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Rings with (x, R, x) in the Left Nucleus

K. Jayalakshmi^1*, S. Madhavi Latha^2**

¹Department of Mathematics, JNTUA College of Engineering, JNTUA University, Ananthapuramu, Andhra Pradesh, INDIA.

²Department of Sciences and Humanities, C. V. Raman Institute of Technology, Tadipatri, Ananthapuramu, Andhra Pradesh, INDIA.

Email: ^*[email protected], ^**[email protected]

Research Article

Abstract If N_l and N_r be the Lie ideals of a nonassociative ring R, then [N_l, R] Í N_l and [N_r, R] Í N_r Also if (x, R, x) is in the left nucleus then N_l[R, R] Í N_l. If R is a prime ring with N_l ¹ 0, and (x, R, x) in the left nucleus then R is either associative or commutative.

Key Word: Nonassociative ring, Left nucleus, Right nucleus, Lie ideals, Associator ideal.

INTRODUCTION

Kleinfeld [1] studied nonassociative rings with (x, R, x) and [R, R] in the left nucleus. Yen [2] considered the rings with the weaker hypothesis that is, rings with (x, R, x) and [N_l, R] in the left nucleus and proved that if R is a semiprime ring, then N_r = N_l. He also proved that if R is a prime ring with N_l ¹ 0 satisfying one additional condition N_l[R, R] Í N_l,then R is either associative or commutative. In this paper by considering N_l and N_r as the Lie ideals of a ring R, we present some properties of R with (x, R, x) in the left nucleus. Using these properties, we show that N_l[R, R] Í N_l. Also we prove that, if R is a prime ring with N_l ¹ 0,then R is either associative or commutative.

PRILIMENARIES

In a nonassociative ring R we define an associator as (x, y, z) = (xy) z – x (yz) and the commutator as [x, y] = xy – yx for all x, y, z Î R. To make the notation more convenient we often use ‘×’ to indicate multiplication as well as juxtaposition. In products, juxtaposition takes precedence, i, e, xy × z º (xy) z. The nucleus of a ring R is defined as N = {n Î R / (n, R, R) = (R, n, R) = (R, R, n) = 0}, the right nucleus as N_r = {n Î R / (R, R, n) = 0} and the left nucleus as N_l = {n Î R / (n, R, R) = 0}. A ring R is said to be prime if whenever A and B are ideals of R such that AB = 0, then either A = 0 or B = 0 and is said to be semiprime if for any ideal A of R, A² = 0 implies A = 0. These rings are also refered to as rings free from trivial ideals. And a ring is said to be simple if whenever A is an ideal of R, then either A= R or A = 0.

Let R be a nonassociative ring satisfying (x, R, x) Í N_l, that is,

(x, y, z) + (z, y, x) Î N_l. (1)

Let N_l and N_r be the Lie ideals of R. Then

[N_l, R] Í N_l (2)

and [N_r, R] Í N_r.

We use Teichmuller identity which is valid in any arbitrary ring.

(wx, y, z) – (w, xy, z) + (w, x, yz) – w(x, y, z) – (w, x, y) z = 0, (3)

for all, w, x, y, z Î R.

Then with w = n Î N_lin (3), we obtain

(nx, y, z) = n(x, y, z).

Since N_l is the Lie ideal from (2), we obtain

(nx, y, z) = n(x, y, z) = (xn, y, z), (4)

for all, n Î N_l.

Thus N_l is the associative subring of R.

MAIN SECTION

Lemma 3.1: Let T = {t Î N_l: t(R, R, R) = 0}, then T is an ideal of R.

Proof: In (4) substituting n = t, we obtain

(tx, y, z) = t(x, y, z) = (xt, y, z) = 0.

Thus tR Ì N_l and Rt Ì N_l.

Also, tw × (x, y, z) = t × w(x, y, z).

Multiplying (3) with t on the left side, we obtain

t × w(x, y, z) = – t × (w, x, y)z

= – t (w, x, y) × z

= 0.

Hence tw × (x, y, z) = 0. Thus TR Í T.

Now using TR Í T, (2), (4), RT Ì N_l and (1), we obtain

wt × (x, y, z) = [w, t] (x, y, z)

= ([w, t]x, y, z)

= ((wt)x, y, z) – ((tw)x, y, z)

= ([wt, x], y, z) + (x(wt), y, z) – (t(wx), y, z)

= ([wt, x], y, z) + (x(wt), y, z)

= – ((x, w, t), y, z) + ((xw)t, y, z)

= – ((x, w, t) + (t, w, x), y, z)

= 0.

Hence RT Í T. Thus T is an ideal of R. From the definition of T, we obtain T(R, R, R) = 0.

This completes the proof of the Lemma.

Let A be the associator ideal of R. We assume that R satisfies (1) and also R is semiprime. Using Lemma 3.1 and equation (3), we obtain T × A = 0 and hence (T Ç A)² = 0. Thus we have T Ç A = 0 and A × T = 0. (5)

From Lemma 3.1 and equation (3), we obtain

(R, T, R) = 0. (6)

Lemma 3.2: Let R be a nonassociative ring satisfying(x, y, z) + (z, y, x) Î N_l. Then(R, R, N_l) = 0.

Proof: Let n Î N_l, then from (1), we obtain

(x, y, n) = (x, y, n) + (n, y, x)Î N_l.

Also from (3), we obtain

z (x, y, n) = (zx, y, n) – (z, xy, n) + (z, x, yn) – (z, x, y)n.

Hence using these, (4) and (1), we obtain

(x, y, n)(z, r, s) = (z(x, y, n), r, s)

= ((zx, y, n), r, s) – ((z, xy, n), r, s) + ((z, x, yn), r, s) – ((z, x, y)n, r, s)

= ((z, x, yn), r, s) – ((z, x, y)n, r, s)

= – ((yn, x, z), r, s) – (n(z, x, y), r, s)

= – (n(y, x, z), r, s) – n((z, x, y), r, s)

= – n((y, x, z), r, s) – n((z, x, y), r, s)

= – n((y, x, z) + (z, x, y), r, s)

= 0.

Hence (x, y, n) Î T.

Since (x, y, n) is also an associator, it is also in A.

Thus from (5), we obtain (x, y, n) = 0.

Hence (R, R, N_l) = 0.

From Lemma 3.2, we obtain N_lÍ N_r. (7)

Let n Î N_r. Then with z = n in (3), we obtain

(w, x, yn) = (w, x, y) n. Thus N_r is an associative subring of R.

Now since N_r is the Lie ideal of R, we obtain

(w, x, yn) = (w, x, y)n = (w, x, ny), (8)

for all n Î N_r and w, x, y Î R.

Lemma 3.3: Let N_r be the Lie ideal of R and let

S = {n Î N_r: (R, R, R)n = 0}, then S is an ideal of R, (R, R, R)S = 0, S Ç A = 0, S × A = A × S = 0 and T Í S.

Proof: Using (1), (3), (5), (7) and (8) and the proof of Lemma 3.1, this Lemma is proved.

Lemma 3.4: If N_r and N_l are the Lie ideals of R, then N_r = N_l and S = T.

Proof: Let us assume that (R, R, n) = 0, then from (1), we obtain (n, x, y) = (n, x, y) + (y, x, n) Î N_l.

Now using (1), (7), (8) and [N_r, R] Í N_r, and since N_r is an associative subring of R, we obtain

(nx, y, z) – n (x, y, z) = {(nx, y, z) + (z, y, nx)} – n{(x, y, z) + (z, y, x)} + [n, (z, y, x)] Î N_r.

From the above equation and (n, R, R) Í N_l Í N_r and with w = n in (3), we obtain

(n, x, y)z = {(nx, y, z) – n(x, y, z)} – (n, xy, z) + (n, x, yz) Î N_r.

Hence using this and (8), we obtain (s, r, z) (n, x, y) = (s, r, (n, x, y) z) = 0, which shows that (n, x, y) Î S Ç A and thus from Lemma 3.3, we have (n, x, y) = 0.

Hence N_r Í N_l. Thus from (7), we have N_r = N_l. From Lemma 3.3 again, S × A = 0 and so S = T. This completes the proof of the Lemma.

Theorem 3.1: If R is a semiprime ring satisfying (x, y, z) + (z, y, x) Î N_l, where N_l is the Lie ideal of R, then T is an ideal of R and (N_l, R, R) = (R, T, R) = (R, R, N_l) = 0. Also, if [N_r, R] Í N_r, then N_r = N_l and S = T Í N.

Proof: From (6) and Lemmas 3.1, 3.2, 3.3 and 3.4 the Theorem is proved.

Lemma 2.5: Let I = {a Î R : N_la = 0}, then I is an ideal of R.

Proof: First we show that ([R, R], R, R) Í I.By taking y = z = x in (1), we obtain (x, x, x) + (x, x, x) = 2(x, x, x) Î N_l. So (x, x, x) Î N_l.

Let S(x, y, z) = (x, y, z) + (y, z, x) + (z, x, y).

Now linearization of (x, x, x) gives (x, y, z) + (y, z, x) + (z, x, y) + (y, x, z) + (z, y, x) + (x, z, y) Î N_l.

i.e., S(x, y, z) + S(y, x, z) Î N_l. (9)

We have D(x, y, z) = [xy, z] – x[y, z] – [x, z]y – (x, y, z) – (z, x, y) + (x, z, y) = 0. (10)

This identity is valid in any arbitrary ring.

Now D(x, y, z) – D(y, x, z) gives

[[x, y], z] + [[y, z], x] + [[z, x], y] = S(x, y, z) – S(y, x, z).

If z Î N_l, we obtain S(x, y, z) – S(y, x, z) Î N_l. (11)

But from (9), S(x, y, z) + S(y, x, z) Î N_l.

i.e., 2S(x, y, z) Î N_l._.

i.e., S(x, y, z) Î N_l.

i.e., (x, y, z) + (y, z, x) + (z, x, y) Î N_l.

But (x, y, z), (z, x, y) Î N_l implies (y, z, x) Î N_l.

i.e., (R, N_l, R) Í N_l implies ((R, N_l, R), R, R) = 0. (12)

Now in (10) substituting x = n and forming the associators with r, s and using (12), we obtain

([ny, z], r, s) = (n[y, z], r, s) + ([n, z] y, r, s) + ((n, y, z), r, s) + ((z, n, y), r, s) – ((n, z, y), r, s)

= (n[y, z], r, s) + ([n, z]y, r, s) + ((z, n, y), r, s).

i.e., ([N_lR, R], R, R) = (N_l[R, R], R, R) + ([N_l, R] R, R, R)+ ((R, N_l, R), R, R).

i.e., (N_l[R, R], R, R) = ([N_lR, R], R, R) – ([N_l, R] R, R, R)

= ((N_lR) R – R (N_lR) – (N_lR) R + (RN_l) R, R, R)

= ((R, N_l, R), R, R)

= 0 from (12).

Thus N_l([R, R], R, R) = (N_l[R, R], R, R) = 0. (13)

Hence we have, ([R, R], R, R) Í I. (14)

Now let a Î I, n Î N_l and x, y, z, w Î R. Thus we obtain

n(ax) = (na)x = 0 implies IR Í I.

Now from (13), we obtain

n(xa) = n[x, a] Î N_l.

Since na = 0 and n Î N_l, we obtain n(a, x, y) = 0. (15)

Using (15), (1) and since N_l is an associative subring of R, we obtain

n((yx)a) – n(y(xa)) = n(y, x, a)

= n((a, x, y) + (y, x, a)) Î N_l. (16)

Applying (16) and n(xa) Î N_l, we obtain

n(y(xa)) Î N_l. (17)

Using (17) and (13), we obtain

(n(xa))y = n((xa)y)

= n[xa, y] + n(y(xa)) Î N_l.

Combining the above with n(xa) Î N_l, we obtain

n(xa)(y, z, w) = ((n(xa))y, z, w)

= 0.

Hence n(xa) Î T and thus n(xa) = 0 implies RI Í I.

Therefore I is an ideal of R and thus NI = 0.

Theorem 3.2: If N_l is the Lie ideal of a prime ring R with N_l ≠ 0 and satisfying (x, y, z) + (z, y, x) Î N_l, then R is either associative or commutative.

Proof: Since R is prime using (5), we obtain either A = 0 or T = 0. If A = 0, then R is associative. Hence we assume that T = 0. Since N_l is the Lie ideal of R, using Lemma 3.2, we see that the ideal of R generated by N_l is N_l + N_lR. Then N_lI = 0 from Lemma 3.5. Hence we obtain

(N_l + N_l R)I Í N_l I + (N_lR)I

= N_l I + (N_l, R, I ) + N_l(RI )

Í N_l I + N_l (RI )

= 0.

Thus [R, R] Í N_l. Now R satisfies Kleinfeld’s hypothesis [1]. Hence it follows that R is either associative or commutative. This completes the proof of the Theorem.

REFERENCES

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Yen, C. T. “Rings with (R, R, R) and [(R, R, R), R] in the left nucleus”, Tamkang Journal of Math., 24, 209-213, (1993).
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Yen, C. T., “Simple rings of characteristic not 2 with associators in the left nucleus are associative”, Tamkang Journal of Math., Vol 33 (1), 93-95, (2002).
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