International Journal of Scientific & Engineering Research, Volume 5, Issue 3, March-2014

ISSN 2229-5518

1286

Simple Antiflexible Rings

M. Hema Prasad, Dr. D. Bharathi, Dr. M. Munirathnam

Abstract— Let R be an antiflexible ring of characteristic ≠ 2,3. In this paper, first we prove that associator is in the center. Using this, we prove that a simple not associative and not commutative antiflexible ring R is right alternative

Index TermsAssociator, antiflexible ring, characteristic, commutator, Lie ring, right alternative

1. INTRODUCTION

Antiflexible albebras were introduced by Kosier [1] , and a subclass of antiflexible rings was studied earlier by Kleinfeld [2]. Thedy [3] proved that simple rings satisfying (a,(b,c,d)) = 0 are either associative or commutative. In [4], K. Survana established the result (R,(R,R,R)) = 0 in (-1,1) ring of characteristic ≠ 2,3. In [5], K. Subhashini proved that a simple not associative and not commutative (1,0) ring R of characteristic ≠ 2,3 is alternative by using (R,(R,R, R)) = 0. In this paper, we prove that a simple not associative and not commutative antiflexible ring R is right alternative.

2. PRELIMINARIES

The associator (x, y, z) and commutator (x, y) are defined by (x, y, z) = (x y) z – x (y z)and (x, y) = x y – y x for all x, y, z in R respectively. The nucleus N of a ring is defined as N = { n ϵ R / (n, R, R) = (R, n, R) = (R, R, n) = 0 }. The center C of R is defined as C = {c ϵ N / (c, R) =0}. A ring is called simple if ≠ 0 and the only nonzero ideal of R is itself. A right alternative ring R is a ring in which y ( x x ) = ( y x ) x, for all x,y in R.
The ring R is said to be Antiflexible if A(x, y, z) = (x, y, z) – (z, y, x) = 0 ----------------------(1)
is an identity in R.
Throughout this paper we assume that R is antiflexible.
A ring R is of characteristic ≠n if nx =0 implies x = 0 for all x ϵ R and n is a natural number when n = 2, 3 and that the third power
associativity condition (x, x, x) = 0 --------------------------(2)
is an identity in R.
With aid of (1), we obtain the identity as a linearization of (2)
B(x, y, z) = (x, y, z) + (y, z, x) + (z, x, y) = 0 ------------------------(3)
We shall also require the Teichmuller identity ( which holds in any ring)
C(w, x, y, z) = (wx, y, z) – (w, xy, z) + (w, x, yz) – w(x, y, z) – (w, x, y)z = 0 ------------------------(4) In any ring, (xy, z) = x(y, z) + (x, z)y + (x, y, z) + (z, x, y) – (x, z, y) holds.
From which we subtract A (x, z, y) + B(x, y, z) = 0
to obtain D(x, y, z) = (xy, z) – x(y, z) – (x, z)y + 2(x, z, y) = 0 -------------------------(5)
We let x o y = x y + y x;
then it can be verified that in any ring
(x o y) o z – x o (y o z) = (x, y, z) + (x, z, y)+ (y, x, z) - (y, z, x) – (z, x, y) – (z, y, x) + (y, (x, y))
so that from (1) we get (x o y) o z – x o (y o z) = (y, (x, z)) -------------------(6)
If we retain the additive group of R but replace the product xy of R by the product xoy, then we obtain a commutative ring , and it follows from (2) and (6) that is a Jordan Ring. Equally, we could get an anticommutative ring by replacing the product xy by the commutator product (x, y) and from
0 = D(x, y, z) – D(y, x, z) and 0 = A(x, z, y).
It would follow that ( (x, y), z) + ( (y, z), x) + ( (z, x), y) = 0 -----------------------(7)
so that would be a Lie ring.
Expanding 0 = C (w, x, y, z) – C (z, y, x, w) and using
0 = A(z, y, xw) = A (z, yx, w) = A(zy, x, w) = z A(y, x, w) = A(z, y, x)w, we get
0 = E(w, x, y, z)=( (w, x), y, z) – (w, (x, y), z) + (w, x, (y, z)) – (w, (x, y, z)) – ((w, x, y), z) -----------------------(8)
Then we expand
0 = E(w, x, y, z) + E(x, y, z, w) + E(y, z, w, x) + E(z, w, x, y) – B( (w, x), y, z) – B( (x, y), z, w) – B( (y ,z), w, x) – B( (z, w), x, y)
To get
0 = F (w, x, y, z) = (w, (x, y), z) + (x, (y, z), w) + (y, (z, w), x) + (z, (w, x), y) ---------------------------(9)

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International Journal of Scientific & Engineering Research, Volume 5, Issue 3, March-2014

ISSN 2229-5518

1287

Now we are able to derive the important identity ( w ,(x, y), z) = 0 --------------------------(10) Expanding
0 = E(x, x, y, x) + E(y, x, x, x) – B(x, x, (y, x)) + ( B(x, x, y), x),
We get, 0 = (x, (x, y, x)),
Hence from 0 = (x, B(x, y, x)) and 0 = (x, A(x, x, y)), we have
0 = (x, (x, y, x)) = (x, (x, x, y)) = (x, (y, x, x)) ----------------------------(11) From (3) and (4), G (w, x, y, z) = (w, (x, y, z)) – (x, (y, z, w)) + (y,(z, w, x)) – (y,(w, x, y)) = 0 ----------------------------(12) G(x, y, x, y) = (x, (y, x, y)) - (y,(x, y, x)) + (x, (y, x, y) – (y,(x, y, x)) = 0

2 (x,(y, x, y)) – 2 (y,(x, y, x) = 0

(x, (y, x, y)) – (y, (x, y, x)) = 0

From (1), A(x, y, y) = 0 & A(y, x, x) = 0 & Thus (x, (y, x, y) ) – (y, (y, x, x)) = 0
Combining this with G(x, y, y, x) = 0
Gives 2 (x, (x, y, y) ) = 0
And therefore (x,(x, y, y)) = 0 ----------------------(13)
Linearize (13) J(x, y, z) = (x, (x, y, z)) + (x, (x, z, y)) = 0 ---------------------(14) Again linearize (13)

K(x, y, z) = (x, (z, y, y)) + (z, (x, y, y)) = 0 ----------------------(15)

Combination of
K(y, x + y, z) – (y, B(y, x, z)) - K(y, x, z) - (y, A(x, y, z)) – K(z, y, y) + (y, A(z, y, x)) – J(y, z, x) – (z, A (y, x, y)) + K (x, y, z) = 0 gives
(y, (z, x + y, x + y)) + (z, (y, x + y, x + y)) – (y, (x, y, z) + (y, z, x) + (z, x, y)) – (y,(z, x, x)) – (z, (y, x, x)) – (y, (x, y, z) – (z, y, x)) –
(z, (y, y, y)) – (y, (z, y, y)) + (y, (z, y, x) –(x, y, z)) – (y, (y, z, x)) – (y, (y, x, z)) – (z, (y, x, y) – (y, x, y)) + (x, (z, y, y)) + (z, (x, y, y)) = 0.

(y, (z, x, x)) + (y, (z, x, y)) + (y, (z, y, x)) + (y, (z, y, y)) + (z, (y, x, x)) + (z, (y, x, y)) + (z, (y, y, x)) + (z, (y, y, y)) – (y, (x, y, z)) –

(y, (y, z, x)) - (y, (z, x, y)) – (y, (z, x, x) – (z, (y, x, x)) – (y, (x, y, z)) + (y, (z, y, x)) – (z, (y, y, y)) – (y, (z, y, y)) + (y, (z, y, x)

– (y, (x, y, z)) – (y, (y, z, x)) – (y, (y, x, z)) + (x, (z, y, y)) + (z, (x, y, y)) = 0
⇒ 2(z, (y, y, x)) + (z, (y, x, y)) + (x, (z, y, y)) – 2(y, (y, z, x)) – (y, (z, x, y)) = 0
⇒ (z, (y, x, y)) + (z, (y, y, x)) – 2 (y, (y, z, x)) – (y, (z, x, y)) = 0
⇒ -(z, (x, y, y)) – 2 (y, (y, z, x)) – (y, (z, x, y)) = 0
⇒Therefore L(x,y,z) = -(z,(x,y,y)) – (y,(y,z,x)) + (y,(x,y,z)) = 0 - ---------------------(16)
Using (1) and (12) can be written as (x, (y, y, x)) = 0 ---------------------(17)
Linearization of this equation gives
(x, (z, y, x)) + (x, (y, z, x)) = 0

(x, (z, y, x)) = - (x, (y ,z, x)) -------------------(18)

Linearize equation (13)
(x, (x, y, z)) + (x, (x, z, y)) = 0

(x, (x, y, z)) = - (x, (x, z, y)) --------------(19)

(x, (x, y, z)) = - (x, (y, x, z)) ( by (18) )
= - (x, (z, x, y)) ( by (1) )
= (x, (z, y, x)) ( by (18) ) -----------------(20)
Commute equation (3) with x
(x, (x, y, z)) + (x, (y, z, x)) + (x, (z, x, y)) = 0

(x, (z, x, y)) = 0 ( By (20) & (18) ) -----------------(21)

From (16) and (19), we have
(z, (x, y, y)) = 0

(R, (x, y, y)) = 0 -------------(22)

Linearize (22), we have
(w, (x, y, z)) + (w, (x, z, y)) = 0

(w, (x, y, z)) = - (w, (x, z, y)) ------------------(23)

(w, (x, y, z)) = (w, (z, y, x)) ( by (1) )
= - (w, (z, x, y)) ( by (23))
= - (w, (y, x, z)) ( by (1) )
= (w, (y, z, x)) ( by (23) )

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International Journal of Scientific & Engineering Research, Volume 5, Issue 3, March-2014

ISSN 2229-5518

1288

Commute equ(3) with w
(w, (x, y, z)) + (w, (y, z, x)) + (w, (z, x, y)) = 0

(w, (x,y,z) ) = 0 -----------------(24)

3. Main Section:

Let us define T ={ t ϵ R / (t, R) = 0 = (t, R, R) = 0}

Lemma 3.1 : T is an ideal of R.

Proof: To prove T is an ideal.

Substitute x = t in (24) that gives
((t,y,z),w) = 0
From this, we have
(ty.z,w) - (t.yz,w) =0
And this becomes
(ty.z,w) = 0 from the definition of T Thus ty ϵ T and so T is a right ideal. However yt ϵ T.
Thus T is a two sided ideal of R.

Theorem3.2 : A simple not associative and not commutative antiflexible ring R of characteristic ≠ 2,3 is right alternative.

Proof: First we prove the identity (r, (z, y, y)w) = 0.

Commute Teichmuller identity C (w, x, y, z) = 0 with r and applying equation (24), we get
-(r, w(x, y, z)) – (r, (w, x, y)z) = 0
Gives (r, w(x, y, z)) = - (r, (w, x, y)z)
If we put y = z = x in this equation, then from (1) it reduces
(r, w(x, x, x)) = - (r, (w, x, x)x)
= 0

(r, (w, x, x)x) = 0 --------------(25)

From (24) & (25) all (w,x,x) are in T Since R is simple and T is an ideal of R, Then either T = R or T = 0.
If T = R then R is commutative.
Since R is not commutative we must have T = 0.
Then (w, x, x) = 0 ------------------(26) This implies R is Right–alternative.

References :

1. Kosier, Frank, On a class of nonflexible algebras, Trans. Am. Math. Soc. 102 (1962) 298-318.

2. Klenfeld. E, Kosier. F, Osborn, J.M., and Rodabaugh.D, The structure of associator dependent rings, Trans. Am. Math.Soc. 110 (1964), 473-483.

3. A. Thedy, On Rings satisfying ((a,b,c),d) = 0 , Proc.Amer.Math. Soc. 29 (1971), 213 – 218.

4. K. Survana and K. Subhashini, A Result on prime (-1,1) rings Actaciencia Indica, VolXXVI M, no.2 (2000), 85-86.

5. Subhashini .K, Simple (1,0) Rings, International Mathematical Forum, Vol 7,2012, no.48, 2407 -2410.

AUTHORS INFORMATION:

M.HEMA PRASAD, Assistant Professor of Mathematics, Dept. of Science of Humanities, SITAMS, CHITTOOR, ANDHRA PRADESH,

INDIA, PH-09440539339,E-mail:mhmprsd@gmail.com.

• Dr. D. BHARATHI, Associate Professor, Dept. of Mathematics, S.V. University, TIRUPATI, ANDHRA PRADESH, INDIA, PH-09440343888,

E-mail: bharathikavali@yahoo.co.in.

• Dr. M. MUNIRATHNAM, Ad-hoc Lecturer, RGUKT, R. K. Valley, KADAPA, ANDHRA PRADESH, INDIA, PH- 9849373260,

E-mail:munirathnam1986@gmail.com.

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