3_dutta.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 5, No. 2, 2012, 116-128

ISSN 1307-5543 – www.ejpam.com

Singular Ideals of Ternary Semirings

Tapan K. Dutta1, Kar Ping Shum 2,∗, Shobhan Mandal3

1 Department of Pure Mathematics, University of Calcutta, 35, Ballygunge Circular Road, Kolkata-

710019,India
2 Institute of Mathematics, Yunnan University, Kunming, 650091, People Republic of China
3 Department of Pure Mathematics, University of Calcutta, 35, Ballygunge Circular Road, Kolkata-

710019, India

Abstract. The notion of singular ideal in a ternary semiring is introduced. The notions of singular

ternary semirings and nonsingular ternary semirings are also defined. Some properties of singular

ideals in ternary semirings are given. Our results obtained can be used to study some radical classes

related to singular ideals.

2010 Mathematics Subject Classifications: 16Y60, 16Y99.

Key Words and Phrases: Singular ideal, Singular ternary semiring and non-singular ternary semirings.

1. Introduction

It was remarked by M. Ferrero and E. R. Puczylowski in [10] “Studying properties of

rings one can usually say more assuming that the considered rings are either singular or non-

singular.” The same remark is equally true in the case of a ternary semiring which was first

introduced by T. K. Dutta and S. Kar in [1]. The notion of ternary semiring was introduced

in 2003. The introduction of ternary algebra dated back to 1932 when Lehmer [13] studied

certain ternary system called triplexes which turn out to be a generalization of abelian groups.

Later on, Banach [cf. Los 15] also studied such algebraic structure and gave some examples

of a ternary semigroup which does not reduce to a semigroup. In addition, W. G. Lister [14]

introduced the notion of ternary ring. Abstractly, a ternary ring T is an abelian group in which

a ternary product tuv is given which is right,center and left distributive and which satisfies

(tuv)x y = t(uv x)y = tu(v x y). In this paper, our ternary semiring is a generalized ternary

ring investigated by W. G. Lister in 1971. Though the notion of ternary semiring generalizes

the notion of semiring but it is not merely a generalization of semiring because there are

certain notions, for example, the lateral ideals which have no analogue in semirings. Some

∗Corresponding author.

Email addresses: duttatapankumar�yahoo.
o.in (T. Dutta), kpshum�ynu.edu.
n (K. Shum),shobhan139�gmail.
om (S. Mandal)

http://www.ejpam.com 116 c© 2012 EJPAM All rights reserved.



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 117

earlier works on ternary semiring may be found in [1, 2, 3, 4, 5, 6, 7, 8, 11].

In this paper, the singular ideals in ternary semirings, singular ternary semirings and non-

singular ternary semirings will be considered. We will investigate the properties of the singular

ideals of a ternary semiring. We will give some examples of singular ternary semirings and

non-singular ternary semirings. We first show that the class of singular ternary semirings with

identity as well as the class of non-singular ternary semirings with identity is closed under

direct products and direct sums. Then we study the image and preimage of singular ternary

semiring and nonsingular ternary semiring under semi-isomorphism. Finally, we show that

the class of semiprime non-singular ternary semirings is hereditary. Our results in this paper

can be applied to study the special radical class of ternary semirings and upper radical class

determined by the above radical classes which are called the singular radical and special

singular radical of ternary semirings respectively. For terminologies and notions not given in

this paper, the reader is referred to W . G. Lister [14].

2. Preliminaries

We first give the following definitions.

Definition 1. [1] A non-empty set S together with a binary operation, called addition and

a ternary multiplication, denoted by juxtaposition, is said to be a ternary semiring if S is an

additive commutative semigroup satisfying the following conditions:

(i) (abc)de = a(bcd)e = ab(cde);

(ii) (a+ b)cd = acd + bcd,

(iii) a(b+ c)d = abd + acd,

(iv) ab(c + d) = abc + abd for all a, b, c, d , e ∈ S.

Definition 2. [1] Let S be a ternary semiring. If there exists an element 0 ∈ S such that 0+x = x

and 0x y = x0y = x y0= 0 for all x , y ∈ S then “0” is called the zero element or simply the zero

of the ternary semiring S. In this case, S is called a ternary semiring with zero.

It is noted that a ternary semiring does not necessarily contain an identity but there are

certain ternary semirings which contain generalized identity in the sense defined below.

Definition 3. [4] A ternary semiring S admits an identity provided that there exist elements

{(ei, fi) ∈ S × S (i = 1,2, . . . , n)} such that
∑n

i=1 ei fi x =
∑n

i=1 ei x fi =
∑n

i=1 xei fi = x for all

x ∈ S. In this case, the ternary semiring S is said to be a ternary semiring with identity

{(ei, fi) : i = 1,2, . . . , n}. In particular, if there exists an element e ∈ S such that

eex = exe = xee = x for all x ∈ S, then “e” is called a unital element of a ternary semiring S.

It is easy to see that x ye = (exe)ye = ex(e ye) = ex y and

x ye = x(e ye)e = xe(yee) = xe y, for all x , y ∈ S. Hence, the following result follows.



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 118

Proposition 1. If e is a unital element of a ternary semiring S, then ex y = xe y = x ye, for all

x , y ∈ S.

We now state the definitions of ternary subsemiring and left (right, lateral) ideals of a

ternary semiring.

Definition 4. [1] An additive subsemigroup T of a ternary semiring S is called a ternary sub-

semiring if t1 t2 t3 ∈ T for all t1, t2, t3 ∈ T.

Definition 5. [1] An additive subsemigroup I of a ternary semiring S is called a left (right,

lateral) ideal of S if s1s2i (respectively is1s2, s1 is2) ∈ I for all s1, s2 ∈ S and i ∈ I . If I is a left, a

right and a lateral ideal of S, then I is called an ideal of S.

In the following proposition, we describe the left(right, lateral) ideal of a ternary semiring.

Proposition 2. [1] Let S be a ternary semiring and a ∈ S. Then the following statements hold:

(i) left ideal generated by “a” is given by 〈a〉l = SSa+ na

(ii) right ideal generated by “a” is given by 〈a〉r = aSS + na

(iii) two-sided ideal generated by “a” is given by 〈a〉t = SSa+ aSS + SSaSS + na

(iv) lateral ideal generated by “a” is given by 〈a〉m = SaS + SSaSS+ na

(v) ideal generated by “a” is given by 〈a〉 = SSa+ aSS+ SaS + SSaSS + na,

where n ∈ Z+
0

(set of all positive integers with zero).

The following definitions are useful in the study of ternary semirings.

Definition 6. An ideal I of a ternary semiring S is said to be a k-ideal if x + y ∈ I; x ∈ S, y ∈ I

imply x ∈ I .

Definition 7. [4] A ternary semiring (ring) S is said to be zero divisor free (ZDF) if for a, b, c ∈ S,

abc = 0 implies a = 0 or b = 0 or c = 0.

Definition 8. [4] A ternary semiring S is said to be commutative if abc = bac = bca for all

a, b, c ∈ S.

Definition 9. [4] A commutative ternary semiring (ring) is called a ternary semi-integral (resp.

integral) domain if it is zero divisor free (ZDF).

Definition 10. [2] A proper ideal P of a ternary semiring S is called a prime ideal of S if for any

three ideals A, B, C of S; ABC ⊆ P implies A⊆ P or B ⊆ P or C ⊆ P.

Definition 11. [2] A ternary semiring S is called a prime ternary semiring if the zero ideal {0}
is a prime ideal of S.

Definition 12. [3] A proper ideal P of a ternary semiring S is called a semiprime of S if for any

ideal A of S; A3 ⊆ P implies A⊆ P.



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 119

Definition 13. [3] A ternary semiring S is called a semiprime ternary semiring if the zero ideal

(0) is a semiprime ideal of S.

Definition 14. [5] An additive commutative semigroup M with a zero element 0M is called a

right ternary semimodule over a ternary semiring S or simply a right ternary S-semimodule if

there exists a mapping M × S × S −→ M (images to be denoted by ms1s2 for all m ∈ M and

s1, s2 ∈ S) satisfying the following conditions:

(i) (m1 +m2)s1s2 = m1s1s2 +m2s1s2

(ii) m1s1(s2 + s3) = m1s1s2 +m1s1s3

(iii) m1(s1 + s2)s3 = m1s1s3 +m1s2s3

(iv) (m1s1s2)s3s4 = m1(s1s2s3)s4 = m1s1(s2s3s4)

(v) 0M s1s2 = 0M = m1s10S = m10Ss2 for all m1, m2 ∈ M and for all s1, s2, s3, s4 ∈ S.

A left ternary S-semimodule can be similarly defined.

Definition 15. [5] A non-empty subset N of a right ternary S-semimodule M is said to be a

ternary subsemimodule of M if (i) a+ b ∈ N and (ii) ast ∈ N for all a, b ∈ N and s, t ∈ S.

Definition 16. A nonzero right ideal I of a ternary semiring S with zero is called an essential

right ideal of S if for any nonzero right ideal J of S, I ∩ J 6= (0).

Definition 17. A class ρ of ternary semirings is called hereditary if I is an ideal of a ternary

semiring S and S ∈ ρ implies I ∈ ρ.

Definition 18. [1] Let S and T be two ternary semirings and f be a mapping which maps S

into T . Then the mapping f : S −→ T is called a homomorphism of S into T if the following

conditions hold:

(i) f (a+ b) = f (a)+ f (b) and (ii) f (abc) = f (a) f (b) f (c) for all a, b, c ∈ S.

Moreover, if both the ternary semirings S and T have zeros 0S and 0T , respectively, then the

following condition: (iii) f (0S) = 0T also holds.

We define ker f = {x ∈ S : f (x) = 0T }.

Throughout this paper, we use S to denote a ternary semiring with zero and S∗ = S \ {0}.
We now use M to denote a right ternary S-semimodule with zero.

3. Singular Ideals

In this section, S is a ternary semiring and M a right ternary S-semimodule. Let m ∈ M .

Then the right annihilator of m in S, denoted by rS(m), is defined by {x ∈ S : mxs = 0M for

all s ∈ S}.
The right annihilator rS(m) has the following property.



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 120

Proposition 3. Let S be a ternary semiring and M a right ternary S-semimodule. Then rS(m) is

a right k-ideal of the ternary semiring S.

Proof. Obviously, the set rS(m) is non-empty, since 0S ∈ rS(m). Let a, b ∈ rS(m). Then

a+b ∈ rS(m). Again let a ∈ rS(m) and s1, s2 ∈ S. Then mas = 0M for all s ∈ S⇒ mas1s2s = 0M

for all s ∈ S. This leads to as1s2 ∈ rS(m). Hence, rS(m) is a right ideal of S.

Let a, a+ b ∈ rS(m). Then mas = 0M = m(a+ b)s for all s ∈ S. This implies mbs = 0M for

all s ∈ S. This shows that b ∈ rS(m). Hence, rS(m) is a right k-ideal of S.

Let S be a ternary semiring and M a right ternary S-semimodule. We define ZS(M) as

follows:

ZS(M) = {m ∈ M : rS(m) is an essential right ideal of S}

Proposition 4. ZS(M) is a ternary subsemimodule of M.

Proof. Clearly, 0M ∈ ZS(M) as rS(0M ) = S is an essential right ideal of S. So ZS(M) is a

non empty subset of M . Let m1, m2 ∈ ZS(M). Then, it follows that rS(m1) and rS(m2) are

two essential right ideals of S. Therefore, rS(m1) ∩ rS(m2) is also an essential right ideal of

S. Let H be a nonzero right ideal of S. Now rS(m1) ∩ rS(m2) ∩ H ⊆ rS(m1 +m2) ∩ H since

rS(m1)∩ rS(m2)⊆ rS(m1+m2). Therefore, rS(m1+m2) is an essential right ideal of S. Hence,

we have m1 +m2 ∈ ZS(M). Let m ∈ ZS(M) and s1, s2 ∈ S. Then, rS(m) is an essential right

ideal of S. Let H be any nonzero right ideal of S. Now if s1s2H = 0, then ms1s2Hs = 0M for all

s ∈ S. This leads to H ⊆ rS(ms1s2). Hence, rS(ms1s2)∩H = H 6= 0. If s1s2H is a nonzero right

ideal of S, then rS(m) ∩ s1s2H 6= 0. Let s1s2h1( 6= 0) ∈ rS(m) ∩ s1s2H for some h1( 6= 0) ∈ H.

Then, ms1s2h1s = 0M , for all s ∈ S. Therefore, h1 ∈ rS(ms1s2)∩H, and so rS(ms1s2)∩ H 6= 0.

Thus, in any case ms1s2 ∈ ZS(M). This Shows that ZS(M) is a subsemimodule of M.

We now give the definition of a singular subsemimodule over a ternary semiring.

Definition 19. The ternary subsemimodule ZS(M) of M is called a singular ternary subsemi-

module of the right ternary S-semimodule M.

The singular ternary subsemimodule ZS(S) is a right ideal of the ternary semiring S and

is called the (right) singular ideal of the ternary semiring S which is denoted by Z(S), i.e.

Z(S) = {t ∈ S : rS(t) is an essential right ideal of S}.
A ternary semiring S is said to satisfy the condition α if for any nonzero element a in S,

rS(a) 6= S or equivalently aSS = 0 implies a = 0.

We now give some examples of ternary semirings satisfying the condition α.

Example 1. Let S be a ternary semi-integral domain. Then S satisfies the condition α.

Example 2. Let S be a prime (semiprime) ternary semiring. Then S satisfies the condition α.

Example 3. Let S be a ternary semiring with a unital element e. Then S satisfies the condition

α.

Example 4. Let S be a ternary semiring with identity. Then S satisfies the condition α.



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 121

In the following proposition, we study the ternary semirings with condition α.

Proposition 5. Let S be a ternary semiring with condition α. Then the (right) singular ideal

Z(S) is a k-ideal of S.

Proof. From Proposition 4, it follows that Z(S) is a right ideal of S. Now let a ∈ Z(S),

s1, s2 ∈ S. Then rS(a) is an essential right ideal of S. and so rS(a) ∩ H 6= 0 for any nonzero

right ideal H of S. Now rS(a) ⊆ rS(s1s2a). It is clear that (0) 6= rS(a) ∩ H ⊆ rS(s1s2a) ∩ H.

Therefore, rS(s1s2a)∩H 6= 0. Thus, s1s2a ∈ Z(S). This shows that Z(S) is a left ideal of S.

Next, let s1, s2 ∈ S and a ∈ Z(S). Now, if s2HS = (0), then H ⊆ rS(s2) ⊆ rS(s1as2).

Thus rS(s1as2) ∩ H = H 6= (0). If s2HS 6= (0), then s2HS is a nonzero right ideal of S. So

rS(a) ∩ s2HS 6= 0. Let s2h1s3( 6= 0) ∈ rS(a) ∩ s2HS for some h1( 6= 0) ∈ H and s3( 6= 0) ∈ S.

So as2h1s3s = 0 for all s ∈ S ⇒ s1as2h1s3st = 0 for all s, t ∈ S. As h1s3s ∈ H for all s ∈ S,

h1s3s ∈ rS(s1as2) ∩ H for all s ∈ S. Now, h1s3s = 0 for all s ∈ S ⇒ s2h1s3st = 0 for all

s, t ∈ S ⇒ S ⊆ rS(s2h1s3) i.e. rS(s2h1s3) = S. Now by condition (α) this implies that

s2h1s3 = 0, a contradiction. Hence h1s3s 6= 0 for some s ∈ S. Thus rS(s1as2) ∩ H 6= (0).
Therefore in any case, s1as2 ∈ Z(S). Consequently, Z(S) is a lateral ideal of S and whence,

Z(S) is an ideal of S.

Finally, let a+ b, a ∈ Z(S). Then rS(a) and rS(a+ b) are both essential right ideals of S.

Therefore, rS(a+ b)∩ rS(a)∩H 6= 0 for any nonzero right ideal H of S.

Suppose that p( 6= 0) ∈ rS(a+ b)∩ rS(a)∩ H. Then, (a+ b)ps = 0= aps, for all s ∈ S and

p ∈ H, which implies bps = 0 for all s ∈ S. This leads to p ∈ rS(b) ∩ H i.e. rS(b) ∩ H 6= 0.

Hence, b ∈ Z(S). Therefore, Z(S) is a k-ideal of S.

Remark 1. The condition “α” is assumed only to show that Z(S) is a lateral ideal of S. In order

to show that Z(S) is an ideal of a commutative ternary semiring, it is not necessary to assume the

condition “α”.

Remark 2. On assuming the condition “α”, we do not lose the generality because our aim is to

study the weakly special radical class and the special radical class of ternary semiring which are

the radical classes of nonsingular semiprime ternary semirings and nonsingular prime ternary

semirings respectively in which the condition α holds which is evident from the above Example 2.

Proposition 6. Let S be a ternary semiring. Then Z(S) = {x ∈ S : x IS = 0 for some essential

right ideal I of S}

Proof. Let Z ′(S) = {x ∈ S : x IS = 0 for some essential right ideal I of S}. Suppose x ∈ Z(S)

and J = rS(x). Then J is an essential right ideal of S. Also xJS = 0. Hence, x ∈ Z ′(S).

Conversely, let x ∈ Z ′(S). Then x IS = 0 for some essential right ideal I of S. Consequently,

I ⊆ rS(x). Let H be any nonzero right ideal of S. Then 0 6= I ∩ H ⊆ rS(x)∩ H which implies

x ∈ Z(S) and hence the result.

Proposition 7. Let S be a ternary semiring. Then Z(S) = {x ∈ S : x IS = 0 for some essential

right k-ideal I of S}



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 122

Proof. Let Z ′′(S) = {x ∈ S : x IS = 0 for some essential right k-ideal I of S}. From

Proposition 6, it follows that Z ′′(S)⊆ Z(S).

Conversely, let x ∈ Z(S) and J = rS(x). Then J is an essential right ideal of S. Now let

a, a + b ∈ J . Then xas = 0 = x(a+ b)s for all s ∈ S, this implies x bs = 0 for all s ∈ S, and

so b ∈ rS(x) = J . Thus, J is a right k-ideal of S. Also, xJS = 0. This leads to x ∈ Z ′′(S) and

hence Z(S)⊆ Z ′′(S). The proposition is hence proved.

Definition 20. A ternary semiring S is said to be singular if Z(S) = S and is said to be non-

singular if Z(S) = 0.

We now study the singular ternary semirings.

Proposition 8. Let S be a non-singular ternary semiring, then S satisfies the condition α.

Proof. Let aSS = 0. Then rS(a) = S is an essential right ideal of S. Hence a ∈ Z(S). Since

S is non-singular, Z(S) = 0 and so a = 0. Thus, S satisfies the condition α.

Definition 21. [5] An element s of a ternary semiring S is said to be nilpotent if for each t ∈ S,

there exists a positive integer n(depending on t) such that (st)ns = 0.

Definition 22. A ternary semiring S is said to be a nil ternary semiring if each a in S is nilpotent.

As examples of singular ternary semirings and non-singular ternary semirings we have the

following propositions.

Proposition 9. Every commutative nil ternary semiring S with a unital element e is singular.

Proof. Pick a( 6= 0) ∈ S. Let H be a nonzero right ideal of S and b( 6= 0) ∈ H. As S is nil, a is

clearly nilpotent. Then, for each t in S, there exists a positive integer n(depending on t) such

that (at)na = 0, or equivalently, we have a(ta)n = 0. Thus, in particular, a(ea)n = 0 for some

positive integer n. This implies a(ea)n bS = 0. Now let m be the least positive integer such

that a(ea)mbS = 0. Then (ea)mb ∈ rS(a). Also (ea)mb ∈ H. Now (ea)m b = a(ea)m−1 be 6= 0

by minimality of m, for a(ea)m−1 be = 0⇒ a(ea)m−1 bese = 0 for all s ∈ S, that is,

a(ea)m−1 bS = 0, a contradiction. Hence, rS(a) ∩ H 6= 0 and so a ∈ Z(S). This proves that

Z(S) = S.

We give below the definition of a right strongly prime ternary semiring.

Definition 23. [12] A ternary semiring S is called right strongly prime if for every nonzero

element a ∈ S, there exist some finite subsets F1, F2, F3 of S such that aF1F2F3 y = {0} ⇒ y = 0

for all y ∈ S.

We study below the properties of a right strongly prime ternary semiring.

Lemma 1. Let S be a right strongly prime ternary semiring with identity. Then every nonzero

ideal of S obtains a finite subset G such that rS(G) = 0, where rS(G) = {t ∈ S : Gts = 0 for all

s ∈ S}.



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 123

Proof. Since S admits identity, there exist elements {(ei, fi) ∈ S × S (i = 1,2, . . . , n)} such

that
∑n

i=1 ei fi x =
∑n

i=1 ei x fi =
∑n

i=1 xei fi = x for all x ∈ S. Suppose I is any nonzero

ideal of S. Let a( 6= 0) ∈ I . Since S is a right strongly prime ternary semiring, there exist

finite subsets F1, F2, F3 of S such that aF1F2F3 y = {0} ⇒ y = 0 for all y ∈ S. Now, set

E = {ei : i = 1,2, . . . , n} and F = { fi : i = 1,2, . . . , n}. Let G = aF1F2F3E. Then G is a finite

subset of I . It is clear that y ∈ rS(G)⇒ aF1F2F3E ys = 0 for all

s ∈ S ⇒ aF1F2F3E yF = 0 ⇒ aF1F2F3

∑n
i=1 ei y fi = 0 ⇒ aF1F2F3 y = 0 ⇒ y = 0. Thus,

rS(G) = 0.

Proposition 10. Every right strongly prime ternary semiring S with identity is non-singular.

Proof. If possible, let Z(S) 6= 0. Then Z(S) is a nonzero ideal of S. Since S is a right

strongly prime ternary semiring, by Lemma 1, there exists a finite subset F = {t1, t2, . . . , tk}
of Z(S) such that rS(F) = 0 i.e. rS(t1)∩ rS(t2)∩ . . . ,∩rS(tk) = 0. Now t i ∈ Z(S) implies that

rS(t i) is an essential right ideal of S for i = 1,2, . . . , k. Therefore rS(F) is an essential right

ideal of S. Consequently, rS(F) 6= 0, a contradiction. The proposition is hence proved.

We now give the definition of a strongly nilpotent element in a ternary semiring.

Definition 24. An element a in a ternary semiring S is said to be strongly nilpotent if there exists

a positive integer n such that a2n+1 = 0.

Definition 25. A ternary semiring S is said to be a reduced ternary semiring if it does not contain

any nonzero strongly nilpotent elements.

In the following proposition, we describe the reduced ternary semirings.

Proposition 11. Every reduced ternary semiring S with identity is non-singular.

Proof. Take any a ∈ S∗. If x ∈ rS(a) ∩ aSS, then for some y1, y2 ∈ S, x = a y1 y2 and

a2 y1 y2s = axs = 0 for all s ∈ S. This implies that (a y1 y2Sa)3 = 0. Hence, since S is reduced,

a y1 y2Sa = 0.

Consequently, xSx = a y1 y2Sa y1 y2 = 0. Thus, xSxSx = 0, xSSxSxS = 0 and

SxSxSSx = 0. Also, (xSSxSSa)3 = 0. Since S is reduced, xSSxSSa = 0. Thus,

xSSxSSx = xSSxSSa y1 y2 = 0. However, the ternary semiring S, being reduced, is semiprime,

and so x = 0.

Consequently, rS(a)∩ aSS = 0. Now,because S admits an identity, aSS is a nonzero right

ideal of S. This means that rS(a) is not an essential right ideal of S, for any a ∈ S∗. This

implies that Z(S) = 0.

Lemma 2. Let {Sα : α ∈ Λ, where Λ is a index set} be a family of ternary semirings. Then,

r∏
α∈Λ Sα

((aα)α∈Λ) =
∏
α∈Λ rSα

(aα).

Proof. Let (xα)α∈Λ ∈ r∏
α∈Λ Sα

((aα)α∈Λ) ⇔ (aα)α∈Λ(xα)α∈Λ(sα)α∈Λ = (0α)α∈Λ for all

(sα)α∈Λ ∈
∏
α∈Λ Sα ⇔ (aαxαsα)α∈Λ = (0α)α∈Λ for all sα ∈ Sα, α ∈ Λ⇔ aαxαsα = 0α for

all sα ∈ Sα,α ∈ Λ⇔ xα ∈ rSα
(aα) for each α ∈ Λ. Hence, the result.



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 124

Lemma 3. Let {Sα : α ∈ Λ where Λ is an index set} be a family of ternary semirings with iden-

tity. Let P be a right ideal of
∏
α∈Λ Sα. Let πα :

∏
α∈Λ Sα → Sα be the projection map. Suppose

Pα = πα(P) for each α ∈ Λ. Then P =
∏
α∈Λ Pα.

Proof. Obviously, P ⊆
∏
α∈Λ Pα. Now let (aα)α∈Λ ∈

∏
α∈Λ Pα. Then, we have

aα ∈ Pα. As Pα is the αth projection map, there exists a (bγ)γ∈Λ−{α} ∈
∏
α∈Λ−{α} Sγ such that

(aα, bγ)γ∈Λ−{α} ∈ P. Let {(eiα
, fiα
) ∈ Sα× Sα (iα = 1,2, . . . , nα)} be the identity of Sα. Now, we

consider the sum of products

nα∑

i=1

(aα, bγ)γ∈Λ−{α}(eiα
, 0γ)γ∈Λ−{α}( fiα

, 0γ)γ∈Λ−{α} =

nα∑

i=1

(aαeiα
fiα

, 0γ)γ∈Λ−{α} = (aα, 0γ)γ∈Λ−{α} ∈ P

as P is a right ideal of
∏
α∈Λ Sα.

Thus, (aα, 0γ)γ∈Λ−{α} ∈ P for each α ∈ Λ. Hence,
∑
α∈λ(aα, 0γ)γ∈Λ−{α} = (aα)α∈Λ ∈ P as P

is a right ideal of
∏
α∈Λ Sα. Hence P =

∏
α∈Λ Sα.

Proposition 12. For every family {Sα : α ∈ Λ where Λ is a index set} of ternary semirings with

identity Z(
∏
α∈Λ Sα) =
∏
α∈Λ Z(Sα) and Z(

⊕
α∈Λ Sα) =
⊕
α∈Λ Z(Sα).

Proof. Suppose that S =
∏
α∈Λ Sα. Let (xα)α∈Λ ∈ Z(S). Then, rS((xα)α∈Λ is an essential

right ideal of S. Let Pα be a nonzero right ideal of Sα, α ∈ Λ. Then
∏
α∈Λ Pα is a nonzero right

ideal of S. This shows that

rS((xα)α∈Λ)∩
∏

α∈Λ

Pα 6= (0α)α∈Λ⇒
∏

α∈Λ

rSα
(xα)∩
∏

α∈Λ

Pα 6= (0α)α∈Λ,

by Lemma 2. We claim that rSα
(xα) ∩ Pα 6= 0α, for each α ∈ Λ. If the assertion is not true.

Then there exists I ⊆ Λ such that rSα
(xα)∩ Pα = 0α for each α ∈ I and rSα

(xα)∩ Pα 6= 0α, for

each α ∈ Λ− I . Now
∏
α∈Λ−I(0α)×
∏
α∈I Pα is a nonzero right ideal of S but

rS((xα)α∈Λ)∩ (
∏

α∈Λ−I

(0α)×
∏

α∈I

Pα) = [
∏

α∈Λ−I

rSα
(xα)∩ (0α)]× [

∏

α∈I

rSα
(xα)∩ Pα] = (0α)α∈Λ,

a contradiction since (xα)α∈Λ ∈ Z(S). Thus, xα ∈ Z(Sα) for each

α ∈ Λ⇒ (xα)α∈Λ ∈
∏
α∈Λ Z(Sα).

Conversely, let (xα)α∈Λ ∈
∏
α∈Λ Z(Sα)⇒ xα ∈ Z(Sα) for each α ∈ Λ. Let P be a nonzero

right ideal of S.

Now, let πα : S→ Sα be the projection map. Then, Pα = πα(P) is a nonzero right ideal of

Sα for at least one α. This shows that P =
∏
α∈Λ Pα by Lemma 3. Now , rSα

(xα)∩ Pα 6= 0α for

at least one

α⇒
∏

α∈Λ

rSα
(xα)∩
∏

α∈Λ

Pα 6= (0α)α∈Λ⇒ rS((xα)α∈Λ)∩ P 6= (0α)α∈Λ⇒ (xα)α∈Λ ∈ Z(S).



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 125

Thus, Z(S) =
∏
α∈Λ Z(Sα).

Similarly, we can prove that Z(
⊕
α∈Λ Sα) =
⊕
α∈Λ Z(Sα).

The following theorem is a main theorem of this paper.

Theorem 1. The class of singular ternary semirings with identity as well as the class of non-

singular ternary semirings with identity is closed under product and direct sum.

Definition 26. A surjective homomorphism of ternary semirings γ : R → S is called a ternary

semi-isomorphism if kerγ = 0.

Lemma 4. If γ : S→ S′ is a ternary semi-isomorphism and Z(S) = S then Z(S′) = S′.

Proof. If possible, let Z(S′) ⊂ S′. Then there exists s′ ∈ S′ but s′ 6∈ Z(S′). Let s ∈ S

such that γ(s) = s′. Since s′ 6∈ Z(S′), there exists a nonzero right ideal H ′ of S′ such that

rS′(s
′) ∩ H ′ = 0. Let H = {b ∈ S : γ(b) ∈ H ′}. Then H is a nonzero right ideal of S. So

rS(s)∩H 6= 0. Therefore there exists a nonzero element h in H such that sht = 0 for all t ∈ S.

Consequently γ(sht) = γ(0) for all t ∈ S ⇒ γ(s)γ(h)γ(t) = 0 for all t ∈ S ⇒ s′γ(h)t′ = 0 for

all t′ ∈ S′ as γ is surjective. But γ(h) ∈ H ′. Therefore γ(h) ∈ rS′(s
′) ∩ H ′ which implies that

γ(h) = 0. Thus h ∈ kerγ = 0. So h= 0, a contradiction. Hence Z(S′) = S′.

Lemma 5. If γ : S→ S′ is a semi-isomorphism and Z(S′) = S′ then Z(S) = S.

Proof. If possible, let Z(S) ⊂ S. So there exists s( 6= 0) ∈ S but s 6∈ Z(S). Therefore there

exists a nonzero right ideal H of S such that rS(s)∩H = 0. Now γ(H) is a nonzero right ideal of

S′. Now γ(s) 6= 0 because γ(s) = 0⇒ s ∈ kerγ = 0⇒ s = 0. Let γ(s) = s′. We now prove that,

rS′(s
′)∩γ(H) = 0. If possible, let rS′(s

′)∩γ(H) 6= 0 and h′( 6= 0) ∈ rS′(s
′)∩γ(H). Let γ(h) = h′,

then h 6= 0. Now s′h′ t′ = 0 for all t′ ∈ S′ and h′ ∈ γ(H). So γ(sht) = γ(s)γ(h)γ(t) = s′h′ t′ = 0

for all t ∈ S. Therefore sht ∈ kerγ = 0 for all t ∈ S. So sht = 0 for all t ∈ S. Thus h ∈ rS(s)∩H,

a contradiction. Therefore rS′(s
′)∩ γ(H) = 0. So s′ 6∈ Z(S′), a contradiction. Hence Z(S) = S.

Lemma 6. If γ : S→ S′ is a semi-isomorphism and Z(S′) = 0 then Z(S) = 0.

Proof. If possible, let Z(S) 6= 0. So there exists s( 6= 0) ∈ S such that s ∈ Z(S). Now γ(s) 6= 0

because γ(s) = 0⇒ s ∈ kerγ = 0⇒ s = 0. Since Z(S′) = 0 , γ(s) 6∈ Z(S′). Hence there exists

a nonzero right ideal H ′ of S′ such that rS′(γ(s)) ∩ H ′ = 0. Let H = {s ∈ S : γ(s) ∈ H ′}.
Then H is a nonzero right ideal of S. Therefore rS(s) ∩ H 6= 0. Therefore there exists a

nonzero element h in H such that sht = 0 for all t ∈ S. Consequently γ(sht) = γ(0) for all

t ∈ S ⇒ γ(s)γ(h)γ(t) = 0 for all t ∈ S ⇒ γ(s)γ(h)t′ = 0 for all t′ ∈ S′ as γ is surjective

⇒ γ(h) ∈ rS′(γ(s)) ∩ H ′ = (0), which implies that γ(h) = 0. Thus h ∈ kerγ = 0. So h = 0, a

contradiction. Hence Z(S) = 0.

Lemma 7. If γ : S→ S′ is a ternary semi-isomorphism and Z(S) = 0 then Z(S′) = 0.



T. Dutta, K. Shum, S. Mandal / Eur. J. Pure Appl. Math, 5 (2012), 116-128 126

Proof. If possible, let Z(S′) 6= 0 and s′( 6= 0) ∈ Z(S′). Since γ is surjective, there exists

a nonzero element s ∈ S such that γ(s) = s′. Since Z(S) = 0, so s 6∈ Z(S). Therefore there

exists a nonzero right ideal H of S such that rS(s) ∩ H = 0. Now γ(H) is a nonzero right

ideal of S′. We now prove that, rS′(s
′) ∩ γ(H) = 0. If possible, let rS′(s

′) ∩ γ(H) 6= 0 and

h′( 6= 0) ∈ rS′(s
′)∩γ(H). Let γ(h) = h′, then h 6= 0. Now s′h′ t′ = 0 for all t′ ∈ S′ and h′ ∈ γ(H).

So γ(sht) = γ(s)γ(h)γ(t) = s′h′ t′ = 0 for all t ∈ S. Therefore sht ∈ kerγ = 0 for all t ∈ S. So

sht = 0 for all t ∈ S. Thus h ∈ rS(s) ∩ H, a contradiction. Therefore rS′(s
′) ∩ γ(H) = 0. So

s′ 6∈ Z(S′), a contradiction. Hence Z(S′) = 0.

Theorem 2. If S and S′ be two semi-isomorphic ternary semirings. Then S is singular (nonsin-

gular) iff S′ is singular (resp. nonsingular).

Proposition 13. If I is an ideal of a ternary semiring S and as a ternary semiring, I is semiprime,

then Z(I) = I ∩ Z(S).

Proof. Let x ∈ Z(I) and P be a nonzero right ideal of S. If PSI = 0, then

(I PS)3 = (I PS)(I PS)(I PS) = I(PSI)(PSI)PS = 0. Now, I is a semiprime ternary semiring

and I PS is an ideal of I . and so, I PS = 0. Consequently P ⊆ rS(I) ⊆ rS(x), since x ∈ Z(I)⊆ I .

So P ∩ rS(x) = P 6= 0. If PSI 6= 0, then PSI is a nonzero right ideal of I . So rI(x)∩ PSI 6= 0.

Let 0 6= a ∈ rI(x) ∩ PSI . Hence, xai = 0 for all i ∈ I ⇒ xaii1s = 0 for all s ∈ S and for

all i, i1 ∈ I . Now if aii1 = 0 for all i, i1 ∈ I , then aI I = 0, aIa = 0 and a3 = 0 as a ∈ I .

Now 〈a〉3 = (aI I + I Ia + IaI + I IaI I + na)3 = (0)⇒ a = 0 as I is semiprime. This arrives a

contradiction as a 6= 0. Thus, aii1 6= 0 for some i, i1 ∈ I ⇒ 0 6= aii1 ∈ rS(x) for some i, i1 ∈ I .

Also aii1 ∈ PSI ⊆ P for all i, i1 ∈ I . Hence, 0 6= aii1 ∈ rS(x)∩ P. This leads to rS(x)∩ P 6= (0).
Thus, in any case, rS(x) is an essential right ideal of S. Hence, x ∈ Z(S). Also, x ∈ I . This

shows that x ∈ I ∩ Z(S). Thus, Z(I) ⊆ I ∩ Z(S).

Conversely, let x ∈ I ∩Z(S) and P be a nonzero right ideal of I . Since the ternary semiring

I is semiprime, PI I 6= 0 as PI I = (0) ⇒ P3 ⊆ PI I = (0)⇒ P = (0), a contradiction. Thus,

PI I is a nonzero right ideal of S. Hence, rS(x)∩ PI I 6= (0). However, PI I ⊆ P and whence,

rS(x)∩ P 6= (0). Consequently, we have 0 6= rS(x)∩ P ⊆ rI (x)∩ P. Thus, x ∈ Z(I). Hence,

I ∩ Z(S)⊆ Z(I). Thus, Z(I) = I ∩ Z(S).

Theorem 3. The class of semiprime non-singular(singular) ternary semirings is hereditary.

Proof. The Proof follows from Proposition 13.

Corollary 1. The class of prime non-singular (singular) ternary semirings is hereditary.

Proof. The proof of this corollary is an immediate consequence of Theorem 3.

Finally, we state the following propositions:

Proposition 14. [7] Let S be a ternary semiring. If Q is a semiprime ideal of S and I is an ideal

of S then Q ∩ I is a semiprime ideal of I.

In closing this paper, we give the following theorem of semiprime non-singular ternary

semirings.



REFERENCES 127

Theorem 4. The class of homomorphic images of semiprime non-singular ternary semirings is

hereditary.

Proof. Let A = {φ(S) : S be a semiprime nonsingular ternary semiring }. Let φ(S) ∈ A
and J a nonzero ideal of φ(S). Then, there exists a nonzero ideal I of S such that

φ(I) = J . Now by Proposition 14, I is semiprime as S is semiprime. Also by Proposition 13, I

is nonsingular. Hence, φ(I) ∈ A , that is, J ∈ A . Thus, the class of homomorphic images of

semiprime non-singular ternary semirings is hereditary.

Remark 3. We make the following observations.

(i) The ternary subsemiring of a semiprime non-singular ternary semiring is not necessarily

semiprime non-singular. If the ring is commutative then the subring of a semiprime ring is

always semiprime. Similar result holds for ternary semirings.

(ii) It has been proved in Proposition 13 that if I is an ideal of a ternary semiring S and I a

semiprime ideal as a ternary semiring, then Z(I) = I ∩ Z(S) .Thus, by the above result,

we can easily deduce that if S is a non-singular ternary semiprime semiring, then I is non-

singular. Similar situation has been investigated by T.K. Dutta and M. L. Das in [9] in the

case of a semiring. In the proof of this result, the properties of ideals have been used. No

result has been proved so far for subsemiring or subring and hence, in the case of ternary

semirings, the above result may not be true.

The reader is invited to provide a counter example.

(iii) It is noted that the direct product or direct sum of semiprime non-singular ternary semirings

with identity is semiprime and non-singular.

In the theory of finite groups and semigroups, the notion of the formation and the heredi-

tary classes of groups and semigroups have been studied by L. A. Shemetkov in [16]. In view

of the above observations and the terminology of formation given by L.A. Shemetkov and A.

N. Skiba [17], we propose the following conjecture. The hereditary semiprime non-singular

ternary semirings form a hereditary formation of ternary semirings.

ACKNOWLEDGEMENTS The third author is thankful to CSIR, India for financial assistance.

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