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A generalization of strongly (−1, 1) rings
JOURNAL
OF ALGEBRA
119, 218~225
A Generalization
(1988)
of Strongly ERWIN
( - 1, 1 ) Rings
KLEINFELD
Department of Mathematics, University of Iowa,Iowa City, Iowa Communicated Received
by Walter April
52242
Feit
28, 1987
I. N-RoDUCTI~N
In the structure theory of (- 1, 1) rings the identity ((x, y), z) =O, becomes a natural stepping stone. Indeed if A is the ideal generated by all associators of R then ((R, R), A) = 0 [2]. Right alternative rings satisfying ((x, y), z) = 0, are called strongly (- 1, l), for they are (- 1, 1). Here we take up the study of right alternative rings which satisfy the weaker identity ((a, b), a) = 0. Under the assumption of semi-prime and characteristc 22, 3 we prove the ring is strongly (- 1, 1). At this point there exist examples of even prime rings which need not be associative, due to Pchelincev [6]. On the other hand there do exist other conditions that imply associativity Cl, 7, 81. We present a non-semi-prime example which satisfies the right alternative identity and ((a, 6), a) = 0, but which is not (- 1, 1).
II. MAIN
RESULTS
We use the usual definitions of assciator and commutator. That is to say (x, y, z) = xy . z - x . yz and (x, y) = xy - yx. Throughout the paper R stands for a ring which is (I) right alternative, (II) satisfies the identity ((a, b), a) = 0, for all a, b E R, and (III) has characteristic # 2, 3. Before we are able to prove the main result we need to establish 32 identities. We shall first list 5 which are valid in right alternative rings and 14 more, which we then prove. After proving Theorem 1 we then list the remaining 13 identities, followed by their proofs. Then we finish up with the main result and an example. First we list without proof the following: (1) (2)
(3) 0021-8693/88
(4 Y2, z) = (x, YYYZ + KY), (wx, y, z) + (4 x7 (Y, z)) = 4-T Y, z) + (w, y, z)x, (XY, z) = X(Y, z) + (x, Z)Y + 2(x, Y, z) + (z, x, Y), 218 $3.00
Copyright c 1988 by Academic Press, Inc. All rights of reproductmn in any form reserved.
ALTERNATIVE
(4) (5)
RINGS
AND
((a,
b),
(4 YT YZ) = (4 Y, Z)Y, ((x, Y), z) + ((Y, z), x) + ((z, xl, y) = 2%
219
a) = 0
YTZ)>
where we define J(x, y, z) = (x, y, z) + ( y, z, x) + (z, x, y). These identities for right alternative rings are well known. In particular proofs may be found in [4].
((4~)~ n(y)), n(z)) = sm 4(x9 Y), ~1, (7) ((w,x), (y,z))=O, (8) (((w xl, Y), z) = 0, (6)
(9) (10) (11) (12)
(13) (14) (15) (16)
(17) (18)
(19) Identity
3((x, Yh z) = 2.0, Y, z), 4x, Y, (4 b)) = 0, ((4 x)3 Y, z) - ((Y, z), w, x) = (WY(x9 Y, 2)) - 6, (WYY, z)), (x7 x9 6, Y)) = 0, (6, Y), x, Y) = -((x, Y), Y, xl = -2(x, Y, (x, Y)) = 2t.K 4 (4
by, (x7Y))
Y)),
= 0,
((Y, 4 Y), x) = (4 YY - 3(X> Y> (xv Y)),
((4 ((4 ((4 b,
YY xl, Y) = ((Y, x7 Y), x)3 Y9 x), x) = 0, Y, xl, Y) = 0, YY = 3(x, Y? (4 Y)).
(6) is just a linearization
of (II).
It follows from
((w,x),(~,z))=-(((~,z),~),w)=(((~,~),z),w)=-((w,z),(y,~)). x(z))) = Sgn n((w, z), (y, 2)). then ((n(w), n(x)), (4~)~
(6) that But
Consequently ((w, x), ( y, z)) = (( y, z), (w, x)). On the other hand it follows from the definition of the commutator that ((w, x), ( y, z)) = -(( y, z), (w, x)), so that 2((w, x), (y, z)) = 0. Then (7) follows because of (III). But then (((w, x), y), z) = -((z, y), (w, x)) using (6), so that (7) now yields (8). As a consequence of (6) we have ((x, y), z) = (( y, z), x) = ((z, x), y). This allows us to rewrite (5) as (9). Using (7) while substituting z = (a, b) in (9) we obtain 2J(x, y, (a, b)) = 0. Then use of (III) leads to (10). As a consequence of (2) we have (wx, y, z) + (w, x, ( y, z)) = w(x, y, z) + (w, y, z)x as well as the last (xw, y, z) + (A w, (Y, z)) = x(w, y, z) + (x, y, z)w. Subtracting equation from the previous, therefore ((w, x), Y, z) + (w & (Y, 2)) - (x, w, (Y, z)) = (w (x, Y, z)) - (XT (w, y, z)). However, it follows from (10) that (w 4 (Y, z)) + (4 (y, z), w) + ((Y, Z)? WYx) =a
220
ERWIN
KLEINFELD
Using (I) on this equation we obtain (w, x, ( y, z)) - (x, w, ( y, z)) = -(( y, z), w, x). With this substitution we have established (11). In a right alternative ring we know (x, x, x) = 0. But since (x, x, x) = (x2, x) we have (x2, x) = 0. But then (6) implies (x2, (x, y)) = - ( y, (x, x2)) = 0. Applying (3) to (x2, (x, y)) we thus have 2(x, x, (x, y)) + ((x, y), x, x) =O, since (x, (x, v)) =0 as a result of (II). But (I) implies ((x, y), x, x) =O. At this point (III) gives (12). Linearize ( 12) by replacing one x by y. Thus (y, x, (x, y)) + (x, y, (x, y)) + (x, x, ( y, y)) = 0. From this follows (x, y, (x, y)) = -(y, x, lx, y)). Using (I) we have (x, Y, (x, Y)) = -(x7 (x, Y), Y) and -(Y, x, (x, Y)) = (Y, (x, Y), x). Now (10) implies 44
Y, (x3 Y)) = 0 = (x, Y? (4 Y)) + (Y, (x3 Y), xl + ((4 Y), x9 Y).
At this point (13) is obvious. It is implied (XY, (x3 Y)) =x(x
by (3) that
6% v)) + (4 (x, Y1J.Y+ 2(x, YT(x3 Y)) + ((xv Yh x> Y).
However, the last two terms cancel each other because of (13). are zero because of (II). Thus (14) follows. Using (6) (by, y), xl = -((x3 Y), v). But -((x9 Y), v) = Cv, (x, v)), definition of the commutator. Thus (14) implies ((xy, y), x) = implies (xy, y) = (x, y)y + ( y, x, y). Substituting this ((x, y) y, x) + (( y, x, v), x) = 0. However, (3) also implies
Two more we have using the 0. But (3) leads to
((4 Y) Y, xl = (4 Y)( YYx) + ((XT Y), x) Y + 2((-% Y), Y9xl + (x9 (4 Y), Y) = -(x9 Y)(X, Y) + 3(x, Y, (XT Y)), using (13). Thus -(x9 y)’ + 3(x, y, (x, y)) + ((y, x, Y), x) = 0. This is equivalent to (15). In (15) interchange x and y. Since ( y, x)’ = (x, y)’ and (Y, x, (y, x)) = -(Y, x, (x, Y)) = (x, Y, 6, Y)) using (13) we obtain (16). Next linearize (16) by replacing y by an x. Thus ((x, x, x), y) + ((x, Y, xl, x) = ((x, x, Y), x) + ((Y, x, x), x). Using (1) thus ((x3 Y, xl, x) = ((x,x, v), x). Since (x, x, y)= -(x, y, x) as a result of (I) thus 2((x, y, x), x) = 0. This implies ((x, y, x), x) = 0, which is (17). In (11) let w=y and z=x. Then ((A xl, YYxl - ((3s xl, Y>xl = (.Y?(x9 YYxl) - (xv (Y, YPx)), so that (Y, (x, Y, x)) = (x, (Y, Y, x)). However, (Y, Y, x) = -(Y, x, v) using (I), thus ( y, (x, y, x)) = -(x, ( y, x, y)). Comparison with (16) now leads to (18). Combining (15) and (18) one obtains (19). We take time out now to prove our first theorem.
ALTERNATIVE THEOREM
RINGS AND (( U,b),u)=O
221
1. Zf e is an idempotent of R then e must be in the center of R.
ProoJ: Put x= e in (12). Thus (e, e, (e, y)) = 0. If we put x= y =e in
(1) we obtain (e, e, z) = (e, e*, z) = (e, e, ez +ze). But (e, e, (e, z)) =0 thus (e, e, ez + ze) = 2(e, e, ez). But then using (4) 2(e, e, ez) = 2(e, e, z)e. Thus (e, e, z)=2(e, e, z)e. If a= (e, e, z) then a=2ae. But then a=2ae=4ae.e. Then (I) implies 4ae . e = 4ae2= 4ae. Thus 2ae= 0, hence a = 0. We have shown 0 = a = (e, e, z). From this, using (I) we obtain (e, z, e) = 0 and also (z, e, e) = 0. This suffices to allow the usual Peirce decomposition of R as RI1 + RIO+ Rol+Roo. If x,,ER,, then (x,,,e)= -x1,,, so that (bm e), e) = x,~. But then (II) implies xi0 = 0, so that Rio= 0. Similarly RO,= 0. At this point it is obvious (e, R) = 0. Also using (1) with y=e and z=zll and also y=e and z=z,,,, then (x, e, z,,) = 2(x, e, zii) while (x, e, zoo) = 0. Thus (x, e, z) = 0, so that (R, e, R) = 0.
But then
(R, R, e) = 0, using (I). Furthermore (e, R) = 0 implies 0 = ((e, y), z) so that J(e, y, z) = 0 using (9). At this point we obtain (e, y, z) = 0, so that (e, R, R) = 0. Thus e is in the center of R. This concludes the proof of the theorem. Define U= {U E R 1(u, R) = O}. U is also known by the name of commutative center. The element u is reserved for a generic element of U. We continue now with the rest of the numbered identities, to be followed by their proofs. (20) (21) (22)
(4 x, Y) = --2(x, y, u) = -2(Y, u, xl = a Y, x9 u) = 2(x, u, Y), (4~)~ (4x), n(y), ~1) = sm n(z, (x9 Y, u)), (x, (Y, z), u) = -2(x, (Y, z, u)),
(23) b(x), (4~), 4~))~ u)=sgne, (24) 3(x, Y, u)=(x, Y)U-(x, ~4, (25) -3(x, (Y,z,u))=(x, (Y,ZW(X,
(Y,z),u),
(26)
((xv Y), zu),
-3(x,
(Y, z, ~1) =(x,
(Y, z))u+
(Y+)),
222
ERWIN
(27) (28)
(29) (30) (31) (32)
KLEINFELD
-3(x, (Y, =, 4) = (4 (Y, 4)~ + ((x3 Y), Z)U- 3((~, Y), =, 4, (4 (Y, =I, u) = 0 = (4 (Y, =9u)), ((x, Y), 42 =o, (R, R, U) c U, CR, U, W = U, (u, R R) c u, (u, R R)(R, R) c u, (U, R, RN@, RI, R)=O.
In (3) let z = U. Thus 2(x, y, U) + (a, x, y) = 0. Also since ((x, y), U) = 0, we obtain from (9) that J(x, y, U) = 0. These identities and (I) suffice to yield (20). As a result of (20) we have (x, y, U) = -(y, x, U) so that (z, (x, y, u)) = -(z, ( y, x, u)). Linearizing (17) by replacing an x by U, we obtain ((u, y, x), x) + ((x, y, u), x) + ((x, y, x), u) = 0. But then (20) implies 3((x, y, u), x) = 0. Thus (III) implies ((x, y, u), x) = 0. Linearizing this yields ((x, y, u), z) = -((z, y, u), x). This suffices to prove (21). Substitute w=u in (11). Thus ((u,x),Y,=)-((y,=),~,x)=(~,(x,y,z))-(x,(u,y,z)). But then -((y, z), U, x) = -(x, (u, y, z)). At this point (20) implies -((y, z), u, xl= -(x, (Y, =I, u) while -(x, (u, Y, 2)) = 2(x, (Y, z, u)), so that we obtain (22). Combining (21) and (22) leads to (23). Also -(x, JJU) = (yu, x), while (3) implies ~Y~,~~=Y~~,~~+~Y,~~~+~~Y,~,~~+~~,Y,~~
= -(x,
y)u + 3(x, y, u)
using (20). This proves (24). Replace y by (y, z) in (24). Thus 3(x, (y, z), U) = (x, (y, z))u- (x, (y, z)u). Also replace x by y and y by z in (24) and commute with x. This yields 3(x, ( y, z, u)) = (x, ( y, z)u) (x, ( y, zu)). Because of (22) we can replace 3(x, ( y, z), U) by -6(x, ( y, z, u)) so that -6(x, ( y, z, u)) = (x, ( y, z))u - (x, ( y, z)u). Adding the last two equations we obtain -3(x, ( y, z, u)) = (x, (y, z))u - (x, ( y, zu)) or (25). As a result of (6) -(x, (y, zu)) = - (ZU, (x, y))= ((x, y), zu). With this replacement in (25) we obtain (26). Now replace x with (x, y) and y with z in (24). Then 3((x, y),z,u)=((x, y),z)u-((x, y),zu). Thus ((x3 Y), =u) = ((4 Y), =)u-
3((x, y), =, u).
Replacing ((x, y), zu) in (26) by this identity we obtain (27). But (6) implies (x, (y,z))u= -((y,z),x)u= -((x, y),z)u. Thus two termsin (27) cancel and we are left with -3(x, (y, z, u)) = -3((x, y), z, u). But now (20), (22), and (23) combined yield -3((x, y), z, U) = 3(z, (x, y), U) = 3(x, (Y, ~1%~1 = -6(x, (Y, =, u)) so that -3(x, (Y, z, u)) = -6(x, (y, z, u)). Then using (III) we obtain (x, (y, z, u)) =O. This combined with (22)
ALTERNATIVE
223
RINGS AND ((a,b),a)=O
implies (28). In (19) replace y by (y. z). Thus (x, (y, z))*= 3(x, (y, z), (x, (y, 2))). Now (x, (y, Z))E U by virtue of (8). But then (x, (y, z), (x, (y, z))) = 0 as a result of (28). This yields (29). Also (28) implies (x, y, u) E U. Then (I) implies (x, u, y) E U. Finally ((x, y), u) = 0 implies J(x, y, u) = 0 using (9). But now (u, x, y) E 17. This proves (30). A linearization of (4) shows that (4 WY(Y, z)x)+ (u, (Y, z), wx) = (4 w, X)(Y, z) + (4 (Y, z), x)w. As a result of (28) and (20) two of those terms drop out and we are left with (u, w, x)( y, z) = (u, w, ( y, z)x). But now (30) implies (3 1). Using (3) we see that (33)
((u, w, X)(Y, z), r) = (u, w, X)((Y, z), r) + ((4 w, x)7 r)(y, z) + a(& w, xl, (Y, z), r) + (r, (u, w, x), (Y, z)).
Use of (31) yields (u, w, x)( y, z) E U so that ((u, w, x)( y, z), r) = 0. Also (30) implies u’ = (u, w, x) E U so that ((u, w, x), r)( y, z) = 0. But then (28) yields (r, ( y, z), u’) = 0. This and (20) imply (u’, ( y, z), r) = 0. Therefore 2((u, w, x), (y,z), r)=O. Similarly, 2(r, u’, (y. t))=(u’(y,z), r)=O, using (20), so that (r, (u, w, x), (y, z)) =0 as a result of (III). Now four of the terms in (33) have been shown to vanish. Thus the remaining term must be zero as well. But then (u, w, x)(( y, z), r) = 0 and we have proven (32). LEMMA 1. IfIisanidealofRandT(I)={t~R~It=O=tI} also an ideal of R and Z(T(Z))=O= (T(I))Z.
thenT(I)is
Prooj Let t E T(Z), XEZ, rE R. Then (t, x, r) =0= (r, x, t). Using (I) we obtain (t, r, x) = 0 = (r, t, x). Observe that (t, x) = 0, so that ((t, x), r) = 0. From (9) it follows that J(t, x, r) = 0. This allows us to conclude that (x, r, t) = 0 = (x, t, r). This is all the associativity one needs to finish the proof of the lemma. DEFINITION.
We define J as all finite sums of elements of the form
(R, R, U) and (R, R, U)R. LEMMA
2. J is an ideal of R.
Proof: Because of (20), (u, x, y) = -2(x, y, u). Thus (u, x, y) (I) implies (x, u, y) = -(x, y, u) E J. Using (30) we have u’ = (x, Thus zu’ = u’z E J. Also u’z . w = (u’, z, w) + u’ . zw E J. Besides -(z, u’, w) + zu’ . w = -(z, u’, w) + u’z . w E J. This completes the the lemma.
E J. Also y, u) E U. z . u’w = proof of
224
ERWIN
KLEINFELD
MAIN THEOREM. If R is semi-prime ((x, y), z) = 0, hence is strongly ( - 1, 1).
then R satisfies
the identity
Proof: As in Lemma 1, form T(J). Define K to be the ideal generated by (((R, R), R), R, R). However, (8) implies ((R, R), R) c U, so that Kc J. From (32) it follows that (U, R, R)((R, R), R) = 0. Then (20) and (III) yield (R, R, U)( (R, R), R) = 0. Using (30) we have (x, y, u) E U so that (x, y, u)z = z(x, y, u). But (28) implies (a, (b, c), a’) = 0. Thus (z, u’, ((b, c), d))=O. Let u’= (x, y, u). Thus z(x, y, u) .((b, c), d) = z . (x, y, u)((b, c), d) = 0. This shows (x, y, u)z . ((b, c), d) = 0. This suffices to show J. ((b, c), d) = 0, so that ((R, R), R) c T(J). Since T(J) is an ideal it follows that (((R, R), R), R, R) c T(J) and so Kc T(J). But now T(J) . J= 0 implies K* = 0. Using semi-prime it follows that K= 0. Thus ((R, R), R) lies in the left nucleus. Using (20) and (III) then ((R, R), R) lies in the nucleus and then by (8) also in the center of R. From (29) we know that ((x, y), z)’ = 0. If L is the ideal generated by ((x, y), z) then L* = 0, so L = 0, using semi-prime. This concludes the proof of the theorem.
Next we present an example of an non-semi-prime ring which satisfies (I), (II), and (III) but which is not ( - 1, 1). It is an algebra of dimension ten over the rationals with basis elements x, y, z, a, b, c, d, r, S, t. All products of basis elements are zero except the following: xy=a, xz= b, zx=c, zy=d, xd=r+s-2t, yb= -t, yc=t, za=r, az=r+2s-2t, by = -s, cy = s, dx = r - s. The only nonzero associators involving basis elements are (x, y,z)=r+2s-2t, (y,z,x)= -t, (z,x, y)= -r+s, (x, z, y) = -r - 2s + 2t, ( y, x, z) = t,. and (z, y,x) = r - s. From this one
can deduce that this algebra is right alternative. All nonzero double commutators of basis elements are ((x, y), z) = 2s - 2t = (( y, z), x) = ((z, x), y) = - (( y, x), z) = - ((x, z), y) = - ((z, y), x). From this one may deduce that the algebra satisfies (II). On the other hand (x, y, z) + ( y, z, x) + (z, x, y) = 3s - 3t, so the algebra fails to be Lie-admissible and thus is not (-1, 1). Pchelincev [6] has constructed prime, strongly ( - 1, 1) rings which are not associative so one needs to assume simplicity if one wants to extend our main theorem to obtain associativity. Hentzel [Z] has shown that simple, strongly ( - 1, 1) rings are associative. We conclude with the following observation. Rings satisfying (I), (II), and (III) as a result of (14) and (9) satisfy the identity J(x, y, xy) = 0. While this is a mild form of locally (- 1, 1) there are some additional results in the literature about such rings.
ALTERNATIVE
RINGS AND ((a, b), a) = 0
225
REFERENCES 1. I. R. HENTZEL, Nil semi-simple (- 1, 1) rings, J. Algebra 22 (1972), 442450. 2. I. R. HENTZEL, The characterization of (- 1, 1) rings, J. Algebra 30 (1974), 236-258. 3. E. KLEINFELD, On a class of right alternative rings, Math. Z. 87 (1965), 12-16. 4. E. KLEINFELD, Right alternative tings, Proc. Amer. Math. Sot. 4 (1953), 939-944. 5. C. MANERI, Simple (- 1, 1) rings with idempotent, Proc. Amer. Math. Sot. 14 (1963), ll@-117. 6. S. V. PCHELINCEV,Prime algebras and absolute divisors of zero (Russian), Izv. Akad. Nuuk SSSR Ser. Mat. 50 (1986), 706720. 7. S. V. PCHELINCEV,Nilpotency of the associator ideal of a free finitely generated (-1, 1) ring (Russian), Algebra i Logika 14 (1975). 543-572. 8. R. E. ROOMELDI,Nilpotency of ideals in a (- 1, 1) ring with minimum condition (Russian), Algebra i Logika 12 (1973), 333-348. 9. N. STERLING, Prime (- 1, 1) rings with idempotent, Proc. Amer. Math. Sot. 18 (1967), 902-909.
OF ALGEBRA
119, 218~225
A Generalization
(1988)
of Strongly ERWIN
( - 1, 1 ) Rings
KLEINFELD
Department of Mathematics, University of Iowa,Iowa City, Iowa Communicated Received
by Walter April
52242
Feit
28, 1987
I. N-RoDUCTI~N
In the structure theory of (- 1, 1) rings the identity ((x, y), z) =O, becomes a natural stepping stone. Indeed if A is the ideal generated by all associators of R then ((R, R), A) = 0 [2]. Right alternative rings satisfying ((x, y), z) = 0, are called strongly (- 1, l), for they are (- 1, 1). Here we take up the study of right alternative rings which satisfy the weaker identity ((a, b), a) = 0. Under the assumption of semi-prime and characteristc 22, 3 we prove the ring is strongly (- 1, 1). At this point there exist examples of even prime rings which need not be associative, due to Pchelincev [6]. On the other hand there do exist other conditions that imply associativity Cl, 7, 81. We present a non-semi-prime example which satisfies the right alternative identity and ((a, 6), a) = 0, but which is not (- 1, 1).
II. MAIN
RESULTS
We use the usual definitions of assciator and commutator. That is to say (x, y, z) = xy . z - x . yz and (x, y) = xy - yx. Throughout the paper R stands for a ring which is (I) right alternative, (II) satisfies the identity ((a, b), a) = 0, for all a, b E R, and (III) has characteristic # 2, 3. Before we are able to prove the main result we need to establish 32 identities. We shall first list 5 which are valid in right alternative rings and 14 more, which we then prove. After proving Theorem 1 we then list the remaining 13 identities, followed by their proofs. Then we finish up with the main result and an example. First we list without proof the following: (1) (2)
(3) 0021-8693/88
(4 Y2, z) = (x, YYYZ + KY), (wx, y, z) + (4 x7 (Y, z)) = 4-T Y, z) + (w, y, z)x, (XY, z) = X(Y, z) + (x, Z)Y + 2(x, Y, z) + (z, x, Y), 218 $3.00
Copyright c 1988 by Academic Press, Inc. All rights of reproductmn in any form reserved.
ALTERNATIVE
(4) (5)
RINGS
AND
((a,
b),
(4 YT YZ) = (4 Y, Z)Y, ((x, Y), z) + ((Y, z), x) + ((z, xl, y) = 2%
219
a) = 0
YTZ)>
where we define J(x, y, z) = (x, y, z) + ( y, z, x) + (z, x, y). These identities for right alternative rings are well known. In particular proofs may be found in [4].
((4~)~ n(y)), n(z)) = sm 4(x9 Y), ~1, (7) ((w,x), (y,z))=O, (8) (((w xl, Y), z) = 0, (6)
(9) (10) (11) (12)
(13) (14) (15) (16)
(17) (18)
(19) Identity
3((x, Yh z) = 2.0, Y, z), 4x, Y, (4 b)) = 0, ((4 x)3 Y, z) - ((Y, z), w, x) = (WY(x9 Y, 2)) - 6, (WYY, z)), (x7 x9 6, Y)) = 0, (6, Y), x, Y) = -((x, Y), Y, xl = -2(x, Y, (x, Y)) = 2t.K 4 (4
by, (x7Y))
Y)),
= 0,
((Y, 4 Y), x) = (4 YY - 3(X> Y> (xv Y)),
((4 ((4 ((4 b,
YY xl, Y) = ((Y, x7 Y), x)3 Y9 x), x) = 0, Y, xl, Y) = 0, YY = 3(x, Y? (4 Y)).
(6) is just a linearization
of (II).
It follows from
((w,x),(~,z))=-(((~,z),~),w)=(((~,~),z),w)=-((w,z),(y,~)). x(z))) = Sgn n((w, z), (y, 2)). then ((n(w), n(x)), (4~)~
(6) that But
Consequently ((w, x), ( y, z)) = (( y, z), (w, x)). On the other hand it follows from the definition of the commutator that ((w, x), ( y, z)) = -(( y, z), (w, x)), so that 2((w, x), (y, z)) = 0. Then (7) follows because of (III). But then (((w, x), y), z) = -((z, y), (w, x)) using (6), so that (7) now yields (8). As a consequence of (6) we have ((x, y), z) = (( y, z), x) = ((z, x), y). This allows us to rewrite (5) as (9). Using (7) while substituting z = (a, b) in (9) we obtain 2J(x, y, (a, b)) = 0. Then use of (III) leads to (10). As a consequence of (2) we have (wx, y, z) + (w, x, ( y, z)) = w(x, y, z) + (w, y, z)x as well as the last (xw, y, z) + (A w, (Y, z)) = x(w, y, z) + (x, y, z)w. Subtracting equation from the previous, therefore ((w, x), Y, z) + (w & (Y, 2)) - (x, w, (Y, z)) = (w (x, Y, z)) - (XT (w, y, z)). However, it follows from (10) that (w 4 (Y, z)) + (4 (y, z), w) + ((Y, Z)? WYx) =a
220
ERWIN
KLEINFELD
Using (I) on this equation we obtain (w, x, ( y, z)) - (x, w, ( y, z)) = -(( y, z), w, x). With this substitution we have established (11). In a right alternative ring we know (x, x, x) = 0. But since (x, x, x) = (x2, x) we have (x2, x) = 0. But then (6) implies (x2, (x, y)) = - ( y, (x, x2)) = 0. Applying (3) to (x2, (x, y)) we thus have 2(x, x, (x, y)) + ((x, y), x, x) =O, since (x, (x, v)) =0 as a result of (II). But (I) implies ((x, y), x, x) =O. At this point (III) gives (12). Linearize ( 12) by replacing one x by y. Thus (y, x, (x, y)) + (x, y, (x, y)) + (x, x, ( y, y)) = 0. From this follows (x, y, (x, y)) = -(y, x, lx, y)). Using (I) we have (x, Y, (x, Y)) = -(x7 (x, Y), Y) and -(Y, x, (x, Y)) = (Y, (x, Y), x). Now (10) implies 44
Y, (x3 Y)) = 0 = (x, Y? (4 Y)) + (Y, (x3 Y), xl + ((4 Y), x9 Y).
At this point (13) is obvious. It is implied (XY, (x3 Y)) =x(x
by (3) that
6% v)) + (4 (x, Y1J.Y+ 2(x, YT(x3 Y)) + ((xv Yh x> Y).
However, the last two terms cancel each other because of (13). are zero because of (II). Thus (14) follows. Using (6) (by, y), xl = -((x3 Y), v). But -((x9 Y), v) = Cv, (x, v)), definition of the commutator. Thus (14) implies ((xy, y), x) = implies (xy, y) = (x, y)y + ( y, x, y). Substituting this ((x, y) y, x) + (( y, x, v), x) = 0. However, (3) also implies
Two more we have using the 0. But (3) leads to
((4 Y) Y, xl = (4 Y)( YYx) + ((XT Y), x) Y + 2((-% Y), Y9xl + (x9 (4 Y), Y) = -(x9 Y)(X, Y) + 3(x, Y, (XT Y)), using (13). Thus -(x9 y)’ + 3(x, y, (x, y)) + ((y, x, Y), x) = 0. This is equivalent to (15). In (15) interchange x and y. Since ( y, x)’ = (x, y)’ and (Y, x, (y, x)) = -(Y, x, (x, Y)) = (x, Y, 6, Y)) using (13) we obtain (16). Next linearize (16) by replacing y by an x. Thus ((x, x, x), y) + ((x, Y, xl, x) = ((x, x, Y), x) + ((Y, x, x), x). Using (1) thus ((x3 Y, xl, x) = ((x,x, v), x). Since (x, x, y)= -(x, y, x) as a result of (I) thus 2((x, y, x), x) = 0. This implies ((x, y, x), x) = 0, which is (17). In (11) let w=y and z=x. Then ((A xl, YYxl - ((3s xl, Y>xl = (.Y?(x9 YYxl) - (xv (Y, YPx)), so that (Y, (x, Y, x)) = (x, (Y, Y, x)). However, (Y, Y, x) = -(Y, x, v) using (I), thus ( y, (x, y, x)) = -(x, ( y, x, y)). Comparison with (16) now leads to (18). Combining (15) and (18) one obtains (19). We take time out now to prove our first theorem.
ALTERNATIVE THEOREM
RINGS AND (( U,b),u)=O
221
1. Zf e is an idempotent of R then e must be in the center of R.
ProoJ: Put x= e in (12). Thus (e, e, (e, y)) = 0. If we put x= y =e in
(1) we obtain (e, e, z) = (e, e*, z) = (e, e, ez +ze). But (e, e, (e, z)) =0 thus (e, e, ez + ze) = 2(e, e, ez). But then using (4) 2(e, e, ez) = 2(e, e, z)e. Thus (e, e, z)=2(e, e, z)e. If a= (e, e, z) then a=2ae. But then a=2ae=4ae.e. Then (I) implies 4ae . e = 4ae2= 4ae. Thus 2ae= 0, hence a = 0. We have shown 0 = a = (e, e, z). From this, using (I) we obtain (e, z, e) = 0 and also (z, e, e) = 0. This suffices to allow the usual Peirce decomposition of R as RI1 + RIO+ Rol+Roo. If x,,ER,, then (x,,,e)= -x1,,, so that (bm e), e) = x,~. But then (II) implies xi0 = 0, so that Rio= 0. Similarly RO,= 0. At this point it is obvious (e, R) = 0. Also using (1) with y=e and z=zll and also y=e and z=z,,,, then (x, e, z,,) = 2(x, e, zii) while (x, e, zoo) = 0. Thus (x, e, z) = 0, so that (R, e, R) = 0.
But then
(R, R, e) = 0, using (I). Furthermore (e, R) = 0 implies 0 = ((e, y), z) so that J(e, y, z) = 0 using (9). At this point we obtain (e, y, z) = 0, so that (e, R, R) = 0. Thus e is in the center of R. This concludes the proof of the theorem. Define U= {U E R 1(u, R) = O}. U is also known by the name of commutative center. The element u is reserved for a generic element of U. We continue now with the rest of the numbered identities, to be followed by their proofs. (20) (21) (22)
(4 x, Y) = --2(x, y, u) = -2(Y, u, xl = a Y, x9 u) = 2(x, u, Y), (4~)~ (4x), n(y), ~1) = sm n(z, (x9 Y, u)), (x, (Y, z), u) = -2(x, (Y, z, u)),
(23) b(x), (4~), 4~))~ u)=sgne, (24) 3(x, Y, u)=(x, Y)U-(x, ~4, (25) -3(x, (Y,z,u))=(x, (Y,ZW(X,
(Y,z),u),
(26)
((xv Y), zu),
-3(x,
(Y, z, ~1) =(x,
(Y, z))u+
(Y+)),
222
ERWIN
(27) (28)
(29) (30) (31) (32)
KLEINFELD
-3(x, (Y, =, 4) = (4 (Y, 4)~ + ((x3 Y), Z)U- 3((~, Y), =, 4, (4 (Y, =I, u) = 0 = (4 (Y, =9u)), ((x, Y), 42 =o, (R, R, U) c U, CR, U, W = U, (u, R R) c u, (u, R R)(R, R) c u, (U, R, RN@, RI, R)=O.
In (3) let z = U. Thus 2(x, y, U) + (a, x, y) = 0. Also since ((x, y), U) = 0, we obtain from (9) that J(x, y, U) = 0. These identities and (I) suffice to yield (20). As a result of (20) we have (x, y, U) = -(y, x, U) so that (z, (x, y, u)) = -(z, ( y, x, u)). Linearizing (17) by replacing an x by U, we obtain ((u, y, x), x) + ((x, y, u), x) + ((x, y, x), u) = 0. But then (20) implies 3((x, y, u), x) = 0. Thus (III) implies ((x, y, u), x) = 0. Linearizing this yields ((x, y, u), z) = -((z, y, u), x). This suffices to prove (21). Substitute w=u in (11). Thus ((u,x),Y,=)-((y,=),~,x)=(~,(x,y,z))-(x,(u,y,z)). But then -((y, z), U, x) = -(x, (u, y, z)). At this point (20) implies -((y, z), u, xl= -(x, (Y, =I, u) while -(x, (u, Y, 2)) = 2(x, (Y, z, u)), so that we obtain (22). Combining (21) and (22) leads to (23). Also -(x, JJU) = (yu, x), while (3) implies ~Y~,~~=Y~~,~~+~Y,~~~+~~Y,~,~~+~~,Y,~~
= -(x,
y)u + 3(x, y, u)
using (20). This proves (24). Replace y by (y, z) in (24). Thus 3(x, (y, z), U) = (x, (y, z))u- (x, (y, z)u). Also replace x by y and y by z in (24) and commute with x. This yields 3(x, ( y, z, u)) = (x, ( y, z)u) (x, ( y, zu)). Because of (22) we can replace 3(x, ( y, z), U) by -6(x, ( y, z, u)) so that -6(x, ( y, z, u)) = (x, ( y, z))u - (x, ( y, z)u). Adding the last two equations we obtain -3(x, ( y, z, u)) = (x, (y, z))u - (x, ( y, zu)) or (25). As a result of (6) -(x, (y, zu)) = - (ZU, (x, y))= ((x, y), zu). With this replacement in (25) we obtain (26). Now replace x with (x, y) and y with z in (24). Then 3((x, y),z,u)=((x, y),z)u-((x, y),zu). Thus ((x3 Y), =u) = ((4 Y), =)u-
3((x, y), =, u).
Replacing ((x, y), zu) in (26) by this identity we obtain (27). But (6) implies (x, (y,z))u= -((y,z),x)u= -((x, y),z)u. Thus two termsin (27) cancel and we are left with -3(x, (y, z, u)) = -3((x, y), z, u). But now (20), (22), and (23) combined yield -3((x, y), z, U) = 3(z, (x, y), U) = 3(x, (Y, ~1%~1 = -6(x, (Y, =, u)) so that -3(x, (Y, z, u)) = -6(x, (y, z, u)). Then using (III) we obtain (x, (y, z, u)) =O. This combined with (22)
ALTERNATIVE
223
RINGS AND ((a,b),a)=O
implies (28). In (19) replace y by (y. z). Thus (x, (y, z))*= 3(x, (y, z), (x, (y, 2))). Now (x, (y, Z))E U by virtue of (8). But then (x, (y, z), (x, (y, z))) = 0 as a result of (28). This yields (29). Also (28) implies (x, y, u) E U. Then (I) implies (x, u, y) E U. Finally ((x, y), u) = 0 implies J(x, y, u) = 0 using (9). But now (u, x, y) E 17. This proves (30). A linearization of (4) shows that (4 WY(Y, z)x)+ (u, (Y, z), wx) = (4 w, X)(Y, z) + (4 (Y, z), x)w. As a result of (28) and (20) two of those terms drop out and we are left with (u, w, x)( y, z) = (u, w, ( y, z)x). But now (30) implies (3 1). Using (3) we see that (33)
((u, w, X)(Y, z), r) = (u, w, X)((Y, z), r) + ((4 w, x)7 r)(y, z) + a(& w, xl, (Y, z), r) + (r, (u, w, x), (Y, z)).
Use of (31) yields (u, w, x)( y, z) E U so that ((u, w, x)( y, z), r) = 0. Also (30) implies u’ = (u, w, x) E U so that ((u, w, x), r)( y, z) = 0. But then (28) yields (r, ( y, z), u’) = 0. This and (20) imply (u’, ( y, z), r) = 0. Therefore 2((u, w, x), (y,z), r)=O. Similarly, 2(r, u’, (y. t))=(u’(y,z), r)=O, using (20), so that (r, (u, w, x), (y, z)) =0 as a result of (III). Now four of the terms in (33) have been shown to vanish. Thus the remaining term must be zero as well. But then (u, w, x)(( y, z), r) = 0 and we have proven (32). LEMMA 1. IfIisanidealofRandT(I)={t~R~It=O=tI} also an ideal of R and Z(T(Z))=O= (T(I))Z.
thenT(I)is
Prooj Let t E T(Z), XEZ, rE R. Then (t, x, r) =0= (r, x, t). Using (I) we obtain (t, r, x) = 0 = (r, t, x). Observe that (t, x) = 0, so that ((t, x), r) = 0. From (9) it follows that J(t, x, r) = 0. This allows us to conclude that (x, r, t) = 0 = (x, t, r). This is all the associativity one needs to finish the proof of the lemma. DEFINITION.
We define J as all finite sums of elements of the form
(R, R, U) and (R, R, U)R. LEMMA
2. J is an ideal of R.
Proof: Because of (20), (u, x, y) = -2(x, y, u). Thus (u, x, y) (I) implies (x, u, y) = -(x, y, u) E J. Using (30) we have u’ = (x, Thus zu’ = u’z E J. Also u’z . w = (u’, z, w) + u’ . zw E J. Besides -(z, u’, w) + zu’ . w = -(z, u’, w) + u’z . w E J. This completes the the lemma.
E J. Also y, u) E U. z . u’w = proof of
224
ERWIN
KLEINFELD
MAIN THEOREM. If R is semi-prime ((x, y), z) = 0, hence is strongly ( - 1, 1).
then R satisfies
the identity
Proof: As in Lemma 1, form T(J). Define K to be the ideal generated by (((R, R), R), R, R). However, (8) implies ((R, R), R) c U, so that Kc J. From (32) it follows that (U, R, R)((R, R), R) = 0. Then (20) and (III) yield (R, R, U)( (R, R), R) = 0. Using (30) we have (x, y, u) E U so that (x, y, u)z = z(x, y, u). But (28) implies (a, (b, c), a’) = 0. Thus (z, u’, ((b, c), d))=O. Let u’= (x, y, u). Thus z(x, y, u) .((b, c), d) = z . (x, y, u)((b, c), d) = 0. This shows (x, y, u)z . ((b, c), d) = 0. This suffices to show J. ((b, c), d) = 0, so that ((R, R), R) c T(J). Since T(J) is an ideal it follows that (((R, R), R), R, R) c T(J) and so Kc T(J). But now T(J) . J= 0 implies K* = 0. Using semi-prime it follows that K= 0. Thus ((R, R), R) lies in the left nucleus. Using (20) and (III) then ((R, R), R) lies in the nucleus and then by (8) also in the center of R. From (29) we know that ((x, y), z)’ = 0. If L is the ideal generated by ((x, y), z) then L* = 0, so L = 0, using semi-prime. This concludes the proof of the theorem.
Next we present an example of an non-semi-prime ring which satisfies (I), (II), and (III) but which is not ( - 1, 1). It is an algebra of dimension ten over the rationals with basis elements x, y, z, a, b, c, d, r, S, t. All products of basis elements are zero except the following: xy=a, xz= b, zx=c, zy=d, xd=r+s-2t, yb= -t, yc=t, za=r, az=r+2s-2t, by = -s, cy = s, dx = r - s. The only nonzero associators involving basis elements are (x, y,z)=r+2s-2t, (y,z,x)= -t, (z,x, y)= -r+s, (x, z, y) = -r - 2s + 2t, ( y, x, z) = t,. and (z, y,x) = r - s. From this one
can deduce that this algebra is right alternative. All nonzero double commutators of basis elements are ((x, y), z) = 2s - 2t = (( y, z), x) = ((z, x), y) = - (( y, x), z) = - ((x, z), y) = - ((z, y), x). From this one may deduce that the algebra satisfies (II). On the other hand (x, y, z) + ( y, z, x) + (z, x, y) = 3s - 3t, so the algebra fails to be Lie-admissible and thus is not (-1, 1). Pchelincev [6] has constructed prime, strongly ( - 1, 1) rings which are not associative so one needs to assume simplicity if one wants to extend our main theorem to obtain associativity. Hentzel [Z] has shown that simple, strongly ( - 1, 1) rings are associative. We conclude with the following observation. Rings satisfying (I), (II), and (III) as a result of (14) and (9) satisfy the identity J(x, y, xy) = 0. While this is a mild form of locally (- 1, 1) there are some additional results in the literature about such rings.
ALTERNATIVE
RINGS AND ((a, b), a) = 0
225
REFERENCES 1. I. R. HENTZEL, Nil semi-simple (- 1, 1) rings, J. Algebra 22 (1972), 442450. 2. I. R. HENTZEL, The characterization of (- 1, 1) rings, J. Algebra 30 (1974), 236-258. 3. E. KLEINFELD, On a class of right alternative rings, Math. Z. 87 (1965), 12-16. 4. E. KLEINFELD, Right alternative tings, Proc. Amer. Math. Sot. 4 (1953), 939-944. 5. C. MANERI, Simple (- 1, 1) rings with idempotent, Proc. Amer. Math. Sot. 14 (1963), ll@-117. 6. S. V. PCHELINCEV,Prime algebras and absolute divisors of zero (Russian), Izv. Akad. Nuuk SSSR Ser. Mat. 50 (1986), 706720. 7. S. V. PCHELINCEV,Nilpotency of the associator ideal of a free finitely generated (-1, 1) ring (Russian), Algebra i Logika 14 (1975). 543-572. 8. R. E. ROOMELDI,Nilpotency of ideals in a (- 1, 1) ring with minimum condition (Russian), Algebra i Logika 12 (1973), 333-348. 9. N. STERLING, Prime (- 1, 1) rings with idempotent, Proc. Amer. Math. Sot. 18 (1967), 902-909.