Communications in Algebra ISSN: 0092-7872 (Print) 1532-4125 (Online) Journal homepage: http://www.tandfonline.com/loi/lagb20 Third power associative, antiflexible rings satisfying (a,b,ac) = a(a,b,c) Dhabalendu Samanta & Irvin Roy Hentzel To cite this article: Dhabalendu Samanta & Irvin Roy Hentzel (2017): Third power associative, antiflexible rings satisfying (a,b,ac) = a(a,b,c), Communications in Algebra, DOI: 10.1080/00927872.2017.1392544 To link to this article: http://dx.doi.org/10.1080/00927872.2017.1392544 Accepted author version posted online: 20 Oct 2017. Submit your article to this journal View related articles View Crossmark data Full Terms & Conditions of access and use can be found at http://www.tandfonline.com/action/journalInformation?journalCode=lagb20 Download by: [UAE University] Date: 25 October 2017, At: 11:10 Third power associative, antiflexible rings satisfying (a, b, ac) = a(a, b, c) 1 Department an us cr ip t Dhabalendu Samanta1 and Irvin Roy Hentzel2 of Applied Sciences & Humanities, PDM College of Engineering, Bhadurgarh, Haryana, India of Mathematics, Iowa State University, Ames, Iowa, USA Abstract M In this paper, we study third power associative, antiflexible rings satisfying the identity (a, b, ac) = a(a, b, c). We prove that third power associative, antiflexible rings satisfying the identity (a, b, ac) = a(a, b, c) with characteristic 6 = 2, 3 are associative of degree five. As a consequence Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 2 Department of this result, we prove that a third power associative semiprime antiflexible ring satisfying the identity (a, b, ac) = a(a, b, c) is associative. KEYWORDS: Antiflexible ring; associative of degree five; semiprime ring; third power asso- ciative Received 2 July 2017; Revised 17 September 2017 Dhabalendu Samanta dhabalendu@gmail.com Department of Applied Sciences & Humanities, PDM College of Engineering, Bhadurgarh, Haryana, India. 1 Ac ce pt ed M an us cr ip t Downloaded by [UAE University] at 11:10 25 October 2017 2010 Mathematics Subject Classification: 17D99; 17A05 2 1. INTRODUCTION A nonassociative ring R is called antiflexible if (a, b, c) = (c, b, a) holds for all a,b,c in R, where the associator is defined by (a, b, c) = (ab)c − a(bc). A number of properties of antiflexible an us cr ip t rings are given by Kosier in [6], by Rodabugh in [7], simple antiflexible rings are characterized to some extent by Anderson and Outcalt in [1]. In [3] Celik has proved that a prime antiflexible ring is either associative or the center is equal to the nucleus. In this paper, we study third power third power associative, antiflexible rings satisfying (a, b, ac) = a(a, b, c) are associative of degree five. That is, any product of degree five (or more) containing an associator is zero. As a consequence of this result, we have proved that a semiprime antiflexible ring satisfying the identity M (a, b, ac) = a(a, b, c) is associative. Thus, we have been able to derive a class of third power associative antiflexible rings which are associative of degree five and associative when they are semiprime. Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 associative, antiflexible rings satisfying the identity (a, b, ac) = a(a, b, c). We have proved that 2. PRELIMINARIES Let R be a nonassociative ring satisfying the following identities (a, b, c) = (c, b, a), (1) (a, a, a) = 0, (2) (3) (a, b, ac) = a(a, b, c). That is, R is a third power associative, antiflexible ring satisfying the identity (3). Throughout this chapter, we assume that characteristic of R is not equal to 2, 3. It is proved in [1, page 312 and 3 (2.10)], that in a ring satisfying (1) and (2) the following identity holds (a, [b, c], d) = 0. (4) (a, b, c) + (b, c, a) + (c, a, b) = 0. an us cr ip t Using (1), linearized (2) and characteristic 6 = 2, we have the cyclic law (5) (a, b, cd) + (c, b, ad) = c(a, b, d) + a(c, b, d). (a, b, a) = −2(a, a, b). M From (5) and (1), we get the following identity (6) (7) Using (4), (7) and characteristic 6 = 2, we obtain Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 Linearization of (3) implies the following identity (a, a, [b, c]) = 0. (8) (T) is the Teichmuller identity. (T) holds in any nonassociative ring (ab, c, d) − (a, bc, d) + (a, b, cd) = a(b, c, d) + (a, b, c)d. 4 (T) From the identity (T), we have (aa, a, b) − (a, aa, b) + (a, a, ab) = a(a, a, b) + (a, a, a)b. Using (2) and (3), we get (aa, a, b) − (a, aa, b) + a(a, a, b) = a(a, a, b). Therefore, we have (9) an us cr ip t (aa, a, b) = (a, aa, b). From (6), we have (a, a, ba) + (b, a, aa) = b(a, a, a) + a(b, a, a). Using (2) and (8), we get (a, a, ab) + (b, a, aa) = a(b, a, a). Using (1) and (3) we obtain a(a,a,b) + (b,a,aa) = a(a,a,b). This (aa, a, b) = (a, aa, b) = 0. (10) (a, b, aa) = 0. Using (3), we get M From (5), we have (aa, a, b) + (a, b, aa) + (b, aa, a) = 0. Therefore, (1) and (10) together imply Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 yields (b, a, aa) = 0. Thus, using (1) and (9) we have a(a, b, a) = 0. (11) (12) Characteristic 6 = 2 and (7) together imply a(a, a, b) = 0. (13) Linearization of (12) implies a(a, b, c)+a(c, b, a)+c(a, b, a)+c(c, b, a)+c(a, b, c)+a(c, b, c) = 0. Since this is a nonhomogeneous identity, we have a(a, b, c) + a(c, b, a) + c(a, b, a) = 0. That is, using (1), we have 2a(a, b, c) + c(a, b, a) = 0. Using (7), we get 2a(a, b, c) − 2c(a, a, b) = 0. Since 5 characteristic 6 = 2, we have (14) a(a, b, c) = c(a, a, b). an us cr ip t From (13) and (3), we have (a, a, ab) = 0. Linearization of this implies (c, a, ab) + (a, c, ab) + (a, a, cb) + (a, c, cb) + (c, a, cb) + (c, c, ab) = 0. Since this is nonhomogeneous identity, we have (c, a, ab) + (a, c, ab) + (a, a, cb) = 0. Using (1), we get (ab, a, c) + (a, c, ab) + (a, a, cb) = 0. (15) (a, ab, c) = (a, a, cb) = (a, a, bc). From (T), we have (aa, b, a) − (a, ab, a) + (a, a, ba) = a(a, b, a) + (a, a, b)a. Using (1) and (3), we M get a(a, b, a)−(a, ab, a)+(a, a, ba) = a(a, b, a)+(a, a, b)a. This implies −(a, ab, a)+(a, a, ba) = (a, a, b)a. Applying (7) and (8), we obtain 3(a, a, ab) = (a, a, b)a. Therefore, by (3) and (13), we have Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 Using (5), we obtain −(c, ab, a) + (a, a, cb) = 0. Using (1) and (8), we have (a, a, b)a = 0. (16) From (7) and (16), we get(a, b, a)a = 0. Linearization of this implies (c, b, a)a + (a, b, c)a + (a, b, a)c + (a, b, c)c + (c, b, a)c + (c, b, c)a = 0. Since this is nonhomogeneous identity, we have (c, b, a)a + (a, b, c)a + (a, b, a)c. Using (1), we have 2(a, b, c)a + (a, b, a)c = 0. Using (7) and characteristic 6 = 2, we get (17) (a, b, c)a = (a, a, b)c. 6 3. RESULTS Lemma 3.1. The identity (a, cd, b) = (c, ab, d) holds in R. an us cr ip t Proof. From (10), we have (a, aa, c) = 0. Linearization of this implies (d, aa, c) + (a, da, c) + (a, ad, c) + (a, dd, c) + (d, ad, c) + (d, da, c) = 0. Since this is nonhomogeneous identity, we have (d, aa, c) + (a, da, c) + (a, ad, c) = 0. Using (4), we get (d, aa, c) + 2(a, da, c) = 0. Applying (1), (c, aa, d) − (a, cd, a) = 0. That is, we have (a, cd, a) = (c, aa, d). Linearization of this implies (a, cd, b) + (b, cd, a) = (c, ab + ba, d). Using (1) and (4), we get 2(a, cd, b) = 2(c, ab, d). Since M characteristic 6 = 2, we have (a, cd, b) = (c, ab, d). This finishes the proof of Lemma 3.1. Lemma 3.2. The identity a(c, b, d) + (c, b, d)a = b(c, a, d) + (c, a, d)b holds in R. Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 (4) and (15), we obtain (c, aa, d) = −2(a, da, c) = −2(a, ad, c) = −2(a, a, cd). Using (7), we get Proof. From (T), we have (c, ab, c) = (ca, b, c) + (c, a, bc) − c(a, b, c) − (c, a, b)c, (18) (c, bc, a) = (cb, c, a) + (c, b, ca) − c(b, c, a) − (c, b, c)a, (19) (c, ca, b) = (cc, a, b) + (c, c, ab) − c(c, a, b) − (c, c, a)b. (20) Adding (18), (19) and (20) and then using (5), we get (c, ab, c)+(c, bc, a)+(c, ca, b) = (ca, b, c)+ (c, a, bc)−(c, a, b)c+(cb, c, a)+(c, b, ca)−(c, b, c)a+(cc, a, b)+(c, c, ab)−(c, c, a)b. Using (4), we get (c, ab, c) + (c, cb, a) + (c, ca, b) = (ca, b, c) + (c, a, bc) − (c, a, b)c + (cb, c, a) + (c, b, ca) − (c, b, c)a + (cc, a, b) + (c, c, ab) − (c, c, a)b. Using (15) and (8), we obtain (c, ab, c) + 2(c, c, ab) = 7 (ca, b, c)+(c, a, bc)−(c, a, b)c+(cb, c, a)+(c, b, ca)−(c, b, c)a+(cc, a, b)+(c, c, ab)−(c, c, a)b. Using (7), we derive (ca, b, c) + (c, a, bc)−(c, a, b)c+(cb, c, a)+(c, b, ca)−(c, b, c)a+(cc, a, b)+ (c, c, ab)−(c, c, a)b = 0. Using (1) and arranging the terms, we get (c, b, ca)+(c, b, ca)+(c, a, bc)+ an us cr ip t (b, a, cc)+(a, c, cb)+(c, c, ab) = (c, a, b)c+(c, b, c)a+(c, c, a)b. Using (6), we obtain c(c, b, a)+ c(c, b, a)+b(c, a, c)+c(b, a, c)+c(a, c, b)+a(c, c, b) = (c, a, b)c+(c, b, c)a+(c, c, a)b. Arranging the terms, we have c((c, b, a) + (b, a, c) + (a, c, b)) + c(c, b, a) + b(c, a, c) + a(c, c, b) = (c, a, b)c + (c, c, a)b. Using (14) and (17), we derive a(c, c, b) + b(c, a, c) + a(c, c, b) = (c, c, a)b + (c, b, c)a + (c, c, a)b. This implies 2a(c, c, b) + b(c, a, c) = 2(c, c, a)b + (c, b, c)a. Applying (7), we have −a(c, b, c) + b(c, a, c) = −(c, a, c)b + (c, b, c)a. Rearranging terms, we get a(c, b, c) + (c, b, c)a = M b(c, a, c) + (c, a, c)b. Linearization of this implies a(c, b, d) + a(d, b, c) + (c, b, d)a + (d, b, c)a = b(c, a, d)+b(d, a, c)+(c, a, d)b+(d, a, c)b. Using (1), we get 2a(c, b, d)+2(c, b, d)a = 2b(c, a, d)+ 2(c, a, d)b. Since characteristic 6 = 2, we have a(c, b, d) + (c, b, d)a = b(c, a, d) + (c, a, d)b. This Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 (c, b, c)a+(c, c, a)b. Using (5), we obtain c(c, b, a)+b(c, a, c)+a(c, c, b) = (c, a, b)c+(c, b, c)a+ completes the proof of Lemma 3.2. Lemma 3.3. The identity (a, b, cd) = (b, a, dc) holds in R. Proof. From (T), we have (a2 , b, d) − (a, ab, d) + (a, a, bd) = a(a, b, d) + (a, a, b)d. Using (15), we get (a2 , b, d) = a(a, b, d) + (a, a, b)d. Using (17), we obtain (a2 , b, d) = a(a, b, d) + (a, b, d)a. (21) 8 Applying (14) and (17), we have (a2 , b, d) = d(a, a, b) + (a, a, b)d. Using (7), we get 2(a2 , b, d) = −d(a, b, a) − (a, b, a)d. Linearization of this implies 2(ac, b, d) + 2(ca, b, d) = −d(a, b, c) − an us cr ip t d(c, b, a) − (a, b, c)d − (c, b, a)d. Using (1), we obtain 2(ac, b, d) + 2(ca, b, d) = −2d(a, b, c) − 2(a, b, c)d. Since characteristic6 = 2, we have (ac, b, d) + (ca, b, d) + d(a, b, c) + (a, b, c)d = 0. Thus using (1), we have From (21), we have (a2 , d, b) = a(a, d, b) + (a, d, b)a. (22) (23) M Adding (21) and (23), we get (a2 , b, d) + (a2 , d, b) = a(a, b, d) + (a, b, d)a + a(a, d, b) + (a, d, b)a. First applying (1) then applying (5), we get (d, a2 , b) = a(d, a, b) + (d, a, b)a. Linearization of this implies (d, ac + ca, b) = a(d, c, b) + c(d, a, b) + (d, a, b)c + (d, c, b)a. Using (4) and Lemma 3.2, Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 d(a, b, c) + (a, b, c)d + (d, b, ac) + (d, b, ca) = 0. we obtain 2(d, ac, b) = 2a(d, c, b) + 2(d, c, b)a. Characteristic 6 = 2 yields (24) (d, ac, b) = a(d, c, b) + (d, c, b)a. Using cyclic law (5), we have (a, b, cd) − (b, a, dc) = −(b, cd, a) − (cd, a, b) − (b, a, dc) = −((b, cd, a) + (cd, a, b) + (b, a, dc)). Using (1), we derive (a, b, cd) − (b, a, dc) = −((b, cd, a) + (b, a, cd) + (b, a, dc)). Applying (22), we get (a, b, cd) − (b, a, dc) = −((b, cd, a) − b(c, a, d) − (c, a, d)b) = −(b, cd, a) + b(c, a, d) + (c, a, d)b. Using Lemma 3.1, we have (a, b, cd)−(b, a, dc) = 9 −(c, ba, d) + b(c, a, d) + (c, a, d)b. Using (24), we get (a, b, cd) − (b, a, dc) = 0. This implies (a, b, cd) = (b, a, dc). This completes the proof of Lemma 3.3. an us cr ip t Lemma 3.4. The following identities hold in R: (ii) (a, [b, c]d, e) = 0. Proof. M (i) From (24), we have (a, b[c, d], e) = b(a, [c, d], e) + (a, [c, d], e)b. Using (4), we obtain (a, b[c, d], e) = 0. This completes the proof of Lemma 3.4(i). Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 (i) (a, b[c, d], e) = 0. (ii) From Lemma 3.4(i) and (4), it follows that (a, [b, c]d, e) = 0. This completes the proof of Lemma 3.4(ii). Lemma 3.5. The identity (a, (b, c, d), e) = 0 holds in R. Proof. (J) is the semi-Jacobi identity. (J) holds in any nonassociative ring [bd, c] = b[d, c] + [b, c]d + (b, d, c) + (c, b, d) − (b, c, d). 10 (J) Applying (5) and (1) to (J), we get 2(b, c, d) = b[d, c] + [b, c]d − [bd, c]. Therefore, we have an us cr ip t (a, 2(b, c, d), e) = (a, b[d, c], e) + (a, [b, c]d, e) − (a, [bd, c], e). Using Lemma 3.4 and (4), we obtain (a, 2(b, c, d), e) = 0. Since characteristic6 = 2, we have (a, (b, c, d), e) = 0. This completes Lemma 3.6. The following identities hold in R: (ii) (RR, RR, R) = (R, RR, RR) = 0. (iii) (RR, R, RR) = 0. M (i) (R, R(RR), R) = (R, (RR)R, R) = 0. Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 the proof of Lemma 3.5. Proof. (i) From (1) and (11), we have (aa, de, a) = 0. Using Lemma 3.1, we get (d, (aa)a, e) = 0. Linearization of this implies (d, (aa)b + (aa)c + (bb)c + (bb)a + (cc)a + (cc)b + (ab)a + (ba)a + (bc)b + (cb)b + (ca)c + (ac)c + (ab)b + (ba)b + (bc)c + (cb)c + (ca)a + (ac)a + (ab)c + (bc)a + (ca)b + (ba)c + (ac)b + (cb)a, e) = 0. Since this is a nonhomogeneous identity, we have (d, (ab)c+(bc)a+(ca)b+(ba)c+(ac)b+(cb)a, e) = 0. Therefore, using Lemma 3.4, Lemma 3.5 11 and (4), we get 6(d, (ab)c, e) = 0. Since characteristic 6 = 2, 3, we have (d, (ab)c, e) = 0. Using an us cr ip t Lemma 3.5, we get (d, a(bc), e) = 0. This finishes the proof of Lemma 3.6(i). (ii) Using Lemma 3.1, we have (ab, cd, e) = (c, (ab)e, d). Using Lemma 3.6(i), we get (ab, cd, e) = (iii) From (5), we have (c, ab, de) + (ab, de, c) + (de, c, ab) = 0. Using Lemma 3.6(ii), we get (de, c, ab) = 0. This finishes the proof of Lemma 3.6(iii). M Lemma 3.7. The following identities hold in R: (i) (a, bc, d)e = 0. Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 0. Similarly using Lemma 3.1 and Lemma 3.6(i), we obtain (a, bc, de) = 0. (ii) a(b, cd, e) = 0. (iii) (a, b, c(de)) = c(a, b, de). Proof. (i) From (T), we have (a(bc), e, a) − (a, (bc)e, a) + (a, bc, ea) = a(bc, e, a) + (a, bc, e)a. Using Lemma 3.6, (1) and (3), we get a(bc, e, a) = a(bc, e, a) + (a, bc, e)a. That is, we have (a, bc, e)a = 0. Using (17), we obtain (a, a, bc)e = 0. (7) and characteristic 6 = 2 together imply (a, bc, a)e = 0. 12 Linearization of this implies (a, bc, d)e + (d, bc, a)e = 0. Using (1), we get 2(a, bc, d)e = 0. Since an us cr ip t characteristic 6 = 2, we have (a, bc, d)e = 0. This finishes the proof of Lemma 3.7(i). (ii) From (24), we have (b, a(cd), e) = a(b, cd, e)+(b, cd, e)a. Using Lemma 3.6(i) and Lemma 3.7 (iii) Again from (T), we have (c(de), b, a) − (c, (de)b, a) + (c, de, ba) = c(de, b, a) + (c, de, b)a. Using Lemma 3.6 and Lemma 3.7 (i), we get (c(de), b, a) = c(de, b, a). Using (1), we obtain M (a, b, c(de)) = c(a, b, de). Lemma 3.8. The identity (a, b, c)d2 = 0 holds in R. Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 (i), we get a(b, cd, e) = 0. This finishes the proof of Lemma 3.7(ii). Proof. Using (5), we have(d2 , a, b) + (a, b, d2 ) + (b, d2 , a) = 0. This implies c(d2 , a, b) + c(a, b, d2 ) + c(b, d2 , a) = 0. Using Lemma 3.7(ii), we getc(d2 , a, b) + c(a, b, d2 ) = 0. Using (1) and Lemma 3.3, we obtain 2c(a, b, d2 ) = 0. Since characteristic 6 = 2, we have c(a, b, d2 ) = 0. (25) Using Lemma 3.7(iii), we obtain (a, b, cd2 ) = 0. (26) 13 Now from (T), we have (ab, c, d2 ) − (a, bc, d2 ) + (a, b, cd2 ) = a(b, c, d2 ) + (a, b, c)d2 . Using an us cr ip t Lemma 3.6, (25) and (26), we get(a, b, c)d2 = 0. This finishes the proof of Lemma 3.8. Lemma 3.9. (R, R, R) is an ideal Proof. Using (3) and (14), we have 2(a, b, ac) = 2a(a, b, c) = 2c(a, a, b). Applying (7), we obtain c(d, b, a). Using (1), we have 2(a, b, dc) + 2(d, b, ac) = −2c(a, b, d). Since characteristic 6 = 2, we get (27) M c(a, b, d) = −(a, b, dc) − (d, b, ac). Thus, (R, R, R) is a left ideal. Therefore, from (T) it follows that (R, R, R) is a two-sided ideal. Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 2(a, b, ac) = −c(a, b, a). Linearization of this implies 2(a, b, dc) + 2(d, b, ac) = −c(a, b, d) − Lemma 3.10. The identity (a, b, (c, d, e)) = 0 holds in R. Proof. Linearization of (26) implies (a, b, c(de + ed)) = 0. (28) Applying Lemma 3.3 and (26), we obtain (a, b, (dd)c) = 0. (29) 14 Linearization of (29) implies (a, b, (de + ed)c) = 0. (30) an us cr ip t Now from (5), we have (a, b, (cd)e+e(cd))+(b, (cd)e+e(cd), a)+((cd)e+e(cd), a, b) = 0. Using Lemma 3.6 and (1), we get (a, b, (cd)e + e(cd)) + (b, a, (cd)e + e(cd)) = 0. Using Lemma 3.3, we obtain (a, b, (cd)e + e(cd)) + (a, b, e(cd) + (cd)e) = 0. This implies 2(a, b, (cd)e + e(cd)) = 0. (a, b, (cd)e + e(cd)) = 0. (31) Now (a, b, (c, d, e)) = (a, b, (cd)e − c(de)). Using (31), we have (a, b, (c, d, e)) = (a, b, −e(cd) + M (de)c). Using (28) and (30), we get (a, b, (c, d, e)) = (a, b, e(dc) − (ed)c) = −(a, b, (e, d, c)). Using (1), we obtain (a, b, (c, d, e)) = −(a, b, (c, d, e)). This implies 2(a, b, (c, d, e)) = 0. Since characteristic 6 = 2, we have (a, b, (c, d, e)) = 0. This completes the proof of Lemma 3.10. Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 Since characteristic 6 = 2, we have Lemma 3.11. The identity (a, b, (cd)e) = 0 holds in R. Proof. Using (30), we have (a, b, (cd)e) = −(a, b, (dc)e). Using Lemma 3.10, we have (a, b, (cd)e) = −(a, b, d(ce)). Using (28), we obtain (a, b, (cd)e) = (a, b, d(ec)). Using (31), we get (32) (a, b, (cd)e) = −(a, b, (ec)d). 15 Applying (32) thrice, we obtain (a, b, (cd)e) = −(a, b, (ec)d) = (a, b, (de)c) = −(a, b, (cd)e). Therefore, we have 2(a, b, (cd)e) = 0. Since characteristic 6 = 2, we have (a, b, (cd)e) = 0. This an us cr ip t completes the proof of Lemma 3.11. Theorem 3.12. Products of degree five are associative. we obtain (a, b, c)d = (ab, c, d) − (a, bc, d) + (a, b, cd) + (b, c, da) + (d, c, ba). (33) M From (27) and (33), it follows that to prove that products of degree five are associative it is sufficient to prove the associators (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2), (1, 2, 2) are zero where the number represents the degree of the terms. We have already proved these associators are zero in Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 Proof. From (T), we have (a, b, c)d = (ab, c, d) − (a, bc, d) + (a, b, cd) − a(b, c, d). Using (27), Lemma 3.6, Lemma 3.10, Lemma 3.11 and (1). This completes the proof of Theorem 3.12. Corollary 3.13. Let R be semiprime. Then R is associative. Proof. In Lemma 3.9, we have already proved that (R, R, R) is an ideal. Now since any product of degree five (or more) containing an associator is zero, we have I 2 = 0. Since R is semiprime, I = 0. This shows that R is associative. 16 REFERENCES 1. Anderson, C. T., Outcalt, D. L. (1968). On simple antiflexible rings. J. Algebra 10:310–320. 2. Celik, H. A. (1971). Commutative associative rings and antiflexible rings. Pacific J. Math. an us cr ip t 38:351–358. 3. Celik, H. A. (1972). On primitive and Prime antiflexible rings. J. Algebra 20:428–440. 4. Jacobs, D. P., Muddana, S. V., Offutt, A. J., Prabhu, K. (0000). Albert 1.0 User’s Guide, 5. Kleinfeld, E. (1955). Primitive alternative rings and semi-simplicity. Amr. J. Math. 77:725–730. 6. Kosier, F. (1962). On a class of nonflexible algebras. Trans. Amer. Math. Soc. 102:299–318. M 7. Rodabaugh, D. (1965). A generalization of flexible law. Trans. Amer. Math. Soc.114:468–487. Ac ce pt ed Downloaded by [UAE University] at 11:10 25 October 2017 Department of Computer Science, Clemson University. 17

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