THE EXTENDED CENTROID OF THE PRIME GAMMA SEMIRINGS

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1 ANALELE ŞTIINŢIFICE ALE UNIVERSITĂŢII AL.I. CUZA DIN IAŞI (S.N.) MATEMATICĂ, Tomul LXII, 2016, f.1 THE EXTENDED CENTROID OF THE PRIME GAMMA SEMIRINGS BY MEHMET ALI ÖZTÜRK Abstract. We define and study the extended centroid of a prime Γ-semiring. We show that the extended centroid is a Γ-semifield and give some properties of the centroid of a prime Γ-semiring. Mathematics Subject Classification 2010: 12K10, 16D30, 16S99, 16Y60, 16Y99. Key words: semiring, quotient semiring, gamma-semiring, quotient gamma-semiring. 1. Introduction Semirings abound in the mathematical world around us. Indeed, the first mathematical structure we encounter-the set of natural numbers-is a semiring. Historically, semirings first appear implicity in [4] and later [10], [11], [12] and [15] in connection with the study of ideals of a ring. They also appear in [7] and [8] in connection with the axiomatization of the natural numbers and nonnegative rational numbers. Over the years, semirings have been studied by various researchers either in their own right, in an attempt to broaden techniques coming from semigroup theory or ring theory, or in connection with applications (see [3]). Nobusawa studied on Γ-ring for the first time in [14]. After this research, Barnes studied on this Γ-ring in [1]. But Barnes approached to Γ-ring in some different way from that of Nobusawa and he defined the concept of Γ-ring and related definitions. After these two papers were published, many mathematicians made good works on Γ-ring in the sense of Barnes and Nobusawa, which are parallel to the results in the ring theory.

2 22 MEHMET ALI ÖZTÜRK 2 In [13] Martindale first constructed for any prime ring R a ring of quotients Q. After, Öztürk and Jun introduced the extended centroid of a prime Γ-ring in the sense of Barnes and obtained some results in Γ-ring M with derivation which was related to Q, and the quotient Γ-ring of M (see [16], [17], [18]) and also, Öztürk and Yazarlı [19] introduced modules over the generalized centroid of semi-prime Γ-rings. In [2] and [9], it was defined the rings of quotients of a prime and semi-prime Γ-ring in the sense of Nobusawa and researched the some properties of it, respectively. Γ-semirings were first studied by Rao[20] as a generalization of Γ-ring as well as of semiring. All definitions and fundamental concepts concerning Γ- semirings can be found in [5], [6], [20], [21], [22] and [23]. Recently, Yazalı and Öztürk [24] considered the extended centroid of a prime semiring. The purpose of this paper is to obtain the extended centroid of a prime Γ- semiring in the sense of Barnes. We show that the extended centroid is a Γ- semifield and give some properties of the centroid of a right multiplicatively cancellable prime Γ-semiring. 2. Preliminaries 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 and α Γ) satisfying the following conditions for all a,b,c S and for all α,β Γ: i) aα(b+c) = aαb+aαc; ii) (a+b)αc = aαc+bαc; iii) a(α+β)b = aαb+aβb; iv) aα(bβc) = (aαb)βc. A Γ-semiring S is said to have a zero element if there exists an element 0 S S such that 0 S +x = x = x+0 S and 0 S γx = 0 S = xγ0 S for all x S and γ Γ. Also, a Γ-semiring S is said to be commutative if xγy = yγx, for all x,y S and γ Γ. Let S be a Γ-semiring with zero. If there exists an element 0 Γ Γsuch that a0 Γ b = b0 Γ a = 0 S for all a,b S and 0 Γ +β = β for all β Γ, then 0 Γ is called the zero of Γ. In order that the context is clear we simply write 0 instead of 0 Γ and 0 S. Let S be a Γ-semiring. An element a S is called a left identity (resp. right identity) of S if x = aγx (resp. x = xγa) for all x S and γ Γ. If

3 3 THE EXTENDED CENTROID OF THE PRIME GAMMA SEMIRINGS 23 a is both a left and right identity, then a is called an identity of S. In this case we say that S is a Γ-semiring with identity. If A and B are subsets of a Γ-semiring S and Γ, we denote by A B, the subset of S consisting of all sums of the form a i α i b i where a i A, b i B and α i. For the singleton subset {x} of S we write x A instead of {x} A. A nonempty subset I of a Γ-semiring S is called a subγ-semiring of S if I is a subsemigroup of (S,+) and aγb I for all a,b I and γ Γ. A right(left)ideal I of a Γ-semiring S is an additive subsemigroup of S such that I ΓS I (SΓI I). If I is both a right and a left ideal of S, then we say that I is a two-sided ideal or simply an ideal of S. Let S be a Γ- semiring. For a S, the principal left ideal (right ideal, ideal) generated by a is denoted by < a (resp. by a >, < a >). Also < a = {na+ x i α i a n Z + {0},x i S,α i Γ}, a >= {na+ aβ j y j n Z + {0},y j S,β j Γ }, < a >= {na+ aγ k x k + y j β j a+ u i α i aδ i v i n Z + {0}, x k,y j,u i,v i S and γ k,β j,α i,δ i Γ}, where Z + is the set of all positive integers. Throughout this paper we consider Γ-semiring with zero. A proper ideal P of S is said to be prime if for any two ideals A and B of S, AΓB P implies that either A P or B P. A Γ-semiring S is called a prime Γ-semiring if < 0 > is a prime ideal of S. Theorem 1 ([5], Theorem 3.6). If P is an ideal of a Γ-semiring S then the following conditions are equivalent: i) If A and B are ideals of S such that AΓB P then either A P or B P. ii) If aγsγb P then either a P or b P where a,b S. iii) For a,b S if < a > Γ < b > P then either a P or b P. iv) If I 1 and I 2 are two right ideals of S such that I 1 ΓI 2 P then either I 1 P or I 2 P.

4 24 MEHMET ALI ÖZTÜRK 4 v) If J 1 and J 2 are two right ideals of S such that J 1 ΓJ 2 P then either J 1 P or J 2 P. Γ-semiring S is said to be zero divisor free (ZDF) if aαb = 0 implies that either a = 0 or α = 0, b = 0 for all a,b S and for all α Γ. A commutative Γ-semiring S is said to be Γ-semifield if for any a( 0) S and for any α Γ there exists b S, β Γ such that aαbβd = d for all d S. Let S be a Γ-semiring. A commutative monoid (M, +) with additive identity 0 M is said to be a right Γ-semiring S-semimodule or simply a ΓSsemimodule, if there exists a mapping M Γ S M (images to be denoted by aαs for a M,α Γ, s S) satisfying the following conditions for all a,b M, for all s,t S and for all α,β Γ: i) (a+b)αs = aαs+bαs, ii) aα(s+t) = aαs+aαt, iii) a(α+β)s = aαs+aβs, iv) aα(sβt) = (aαs)βt, v) 0 M αs = 0 M = aα0 S. One defines a left Γ-semiring S-semimodule in an analogous fashion. Let R and S both be Γ-semirings, and f a map of R into S. Then f is a Γ-homomorphism if and only if f (r 1 +r 2 ) = f (r 1 )+f (r 2 ) and f (r 1 γr 2 ) = f (r 1 )γf (r 2 ) for all r 1,r 2 R and for all γ Γ. A Γ-homomorphism of semirings which is both injective and surjective is called isomorphism. If there exists isomorphism between Γ-semirings R and S we write R = S. If f : R S is Γ-homomorphism of semirings, then the kernel of f, Kerf = {r f(r) = 0 S } is immediately seen to be an ideal of R and also Im(f) = {f (r) r R} is Γ-subsemiring of S. Let S be Γ-semiring, M and N be ΓS-semimodule. Then a function f from M to N is a right ΓS-semimodule homomorphism if and only if the following conditions are satisfied: i) f (m 1 +m 2 ) = f(m 1 )+f(m 2 ) for all m 1,m 2 M, ii) f(mαs) = f (m)αs for all m M, for all s S and for all α Γ.

5 5 THE EXTENDED CENTROID OF THE PRIME GAMMA SEMIRINGS Extended centroid Let S be a prime Γ-semiring such that SΓS S. We regard ideals U of S as right (left) ΓS-semimodules. Denote M: = {f : U S < 0 S > U is ideal of S, f is right ΓS-semimodule homomorphism}. Define a relation on M by f g M K( < 0 >) U V such that f = g on ideal K of S where Uand V are domains of f and g respectively. Since S is a prime Γ-semiring, it is possible to find a non-zero K and so is an equivalence relation. This gives a chance for us to get a partition of M. Then we denote the equivalence class by f = [U,f], where f := {g : V S f g} and denote by Q r set of all equivalence classes. That is, Q r = { f f : U S is right ΓS-semimodule homomorphism and < 0 S > U is ideal of S}. Now we define an addition + on Q r as follows f +ĝ = f +g, for all f,ĝ Q r.let f,ĝ Q r where U and V are domains of f and g respectively. Therefore f + g : U V S is a right S homomorphism. Assume that f 1 f 2 and g 1 g 2 where U 1,U 2, V 1 and V 2 are domains of f 1,f 2,g 1 and g 2 respectively. Then K 1 ( < 0 >) U 1 U 2 such that f 1 = f 2 on K 1 and K 2 ( < 0 >) V 1 V 2 suchthatg 1 = g 2 onk 2.TakingK = K 1 K 2. Then K < 0 > andk = K 1 K 2 (U 1 U 2 ) (V 1 V 2 ) = (U 1 V 1 ) (U 2 V 2 ) For any x K, we have (f 1 +g 1 )(x) = f 1 (x) +g 1 (x) = f 2 (x) +g 2 (x) = (f 2 +g 2 )(x), andsof 1 +g 1 = f 2 +g 2 onk. Thereforef 1 +g 1 f 2 +g 2 where f 1 +g 1 : U 1 V 1 S and f 2 +g 2 : U 2 V 2 S are right ΓS-semimodule homomorphisms. That is, addition + is well-defined. Now we prove that Q r is a commutative monoid. Let f,ĝ,ĥ Q r where U, V and W are domains of f, g and h respectively. Since U (V W) = (U V) W, we get for all x U (V W), [f +(g+h)](x) = f (x)+(g +h)(x) = f (x)+[g(x)+h(x)] = [f (x)+g(x)]+h(x) = (f +g)(x)+h(x) = [(f +g)+h](x). Hence f +(g +h) = (f +g) +h on U (V W). That is, f +(ĝ + ĥ) = ( f +ĝ)+ĥ. Taking θ Q r whereθ : S S, x 0 S for all x S. Let f Q r, where U is domainof f. SinceU U S, weget forall x U,(f +θ)(x) = f (x)+ θ(x) = f (x)+0 S = f (x)and(θ+f)(x) = θ(x)+f (x) = 0 S +f (x) = f (x). Thus, f + θ = θ + f = f. Hence θ is the additive identity in Q r.

6 26 MEHMET ALI ÖZTÜRK 6 Finally, for any elements f,ĝ Q r where U and V are domains of f and g respectively, we have for all x U V = V U, (f +g)(x) = f (x)+g(x) = g(x)+f (x) = (g +f)(x). That is, f +ĝ = ĝ + f. Therefore (Q r,+) is commutative monoid. Since SΓS S and S is a prime Γ-semiring, SΓS( < 0 >) is an ideal of S. We can take the homomorphism 1 SΓ : SΓS S as a unit ΓSsemimodule homomorphism. Note that SβS < 0 > for all < 0 > β Γ so that 1 Sβ : SβS S is non-zero ΓS-semimodule homomorphism. Denote N: = {1 Sβ : SβS S 0 β Γ}, and define a relation on N by 1 Sβ 1 Sγ W := SαS( < 0 >) SβS SγS such that 1 Sβ = 1 Sγ on W where SβS and SγS are domains of 1 Sβ and 1 Sγ respectively. We can easily check that is an equivalence relation on N. Denote by β = [SβS,1 Sβ ], the equivalence class containing [SβS,1 Sβ ] and by Γ the set of all equivalence classes of N with respect to, that is β: = {1 Sγ : SγS S 1 Sβ 1 Sγ } and Γ := { β 0 β Γ}. Define an addition + on Γ as follows β + γ = β +γ, for all β( 0),γ( 0) Γ. Then it is routine to check that ( Γ,+) is commutative monoid. Now we define a (,, ) : Q r Γ Q r Q r,( f, β,ĝ ) f β ĝ, as follows f β ĝ = fβg, where U,V and SβS are domains of f, g and 1 Sβ respectively. Therefore f1 Sβ g : VΓSβSΓU S is a right ΓS-semimodule homomorphism where VΓSβSΓU = { v i γ i s i βr i α i u i v i V,u i U,s i,r i S and γ i,α i Γ }, an ideal of S. Assume that f 1 f 2, g 1 g 2 and 1 Sβ 1 Sβ where U 1,U 2, V 1,V 2,SβS and Sβ S are domains of f 1,f 2,g 1, g 2,1 Sβ and 1 Sβ respectively. Then K 1 U 1 U 2 such that f 1 = f 2 on K 1, K 2 V 1 V 2 such that g 1 = g 2 on K 2 and W SβS Sβ S such that 1 Sβ = 1 Sβ on W. Also V 1 ΓSβSΓU 1 V 2 ΓSβ SΓU 2 (U 1 SβS V 1 ) (U 2 Sβ S V 2 ) = (U 1 U 2 ) (V 1 V 2 ) (SβS Sβ S) and there exists < 0 S > K is an ideal of S such that K V 1 ΓSβSΓU 1 V 2 ΓSβ SΓU 2. For any x K, x V 1 ΓSβSΓU 1 V 2 ΓSβ SΓU 2 so that x V 1 ΓSβSΓU 1 and x V 2 ΓSβ SΓU 2. Then, x = v iγ i s i βr i α i u i ; v i V 1 V 2, u i U 1 U 2, s i,r i S and γ i,α i Γ.

7 7 THE EXTENDED CENTROID OF THE PRIME GAMMA SEMIRINGS 27 Therefore (f 1 1 Sβ g 1 )(x) = f 1 (1 Sβ (g 1 ( v i γ i s i βr i α i u i ))) = f 1 (g 1 ( v i γ i s i βr i α i u i )) = f 1 ( g 1 (v i )γ i s i βr i α i u i ) = f 1 ( g 2 (v i )γ i s i βr i α i u i ) = f 2 ( g 2 (v i )γ i s i βr i α i u i ) = f 2 (g 2 ( v i γ i s i βr i α i u i )) = f 2 (1 Sβ (g 2 ( v i γ i s i βr i α i u i ))) = (f 2 1 Sβ g 2 )(x) and so f 1 1 Sβ g 1 = f 2 1 Sβ g 2 on K. Hence, f 1 β ĝ1 = f 2 β ĝ 2. That is,. is well-defined. Now we will prove that Q r is a Γ-semiring with identity. Let f,ĝ,ĥ Q r whereu,v andw aredomainsoff,gandhrespectively and γ Γ where SγS is domains of 1 Sγ. Since (V W)ΓSγSΓU VΓSγSΓU WΓSγSΓU, we get for all x (V W)ΓSγSΓU, [f1 Sγ (g +h)](x) = f(1 Sγ (g +h)(x)) = f(1 Sγ (g(x)+h(x))) = f(1 Sγ (g(x))+1 Sγ (h(x))) = f(1 Sγ (g(x))) +f(1 Sγ (h(x))) = (f1 Sγ g)(x)+(f1 Sγ h)(x) = [f1 Sγ g +f1 Sγ h](x). Hence f1 Sγ (g+h) = f1 Sγ g+f1 Sγ h on (V W)ΓSγSΓU. That is, f γ(ĝ+ ĥ) = f γĝ + f γĥ. Similarly, the equalities ( f + ĝ) γ ĥ = f γ ĥ + ĝ γ ĥ and f( γ + β) ĝ = f γ ĝ + f β ĝ are proved in analogous way. Also, let f,ĝ,ĥ Q r where U, V and W are domains of f, g and h respectively and γ, β Γ where SγS,SβS are domains of 1 Sγ,1 Sβ respectively. Since WΓSβSΓ(VΓSγSΓU) = (WΓSβSΓV)ΓSγSΓU, we get for all x WΓSβSΓ(VΓSγSΓU), [(f 1 Sγ g)1 Sβ h](x) = ((f 1 Sγ g)1 Sβ )(h(x)) = f (1 Sγ g(1 Sβ h(x))) = f (1 Sγ (g1 Sβ h)(x)) = [f1 Sγ (g1 Sβ h)](x). Hence (f 1 Sγ g)1 Sβ h = f1 Sγ (g1 Sβ h) on WΓSβSΓ(VΓSγSΓU). That is, ( f γĝ) βĥ = f γ(ĝ βĥ). Next we will show that Q r has an identity. Taking Î Q r where I : S S, s s for all s S. Let f Q r, where U is domain of f and γ Γ where SγS is domains of 1 Sγ Since SΓU U, we get for all x SΓSγSΓU, (f1 Sγ I)(x) = f(1 Sγ ( I(x))) = f (x) and

8 28 MEHMET ALI ÖZTÜRK 8 (I1 Sγ f)(x) = I(1 Sγ (f (x))) = f (x). Thus, f γ 1 = 1 γ f = f. Hence Î is the multiplicative identity in Q r. Therefore (Q r,+,.) is a Γ-semiring with identity. Moreover we have that θ Î. Finally, noticing that the mapping φ : Γ Γ defined by φ(γ) = γ for every 0 γ Γ is an isomorphism, we know that the Γ-ring Q r is a Γ- semiring. Thus, (Q r,+,.) be a Γ-semiring. One can, of course, characterize Q l, the left quotient Γ-ring of S in a similar manner. Definition 1. A Γ-semiring S is said to be right (left) multiplicatively cancellative if xγy = zγy; (resp. xγy = xγz), for all x,y,z S and for all γ Γ implies that x = z (resp. y = z). Theorem 2. Let S is a prime Γ-semiring. If S is right multiplicatively cancellable semiring, then S may be embedded in Q r as a subγ-semiring. Proof. Let a S. Define λ aγ : S S by λ aγ (s) = aγs, for all s S and for all γ Γ. It is clear that λ aγ is a right ΓS-semimodule homomorphism, so that λ aγ defines element λ aγ of Q r. Hence, we may define ψ : S Q r by ψ(a) = λ aγ for a S. Clearly ψ is well-defined. Let for any a,b S, ψ(a) = ψ(b). Then, λ aγ = λ bγ, i.e., λ aγ = λ bγ on S S S. Thus, we get that λ aγ (s) = λ bγ (s), for all s S, i.e., aγs = bγs. Since R is right multiplicatively cancellable semiring, a = b. That is, ψ is injective mapping. In order to prove ψ is a monomorphism, let a, b S and γ,β Γ. Then λ (a+b)γ (s) = (a+b)γs = aγs + bγs = λ aγ (s)+λ bγ (s) = (λ aγ +λ bγ )(s) and λ (aβb)γ (s) = (aβb)γs = aβ(bγs) = λ aβ (bγs) = λ aβ (1 Sβ (λ(bγs))) = λ aβ (1 Sβ (λ bγ (s))) = (λ aβ 1 Sβ λ bγ )(s), for all s S. It follows that λ (a+b)γ = λ aγ + λ bγ and λ (aβb)γ = λ a 1 Sβ λ b. Hence, ψ(a+b) = λ (a+b)γ = λ aγ +λ bγ = λ aγ + λ bγ = ψ(a) + ψ(b) and ψ(aβb) = λ (aβb)γ = λ aβ 1 Sβ λ bγ = λ aβ β λbγ = ψ(a)βψ(b) [Γ = Γ]. Thus S is sub Γ-semiring of Q r. Therefore S is subγ-semiring of Q r. We call Q r the right quotient Γ- semiring of S. For purposes of convenience, we use q instead of ˆq Q r. Lemma 1. Let S be a prime Γ-semiring. For each nonzero q Q r, there is a nonzero ideal U of S such that q(u) S. Proof. Straightforward. Lemma 2. Let S be a prime Γ-semiring. Then the quotient Γ-semiring Q r of S is a prime Γ-semiring.

9 9 THE EXTENDED CENTROID OF THE PRIME GAMMA SEMIRINGS 29 Proof. Let p, q Q r be such that pγq r Γq = θ. If p θ q, then there exist nonzero ideals U and V of S such that p(u) S and q(v) S. Since p θ q, then there exist nonzero elements u U and v V such that p(u) 0 S q(v). Noticing that S is subγ-semiring of Q r, we have p(u)γsγq(v) p(u)γq r Γq(v) = 0 S and so p(u)γsγq(v) = 0 S. This is a contradiction. Hence p = θ or q = θ, ending the proof. Definition 2. The set C Γ := {q Q r qγp = pγq for all p Q r and for all γ Γ} is called the extended centroid of a Γ-semiring S. It is clear that C Γ is a subγ-semiring of Q r. Let θ c C Γ. Assume that cγq = θ for any q C Γ and for all γ Γ. Then, pβcγq = θ for all p Qr and for all γ,β Γ, and so cβpγq = θ for all p Qr and for all γ,β Γ, i.e., cγpγq = θ. Since θ c and S is a prime Γ-semiring, we get q = θ. Thus, C Γ is a zero divisor free (ZDF) Γ-semiring. Theorem 3. The center C Γ of Q r is a Γ-semifield. Proof. We prove that C Γ is a Γ-semifield in order to every θ c C Γ and for any α Γ there exists a S and α,β Γsuch that cαaβd = d for all d S. Let θ c C Γ. Since c Q r, there is a nonzero ideal U of S such that c(u) S. Since S is a prime Γ-semiring, cγ(u) 0 S. Taking, 0 S V = cγu ideal of S. Moreover, mapping a : V S, a(cβd) = d is a right ΓS-semimodule homomorphism.then, a = [V,a] Q r. Also, since a(cβd) = d = I(d) for all a U where I : S S,s s for all s S (I = [S,I]), and so aαc = I for all α Γ. Therefore, aαcβd = d for all d S. That is, C Γ is a Γ-semifield. We now let R = SΓC Γ, a subγ-semiring of Q containing S. We shall call R the central closure of S. The same proof used in showing that Q r was prime may be employed to show that R is prime Γ-semiring. Also, if S is right (left) multiplicatively cancellable prime Γ-semiring, then R is a right (left) multiplicatively cancellable prime Γ-semiring. Theorem 4. Let S be multiplicatively cancellable prime Γ-semiring. If a and b are nonzero elements in R such that aγxβb = bβxγa for all x S and for all γ,β Γ, then there exists q C Γ and γ Γ such that qγa = b. Proof. We may assume that a θ and b θ. Let U be a nonzero ideal of S such that aγu S and bγu S, and V = UΓaΓU. We define a mapping f : V S, f( i x iγ i aβ i y i ) = i x iγ i bβ i y i, x i, y i U

10 30 MEHMET ALI ÖZTÜRK 10 and γ i,β i Γ. Let i x iγ i aβ i y i = i x i γ i aβ i y i ; x i, y i, x i, y i U and γ i,β i,γ i,β i Γ. Then ( x i γ i aβ i y i )αb = ( x iγ iaβ iy i)αb, i i x i γ i (aβ i y i αb) = x iγ i(aβ iy iαb), i i x i γ i (bβ i y i αa) = x iγ i(bβ iy iαa), i i ( i x i γ i bβ i y i )αa = ( i x i γ i bβ i y i )αa. Since R is multiplicatively cancellable, we get i x iγ i bβ i y i = i x i γ i bβ i y i. Therefore, f is well-defined. f is a right ΓS-semimodule homomorphism. Because f ((xγaβy)αs) = f (xγaβ(yαs)) = xγbβ(yαs) = (xγby)αs = f (xγaβy)αs for all x,y U, s S and γ,β,α Γ. Let q denote the element of Q r determined by f and let p be any element of Q r, with p(v) S for some nonzero ideal V of S by Lemma 1. For x,y U, z V and γ,β,α,δ,σ Γ, we have (qγp)(zαxδaσy) = q((pβz)αxδaσy) = (pβz)αxδbσy = p(zαxδbσy) = pγq(zαxδaσy). Since VΓSγSΓU V U and pγq = qγp on VΓSγSΓU, we get, pγq = qγp for all p Q r, and for all γ Γ, and so q C Γ. For γ,β,α Γ,qγ(xαaβy) = q(xαaβy) = xαbβy and so xαbβy = qγ(xαaβy) = (xαaβy)γq = xα(aβy)γq = xαqγ(aβy) = (xαq)γaβy = xα(qγa)βy. Since R is multiplicatively cancellable, we obtain qγa = b. Acknowledgement. The author is deeply grateful to the referee for the valuable suggestions. REFERENCES 1. Barnes, W.E. On the Γ-rings of Nobusawa, Pacific J. Math., 18 (1966),

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12 32 MEHMET ALI ÖZTÜRK Sardar, S.K.; Dasgupta, U. On primitive Γ-semirings, Novi Sad J. Math., 34 (2004), Sardar, S.K. A note on Γ-semifield, Int. Math. Forum, 6 (2011), Yazarli, H.; Öztürk, M.A. On the centroid of prime semirings, Turkish J. Math., 37 (2013), Received: 11.VIII.2012 Revised: 24.IV.2013 Accepted: 13.V.2013 Adıyaman University, Faculty of Arts and Sciences, Department of Mathematics, Adıyaman, TURKEY maozturk@adiyaman.edu.tr