Asian Journal of Mathematics & Statistics

Year: 2010 | Volume: 3 | Issue: 4 | Page No.: 244-248
DOI: 10.3923/ajms.2010.244.248

Simple (-1, 1) Rings

V.M. Rao and K. Suvarna

Abstract: This study proves that a simple (-1, 1) ring of characteristic ≠2,3 is a derivation alternator ring.

Keywords: associator, derivation alternator rings, Simple rings, commutator and nucleus

INTRODUCTION

Algebras of type (γ, δ) were first defined by Albert (1949). They are algebras (non- associative) satisfying the following identities:

and

where γ²-δ² + δ = 1 and where the associator (x, y, z) = (xy)z- x(yz).

Simple rings of type (γ, δ) have been studied by Kokories (1958) and Kleinfeld (1959). Their study show that except for type (-1, 1) and (1, 0), all simple (γ, δ) rings with an idempotent e which is not the unity element are associative. Thedy (1971) proved that a simple non-associative ring with ((a, b, c), d) = 0 is either associative or commutative. Hentzel et al. (1980) studied derivation alternator rings. These rings are a generalization of alternative rings. In this study it is proven that in a (-1, 1) ring R every associator commutes with every element of R, i.e., (R, (R, R, R)) = 0. Using this it is proven that a simple (-1, 1) ring of characteristic ≠2, 3 is a derivation alternator ring. At the end of this study an example of (-1, 1) ring which is not a derivation alternator ring is provided.

PRELIMINARIES

A non-associative ring is said to be (-1, 1) ring if it satisfies the following identities:

is an immediate consequence of Eq. 1 and 2 since 0 = A(x, x, y)-B(x, x, y).

A non-associative ring R is called a derivation alternator ring if it satisfies the following identities:

and

Throughout this study R will represent a (-1, 1) ring of characteristic ≠2, 3. The commutator (x, y) of two elements x and y in a ring is defined by (x, y) = xy-yx. A ring R is said to be simple if whenever A is an ideal of R, then either A = 0 or A = R. A ring R is said to be of characteristic ≠n if nx = 0 implies x = 0, x∈R and n is a natural number.

In any ring, we have the following Teichmuller identity:

The following identity also holds in any ring:

By forming C(x, y, y, z)-C(x, z, y, y) + C(x, y, z, y) = 0, we obtain 2(x, y, yz) = 2(x, y, z)y. This implies that:

In C(x, z, y, y) = 0, we make use of Eq. 6, so that

By linearzing Eq. 6 (replace y with w+y), we obtain the identity:

From C(w, x, y, z)-F(w, z, x, y) = 0, it follows that:

In a (-1, 1) ring Eq. 5 becomes H(x, y, z) = (xy, z)-x(y, z)-(x, z)y-2(x, y, z)-(z, x, y) = 0 because of Eq. 2.

The combination of Eq. 1 and 4 as Kleinfeld (1959) gives:

From J(x, x, x, y) + (x, B(x, y, x)) = 0 it follows that 2(x, (x, x, y)) = 0. From this and the fact that (x, y, x) = -(x, x, y) we obtain:

Now J(y, x, y, x) = 0 gives 2(y, (x, y, x))-2 (x, (y, x, y)) = 0 and thus (y, (x, y, x))-(x, (y, x, y) = 0. From B(x, x, y) = 0 and B(y, y, x) = 0, we have (y, (x, x, y))-(x, (y, y, x)) = 0. Combining this with G(y, x, x, y) = 0 gives 2(y, (x, x, y)) = 0 and therefore:

Using the right alternative property of R, identity (10) can be written as:

Lemma 1
If R is a (-1, 1) ring of characteristic ≠2, 3, then (R, (R, R, R)) = 0.

Proof
By linearizing the identity (11) and (10), we have:

and

From Eq. 2, 12, and 13 and again 2 we get:

Commuting Eq. 1 with y, we have:

From Eq. 14, this equation becomes 3(y, (x, y, z))= 0. Since R is of characteristic ≠3;

The following identity holds in any (-1, 1) ring as in Hentzel (1972):

From Eq. 15 the identity K(x, y, z) = (x, (y, y, z)) - 3 (y, (x, z, y) = 0 becomes (x, (y, y, z)) = 0 Thus:

By linearizing equation Eq. 16, we obtain:

Applying Eq. 2 and 17 repeatedly, we get:

Commuting Eq. 1 with w and applying the above equation, we obtain 3(w, (x, y, z)) = 0. Since R is of characteristic ≠3, we have

The identity Eq. 18 completes the proof of the Lemma.

Next we prove the identity (r, (y, y, z)w) = 0.Commuting Teichmuller identity C(w, x, y, z) = 0 with r and applying lemma 1, we get (r, (x, y, z)w) = -(r, (w, x, y)z).

If we put x = y in this equation, then it reduces to:

Lemma 2
If R is a (-1, 1) ring of characteristic ≠2, 3, then T = {t∈R/(t, R) = 0 = (tR, R)} is an ideal of R.

Proof
By substituting x = t in Eq. 18, we get ((t, y, z), w) = 0. From this equation it follows that (ty.z, w) = 0. Thus ty∈T and so T is a right ideal. However yt = ty. Thus T is a two sided ideal of R.

MAIN RESULT

Theorem
A simple (-1, 1) ring of characteristic ≠2, 3 is a derivation alternator ring.

Proof
From Eq. 16 and 19, we have:

and

So, (x, x, yz)-y(x, x, z)-(x, x, y)z∈T.

Since R is simple and T is an ideal of R, either T = R or T = 0. If T = R, then R is commutative. But R is not commutative.

Thus T = 0 and (x, x, yz) –y(x, x, z)-(x, x, y)z.
That is, (x, x, yz) = y(x, x, z) + (x, x, y)z.
Similarly, (x, yz, x) = y(x, z, x) + (x, y, x)z.
By taking y = x in Eq. 3, we get (x, x, x) = 0.
Hence, R is a derivation alternative ring.

The following example illustrates that a (-1, 1) ring, which is not derivation alternator ring.

Example
Consider the algebra having basis elements x, y and z over an arbitrary field. We define x² = y, yx = z and all other products of basis elements equal to zero. It clearly satisfies (1) and (2). Hence it is a (-1, 1) ring, but not a derivation alternator ring since (x, x, x) = z.

REFERENCES