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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.
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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
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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
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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
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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.
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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
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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
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(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.
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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.
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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).
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From (5) and (1), we get the following identity
(6)
(7)
Using (4), (7) and characteristic 6 = 2, we obtain
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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.
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(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)
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(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
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From (5), we have (aa, a, b) + (a, b, aa) + (b, aa, a) = 0. Therefore, (1) and (10) together imply
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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
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characteristic 6 = 2, we have
(14)
a(a, b, c) = c(a, a, b).
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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
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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
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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.
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3. RESULTS
Lemma 3.1. The identity (a, cd, b) = (c, ab, d) holds in R.
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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
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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.
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(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) =
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(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)+
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(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 =
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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
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(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)
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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) −
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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)
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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,
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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) =
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−(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.
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Lemma 3.4. The following identities hold in R:
(ii) (a, [b, c]d, e) = 0.
Proof.
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(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).
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(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).
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(J)
Applying (5) and (1) to (J), we get 2(b, c, d) = b[d, c] + [b, c]d − [bd, c]. Therefore, we have
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(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.
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(i) (R, R(RR), R) = (R, (RR)R, R) = 0.
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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
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and (4), we get 6(d, (ab)c, e) = 0. Since characteristic 6 = 2, 3, we have (d, (ab)c, e) = 0. Using
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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).
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Lemma 3.7. The following identities hold in R:
(i) (a, bc, d)e = 0.
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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.
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Linearization of this implies (a, bc, d)e + (d, bc, a)e = 0. Using (1), we get 2(a, bc, d)e = 0. Since
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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
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(a, b, c(de)) = c(a, b, de).
Lemma 3.8. The identity (a, b, c)d2 = 0 holds in R.
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(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)
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Now from (T), we have (ab, c, d2 ) − (a, bc, d2 ) + (a, b, cd2 ) = a(b, c, d2 ) + (a, b, c)d2 . Using
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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)
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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.
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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)
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Linearization of (29) implies
(a, b, (de + ed)c) = 0.
(30)
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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) +
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(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.
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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).
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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
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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)
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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
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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.
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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.
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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.
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7. Rodabaugh, D. (1965). A generalization of flexible law. Trans. Amer. Math. Soc.114:468–487.
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Department of Computer Science, Clemson University.
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