Download A NOTE ON SIMPLE (-1,1) RINGS

Int. J. of Mathematical Sciences and Applications,
Vol. 1, No. 3, September 2011
Copyright  Mind Reader Publications
www.journalshub.com
A NOTE ON SIMPLE (-1,1) RINGS
K.SUBHASHINI
Department of Basic Sciences, G.Pulla Reddy Engineering College (Autoomus)
Nandyala Road, Kurnool-518002, Andhra Pradesh.
[email protected]
ABSTRACT
In this paper I show that a simple non -associative and not commutative (-1,1) ring of characteristic ≠ 2,3
is alternative.
Key words: Associator, Commutator, Characteristic, Alternative ringc Simple,Ideal,idempotent
AMS subject classification: 2010 MSC 17AXX
1.INTRODUCTION:
A (-1,1) ring R is a non-associative ring in which the following identities hold:
(x,y,z) + (x,z,y) = 0
………..(1)
(x,y,z) + (y,z,x) + (z,x,y) = 0 ………. (2)
for all z,y,z in R. The associator (x,y,z) is defined by (x,y,z) = xy.z – x.yz and the commutator (x,y) = xy-yx. The ring R is
called alternative, if R satisfies identity (1)
and
(x,x,y) = 0
………...(3)
If there exists a positive integer n such that na = 0 for every element a of the ring R, the smallest such positive integer is called
the characteristic of R. Thedy [3] proved that a simple non-associative ring with ((a,b,c),d) = 0 is either associative or
commutative. Maneri[1] has shown that a simple A (-1,1) ring of characteristic ≠ 2,3with idempotent e ≠ 0,1 is associative.
Here without using any additional conditions, we prove that a simple non –associative and not commutative (-1,1) ring R of
characteristic ≠ 2,3 is alternative.
Throughout this paper, we assume R to be a (-1,1) ring of characteristic ≠ 2,3.
2. PRELIMINARIES: −
In R we have the following identity (identity (4) in [1]):
(y,(x,y,x)) = 0.
……………..(4)
From (1) this implies that
(y,(x,x,y)) = 0
…………… (5)
By linearizing the identies (4) and (5) , we have
(y,(x,y,z)) = - (y,(z,y,x)),
…………….(6)
(y,(x,z,y)) = - (y,(z,x,y))
…………….(7)
From equations (1),(6),(7) and again using (1)
(y,(y,z,x)) = - (y,(z,x,y)) = - (y,(x,z,y)) = (y,(x,y,z))
…………… (8)
Commuting (2) with y,we have (y,((x,y,z) + (y,z,x) + (z,x,y))) = 0.
From (8) this equation becomes 3(y,(x,y,z)) = 0. Since R is of char ≠ 3
(y,(x,y,z)) = 0.
……………(9)
1483
K.SUBHASHINI
The following identity L in [2] holds in R: (x,(y,y,z))- 3(y,(x,z,y)) = 0.
From (8) and (9) this equation becomes (x,(y,y,z)) = 0. Thus
(R,(y,y,z)) = 0.
…………… (10)
Linearise equation (10), we obtain
(w,(x,y,z)) = - (w,(y,x,z)
……………(11)
Applying equations (1) and (11) repeatedly, we get
(w,(x,y,z)) = - (w,(y,x,z) = (w,(y,z,x)) = - (w,(z,y,x) = (w,(z,x,y)) …(12)
Commute equation (2) with w and apply (12).
Then 3(w,(x,y,z)). Since char ≠ 3 ,
(w,(x,y,z)) = 0 .
……………(13)
We need the following identity which holds in any ring known as Teichmuller
Identity
(wx,y,z) – (w,xy,z) + (w,x,yz) = w(x,y,z) + (w,x,y)z. Commute this equation
with any element of r of R and apply (13). Now (r,(x,y,z)w) = - (r,(w,x,y)z). If we put x = y in this equation, then it reduces
to (r,(y,y,z)w) = 0.
…………….(14)
3. MAIN THEOREM:
To prove the main theorem we need the following Lemma.
Lemma:
If R is a (-1,1) ring of char ≠ 2,3 , then T = {t € R/(t,R)= 0 = (tR,R)} is an ideal of R.
Proof:
By substituting x = t in (13), we get ((t,y,z),w) = 0. From this equation it follows
and so T is a right ideal. However yt = ty. Thus T is a two sided ideal of R.
that (ty,z,w) = 0. Thus ty € T
Theorem: A simple not associative and not commutative (-1,1) ring R of char ≠ 2,3 is alternative.
Proof: From equations (13) and (14) all (y,y,z) are in the ideal T. Since R is simple and T is an ideal of R,either T = R
or T = 0.
If T = R, then Ris commutative. Since R is not commutative, we must have T = 0.
Then (y,y,z) = 0.
This implies R is left alternative. And hence from (1) R is alternative.
4 . EXAMPLE:
Take the example of Maneri [1].
Let F be a field and A an algebra with basis elements e,u,v,w,z . We define multiplication of basis elements as follows :
e.e = e = e, eu = v, ev = 0,ue = u, ew = w-z, ez = ze = z and all other products are zero. R be a ring defined by the above
multiplication table with all finite sums of e,u,v,w,z . It is easy to check that R is a (-1,1) ring Since (e,u) = eu-ue = v-u ≠
0. So R is not commutative. Similarlly (e,e,u) ≠ 0. Identity (13) was not proved by Maneri but it holds good in R., since
(e,(e,e,w)) = 0.
BIBLIOGRAPHY:
(1)
(2)
(3)
C. Maneri, Simple (-1,1) rings with idempotent, Proc.Amer.Math.Soc.14,110-117 (1963)
I.R Hentzel, The Characterization of (-1,1) rings ,J.Algebra 30 (1974),236-258
A.Thedy, On Rings satisfying ((a,b,c),d) = 0, , Proc.Amer.Math.Soc. 29 (1971),213
1484