Int. J. Contemp. Math. Sciences, Vol. 6, 2011, no. 13, 619 - 635

Boolean Like Semirings

K. Venkateswarlu

Department of Mathematics, P.O. Box. 1176 Addis Ababa University, Addis Ababa, Ethiopia [email protected]

B. V. N. Murthy

Department of Mathematics Simhadri Educational Society group of institutions Sabbavaram Mandal, Visakhapatnam, AP, India [email protected]

N. Amarnath

Department of Mathematics Praveenya Institute of Marine Engineering and Maritime Studies Modavalasa, Vizianagaram, AP, India [email protected]

Abstract

In this paper we introduce the concept of Boolean like semiring and study some of its properties. Also we prove that the set of all nilpotent elements of a Boolean like semiring forms an ideal. Further we prove that Nil radical is equal to Jacobson radical in a Boolean like semi ring with right unity.

Mathematics Subject Classification: 16Y30,16Y60 Keywords: Boolean like semi rings, Boolean near rings, prime ideal , Jacobson radical, Nil radical

1 Introduction

Many ring theoretic generalizations of Boolean rings namely , regular rings of von- neumann[13], p- rings of N.H,Macoy and D.Montgomery [6 ] ,Assoicate rings of I.Sussman k [9], p -rings of Mc-coy and A.L.Foster [1 ] , p1, p2 – rings and Boolean semi rings of N.V.Subrahmanyam[7] have come into light. Among these, Boolean like rings of A.L.Foster [1] arise naturally from general ring dulity considerations and preserve many of the formal

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properties of Boolean ring. A Boolean like ring is a commutative ring with unity and is of characteristic 2 with ab (1+a) (1+b) = 0. It is clear that every Boolean ring is a Boolean like ring but not conversely (see example 1.9 ).Later Swaminathan [ 10,11,12 ] has demonstrated that every Boolean like ring is a direct product of Boolean ring and a module over a Boolean ring. Further he investigated the study on algebraic and geometry of Boolean like rings extensively.

Subrahmanyam N.V.[ 7 ] introduced the concept of Boolean semi rings and studied many geometric properties. In fact Boolean semi rings are nothing but a special class of near rings. In this paper our objective is to introduce the concept of Boolean like semi ring which is a generalization to Boolean like rings of A.L. Foster[ 2] and also a special class of near rings. Further we study its properties and also furnish examples that the two systems Boolean semirings of Subrahmanyam and Boolean like semi rings of our system are independent.

1. PRELIMINARIES

In this article we recall certain definitions and results concerning Boolean semi rings and Boolean like rings. Further we prove some motivating results to introduce the concept of Boolean like semi rings.

Definition [7] A system ( R,+, . ) a Boolean semi ring if and only if the following properties hold : 1. (R,+) is an additive (abelian) group ( whose “zero” will be denoted by “0”.) 2. (R, .) is a semi group of idempotents in the sense, aa=a ,for all a R 3. a(b+c) = ab + ac and 4. abc = bac , for all a,b,c R.

Example [7]. Let (G,+) be any abelian group and define ab = b , for all a,b G. Then (G,+,.) is a Boolean semi ring .

Here we give some examples which motivates us to introduce the concept of Boolean like semi rings

Example 1.1. Consider the near ring defined on the Klein’s four group with N= {0,a,b,c}. “ + ” and “. “ defined by

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+ 0 a b c . 0 a b c

0 0 a b c 0 0 0 0 0

a a 0 c b a 0 a b c b b c 0 a b 0 a b c

c c b a 0 c 0 a b c

Then (N,+,.) is a Boolean semi ring. Note that (a+b)c ≠ ac + bc, ab ≠ ba , Characteristic of N = 2 In the sense that N is not commutative and right distributive law fails Also ab(a+b+ab) = ab is not true for all a,b N , since ac(a+c+ac) = a ≠ ac

Example 1.2 Let M = {0,1,2} , “+” denotes addition modulo 3 and “.” Defined as follows

+ 0 1 2 . 0 1 2

0 0 1 2 0 0 1 2 1 1 2 0 1 0 1 2 2 2 0 1 2 0 1 2

Then (M,+, . ) is a Boolean semi ring. Note that Chr M ≠ 2 and ab(a+b+ab) ≠ ab as 0.1(0+1+0.1) = 2 ≠ 0.1

Example 1.3 .Let R = {0,x,y,z } , + and . are defined by

+ 0 x y z . 0 x y z

0 0 x y z 0 0 0 0 0

x x 0 z y x 0 x 0 x y y z 0 x y 0 0 0 0 z z y x 0 z 0 z 0 z

Then clearly (R,+,.) is a (Left) near ring in which Chr = 2 and

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ab(a+b+ab) =ab ,for all a,b R. Clearly it is not a Boolean semi ring since y2 ≠ y

Example 1.4. Let K = {0,1,2,3} , “+” denotes addition modulo 4 and “.” defined by

+ 0 1 2 3 . 0 1 2 3 0 0 1 2 3 0 0 0 0 0 1 1 2 3 0 1 0 1 2 3 2 2 3 0 1 2 0 1 2 3 3 3 0 1 2 3 0 0 0 0

Then (M,+, .) is a(left) near ring , but not Boolean semi ring and Chr M ≠ 2 and also ab(a+b+ab) = ab is not true for all a,b M , as 1.2 (1+2+1.2) =1 ≠ 1.2

Example 1.5. Let L = {0,1,2,3} , “+” denotes addition modulo 4 and “.” Defined by

+ 0 1 2 3 . 0 1 2 3 0 0 1 2 3 0 0 0 0 0 1 1 2 3 0 1 0 1 2 3 2 2 3 0 1 2 0 2 0 2 3 3 0 1 2 3 0 1 2 3

Clearly (L,+, .) is a (left) near ring in which ab(a+b+ab) = ab , for all a,b L and Chr L ≠ 2 and which is not a Boolean semi ring and also x2 = x is not true for all x L , as 2.2 =0

Definition [ 4 ]. A near ring R is called Boolean near ring if x2 = x ,for all x R

Example [ 4 ]. Let (Z, +) be a group and define ab = b, for all a,b Z. Then (Z,+,.) is a Boolean Near ring.

We have the following in the Theorem 1.6. If R is a Boolean Near ring with ab(a+b+ab) = ab for all a R then characteristic R is 2 Proof. Let a = b in ab(a+b+ab) =a. Then aa(a+a+aa) = aa .Since a2 = a, we have that a( a+a+a ) = a ⇒ aa + aa + aa = a ⇒ a+a+a = a ⇒ a+a = 0

We observe an important fact in the following

Remark 1.7 . If R is a near ring with characteristic 2 in which ab(a+b+ab)=ab holds for all a,b in R but still R may not be a Boolean near ring.

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For instance, look into the example 1.3 , it is clear that ab(a+b+ab)=ab holds for all a,b in R but y2 ≠ y

Definition [2 ]. A Boolean like ring R is a commutative ring with unit element in which for all elements a,b in R, a+a =0, ab(1+a)(1+b)=0.

Remark 1.8[2]. Clearly every Boolean ring with 1 is a Boolean like ring , but not conversely.

Example [10] Let R = {0,1,p,q} , “+”and “ .” are defined by

+ 0 1 p q . 0 1 P q 0 0 1 P q 0 0 0 0 0 1 1 0 q p 1 0 1 p q p p q 0 1 p 0 p 0 p q q p 0 1 q 0 q p 1

Then (R,+,.) is a Boolean Like ring but not Boolean ring as o = p2 ≠ p.

Motivated from the above examples with different properties we are in a position to introduce the concept of Boolean like semi rings.

2 Boolean like semi rings and its properties

We introduce the concept of Boolean like semi ring and give some examples of a Boolean like semi ring .we obtain certain properties of Boolean like semi rings .Also we introduce the concept of ideal in a Boolean like semi ring and obtain the various properties of ideals. Further we prove that every Jacobson radical is a nil radical in any Boolean like semi ring.

Now we begin with the following

Definition 2.1 . A non empty set R together with two binary operations + and . satisfying the following conditions is called a Boolean like semi ring 1.( R,+ ) is an abelian group 2. (R , . ) is a semi group 3. a.( b+c) = a.b +a.c for all a, b, c ∈ R 4. a + a = 0 for all a in R 5. ab( a+b+ab ) =ab for all a, b ∈ R.

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Remark 2.2 1. Condition 3 is called left distributive law and hence a system (R,+,.) satisfying Conditions 1 , 2 and 3 is called a (left) near ring 2. From condition 4 of the above definition , it is clear that the near ring R is of characteristic 2 3. Conditions 4 and 5 of the above definition are independent as it can be observed from example 1.1 , that condition 4 holds but not condition 5 since Chr R = 2 and ab(a+b+ab) ≠ ab and further from example 1.5 , it is clear that condition 5 holds but not condition 4 as ab(a+b+ab) = ab and Chr ≠2

Let R be a Boolean like semi ring . Then we have the following easy consequences of the definition. Lemma 2.3 For a R , a.0 =0 Lemma 2.4 For a R , a4 = a2

Proof . Substituting b = a in (5) of the definition 2.1, then aa(a + a +aa) = aa ⇒ aa(0+aa) = aa ⇒ aaaa = aa ⇒ a4 = a2 Remark 2.5. (a) a4 = a2 is called weak idempotent law (b) From the above lemma 2.3, we have a2n = a2 , for any integer n > 0

Lemma 2.6. If R is a Boolean like semi ring then, an = a or a2 or a3 for any integer n > 0 Proof , Clearly a2n+1 = a3 and a2n = a2 for all n follows from Lemma 2.4

We furnish few examples of Boolean like semi rings here under.

Example 2.7 Let R ={0,a,b,c } , + and . are defined as follows

+ 0 a b c . 0 a b c 0 0 a b c 0 0 0 0 0 a a 0 c b a 0 0 a a b b c 0 a b 0 0 b b c c b a 0 c 0 a b c

Then R is a Boolean like semi ring. Observe that cab ≠ cba .

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Remark 2.8 The above example is not a Boolean semi ring as a2 a and example 1.1 shows that R is a Boolean semi ring but not Boolean like semi ring as ab(a+b+ab) ≠ ab

Example 1.1 is a Boolean semi ring but not Boolean like ring Thus the two systems Boolean semi ring of subrahmanyam and Boolean like semirings of our system are independent.

Example 2.9. Let R ={ 0,x,y,z } , + and . are defined by

+ 0 x y z . 0 x y z 0 0 x y z 0 0 0 0 0 x x 0 z y x 0 x 0 x y y z 0 x y 0 0 0 0 z z y x 0 z 0 z 0 z

Then ( R,+,. ) is a Boolean like semi ring.

We note that abc = acb for all a,b,c R.

Example 2.10. Every Boolean like ring is a Boolean like semi ring but not conversely

For instance Example 2.9 , is a Boolean like semi ring but not Boolean like ring as y2 ≠ y

Definition 2.11. A Boolean like semi ring R is said to be weak commutative if abc=acb ,for all a,b,c R. Example 2.9 is weak commutative Boolean like semi ring

Lemma 2.12. If R is a Boolean like semiring with weak commutativity then 0.a = 0.for all a R Proof. Now 0.a = a.0.a = a.a.0 = a(a.0) =a0 = 0

Lemma 2.13 Let R be Boolean like semi ring then for any a,b R and for any integers m, n 1. aman = am+n 2. ( am) n = amn 3. (ab)n = an bn if R is weak commutative. Proof. Follows by mathematical induction and weak commutativity

Lemma 2.14. For any a, b R , ( a+b )a = (a+b)ab If a2=0 Proof. ( a+b )a = (a+b)a( a+b+a+(a+b)a) = (a+b)a( b+(a+b)a)

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= (a+b)ab + (a+b)a (a+b)a = (a+b)ab + (a+b)aa(a+b) = (a+b)ab + (a+b)a2(a+b) = (a+b)ab + (a+b)0(a+b) = (a+b)ab

3 Ideals of a Boolean like Semiring

We begin with the following

Definition 3.1. A non empty subset I of R is said to be an ideal if 1 (I,+) is a sub group of (R,+) ,i.e , for a,b R ⇒ a + b R 2 ra R for all a I , r R ,i.e RI ⊆ I 3 (r+a)s + rs I for all r, s R , a I

Remark 3.2. If R is weak commutative, By definition 3.1(3) 0,r R , aI ⇒ (0+a)r + 0r I ⇒ ar I . Thus If R is weak commutative r R, aI ⇒ ar I and ra I

We prove certain lemmas here under which we use in the sequel. . Lemma 3.3 For any a , b R , ( a + b )2 = ( a + b )2 ( a 2 + b2 ) Proof. ( a + b )2 = (a + b) ( a + b ) = (a+b)a + (a+b)b = (a+b)ab + (a+b)2a2 + (a+b)ba + (a+b)2 b2 = (a+b)2a2 + (a+b)2 b2 = ( a + b )2 ( a 2 + b2 )

Lemma 3.4. For any a R, a + a2 is nilpotent element. Proof. Let b = a in lemma 3.3 then we have ( a + a2 )2 = ( a + a2 )2( a 2 + a2 ) = ( a + a2 )2 0 = 0

Remark. 3.5 Every element a R can be expressed as a sum of nilpotent and idempotent element in the sense that a = (a+ a2 )+ a2 where a2 is a Boolean element and (a+ a2 ) is a nilpotent element. How ever this representation is not unique. For instance in example 2.8, z = 0 + z = x + y with two different representations

Lemma 3.6 For any a , b R , ( a + b )3 = ( a + b )2 (a3+ab2+ba2+b3) Proof . ( a + b )3 = ( a + b )2 ( a+b) = ( a + b )2 ( a 2 + b2 )( a+b ) ( Lemma 3.3) = ( a + b )2 [ (a 2 + b2 )a + ( a 2 + b2 )b ] =( a + b )2 ( a 2 + b2 )a + ( a + b )2 ( a 2 + b2 )b =( a + b )2 a(a 2 + b2 ) + ( a + b )2 b( a 2 + b2 ) (by Weak Commutativity)

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= ( a + b )2 a 3 +( a + b )2 ab2 + ( a + b )2 ba 2 + ( a + b )2 b3 = ( a + b )2 [ a 3 +ab2 + ba2+ b3] Lemma 3.7 For any a,r and s R , [ (r+a)s+rs]2 = [(r+a)s+rs]sa Proof. [(r+a)s+rs]2 = [(r+a)s+rs] [(r+a)s+rs] =[(r+a)s+rs](r+a)s + [(r+a)s+rs]rs =[(r+a)s+rs]s(r+a) + [(r+a)s+rs]rs =[(r+a)s+rs](sr+sa) + [(r+a)s+rs]rs =[(r+a)s+rs]sr+ [(r+a)s+rs]sa +[(r+a)s+rs]rs =[(r+a)s+rs]rs + [(r+a)s+rs]sa + [(r+a)s+rs]rs (by weak commutativity) = [(r+a)s+rs]sa (since Chr R = 2). Lemma 3.8. For any a,r and s R , [(r+a)s+rs]4 =[(r+a)s+rs]2s2a2 Proof. Follows from lemma 3.7 and weak commutativity

Thus from the above two lemmas we have the following Lemma 3.9. For any a,r,s R and for any integer n , [(r+a)s+rs]2n = [(r+a)s+rs]nsnan

Theorem 3.10. The set of all nilpotent elements of a week commutative Boolean like semi ring form an ideal. Proof. Let N = { x R / xn = 0 for some positive integer n }. Clearly, 0 N and hence N is non empty sub set of R. Let a, b N .Then an=0 and bm=0 for some positive integers n,m. Then from lemma 2.6 a=0 or a2=0 or a3=0 and b=0 or b2=0 or b3=0 Now the following cases arise , Case 1. a=0 , b=0 ⇒ a+b =0 ⇒ a + b N 2. a=0 , b2=0 ⇒ (a + b)2= b2=0 ⇒ a + b N 3. a=0 , b3=0 ⇒ (a+b)3 = 0 ⇒ a + b N 4. a2 = 0 , b2 = 0 then from lemma 3.3 ( a + b )2 = ( a + b )2 ( a 2 + b2 ) = ( a + b )2 ( 0 +0 )= 0 ⇒ a + b N 5. a2=0 , b3=0 , consider (a+b)6 = {(a+b)2}3 = [( a + b )2 a 2 +( a + b )2 b2 ]3 = [ ( a + b )2 0 + ( a + b )2 b 2 ]3 =[ ( a + b )2 b 2 ]3 = [( a + b )2 ]3 (b2 )3 (by 2.12(iii) ) = ( a + b )6(b3)2 = ( a + b )6 0 = 0. Hence a + b N 6. a3=0 , b3=0 Consider, ( a + b )3 = ( a + b )2 (a3+ab2+ba2+b3) (by lemma 3.6) = ( a + b )2 (ab2+ba2)= ( a + b )2 ab2+ ( a + b )2 ba2 = (a+b)( a + b )ab2 + (a+b)(a + b)ba2 = (a+b)( a + b )ab2 + (a+b) (a + b ) ba2 =(a+b)ab2 (a+b) + (a+b) ba2 (a + b)

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= (a+b)(ab2a+ ab3 ) +(a+b) (ba3 +ba2b) = (a+b)a2b2 + (a+b) b2a2 = (a+b)a2b2+ (a+b)a2b2 = 0 . Thus , a + b N. In all cases, for all a,b N ,we have that a + b N Now we prove that ra N whenever r R , a N Let a N. Then a n = 0 ,for some positive integer n Consider , (ra )n = r n an ( by lemma 2.12(iii) ) = r n 0 = 0 , Thus for r R , a N we have that ra N which shows that ,N is a left ideal. Our next claim is if r,s R , a N then (r+a)s + rs N. Let a N . Then a n = 0 ,for some positive integer From lemma 2.6 , we have that either a=0 or a2=0 or a3=0. Suppose a=0 then (r+a)s+rs = (r+0)s+rs = rs+rs = 0 If a2 =0 then [(r+a)s+rs]4 =[(r+a)s+rs]2s2a2 ( by lemma 3.8 ) = [(r+a)s+rs]2s20 = 0.Hence (r+a)s + rs N. If a3=0 then [(r+a)s+rs]6 = [(r+a)s+rs]3s3a3 (by lemma3.9 ) = [(r+a)s+rs]3s30 = 0 Therefore , (r+a)s + rs N. Thus in all the above cases (r+a)s + rs N. Hence the theorem

Remark 3.11. The ideal N in the above theorem is denoted by N(R)

Theorem 3.12. If I is an ideal of a weak commutative Boolean like semi ring R then radical of I, denoted by r(I) = { x R / xn I, for some positive integer n } is an ideal of R.

Proof. Clearly r(I) ≠ .Let a, b r(I). Then an , bm I. Hence from lemma 2.6, we have that a I or a2 I or a3 I and b I or b2 I or b3 I . we discuss the following cases. (i) a I ,b I . Clearly a + b I . Hence a+b r(I) (ii) a I ,b2 I ⇒ (a+b)4 = (a + b)2 = (a + b)2 ( a 2 + b2 ) (by lemma 3.3 ) = ( a + b )2 a2+ ( a + b )2 b2 = ( a + b )2 aa + ( a + b )2 b2 I Hence a+b r( I ). (iii) a I ,b3 I ⇒ ( a + b )3 = (a + b)2(a3+ab2+ba2+b3)(by lemma 3.6 )

= ( a + b )2 a3 + ( a + b )2 ab2 + ( a + b )2 ba2 + ( a + b )2 b3

= ( a + b )2 a3 + ( a + b )2 b2 a + ( a + b )2 ba2 + ( a + b )2 b3

= ( a + b )2 a2 a + ( a + b )2 b2 a + ( a + b )2 ba2 + ( a + b )2 b3 I Therefore a+b r( I ). (iv) a2 I ,b2 I ⇒ (a+b)4 = ( a + b )2( a + b )2 ( a 2 + b2 ) (by lemma 3.3)

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= ( a + b )2 a 2 +( a + b )2 b2 I Therefore a+b r( I ). (v) a2 I , b3 I .Then ( a + b )3 = (a + b)2(a3+ab2+ba2+b3)(by lemma 3.6) = ( a + b )2 a3 + ( a + b )2 ab2 + ( a + b )2 ba2 + ( a + b )2 b3 = ( a + b )2 a3 + ( a + b )2 b2 a + ( a + b )2 ba2 + ( a + b )2 b3 = ( a + b )2 a3 + ( a + b ) b2 a (a+b)+ ( a + b )2 ba2 + ( a + b )2 b3 = ( a + b )2 a3 + ( a + b )(b2a2+b2ab) + ( a + b )2 ba2 + ( a + b )2 b3 = ( a + b )2 a3 + ( a + b )(b2a2+b3a) + ( a + b )2 ba2 + ( a + b )2 b3 = ( a + b )2 a3 + ( a + b )b2a2+(a+b)b3a + (a + b)2 ba2 + ( a + b )2 b3 = ( a + b )2aa2 + ( a + b )b2a2 + (a+b)ab3 + (a + b)2 ba2 + ( a + b )2 b3 I Thus a+b r( I ). In all the cases a + b r(I) whenever a, b r(I) We now prove s R , a r(I) ⇒ sa r(I) Let a r(I) .Then an I for some positive integer n. Consider (sa)n = snan I. Hence sa r(I) We now prove s, t R , a r(I) ⇒ (s+a)t + st r(I) Let a r(I) then an I for some positive integer n. This implies either a I or a2 I or a3 I. (i) If a I then (s+a)t + st I (by def of ideal ) ⇒ (s+a)t + st r(I)

(ii) If a2 I then consider [ (s+a)t + st ] 4 =[ (s+a)t + st ] 2 t2a2 I (by lemma 3.8)

Hence (s+a)t + st r(I)

(iii) If a3 I , consider [(s+a)t+st]6 = [(s+a)t+st]3t3a3 (by lemma 3.9) Therefore (s+a)t + st r(I). Hence in any case (s+a)t + st r(I) Thus r(I) is an ideal of R.

Theorem 3.13. Let R be a weak commutative Boolean like semi ring and I be an ideal of R. Then the set Ann I = { a R sa 0 for all s I } is an ideal of R

Proof. Clearly Ann(I) ≠ Ø. Let a ,b Ann(I). then sa =0 = sb, for all s I Consider s(a+b) = sa + sb = 0 + 0 = 0 , therefore a + b Ann(I). Let x Ann(I) , r R. Then sx = 0 , for all s I. Now s(rx) = srx = sxr ( by weak commutativity) = = 0r = 0 .Hence rx Ann(I) Finally, let u , v R , x Ann(I) then sx = 0 for all s I consider s[ (u + x )v +uv ) ] = s (u + x )v + suv = sv( u + x ) + suv ( by weak commutativity)

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= svu + svx + svu = svx ( since Char R = 0 ) = sxv (by weak commutativity) = 0. Hence (u + x)v + uv Ann(I). Thus Ann(I) is an ideal of R.

The following are in similar lines

Theorem 3.14. Let R be a weak commutative Boolean like semi ring and I be an ideal of R. Then the set K = { r R ra 0 , for all a I} is a left ideal of R

Remark 3.15. Even Ideal S is not necessary to show that Ann (S) to become an ideal.

We have this in the following

Theorem 3.16. If R is a weak commutative Boolean like semi ring R and S be a non empty sub set of R then the set Ann(S) = { x R / sx = 0 for all s S } is an ideal of R , which is called right annihilator of S.

Proof. Since s0 =0 ,for all s S ,we have that 0 Ann(S) and hence Ann(S) is non empty subset of R. We now prove x ,y Ann(S) ⇒ x + y Ann(S). Let x ,y Ann(S) ⇒ sx =0 = sy for all s S Consider s(x + y) = sx + sy = 0 + 0 = 0 , therefore x + y Ann(S). Let x Ann(S) , r R. Then sx = 0 , for all s S. consider s(rx) = srx = sxr ( by weak commutativity) = 0r ( by hypothesis). = 0 Hence rx r(S). Finally we prove that for any u , v R , x Ann(S) ⇒ (u + x)v + uv Ann(S). Now x Ann(S) ⇒ sx = 0 for all s S. consider s[ (u + x )v +uv ) ] = s (u + x )v + suv = sv( u + x ) + suv ( by weak commutativity) = svu + svx + svu = svx ( since a +a = 0 ) = sxv (by weak commutativity) = 0. (by hyp) Hence (u + x)v + uv Ann(S). Thus Ann(S) is an ideal of R. Theorem 3.17. Let R be a weak commutative Boolean like semi ring and x R, then the Annihilator of x denoted by Ann x = { a R / xa = 0 } is an ideal of R

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Proof. Clearly Ann x ≠ ∅. Let a, b Ann x .Then xa =0 = xb. Now x( a + b ) = xa + xb = 0 + 0 = 0 .Thus a + b Ann x. Let r R ,a Ann x then x(ra) = xra = xar = 0r( by weak commutativity ) = 0.Thus ra Ann x Finally let r, s R a Ann x then x[ (r + a )s + rs ) ] = x(r + a)s + xrs = xs(r+a) + xsr ( by weak commutativity ) = xsr + xsa + xsr = xsa = xas ( by weak commutativity ) = 0s = 0 Hence ( r+a )s + rs Ann x . Thus Ann x is an ideal of R.

Theorem 3.18 If R is a weak commutative Boolean like semi ring and I ideal of R then the set [R:I] = { x R/ rx I , for all r R} is a left ideal of R and I ⊆ [R:I].

Proof. Routine verification

We state the following trivial facts :

Theorem 3.19. If I and J are ideals of a Boolean like semi ring R then I +J is an ideal of R.

Theorem 3.20. If I and J are ideals of a Boolean like semi ring R then I ∩ J is an ideal of R.

Theorem 3.21. If I is an ideal of a Weak commutative Boolean like semi ring R then (a) I ⊆ r(I) , (b) r (I∩J) = r(I) ∩ r(J) (c) If J is an ideal , I ⊆ J ⇒ r(I) ⊆ r(J) (d) r(r(I)) = r(I) Proof. (a) , (c) and (d) are routine. For (b) , let x r( I ∩ J ) then xn I ∩ J ⇒ xn I , xn J ⇒ x r(I) , x r(J) ⇒ x r( I) ∩ r(J) ⇒ r( I ∩ J) ⊆ r( I) ∩ r(J) ………..(i) If x r(I) ∩ r(J) ⇒ x r(I) , x r( J) ⇒ xn I , xm J , for some positive integers n,m Now xn I ⇒ xmxn I and xm J ⇒ xnxm J ⇒ x m+n I and x n+m J ⇒ x m+n I ∩ J ⇒ x r( I ∩ J) ⇒ r( I) ∩ r(J) ⊆ r(I ∩J) …………………(ii) from (i) and (ii) , r( I ∩ J ) = r(I) ∩ r(J).

Theorem 3.22. If I and J are left ideals of a Boolean like semi ring R then the product IJ = { a1b1+a2b2+----+anbn /ai I , bi J } is a left ideal of R. Proof. Routine

Theorem 3.23. If I , J and K are ideals of R then I ∩ (J + K) = ( I ∩ J ) + ( I ∩ K ) ,if J⊆ I or K⊆ I

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Proof. Without loss of generality, we may assume that K ⊆I , then J⊆ J+K ⇒ I ∩ J I ∩ (J+K) Also K ⊆ J+K ⇒ I ∩ K ⊆ I ∩ (J+K) , so ( I ∩ J) + I ∩ K ⊆ I ∩ (J+K) ………………..(i) Conversely, x I ∩ (J+K) ⇒ x I and x J+K x J+K ⇒ x = j + k , where k K , j J k K , x I ⇒ x+ k I ⇒ j+k+k I ⇒ j I ⇒ j I ∩ J x = j +k (I ∩ J ) + K = (I∩ J ) + (I ∩K ) ⇒ x ( I ∩ J ) + ( I ∩ K ) therefore I ∩ (J + K) ⊆ ( I ∩ J ) + ( I ∩ K ) ………………….(ii) From (i) ,(ii) I ∩ (J + K) = ( I ∩ J ) + ( I ∩ K ) Theorem 3.24. If the set of right zero divisors of R =D = { a R/ Ǝ xR such that xa = 0 }

, then D = ∪ x≠0 Ann x. Proof. If a D then a is a right zero divisor .Hence there exists 0 ≠y R such that ya = 0 ⇒ a Ann y ⇒ a ∪ x≠0 Ann x , conversely , a ∪ x≠0 Ann x ⇒ a Ann x ,for some 0 ≠x R ⇒ xa = 0 ⇒ a is a right zero divisor ⇒ a D. Hence the theorem.

4 Homomorphism of Boolean like semi rings and quotient Boolean like semi rings

Theorem 4.1. A mapping f: R→ R’ is a homomorphism of a Boolean like semi rings R into R’ then f(0) =0’ and f(R) is weak commutative if R is weak commutative. Theorem 4.2. If R and R’ are Boolean like semi rings and also f : R→ R’ is homomorphism of R into R’ then f(R) is a sub ring of R’. i.e , every homomorphic image of a Boolean like semi ring is Boolean like semi ring. Theorem 4.3. If I is an ideal of a Boolean like semi ring R then R/I is also Boolean like semi ring. Proof. Define a map g : R→ R/I by g(x) = x + I , for all x R. Clearly which is an onto homomorphism. So the homomorphic image of R is Boolean like semi ring ( by Theorem 4.2). R/I = g(R) = { g(x) / x R } = { x +I / x R } is a Boolean like semi ring. Also Chr of R/I is 2 , since (x+I) + (x+I) = (x+x) + I = 0 +I and For any x,y R , (x+I) (y+I)[ (x+I) + (y+I) +(x+I)(y+I) ] = (xy +I)[ (x+y+xy) +I ] = (xy(x+y+xy)) +I = xy+I = (x + I )(y + I)

Definition 4.4. If R is a Boolean like semi ring , the set R/I ={ x+I /x R } is called a quotient Boolean like semi ring of R.

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Definition 4.5. If R and R’ are Boolean like semi rings and f : R→ R’ is a homomorphism of R into R’ , the set { x R / f(x) = 0’ , 0 R’ is the additive identity } called kernel of the homomorphism f. It is denoted by Ker f. Theorem 4.6. If R and R’ are Boolean like semi rings and f : R→ R’ is a homomorphism of R into R’ then Ker f is an ideal of R. Proof. Ker f = { x R / f(x) = 0’ , 0 R’ is the additive identity } is a non empty and subset of R is clear. We now prove that x,y Ker f ⇒ x + y Ker f Clearly x+y and rx Ker f whenever x ,y Ker f , r R Let r ,s R , x Ker f. Consider f[ (r+x)s + rs ) ] = f [ ( r + x )s ]+f(rs) = f(r+x)f(s) + f(rs) = [ f(r)+f(x) ]f(s) +f(rs) = [ f(r) + 0’ ]f(s) + f(r)f(s) = f(r)f(s) + f(r)f(s) = 0’ (since Chr R’ =2 ) Therefore (r+x)s +rs Ker f. Hence Ker f is an ideal of R.

Theorem 4.7. If R and R’ are Boolean like semi rings and f : R→ R’ is a homomorphism of R into R’ then Ker f ={0} iff f is one-one. Proof. Suppose Ker f = {0}. Then for x , y R , let f(x) = f( y) ⇒ f(x) + f(y) = f(y) + f(y) ⇒ f(x+y) = f(y +y) = f(0) = 0’ ⇒ x+y Ker f = {0} ⇒ x+y = 0 ⇒ x = y. Hence f is 1-1. Converse is clear. Theorem 4.8. If R and R’ are Boolean like semi rings and f : R→ R’ is a homomorphism of R onto R’ with kernel f = K then R/ K is isomorphic to R’. Proof. Define a map g : R/K →R’ by g( x+K) = f(x) , for all x R. For any x, y R such that g(x+K) = g(y+K) ⇔ f(x) = f(y) ⇔f(x) + f(y) = 0 ⇔f(x+y) = 0 ⇔x+y kerf =K ⇔ x+y K ⇔ (x+y)+K = 0+K ⇔ (x+K)+(y+K) = 0+K ⇔ x+K = y+K. Hence g is well- defined and is 1-1. It is clear that g is an onto homomorphism Thus g: R/K→ R’ is an isomorphism .

We recall that

Definition[3]. An ideal I of a near ring is Prime if and only if ab P then either a P or b P Definition [3]. The intersection of prime ideals of R is called a nilradical.

Definition . An ideal M of a near ring is maximal if M R and there is no ideal I such that M⊆ I⊆ R.

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Definition. The intersection of maximal ideals of R is called Jacobson radical.

Theorem 4.9. If I is an ideal of a Boolean like semi ring R then I is prime iff R/I has no nonzero nilpotents and R/I has no zero divisors. Proof. Suppose I is a prime ideal . If x+I R/I such that ( x+I )n = 0+I ⇒ xn +I = 0+I ⇒ xn I ⇒ x I (I is prime ideal ) ⇒ x + I = 0 + I. Therefore R/I has no zero divisors. If (x+I)(y+I) = 0+I ⇒ xy +I = 0+I ⇒ xy I ( I is prime ideal ) ⇒ x I or y I ⇒ x+I = 0+I or y+I = 0+I Thus R/I has no zero divisors. Conversely , if ab I such that a ∉ I then a2b I ⇒ a2b +I = 0+I ⇒ (a2+I )(b + I) = 0+I ⇒ b+I = 0+I (since a∉ I ⇒ a2∉ I and R/I has no zero divisors ) ⇒ b I Theorem 4.10. If R is a weak commutative Boolean like semi ring and N is the nilradical of R then R/N is Boolean near ring. Proof. we have x+x2 is a nilpotent element from lemma 3.4 and hence (x+x2 ) + N = 0 , zero in R/N.Consider (x+N)2 = x2 + N = (x+x+x2)+N = (x+N) + ( x+x2+N) = x+N Hence R/N is Boolean near ring.

Theorem 4.11. Let R be a Boolean like semi ring with unity 1. If I is an Ideal of R such that 1 I then I = R. Proof. If x R then x = x1 I ( by 3.1(2) ) ⇒ x I . Hence I = R Now we are in a position to prove the following

Theorem 4.12. Every prime ideal in a Boolean like semi ring R with right unity is maximal. Proof. Let P be a prime ideal of R. Let J be an ideal of R such that P ⊊ J ⊆ R.Then there exists a J such that a ∉ P . By 1.1.2 a4 = a2 ⇒ a4 + a2 = 0 ⇒ a2a2 + a2 1 = 0 ⇒ a2(a2 + 1 ) = 0 P and a2∉ P ⇒ a2 + 1 P ⊊ J ⇒ a2 +a2 +1 J ( a J ⇒ a2 J) ⇒ 1 J ⇒ J = R ( by Theorem 3.35 ) . Hence P is maximal. Remark [3] .Every ideal I in a near ring R, is Maximal either I is prime or R2 I.

Theorem4.13. Every maximal ideal in a Boolean like semi ring R with right unity is prime. Proof. Let M be a maximal ideal in R. clearly M ≠ R. From the above remark it is sufficient to prove that R2 is not contained in I. Clearly 1 R , hence 1 I which in turn makes R = I , a contradiction. Hence I is prime ideal. From theorems 4.12 & 4.13, we conclude this article in the following

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Theorem 4.14. In a Boolean like semi ring with right unity, Jacobson radical is equal to Nil radical

References

1. Foster .A.L: Pk rings and ring logice, Ann.Scu.Norm. pisa , 5, 1951,279-300 2. Foster .A.L: The theory of Boolean like rings, Trans.Amer.Math.Soc. Vol.59, 1946, 3. Gunter Pilz: Near-Rings, The theory and its applications (North-Holland) 1983. 4. James R.Clay and A. Lawyer: Boolean Near-Rings, Canada. Math. Bull. vol.12,no.3,1969 265-273 5. Macoy N.H : Sub rings of Direct sums, Amer.J.Math,60, 1938, 374-382 6. Macoy N.H: and Montgomery’s: A representation of generalized Boolean rings, Duke Math. J. 3, 1937,455-459 7. Subrahmanyam. N.V: Boolean semi rings, Math. Annalen 148, 395-401,1962 8. Subrahmanyam. N.V : Lattice theory for certain classes for rings, Math,Ann.139,1960 9. Sussman I : A generalization of Boolean rings, Math.Ann,136,1958,326-338 10. Swaminathan V : Boolean- like rings, PhD dissertation ,, Andhra University, India, 1982 11. Swaminathan V: Geometry of Boolean- like rings, Math. Seminar Notes, Kobe University, Japan, Vol 9, 1981, 195-214. 12. Swaminathan V: On Foster’s Boolean- like rings, Math. Seminar Notes, Kobe University, Japan, Vol 8, 1980, 347-367. 13. Von-Neumann: On Regular rings proc.Nat..Acc.Sci, U.S.A 22, 1936, 707-713.

Received: September, 2010