideal A of a Γ-semiring S is called a completely prime
ideal provided xΓy ⊆ A; x, y ∈ S implies either x ∈ A or y ∈ A.
118
Definition 5.1.14. An ideal A of a Γ-semiring S is called a semiprime ideal provided
xΓSΓx ⊆ A; x ∈ S implies x ∈ A. An ideal A of a Γ-semiring S is called a completely
semiprime ideal provided (xΓ)n−1 x ⊆ A; x ∈ S for some natural number n implies
x ∈ A.
Definition 5.1.15. A Γ-semiring S is said to be a left(right) duo Γ-semiring provided
every left(right) ideal of S is a two sided ideal of S. A Γ-semiring S is said to be a
duo Γ-semiring provided it is both a left and a right duo Γ-semiring.
Definition 5.1.16. An ideal A of a Γ-semiring S is said to be left(right) primary
provided that the following conditions hold :
(i) If X,Y are ideals of S such that XΓY ⊆ A and Y * A (X * A), then X ⊆
MA (S) (Y ⊆ MA (S)), MA (S) is the intersection of all prime ideals of S containing A.
(ii) MA (S) is a prime ideal of S.
Theorem 5.1.1. Every (completely)semiprime ideal of a Γ-semiring S is the intersection of all minimal (completely)prime ideals of S containing it.
Definition 5.1.17. An element ‘a’ of a Γ-semiring S is said to be semisimple provided
a ∈ (hai Γ)1 hai, that is (hai Γ)1 hai = hai.
Definition 5.1.18. An ideal A of a Γ-semiring is called a maximal ideal provided A
is a proper ideal of S and is not properly contained in any proper ideal of S.
5.2
Pseudo Symmetric Ideals
Definition 5.2.1. An ideal A in a Γ-semiring S is said to be pseudo symmetric
provided xΓy ⊆ A ; x, y ∈ S implies xΓsΓy ⊆ A, for all s ∈ S.
119
Definition 5.2.2. A Γ-semiring S is said to be pseudo symmetric provided every
ideal is pseudo symmetric ideal.
We remark that every commutative Γ-semiring is a pseudo symmetric Γ-semiring
and the converse need not necessarily be true.
Example 5.2.1. Let S = {a, b, c} and Γ = {a, b, c}. We define binary operations 0 +01
in S and 0 +02 in Γ as shown in the following tables :
+1
a
b
c
+2
a
b
c
a
a
a
a
a
a
b
a
b
a
a
a
b
a
b
a
c
a
b
c
c
a
b
c
Now, clearly (S,+1 ) and (Γ,+2 ) are semigroups. Define a mapping S×Γ×S →
S by aαb = ab, for all a,b ∈ S and α ∈ Γ. It is easy to see that S is a Γ-semiring.
Now clearly S is pseudo symmetric Γ-semiring which is not commutative Γ-semiring.
We now discuss the relationships among prime, completely prime and pseudo
symmetric ideals.
Theorem 5.2.2. Let S be a Γ-semiring. Then the following statements hold :
(i)
Every completely prime ideal is both prime and pseudo symmetric.
(ii) Let A be a pseudo symmetric ideal of S. Then A is prime ⇔ A is completely
prime.
(iii) Let A be a prime ideal of S. Then A is pseudo symmetric ⇔ A is completely
prime.
120
Proof.
(i) This statement is easy to observe. We hence omit the proof.
(ii) This result follows from lemma 1 in [5].
(iii) Let A be a pseudo symmetric ideal of S. If xΓy ⊆ A, for some x, y ∈ S, then
xΓSΓy ⊆ A. Since A is prime, we have x ∈ A or y ∈ A. This shows that A is
completely prime. Conversely if A completely prime ideal, then from (i) A is pseudo
symmetric ideal.
Theorem 5.2.3. The following statements are equivalent for an ideal A in a Γsemiring S.
(i) A is a pseudo symmetric ideal.
(ii) Ar (a) = {x ∈ S; aΓx ⊆ A} is an ideal of S for all a∈S.
(iii) Al (a) = {x ∈ S; xΓa ⊆ A} is an ideal of S for all a∈ S.
Proof.
(i) ⇒ (ii) :
Let x ∈ Ar (a). Then aΓx ⊆ A. Since A is a pseudo symmetric ideal, aΓsΓx ⊆ A and
clearly aΓxΓs ⊆ A. Therefore sΓx, xΓs ⊆ Ar (a) and hence Ar (a) is an ideal in S.
(ii) ⇒ (i) :
Let xΓy ⊆ A. Then y ∈ Ar (x). Since Ar (x) is an ideal, we have sΓy ⊆ Ar (x) and
hence xΓsΓy ⊆ A, for all s ∈ S. Therefore A is a pseudo symmetric ideal.
(i) ⇒ (iii) :
Let x ∈ Al (a). Then xΓa ⊆ A. Since A is a pseudo symmetric ideal, xΓsΓa ⊆ A and
clearly sΓxΓa ⊆ A. Therefore xΓs, sΓx ⊆ Al (a) and hence Al (a) is an ideal in S.
121
(iii) ⇒ (i) :
Let xΓy ⊆ A. Then x ∈ Al (y). Since Al (y) is an ideal of S, xΓs ⊆ Al (y), for all s ∈ S
and hence xΓsΓy ⊆ A, for all s ∈ S. Therefore A is a pseudo symmetric ideal of S.
We now describe the relationship between the one-side duo Γ-semiring and the
pseudo symmetric Γ-semiring.
Corollary 5.2.4. Every left(right)duo Γ-semiring S is a pseudo symmetric Γ-semiring.
Proof. Let A be any ideal in S. Since for all a ∈ S, Al (a) is a left ideal and hence by
theorem 5.2.3, A is a pseudo symmetric ideal. Therefore S is a pseudo symmetric Γsemiring. Similarly every right duo Γ-semiring is a pseudo symmetric Γ-semiring.
Remark 5.2.1. In fact pseudo symmetric Γ-semirings are in abundance.
(1) Every left(right) pseudo commutative Γ-semiring is a pseudo symmetric Γsemiring.
(2) Every quasi commutative Γ-semiring is a pseudo symmetric Γ-semiring.
(3) Every generalized commutative Γ-semiring is a pseudo symmetric Γ-semiring.
(4) Every normal Γ-semiring is a pseudo symmetric Γ-semiring.
Lemma 5.2.5. Every idempotent Γ-semiring S is a pseudo symmetric Γ-semiring.
Proof. Let A be any ideal in S and let aΓb ⊆ A. Then bΓa = bΓaΓbΓa ⊆ A and
hence aΓsΓb = aΓsΓbΓaΓsΓb ⊆ A. Therefore A is a pseudo symmetric ideal.
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Lemma 5.2.6. If S is a Γ-semiring in which every element is a mid unit, then S is
a pseudo symmetric Γ-semiring.
Proof. Let A be an ideal in S and let aΓb ⊆ A. Now, for any s ∈ S, aΓsΓb = aΓb ⊆ A.
So, A is a pseudo symmetric ideal.
Lemma 5.2.7. Every completely semiprime ideal A in a Γ-semiring S is a pseudo
symmetric ideal and the converse is not true.
Proof. Let xΓy ⊆ A. Then (yΓxΓ)1 yΓx = yΓxΓyΓx ⊆ A. Since A is a completely
semiprime ideal, yΓx ⊆ A. Now, (xΓsΓyΓ)1 xΓsΓy = xΓsΓyΓxΓsΓy ⊆ A, for all
s ∈ S and hence xΓsΓy ⊆ A. Therefore A is a pseudo symmetric ideal.
Remark 5.2.2. Let S be a Γ-semiring. An element a ∈ S is called a left identity(resp.
right identity) of S if x = aαx(resp. x = xαa) for all x ∈ S and α ∈ Γ. If ‘a’ is both
a left and right identity, then ‘a’ is called an identity of S.
Let S be a Γ-semiring. If S has an identity element 1, set S 1 = S and if S does
not have an identity element 1, let S 1 be the Γ-semiring S with an identity element
1 adjoined.
The following theorem gives a characterization for a pseudo symmetric ideal to be
a one sided primary ideal.
Theorem 5.2.8. Let S be a Γ-semiring and A be a pseudo symmetric ideal of S.
Then A is left primary if and only if the following condition holds :
For all x, y ∈ S, xΓy ⊆ A and y ∈
/ A imply (xΓ)n−1 x ⊆ A, for some n > 0−−−(∗)
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Proof. Suppose that A is a left primary ideal of S and xΓy ⊆ A with y ∈
/ A. Then,
since A is pseudo symmetric and xΓy ⊆ A, we have hxi Γ hyi = S 1 Γ(xΓS 1 ΓS 1 Γy)ΓS 1 ⊆
S 1 ΓAΓS 1 ⊆ A. Thus, we have hxi ⊆ MA (S), where MA (S) is the intersection of all
prime ideals of S containing A. Then we have MA (S) = NA (S), where NA (S) is the
set of all elements of S nilpotent with respect to A. This implies that x ∈ NA (S) and
so (xΓ)n−1 x ⊆ A, for some n ≥ 1. Hence (∗) holds.
Suppose that (∗) holds. Then we have the following situations :
(i) X and Y are ideals of S with XΓY ⊆ A but Y * A. Then there exists an
element y ∈ Y but y ∈
/ A such that for all x ∈ X, xΓy ⊆ XΓY ⊆ A. By (∗), we
immediately obtain that x ∈ MA (S), for all x ∈ X. This implies that X ⊆ MA (S).
(ii) Assume that xΓy ⊆ MA (S). Then we have xΓy ⊆ NA (S) and hence we find
a smallest positive integer n such that (xΓyΓ)n−1 xΓy ⊆ A. If n = 1, then xΓy ⊆ A.
By (∗), we have (xΓ)k−1 x ⊆ A, for some integer k > 0 or y ∈ A. This means that
x ∈ MA (S) or y ∈ MA (S). Now, we assume that n > 1.
We have the following cases :
Case (i): If yΓ(xΓyΓ)n−2 xΓy * A, then by (∗) and xΓ(yΓ(xΓyΓ)n−2 xΓy) =
(xΓyΓ)n−1 xΓy ⊆ A, we have (xΓ)n−1 x ⊆ A, for some n > 0. This implies that
x ∈ NA (S) = MA (S).
Case (ii): If yΓ(xΓyΓ)n−2 xΓy ⊆ A, then since (xΓyΓ)n−2 xΓy * A, we have
y ∈ NA (S) = MA (S).
Hence, in all cases we must have x ∈ MA (S) or y ∈ MA (S). This shows that
MA (S) is completely prime, and so MA (S) is prime.
Corollary 5.2.9. If S is a one-side duo Γ-semiring, then an ideal A of S is left
primary if and only if (∗) holds.
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We now make an attempt to characterize pseudo symmetric ideals in Γ-semirings.
Lemma 5.2.10. Let A be any pseudo symmetric ideal in a Γ-semiring S. Then
a1 α1 a2 α2 ......an−1 αn−1 an ∈ A if and only if ha1 i Γ ha2 i ......Γ han i ⊆ A.
Proof. Let A be any pseudo symmetric ideal in a Γ-semiring S. Clearly if
ha1 i Γ ha2 i ......Γ han i ⊆ A,
then a1 α1 a2 α2 ......an−1 αn−1 an ∈ A.
Conversely if a1 α1 a2 α2 ......an−1 αn−1 an ∈ A, then for any t ∈ ha1 i Γ ha2 i ......Γ han i, we
have t = s1 α1 a1 β1 s2 α2 a2 β2 ......αn an βn sn+1 , where si ∈ S 1 and αi , βi ∈ Γ. Since A is
a pseudo symmetric ideal, we have t ∈ A. Therefore ha1 i Γ ha2 i ......Γ han i ⊆ A.
Corollary 5.2.11. If A is a pseudo symmetric ideal in a Γ-semiring S, then for any
natural number n, (aα)n−1 a ∈ A; α ∈ Γ implies (hai Γ)n−1 hai ⊆ A.
Proof. The proof of this corollary follows from lemma 5.2.10 by taking a1 = a2 =
a3 = ... = an = a.
Theorem 5.2.12. Every prime ideal P minimal relative to containing a pseudo symmetric ideal A in a Γ-semiring S is completely prime.
Proof. Let T be the sub Γ-semiring generated by S\P . First we show that A ∩ T = φ.
If A ∩ T 6= φ, then there exist x1 , x2 , x3 , ..., xn ∈ S\P such that x1 α1 x2 ...αn−1 xn ∈ A.
By lemma 5.2.10, we have hx1 i Γ hx2 i Γ......Γ hxn i ⊆ A ⊆ P . Since P is a prime ideal,
we have hxi i ⊆ P for some i, a contradiction. Thus A ∩ T = φ. Consider the set
Σ = {B; B is an ideal in S containing A such that B ∩ T = φ}. Since A ∈ Σ, Σ
is not empty. Now, Σ is a poset under set inclusion and satisfies the hypothesis of
Zorn’s lemma. Thus by Zorn’s lemma, Σ contains a maximal element, say M . Let X
and Y be two ideals in S such that XΓY ⊆ M . If X * M and Y * M , then M ∪ X
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and M ∪ Y are ideals in S containing M properly and hence by the maximality of
M , we have (M ∪ X) ∩ T 6= φ and (M ∪ Y ) ∩ T 6= φ. Since M ∩ T = φ, we have
X ∩ T 6= φ and Y ∩ T 6= φ. So there exists x ∈ X ∩ T and y ∈ Y ∩ T . Now,
xΓy ⊆ XΓY ∩ T ⊆ M ∩ T = φ, a contradiction. Therefore M is a prime ideal
containing A. Now, A ⊆ M ⊆ S\T ⊆ P . Since P is a minimal prime ideal relative
to containing A, we have M = S\T = P . Therefore P is a completely prime ideal.
Corollary 5.2.13. Every prime ideal P minimal relative to containing a completely
semi prime ideal A in a Γ-semiring S is completely prime.
Proof. By lemma 5.2.7, A is a pseudo symmetric ideal and hence by theorem 5.2.12,
P is a completely prime ideal.
5.3
Semipseudo Symmetric Ideals
Definition 5.3.1. For any ideal A in a Γ-semiring S, we call A1 , the complete prime
radical or simply complete radical of the ideal A and denote it by c.rad A.
Definition 5.3.2. For any ideal A in a Γ-semiring S, we call A3 , the prime radical
√
or simply radical of the ideal A and denote it by rad A or A.
We now state without proof a few properties of the complete prime and prime
radicals.
Proposition 5.3.1. For any two ideals A and B of a Γ-semiring S, the following are
true :
(i)
rad(rad A) = rad A and c.rad(c.rad A) = c.rad A.
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