J. Comp. & Math. Sci. Vol.2 (4), 613-616 (2011)
Semi-Prime (-1,1) Rings K. SUBHASHINI 1, K. SUVARNA2 , M. RAVI KUMAR1 and K. MALLIKARJUNUDU1 1
Department of Mathematics, G.Pulla Reddy Engineering College (Autonomus), Kurnool - 518002, Andhra Pradesh. India 2 Department of Mathematics, Sri Krishnadevaraya University, Anantapur-515003, Andhra Pradesh. India ABSTRACT In this paper, we prove the result that every associater commutes with every element of a (-1,1) ring R of characteristic ≠ 2,3. From this we derive an identity ((R,R), (R,R,R)) = 0. Using this we establish the property that in a semi - prime (-1,1) ring R of characteristic ≠ 2 ,3 the double commutator ((R,R),R) is in the nucleus. With this we show that R is strongly (-1,1) ring. Keywords: Associator, commutator, (-1,1) ring, characteristic, semi-prime, nucleus, strongly (-1,1) ring.
1. INTRODUCTION A (-1,1) ring R is a non-associative ring in which the following identities hold: (a,b,c) + (a,c,b) = 0
(1)
(a,b,c) + (b,c,a) + (c,a,b) = 0. ∀ a,b,c ∈ R.
(2)
Where the associater (a,b,c) = ab.c - a.bc. And the commutator (a,b) = ab - ba. The ring R is called a semi-prime, if R has no non zero ideal squaring to zero. If there exists a positive integer n such that na = 0 for every element a in R, the smallest such positive integer is known as the characteristic of R. A a non-associative ring R is said to be strongly (-1,1) ring, if it satisfies the identities (1) and ((a,b),c) = 0.
Kleinfeld2 proved that a semi-prime right alternative ring with characteristic ≠ 2,3 under the assumption of ((a,b),a) = 0 is strongly (-1,1) ring. Paul [5] proved that a semi-prime (-1,1) ring R of characteristic ≠ 2,3 is associative with the additional assumption of commutators are in the left nucleus that is ((R,R),a,b) = 0. Hentzel3 proved an identity ((R,R), (R,R,R)) = 0 if R is a semi-prime (-1,1) ring. But we will prove that this identity holds good in any (-1,1) ring of characteristic ≠ 2,3. Using this we establish the property that in a semi-prime (-1,1) ring R of characteristic ≠ 2,3 the double commutator ((R,R),R) is in the nucleus. Without any additional assumption here we prove a semi-prime (-1,1) ring R of characteristic ≠ 2,3 is strongly (-1,1) ring.
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K. Subhashini, et al., J. Comp. & Math. Sci. Vol.2 (4), 613-616 (2011)
2. BASIC IDENTITIES
The following identity L in3 holds in R:
Throughout this paper, we assume R to be a (-1,1) ring of characteristic ≠ 2,3 . From identity (4) of Maneri1 , we have
(a,(b,b,c)) - 3(b,(a,c,b)) = 0
(10)
From (9) and (10) it becomes (a,(b,b,c)) = 0
(11)
(b,(a,a,b)) = 0. Using (1) this can be written as (b,(a,b,a)) = 0.
(3)
(4)
The following identity holds in any ring known as Teichmuller identity (ab,c,d) – (a,bc.d) + (a,b,cd) = a(b,c,d) + (a,b,c)d
(5)
3. PRELIMINARIES Lemma 1: If R is a (-1,1) ring R of characteristic ≠ 2,3, then ((a,b,c),d) = 0. Proof: Replace a by a+c in identity (3) . (b,(a+c,a+c,b)) = 0. ⇒ (b,(a,a,b)) + (b,(a,c,b)) + (b, (c,c,b)) + (b,(c,a,b)) = 0. From (3), we have (b,(a,c,b)) = - (b, (c,a,b)) (6) Again replace a by a+c in (4) and using (4) , we get (b,(a,b,c)) = - (b,(c,b,a)) (7) From equations (1), (6), (7) and again using (1) (b,(b,c,a) = - (b,(b,a,c)) = (b,(c,a,b)) = (b,(a,c,b)) = (b,(a,b,c)) (8) Commuting equation (2) with b, we have (b,(a,b,c) + (b,(b,c,a)) + (b,(c,a,b)) = 0. From (8) this equation becomes 3(b,(a,b,c)) = 0. Since of characteristic ≠ 3, (b,(a,b,c)) = 0 (9)
Consider an ideal A of R which is generated by {(a,b,c)/ a,b,c ∈ R}. From Lemma 4 in4 We have A ⊆ {(b,b,r) / b ∈ R}
(12)
Using (11) and (12) we get (R,(a,b,c)) = 0 . ⇒ (d,(a,b,c)) = 0 . ∀ d ∈ R. And hence ((a,b,c),d) = 0 ∀ a,b,c,d ∈ R (13) Lemma 2: If R is a (-1,1) ring R of characteristic ≠ 2,3 then ((R,R), (R,R,R)) = 0. Proof: Replace d by a commutator (R,R) in (13) , we have ((a,b,c),(R,R)) = 0. This gives ((R,R,R),(R,R)) = 0 ∀ a,b,c in R And hence ((R,R),(R,R,R)) = 0. (14) From Lemma 7 of Hentzel3 (a,a,(R,R)) = 0 holds good in semi-prime (-1,1) rings. This implies that (a,a,(b,c)) = 0 .
(15)
Linearizing (15) gives us (a,d,(b,c)) + (d,a,(b,c)) = 0
(16)
From (1) and (2) it follows that (a,x,x1) = 0 for x, x1 ∈ (R,R)
(17)
Let S = {(x,(b,c))/ ∀ a,b,c ∈ R}.be an ideal generated by double commutators
Journal of Computer and Mathematical Sciences Vol. 2, Issue 4, 31 August, 2011 Pages (581-692)
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K. Subhashini, et al., J. Comp. & Math. Sci. Vol.2 (4), 613-616 (2011)
The nucleus N of R = {x ∈R/ (a,b,x) = (a,x,b) = (x,a,b) = 0 ∀ a,b ∈R}. Lemma 3: If R is a semi-prime (-1,1) ring R of characteristic ≠ 2,3 then S is contained in the nucleus N. Proof: The following identity O in Hentzel3 holds in any (-1,1) ring R. ((a,b),c,d) – ((c,d),a,b) – (a,(b,c,d)) + (b,(a,c,d)) = 0 (18) Apply (13) to the equation (18), we get ((a,b),c,d) = ((c,d),a,b) Replace b by (R,R) ((a,(R,R)),c,d) = ((c,d),a,(R,R)) = - (a,(c,d),(R,R)) from (16) = 0 from (17) Thus (a,(R,R)) is in the nucleus N. Hence S ⊆ N. Throrem: If R is a semi-prime (-1,1) ring R of characteristic ≠ 2,3, then R is a strongly (-1,1) ring. Proof: Let A = (R,R,R) + (R,R,R)R be the associator ideal of R. From Lemma 2, S ⊆ N. Implies that (S,R,R) = (R,S,R) = (R,R,S) = 0. For any s in S and a,b,c in R, by (5) (sa,b,c) – (s,ab,c) + (s,a,bc) – s(a,b,c) – (s,a,b)c = 0.Since S is an ideal then the terms (sa,b,c),(s,ab,c),(s,a,bc),(s,a,b) ∈ (S,R,R) = 0.Therefore s.(a,b,c) = 0. This implies that S((R,R,R) + (R,R,R)R) =0.Thus SA = 0. Now R(S+RS) ⊆ RS + R(RS) ⊆ RS + (RR)S – (R,R,S) ⊆ S+RS. Also (S+RS)R ⊂ SR + (RS)R ⊆ S+R(SR) ⊆ S+RS. Thus if <S> is the ideal generated
by S then S+RS = <S> . Therefore <S>A = (S+RS)A ⊆ SA + (RS)S ⊆ R(SA) ⊆ (0). More over S ⊆ A and A is an ideal implies that <S> ⊆ A. Thus <S><S> = 0. For any s in S and a,b,c in R ,it is easy to prove that AS = 0. Now (S+SR)R ⊆ SR + (SR)R ⊆ SR + S(RR) + (S,R,R) ⊆ S+SR. Also R(R+SR) ⊆ RS + R(SR) ⊆ S+(RS)R ⊆ S+SR. Thus if <S> is the ideal generated by S then S+SR = <S>. Therefore A<S> = A(S+SR) ⊆ AS + A(SR) ⊆ (AS)R ⊆ (0). More over S ⊆ A and A is an ideal implies that <S> ⊆ A. Thus <S><S> = 0. Since R is semi-prime and <S> is an ideal of R, this implies that <S> = 0. Hence S = 0. Therefore (a,(b,c)) = 0 and thus R is a strongly (-1,1) ring. 4. EXAMPLE Take the example of Paul6
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Journal of Computer and Mathematical Sciences Vol. 2, Issue 4, 31 August, 2011 Pages (581-692)
616
K. Subhashini, et al., J. Comp. & Math. Sci. Vol.2 (4), 613-616 (2011)
n this example it was shown that the commutators do not lie in the left nucleus. Also it was said that the ring R is a (-1,1) ring. But the identities (1) and (2) are not satisfied when a,b,c are replaced by a,b,e of this example. (a,b,e) + (a,e,b) = (ab.e –a.be) + (ae.b – a.eb) = (a.e – a.0) + (h.b – a.0) = h ≠ 0. Also (a,b,e) + (b,e,a) + (e,a,b) = h ≠ 0. Hence R given in this example is not a (-1,1) ring.
good in any (-1,1) ring of characteristic ≠ 2,3. To show (13) it is enough to check ((a,b,e),e) = 0 and ((a,b,a),e) = 0. Take ((a,b,e),e) = ((ab.e – a.be),e) = ((a.e – a.e),e) = (0,e) = 0. And ((a,b,a),e) = ((ab.a – a.ba),e) = ((a.a – a.0),e) = (c,e) = (ce – ec) = 0. Hence ((R,R,R),R) = 0 .
We modify Paul’s example6 in which the commutators do not lie in the left nucleus , every associator commutes with every element of R and R is a (-1,1) ring. Let R be defined by the following multiplication table together with all finite sums of e,a,b,c,d,h. R is a non associative and non commutative ring. e
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0
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Now, ((a,b),a,b) = (ab –ba,a,b) = (a,a,b) – (0,a,b) = aa.b – a.ab = cb –aa = 0-c = c ≠ 0. Thus commutators do not lie in the left nucleus. Paul5 proved an identity (13) by assuming the commutators are in the left nucleus of a (-1,1) ring. But (13) holds
Finally R is a (-1,1) ring since (a,b,e) + (a,e,b) = (ab.e –a.be) + (ae.b – a.eb) = (a.e – a.e) + (h.b – a.0) = 0. Also (a,b,e) + (b,e,a) + (e,a,b) = 0 , ∀ elements in R. REFERENCES 1. C. Maneri, Simple (-1,1) rings with idempotent, Proc. Amer. Math. Soc. 14, 110-117 (1963). 2. E. Kleinfeld, A generalization of strongly (-1,1) rings, J. Algebra, 218225 (1988). 3. I. R. Hentzel, Nil semi simple (-1.1) rings, J. Algebra 22, 442-450 (1972). 4. I. R. Hentzel, The Characterization of (-1,1) Rings, J, Albebra 30, 236-258 (1974). 5. Y. Paul, A note on (-1,1) rings, Jnanabha, Vol. 11, 107-109 (1981). 6. Y. Paul, A note on radicals in Lattice ordered (-1,1) rings with commutators in the left nucleus Procedings of the Fourth Ramanujan Symposium on Algebra and its Applications University of Madras, Madras1-3, 121-126, FEB.(1995).
Journal of Computer and Mathematical Sciences Vol. 2, Issue 4, 31 August, 2011 Pages (581-692)