Extensions of Boolean Rings and Nearrings Текст научной статьи по специальности «Математика»

Extensions of Boolean Rings and Nearrings

Hamsa Nayak Syam P. Kuncham Babushri S. Kedukodi*

Department of Mathematics Manipal Institute of Technology Manipal Academy of Higher Education Manipal Karnataka, 576104 India

Received 15.06.2018, received in revised form 06.09.2018, accepted 01.12.2018 In this paper, we introduce the notions of left (right) Boolean rings and nearrings. We give examples to show that left (right) Boolean rings are not commutative in general. We obtain interrelations among these algebraic structures and get conditions under which the structures are commutative. Finally, we study the concept of derivations on left (right) Boolean rings and nearrings and obtain commutativity results.

Keywords: ring, nearring, Boolean ring, Boolean nearring. DOI: 10.17516/1997-1397-2019-12-1-58-67.

Introduction

A nearring (N, +, ■) is an algebraic system with binary operations addition and mutiplication satisfying the axioms of a ring, except commutativity of addition and one of the distributive laws. Right nearrings satisfy the right distributive law and left nearrings satisfy the left distributive law. A natural example of right nearring is the set of all mappings from a group (G, +) to itself under addition and composition of mappings. In this sequel, N denotes a right nearring.

A Boolean ring R is a ring for which x2 = x for all x G R. An example of Boolean ring is the ring of integers modulo 2. If all the elements in a nearring are idempotent, then such a nearring is called a Boolean nearring. Clay and Lawyer [7] showed that a distributively generated Boolean nearring is multiplicatively commutative by using subdirect sum representations of nearrings. Reddy [27] studied recent developments in Boolean nearrings. Bhavanari and Kuncham [2] introduced the concepts of linearly independent elements and u-linearly independent elements in N-group. Bhavanari, Godloza, Babu and Kuncham [6] introduced zerosquare nearring of type-1/type-2, their ideals, zero square dimension of a nearring and obtained several results. Bhavanari, Kuncham and Kedukodi [5] studied graph theoretic aspects of nearrings. Kedukodi, Kuncham and Bhavanari [12,13] studied equiprime, 3-prime and c-prime fuzzy ideals of nearrings. Jagadeesha, Kedukodi, Kuncham [9,16,19] studied homomorphic images of interval valued L-fuzzy ideals and proved isomorphism theorems. Molkhasi [22] defined the notion of strongly algebraically closed lattices.

For recent developments in nearrings and Boolean nearrings, we refer Kuncham, Kedukodi, Panackal and Bhavanari [20], Nayak, Kuncham and Kedukodi [23], Koppula, Kedukodi and Kuncham [17] and Kedukodi, Kuncham and Bhavanari [14,15]. We refer Bhavanari and Kuncham [3],

* [email protected] © Siberian Federal University. All rights reserved

Kuncham [18], Pilz [24], Bhavanari [4], Lafuerza-Guillen and Panackal [21], Panackal, Lafuerza-Guillen and Ravindran [25] and Jagadeesha, Kuncham and Kedukodi [8] for basic definitions. The paper is divided into two sections. In Section 1, we introduce left, right Boolean rings and nearrings. We give conditions under which the structures are weak commutative with respect to an ideal. In Section 2, we study derivations on left and right Boolean ring and nearring.

1. Left and right Boolean rings and nearrings

Definition 1.1. Let N be a nearring. N is called a left (resp. right) Boolean nearring if there exists n £ N such that x2 = nx (resp. x2 = xn) for all x £ N. If N is a ring satisfying x2 = nx (resp. x2 = xn) for all x £ N then N is called a left (resp. right) Boolean ring.

We begin with an example of left (resp. right) Boolean ring which is not a Boolean ring. Example 1.2. Let R = {0, a, b, c} with operations + and • 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 0 0

b b c 0 a b 0 0 b b

c c b a 0 c 0 0 b b

Then R is a left as well as right Boolean ring for n £ {b, c}. We have a2 = 0 = a. Hence R is not a Boolean ring.

Example 1.3. Let R be a left(resp. right) Boolean ring. In R3, define addition componentwise and multiplication by (xi, yi, zi)(x2, y2, z2) = (0,0, x^2 — x2yi). Then R3 is a left(resp. right) Boolean ring.

Verification: Let (xi,yi,zi), (x2,y2,Z2) £ R3. (xi,yi,zi)(x2,y2,z2) = (0, 0,xiy2 — x2yi) and (x2,y2,Z2)(xi,yi,zi) = (0, 0, x2yi —xiy2). Clearly, (xi, yi, zi)(x2, y2, z2)+(x2, y2, z2)(xi, yi, zi) =

= (0,0,0).

We show that distributive properties are satisfied: Let (xi,yi,zi), (x2,y2,z2), (x3,y3,z3) £ R3.

(xi,yi,zi)((x2 ,y2,z2) + (x3,y3,z3)) = (xi,yi,zi)(x2 + x3,y2 + y3,z2 + z3) =

= (0,0,xi(y2 + y3) — (x2 + x3)yi) = (0, 0,xiy2 + xiy3 — x2yi — x3yi).

Now,

(xi, yi, zi)(x2,y2,z2) + (xi,yi,zi)(x3,y3, z3) = (0, 0, xiy2 — x2yi) + (0, 0, xiy3 — x3yi) =

= (0, 0, xiy2 — x2yi + xiy3 — x3yi).

Hence (xi, yi, zi)((x2, y2, z2) + (x3, y3, z3)) = (xi, yi, zi)(x2, y2, z2) + (xi, yi, zi)(x3, y3, z3). Similarly we can verify that

((xi,yi,zi) + (x2,y2 ,z2))(x3,y3,z3) = (xi,yi,zi)(x3,y3,z3) + (x2, y2, z2)(x3, y3, z3).

Now, we show that the associative law is satisfied.

(xi,yi,zi)((x2,y2,z2)(x3,y3,z3)) = (xi,yi,zi)(0,0,x2y3 — x3y2) = (0,0, 0).

Hence ((xi,yi,zi)(x2,y2,z2))(x3,y3, z3) = (0,0,xiy2 — x2yi)(x3, y3, z3) = (0,0,0). Now, we show that R3 is a left(resp. right) Boolean ring. We have (x,y, z)(x,y, z) = (0,0, xy — xy) =

= (0, 0, 0) = (x,y,z)(0, 0, n) = (0, 0,n)(x,y,z).

Example 1.4. Let M

on N as follows:

x 0 0 x

I

\x G N >, where N = {0, a, b, c} and + and • are defined

+ 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 0 0

b b c 0 a b 0 0 0 a

c c b a 0 c 0 0 0 a

Then M is a left Boolean nearring with n =

c0 0c

because

x 0 x 0 2 x2 0 cx 0 and c 0 x 0 cx 0

0 x 0 x = 0 2 x2 = 0 cx 0 c 0 x = 0 cx

for every x G N.

Note that M is not a right Boolean nearring. Also, M is not commutative because

b 0 c0 a0 = 0 0 c0 b 0

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

Example 1.5. Let M

on N as follows:

x0 0x

\x G N >, where N = {0, a, b, c} and + and • are defined

+ 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 0 0

b b c 0 a b 0 a b b

c c b a 0 c 0 a b b

Then M is a right Boolean nearring with n

b0 0b

because

x0 x0 x2 0 xb 0 and x0 b 0 xb 0

0x 0x = 0 x2 = 0 xb 0x 0 b = 0 xb

for every x G N. Note that M is not a left Boolean nearring. Also, M is not commutative because

a0 b 0 0 0 = a0 b 0 a0

0a 0 b = 0 0 0a = 0 b 0a

Proposition 1.6. Let R be a left(right) Boolean ring. If n is not a zero divisor then R is commutative.

Proof. Let R be a left Boolean ring. We have (x + y)2 = n(x + y). Then

(x + y)(x + y) = nx + ny ^ x(x + y) + y(x + y) = nx + ny ^ x2 + xy + yx + y2 =

= nx + ny ^ nx + xy + yx + ny = nx + ny ^ xy + yx = 0 ^ xy = -yx. (1)

Also,

(x + x)2 = n(x + x) ^ x2 + x2 + x2 + x2 = nx + nx ^ n(x + x) = 0 ^ x = -x. (2) By (1) and (2) we get, xy = yx. The proof is similar for right Boolean ring. □

Proposition 1.7. If N is a Boolean nearring with left (resp. right) identity then N is a left (resp. right) Boolean nearring.

Proof. Let e be a left identity of N. Then x2 = x = ex. Hence N is a left Boolean nearring. Now, let e be a right identity of N. Then x2 = x = xe. Hence N is a right Boolean nearring. □

Example 1.8. Let N = {0, a, b, c} with operations + and • 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 a a a

b b c 0 a b 0 b b b

c c b a 0 c 0 c c c

Then N is a non-commutative Boolean nearring. Note that a, b, c are right identities of N. By Proposition 1.7, N is a right Boolean nearring for n £ {a,b,c}. Note that a,b,c are not zero divisors of N. This example also shows that Proposition 1.6 is not true in general for left (resp. right) Boolean nearrings.

Proposition 1.9. Let R be a left(resp. right) Boolean ring with |R| ^ 3. If 0 = n is not a zero divisor then R has a proper zero divisor.

Proof. Let x,y £ R such that 0 = x = y = 0. Note that x + y = 0. If xy = 0 then x is a proper zero divisor. Let xy = 0. (xy)(x + y) = xyx + xyy = yx2 + xy2 (Since R is commutative when n is not a zero-divisor) = ynx + xny = xny + xny = 0. Thus xy is a proper zero-divisor. The proof is similar for right Boolean ring. □

Proposition 1.10. Let f : Ri ^ R2 be a onto ring homomorphism. If Ri is a left(resp. right) Boolean ring then R2 is a left(resp. right) Boolean ring.

Proof. Let Ri be a left Boolean ring. Let y £ R2. Then y = f (x) for some x £ Ri. Now, y2 = f (x)f (x) = f (x2) = f (nx) = f (n)f (x) = f (n)y. Hence R2 is a left Boolean ring. Similarly, if Ri is a right Boolean ring then R2 is a right Boolean ring. □

Proposition 1.11. Let R be a field and a left(resp. right) Boolean ring. Then n = 1.

Proof. Let n =1. Now, x2 = nx ^ x2 — nx = 0 ^ x(x — n) = 0 ^ x — n = 0 ^ x = n. This shows that n varies with each x, a contradiction. Hence n = 1. The proof is similar for right Boolean ring. □

Theorem 1.12. Let N be a left(right) Boolean nearring and P be a c-prime ideal of N. If n and m are not zero divisors then P is maximal.

Proof. Let N be a left Boolean nearring. Then we have, x2 = nx. This implies x2 — nx = 0 £ P ^ x(x—nx) £ P ^ x £ P or x—nx £ P ^ x £ P or x £ nx+P ^ N/P = {P, nx+P} ^ N/P is a field. Hence P is maximal. The proof is similar for right Boolean nearring exists. □

Definition 1.13 (Pilz [24]). A nearring N is said to have insertion of factors property(IFP) if for all a, b £ N, ab = 0 implies anb = 0 for all n £ N.

Corollary 1.14. Let N be a left(right) Boolean nearring and P be an equiprime ideal that has IFP. If n and m are not zero divisors and P be an equiprime then P is maximal.

Proof. The proof follows from Theorem 2.21 of Kedukodi, Kuncham and Bhavanari [11] and Theorem 1.12. □

Definition 1.15. Let I be an ideal of N. I is called left(resp. right) Boolean type if there exists n £ N such that x2 — nx £ I (resp. x2 — xn £ I) for all x £ N.

Proposition 1.16. Let I be an ideal of N. Then N/I is a left(resp. right) Boolean ring if and only if I is left(resp. right) Boolean type.

Proof. I is left Boolean type ^ x2 — nx £ I ^ x2 +1 = nx +1 ^ (x + I)2 = (n + I)(x +1). Hence N/I is a left Boolean ring. Similarly, we can prove that N/I is a right Boolean ring if and only if I is right Boolean type. □

Note 1.17. If I = {0} is left (resp. right) Boolean type then N is a left (resp. right) Boolean nearring.

Example 1.18. Let R = {0, a, b, c} with operations + and • defined as follows:

+ 0 a b c

0 0 a b c

a a 0 c b

b b c 0 a

c c b a 0

0 a b c

0 0 0 0 0

a 0 a 0 a

b 0 0 b b

c 0 a b c

Then R is a left (resp. right) Boolean ring with n = c and left identity c.

Theorem 1.19. Let R be a left(resp. right) Boolean nearring. If n is a distributive element and has a left identity e, then R is a zero symmetric nearring. Further, if there exists a left ideal I such that e £ I and (i) nx £ I ^ x £ I (resp. xn £ I ^ x £ I); (ii) [X, Y] n I = {0} then R is a commutative ring.

Proof. As e is left distributive, (e + e)2 = n(e + e) = ne + ne. Now, (e + e)2 = (e + e)(e + e) = = e(e+e)+e(e+e) = e2+e2 +e2+e2 = ne+ne+ne+ne ^ ne+ne = 0 ^ n(e+e) = 0 ^ e+e = 0. If x is in R, then x + x = (e + e)x = 0.x = 0 ^ x + x = 0. Let w be arbitrary element in R. Then (e + w)2 = n(e + w) ^ (e + w)(e + w) = ne + nw ^ e(e + w) + w(e + w) = ne + nw ^ e2 + w + w(e + w) = ne + nw ^ w(e + w) = —w + nw = w + w2 = (e + w)w.

w(e + w) — (e + w)w = 0. (3)

Now, w(e + w)0 = (e + w)w0 ^ w(e0 + w0) = ew0 + ww0 ^ ww0 = ew0 + ww0 ^ w0 = 0. Thus R is a zero symmetric nearring. Replacing w = ab and w = ba in equation (3), we get (e + ab)ab = ab(e + ab) ^ ab + (ab)2 = ab(e + ab). Now, we have

ab = ab( e + ab) — nab. Similarly, ba = ba(e + ba) — nba.

Hence ab — ba = ab(e + ab) — nab — [ba(e + ba) — nba] £ I. We have ab — ba £ [X, Y] = {xy — yx\x £ X,y £ Y} and ab — ba £ I ^ ab — ba = 0 ^ ab = ba. Proof is similar for right Boolean ring. □

Corollary 1.20. Let R be a left(resp. right) Boolean nearring. Let n be a distributive non zero divisor of R and I be a left ideal such that [X, Y] n I = {0}. Then (R, is a partially ordered set with ^ defined by x ^ y if xy = nx (resp. xy = xn). Further, if nx £ I ^ x £ I (resp. xn £ I ^ x £ I) and R has a left identity e such that e £ I then (R, is a lattice with meet and join operations given respectively by x A y = xy and x V y = x + y + xy.

Proof. It is straightforward to verify that (R, <) is a partially ordered set. The rest of the proof follows from Theorem 1.19. □

Definition 1.21. Let N be a left(resp. right) Boolean nearring and I be an ideal of N. N is said to satisfy weak commutative property with respect to ideal I if for all a,b,c £ N, abc — acb £ I.

Lemma 1.22. Let N be a left(resp. right) Boolean nearring. If there exists an ideal I of N such that (i) nx G I ^ x G I (resp. xn G I ^ x G I), and (ii) xIx C I for all x G N, then ab — aba G I for all a,b G I.

Proof. We have (ab — aba)2 = n(ab — aba). Now, (ab — aba)2 = (ab — aba)(ab — aba) = = ab(ab — aba) — aba(ab — aba) = ab(ab — ¿i) — ¿2 [where ¿1 = aba and ¿2 = aba(ab — aba)] = = ¿3 + abab — i4 G I. Hence we get n(ab — aba) G I. This implies ab — aba G I. The proof is similar for right Boolean ring. □

Theorem 1.23. Let N be left(resp. right) Boolean nearring. If there exists an ideal I in N such that (i) nx G I ^ x G I (resp. xn G I ^ x G I) and (ii) xIx C I for all x G N, then N satisfies weak commutative property with respect to I.

Proof. We have abc — acb = abc — a(cbc + i1) [because cb — cbc G I implies cb = cbc + i1] = abc — i,2 — acbc = abc — acbc — i2 = (a — ac)bc — i,2 = ((a — ac)b(a — ac) + ¿3)c — i2 G I. □

Definition 1.24 (Plasser [26]). A nearring N has strong IFP if and only if for all ideals I of N and for all a,b,n G N, ab G I implies anb G I.

Corollary 1.25. Let N be a left (resp. right) Boolean nearring with an ideal I such that (i) nx G I ^ x G I (resp. xn G I ^ x G I) and (ii) xIx C I for all x G N. Then N has strong IFP.

Proof. Let ab G I. By Theorem 2.4, axb = abx + i G I. This implies N has strong IFP. □

Theorem 1.26. Let N be a left(resp. right) Boolean nearring and I be an ideal of N such that (i) nx G I ^ x G I (resp. xn G I ^ x G I); (ii) xIx C I for all x G N. If L is any left ideal of N containing I then L is an ideal of N.

Proof. Let L be a left ideal of N. To show that L is an ideal it suffices to show that LN C L. Let l G L,n G N. Then we have l = lo + lc and n = no + nc as Pierce decompositions, where l0,n0 G N0; lc,nc G Nc—(4). As L is a left ideal, we have N0L C L, for m0l = = m0(0 +1') — m00 G L for all m0 G N0, l G L—(5). Now, ln = l2 = (l0 + lc)l = l0l + lcl = l0l + lc. By (5), l, l0l G L and it follows that lc G L and hence l0 G L. We have, ln = (l0 + lc)n = = lon + lcn = lon + lc = lo(no + nc) + lc = lo(no + nc)lo + i + lc = lo(nolo + nc) + i + lc. We have lonc = lonc0 = lo0nc + i = 0 + i G I. As no, lo G L, we have lo(nolo + nc) = lonc + ¿2 G I C L. Hence ln = lo(nolo + nc) + lc G L. Thus L is an ideal of N. □

2. Derivations

Definition 2.1 (Bell [1]). A derivation on N is defined to be an additive endomorphism satisfying the product rule D(xy) = xD(y) + D(x)y for all x,y G N.

Proposition 2.2. Let N be a 3-prime, 2 torsion left(resp. right) Boolean nearring. If D is a commuting derivation on N, then either D(n) = 0 or D(n) = n.

Proof. Let N be 3-prime left Boolean nearring. Then x2 = nx. We have D(x2) = D(nx) ^ xD(x) + D(x)x = nD(x) + D(n)x. As D is commuting and x + x = 0, we get nD(x) + D(n)x = 0. Put x = n, we have nD(n) + D(n)n = 0 ^ (D(n))2 + D(n)n = 0 ^ D(n)(D(n) + n) =0. Hence D(n) = 0 or D(n) = n. The proof is similar for right Boolean nearring. □

Proposition 2.3. Let N be 2 torsion left(resp. right) Boolean nearring. If D is a commuting derivation on N with D(n) = n, then x2 + y2 = (D(x))2 + (D(y))2 for all x,y G N.

Proof. Let N be a left Boolean nearring. Then we have (x + y)2 = n(x + y). This gives D(x + y)(x + y) = D(n(x + y)) ^ (x + y)D(x + y) + D(x + y)(x + y) = D(n(x + y)) ^ (x + y)D(x + y) + D(x + y)(x + y) = nD(x + y) + D(n)(x + y) ^ ((x + y) + (x + y) — n)D(x + y) = = n(x + y) ^ nD(x + y) = x2 + y2 ^ x2 + y2 = nD(x) + nD(y) = (D(x))2 + (D(y))2. The proof is similar for right Boolean nearring. □

Theorem 2.4. Let R be a left(resp. right) Boolean ring and D be the derivation on R with D(nx) = 0( resp. D(xn) = 0) for all x £ R. Then D(xy) = —D(yx).

Proof. Let R be left Boolean ring. Then x2 = nx. We have D(x2) = D(nx) =0. ^ xD(x) + D(x)x = 0 ^ D(x)x = —xD(x). Now we have (x + y)2 = n(x + y). Then D(x + y)2 = = D(n(x + y)) = 0 ^ D((x + y)(x + y)) = 0 ^ (x + y)D(x + y) + D(x + y)(x + y) = 0 ^ xD(x) + xD(y) + yD(x) + yD(y) + D(x)x + D(y)x + D(x)y + D(y)y = 0 ^ xD(y) + yD(x)+ +D(y)x + D(x)y = 0 ^ D(xy) = —D(yx). The proof is similar for right Boolean nearring. □

Corollary 2.5. 1. If x + x = 0, then D(xy) = D(yx). 2. If R is a zero square ring, then D(xy) = —D(yx).

Proposition 2.6. Let N be left Boolean nearring without zero divisors. Let D be the derivation on N. Then D(n) = n or D(n) = 0.

Proof. We have x2 = nx. Then (D(n))2 = nD(n) ^ D(n)D(n) = nD(n). Then (D(n) — —n)D(n) =0 ^ D(n) = n or D(n) =0. □

Proposition 2.7. Let N be right Boolean nearring without zero divisors. Let D be the commuting derivation on N. Then D(n) = n or D(n) = 0.

Proof. We have x2 = xn. Then (D(n))2 = D(n)n ^ D(n)D(n) = D(n)n. Then (D(n) — —n)D(n) =0 ^ D(n) = n or D(n) =0. □

Proposition 2.8. Let N be a left Boolean nearring. If there exists a nilpotent element in N, then n0 = 0.

Proof. Let x be a nilpotent element in N. Then we have xk =0 for some k £ N. As N is left Boolean nearring, we have x2 = nx for n £ N. Now, (xk)2 = nxk = n0 ^ n0 = (xk)2 =0. □

Proposition 2.9. Let D be the derivation on left Boolean nearring N such that Dk(x) = kx. If n is not a zero divisor then Dk (x) = 0.

Proof. We have Dk (x2) = kx2 = knx ^ Dk (nx) = knx ^ nDk (x) + Dk(n)x = knx ^ nkx + knx = knx. Hence nkx = 0. As n is not a zero divisor, we get kx = Dk (x) = 0. □

Proposition 2.10. Let N be a nearring. If Dk (x) = kx is a nonzero derivation on N then NkN = {0}.

Proof. As D is a derivation on N, we have Dk (xy) = xDk (y) + Dk (x)y. Hence kxy = = xky + kxy ^ xky = 0 ^ NkN = {0}. □

Proposition 2.11. Let N be a left Boolean nearring such that xNy = 0 for all x and y. If n is a distributive element then D(x) = x2 is a derivation on N.

Proof. We have D(x) = x2 = nx. Now, D(x + y) = n(x + y) = nx + ny = D(x) + D(y). We have D(xy) = nxy. Now xD(y) + D(x)y = xy2 + x2y = xny + nxy = 0 + nxy = nxy ^ D(xy) = = xD(y) + D(x)y. Hence D is a derivation on N. □

Proposition 2.12. Let N be a right Boolean nearring. If xNy = 0 for all x and y then D(x) = x2 is a derivation on N.

Proof. We have D(x) = x2 = xn. Now, D(x + y) = (x + y)n = xn + yn = D(x) + D(y). We have D(xy) = xyn. Now xD(y) + D(x)y = xy2 + x2y = xyn + xny = xyn + 0 = xyn ^ D(xy) = = xD(y) + D(x)y. Hence D is a derivation on N. □

Theorem 2.13. Let R be a ring, D is a derivation on R and k G R. Then Dk(x) = D(x) + kx is a derivation on R if and only if RkR = {0}.

Proof. Let RkR = {0}. Now, Dk(x+y) = D(x+y)+k(x+y) = D(x)+D(y)+kx+ky = D(x)+ +kx + D(y) + ky = = Dk (x) + Dk (y). We have Dk (xy) = D(xy) + kxy = xD(y) + D(x)y + kxy. Now, xDk(y) + Dk(x)y = x(D(y) + ky) + (D(x) + kx)y = xD(y) + xky + D(x)y + kxy = = xD(y) + D(x)y + kxy ^ Dk(xy) = xDk(y) + Dk(x)y. Hence Dk(x) = D(x) + kx is a derivation on R.

Conversely, let Dk (x) = D(x) + kx be a derivation on R. We have Dk (xy) = D(xy)+ +kxy = xD(y) + D(x)y + kxy. Now, xDk(y) + Dk(x)y = x(D(y) + ky) + (D(x) + kx)y = = xD(y) + xky + D(x)y + kxy. As Dk (xy) = xDk (y) + Dk (x)y, we have xD(y) + D(x)y + kxy = = xD(y) + xky + D(x)y + kxy ^ xky = 0. Hence RkR = {0}. □

Theorem 2.14. Let R be a 3-prime left(resp. right) Boolean ring. If D is the derivation on R with D(nx) = 0 (resp. D(xn) =0) for all x G R then R is a commutative ring.

Proof. The proof follows from Corollary 3.4 (ii) of Kamal and Al-Shaalan [10] and Theorem 2.4. □

Corollary 2.15. Let R be a 3-prime left(resp. right) Boolean ring with x + x = 0 for all x G R. If D is the derivation on R such that D(nx) = 0 (resp. D(xn) =0) then R is a commutative ring.

Proof. The proof follows from Corollary 2.2 of Kamal and Al-Shaalan [10] and (1) of Corollary 2.5. □

The authors thank the anonymous reviewers for their comments and suggestions. All authors acknowledge Manipal Institute of Technology, Manipal Academy of Higher Education for their encouragement. The first author acknowledges Manipal Academy of Higher Education for Dr. T. M.A. Pai PhD Scholarship.

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Расширения булевых колец и почти колец

Хамса Наяк Сям П. Кунчам Бабушри С. Кедукоди

Математический факультет Манипальский технологический институт Манипальская Академия высшего образования Манипал, Карнатака, 576104 Индия

В этой статье мы вводим понятия левых (правых) булевых колец и почти колец. Мы приводим примеры, чтобы показать, что левые (правые) булевы кольца вообще не коммутативны. Мы получаем взаимосвязи между этими алгебраическими структурами и получаем условия, при которых структуры являются коммутативными. Наконец, мы изучаем концепцию дифференцирования на левых (правых) булевых кольцах и почти кольцах и получаем результаты коммутативности.

Ключевые слова: кольцо, почти кольцо, булево кольцо, булево почти кольцо.