Paul, Yash

English

Paul Yash

University of Szeged

Acta Sci. Math.; 56 (1992), 47—49 On rings satisfying (x, y, z)=(y, z, x) YASH PAUL Let R be an arbitrary nonassociative ring having a Peirce decomposition, into a direct sum of Z modules, relative to nonzero idempotent e in the nucleus of R. If R satisfies the identity (x, y, z)—(y, z, x) then (i) under certain conditions on the Z modules, R is associative and (ii) if R is prime then e is the identity element of R. As usual, the associator (x,y,z) denotes (xy)z-x(yz) and the commutator (x, y)—xy—yx. STERLING [3] has shown that semiprime rings satisfying (1) (x, y, z) = (y, z, x) are alternative. The nucleus N(R) of an arbitrary nonassociative ring R consists of all elements n in R such that (n, r,s) = 0 = (r, s, n) — (s, n, r) for all r, s in R. It is well known, see SCHAFER[2] p. 18, that N(R) is an associative subring of R. We need an identity valid in all rings, the so-called Teichmiiller identity (2) (ivx, y, z) - (w, xy, z) + (w, x, yz) = vc(x, y, z) + (w, x, y) z. We now take turns letting one of the four elements in (2) be in the nucleus. Thus (3) (4) (5) (6) (fix, y, z) = n(x, y, z) (wn, y, z) = (w, ny, z) (w, xn, z) = (w, x, nz) (w, X, yn) = (W\ X , y)n for any element n in N(R) and the rest of the elements arbitrary in R. Received June 7, 1990 and in revised forms January 16 and July 24, 1991. 48 Yash Paul Lemma. Let R be a ring satisfying (1). Then (R, N(R)) Q N(R). Proof . Let x, y, z be arbitrary elements of R and n be an element of N(R). Using (1), (3), (5) and (6) we get n(x, y, z) = (nx, y, z) = (y, z, nx) = (y, zn, x) = (x, y, zn) = (x, y, z)n. Thus (7) n(x,y,z) = (x,y,z)n. Again using (1), (3), (6) and (7) we get (nx, y,z) = n (x, y, z) = (x, y, z)n = (y, z,x)n = (y, z, xri) = (xn, y, z). Thus ((x, n),y, z)—0. This implies that (R, N(R))QN(R). A ring R is said to have a Peirce decomposition relative to the idempotent e£R if J? can be decomposed into a direct sum of Z modules RtJ (i,j=0, 1) where RIJ = {x£R: xe = jx and ex = ix}. It is known, see JACOBSON [1], that if J? is an associative ring and if e is an idempotent in R then R has a Peirce decomposition relative to e. Also, if R has an identity ele-ment 1 and if we write et=e and e 0 =l—e then RlJ=eiRej. Let e£N(R). Embed R into the ring R'=Z+R which contains an identity element 1. Clearly, e and 1 — e are in N(R'). It follows that R=(BRjj and Rij—eiRej for i,j=0, 1. Thus RijRu = (<>;Rej)(ekRe,) = e,.jR(e,£gRe, 8jkeiRel = 5JkRn for i,j, k,l—0,1 (5 denotes the Kronecker delta). T h e o r e m . Let R be a ring satisfying (1) with an idempotent e^O in N(R). (i) If R satisfies the condition Rij Rji = Ru when i ^ j then R is associative. (ii) If R is prime then e is the identity element of R. Proof , (i) R10=(e,R10) = -(R10,e) and R01=(R01, e). Since e£N(R), by the above Lemma, R10 and RmQN(R). N(R is an associative subring of R. So R10Rqx and R01R10QN(R). By the given condit ion R1± and R00 <gN(R). I t fol-lows that R — Rn + Rw + *<» + Roo i Hence R is associative. Rings satisfying (x, y, z)=(y, z, x) 49 (ii) RwRn+RfuRwQNiR). This, together with the property RijRk lQdjkRu , allows us to conclude that B=R10 R01++Rox+R0i Rw is an ideal of R con-tained in N(R). All rings R have an ideal A, called the associater ideal. It is defined as the smallest ideal which contains all associators. It actually consists of all finite sums of associators and right multiples of associators. The associator ideal is never zero, except when R is associative. We shall show that BA=(0). Let b£B. Then using Teichmuller identity (2) we get (bx, y, z) — (b, xy, z) + (b, x, yz) = b (x, y, z) + (b,x,y)z. Since B is an ideal contained in N(R) and b£B we have (bx, y, z) = (b, xy, z) = (b, x, yz) = (b, x, y) = 0. Thus, from the above equation, we get b(x,y,z) = 0. Also, since b£N(R), b((x,y, z)w) = (b(x,y, z))w = 0. Thus we have proved that bA=(0) for all b in B. Hence BA=(0). But R is prime and nonassociative. This implies that B—(0). So we have R = © Roo • Thus, i? n and R00 are ideals of R such that = (0). From the primeness of R again Ru=(0) or i?00=(0). But 0¿¿e£Ru so that -^11^(0)- We must have i?00=(0). This implies that e is the identity element of R. References [1] N. JACOBSON, Structure of rings, Amer. Math. Soc. Colloquium Publications 37, Amer. Math. Soc. (Providence, R.I. 1964). [2] R. D. SCHAFER, An introduction to nonassociative algebras, Pure and Appl. Math. 22, Academic Press (New York—London, 1966). [3] N. J. STERLING, Rings satisfying (x, y, z)=(y, z, x), Can. J. Math., 20 (1968), 913—918. INDIAN INSTITUTE OF TECHNOLOGY NEW DELHI-110016 INDIA 2 

On rings satisfying (x, y, z) = (y, z, x)

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On rings satisfying (x, y, z) = (y, z, x)

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