A short characterization of the Octonions
aa r X i v : . [ m a t h . R A ] F e b ALTERNATIVE RINGS WHOSE COMMUTATORS ARENOT ZERO-DIVISORS
ERWIN KLEINFELD YOAV SEGEV
Abstract.
The purpose of this short paper is to prove that an alterna-tive ring R whose commutators are not zero-divisors, in characteristic = 2 , , is an octonion algebra over its center. We also give other hy-potheses for R to be an octonion algebra. Introduction
In this paper we charaterize the octonions in charateristic not 2 as inTheorems 1.1 below. This differs from previous characterization as we donot emphasize ideals. For historical information on the octonions see [Bl, E],and for their connection with algebraic groups see [SpV].The second author would like to thank the first author for contacting him,and for including him in this project.
Theorem 1.1.
Let R be an alternative, not associative ring. Assume that (1) a non-zero commutator in R is not a divisors of zero; (2) 2 x = 0 or x = 0 = ⇒ x = 0 , for all x ∈ R. Then R contains no divisors of zero. As a corollay we get:
Theorem 1.2.
Let R be as in Theorem 1.1. Then the localization of R atits center is an -dimensional octonion algebra over its center–the fractionfield of the center of R. Proof.
This follows from Theorem 1.1 together with [BK, Theorem A]. (cid:3)
In a commutative alternative ring ( x, y, z ) = 0 , for each associator ( x, y, z )(see [S, Lemma 9, p. 706]). Hence hypothesis (2) of Theorem 1.3 below is anecessay hypothesis and is implied by the hypothesis 3 x = 0 = ⇒ x = 0 inTheorem 1.1. Indeed this holds by Remark 1.4(1) below.The hypothesis 3 x = 0 = ⇒ x = 0 in Theorem 1.1, is not a necessayhypothesis. Theorem 1.3.
Let R be an alternative, not associative ring. Assume that Date : March 1, 2021.2000
Mathematics Subject Classification.
Primary: 17D05 ; Secondary: 17A35.
Key words and phrases.
Octonion algebra, Alternative ring, commutator, associator. (1) a non-zero commutator in R is not a divisors of zero; (2) commutative subrings of R are associative; (3) 2 x = 0 = ⇒ x = 0 , for all x ∈ R. Then the localization of R at its center is an -dimensional octonion algebraover its center–the fraction field of the center of R. Remarks 1.4. (1) Let R be a commutative alternative ring. It is wellknown that 3( x, y, z ) = 0 , for each associator ( x, y, z ) ∈ R (see, e.g.,[K2, equation (6), p. 131]). Hence the hypothesis 3 x = 0 = ⇒ x = 0 , implies hypothesis (2) of Theorem 1.3.(2) The hypothesis that 2 x = 0 = ⇒ x = 0 in the above theoremsis used in the proof of Lemma 3.1(3&4). We do not know whathappens without this hypothesis.By Lemma 2.3 below, hypotheses (1) and (2) of Theorem 1.5 below implyhypothesis (1) of Theorem 1.1. Hence Theorems 1.1 and 1.3 imply Theorem 1.5.
Let R be an alternative, not associative ring. Assume that (1) no non-zero commutator of R is nilpotent; (2) non-zero elements of the nucleus of R are not divisors of zero in R ;(3) at least one of the following two hypotheses hold: (i) commutative subrings of R are associative. (ii) 3 x = 0 implies x = 0 , for all x ∈ R. (4) 2 x = 0 implies x = 0 , for all x ∈ R. Then the localization of R at its center is an -dimensional octonion algebraover its center–the fraction field of the center of R. Finally, hypotheses (1)–(3) of Theorem 1.6 below imply hypotheses (1)&(2) of Theorem 1.5. This is used in the proof of
Theorem 1.6.
Let R be an alternative, not associative ring such that (1) no non-zero commutator of R is nilpotent; (2) R has an identity; (3) the nucleus of R is a simple ring; (4) hypothesis (3) of Theorem 1.5 holds. (5) 2 x = 0 implies x = 0 , for all x ∈ R. Then the center C of R is a field and R must be an -dimensional octonionalgebra over C. Preliminaries on alternative rings
Our main references for alternative rings are [K1, K2]. Let R be a ring,not necessarily with and not necessarily associative. Definitions 2.1.
Let x, y, z ∈ R. (1) The associator ( x, y, z ) is defined to be( x, y, z ) = ( xy ) z − x ( yz ) . (2) The commutator ( x, y ) is defined to be( x, y ) = xy − yx. (3) R is an alternative ring if( x, y, y ) = 0 = ( y, y, x ) , for all x, y ∈ R. (4) The nucleus of R is denoted N and defined N = { n ∈ R | ( n, R, R ) = 0 } . Note that in an alternative ring the associator is skew symmetric inits 3 variables ([K2, Lemma 1]). Hence (
R, n, R ) = (
R, R, n ) = 0 , for n ∈ N. (5) The center of R is denoted C and defined C = { c ∈ N | ( c, R ) = 0 } . In the remainder of this paper R is an alternative ring which is notassociative . N denotes the nucleus of R and C its center. Facts 2.2.
Let R be an alternative ring. (1) If v ∈ R is a commutator, then v ∈ N. (2) If v ∈ R is a commutator, then [ v , R, R ] v = 0 . (3) Any subring of R generated by two elements is associative.Proof. For (1)&(2) see [K1, Theorem 3.1]. Part (3) is a theorem of E. Artin. (cid:3)
We also need the following easy lemma.
Lemma 2.3.
Assume that (i)
No non-zero commutator of R is nilpotent. (ii) non-zero elements of the nucleus of R are not divisors of zero in R ; Then a non-zero commutator in R is not a divisors of zero.Proof. Let v be a non-zero commutator. Then, by Facts 2.2(1), 0 = v ∈ N. If t ∈ R is such that vt = 0 , then since the subring generated by v and t isassociative, v t = 0 , so t = 0 . (cid:3) The following lemma gives some properties of R which we will requirelater. Lemma 2.4.
Let w, x, y, z ∈ R, and n, n ′ ∈ N, then (1) [ N, R ] ⊆ N. (2) Nuclear elements commute with associators. (3) ( w, n )( x, y, z ) = − ( x, n )( w, y, z ) . In particular, ( n ′ , n )( x, y, z ) = 0 . (4) If N contains no zero divisors, then N is commutative. (5) ( w, n )( w, n )( x, y, z ) = 0 . KLEINFELD AND SEGEV
Proof. (1)&(2) By [K2, equation (2), p. 129],(2.1) ( nx, y, z ) = n ( x, y, z ) . Also, by the beginning of the proof of Lemma 4, p. 132 in [K2],(2.2) ( nx, y, z ) = ( x, y, z ) n and ( xn, y, z ) = n ( x, y, z ) . So ( nx, y, z ) = ( xn, y, z ) , hence (1) holds, and also n ( x, y, z ) = ( x, y, z ) n, so(2) holds.(3) By [K2, equation (6), p. 131],(2.3) ( wx, n ) = w ( x, n ) + ( w, n ) x. Now ( w, n )( x, y, z ) = (( w, n ) x, y, z ) (by (1) and equation (2.1))= − ( w ( x, n ) , y, z ) (by equation (2.3) and (1))= − ( x, n )( w, y, z ) (by equation (2.2) and (1)) . Taking w = n ′ ∈ N, we get the last statement of (3).(4) This is immediate from the last statement of (3) and the hypothesisthat R is not associative.(5) We have( w, n )( w, n )( x, y, z ) = − ( w, n )( x, n )( w, y, z ) (by (3))= − ( w, n )( w, y, z )( x, n ) (by (2))= ( w, n )( y, w, z )( x, n ) (commutators are skew-symmetric)= − ( y, n )( w, w, z )( x, n ) (by (3))= 0 (because R is alternative) . (cid:3) Proof of the main results.
In this section R is an alternative ring which is not associative. We assumethat(1) 2 x = 0 = ⇒ x = 0 , for all x ∈ R. In view of Lemma 2.3, to prove Theorems 1.1, 1.3 and 1.5, we may (and wedo) assume that(2) non-zero commutator are not zero-divisors in R.We further assume that(3) commutative subrings of R are associative.Recall from Remark 1.4(1), that hypothesis (3) is implied by the hypothesis3 x = 0 = ⇒ x = 0 . In the proof of Theorem 1.6, at the end of this section, we show thatthe hypotheses of that theorem together with Lemma 2.3 also imply thatnon-zero commutator are not zero-divisors in R. Lemma 3.1. (1) N is contained in the center of R. So N = C. (2) v ∈ N, for any commutator v ∈ R, hence v ∈ C. (3) u = 0 , and u ∈ C, for any non-zero associator u ∈ R. (4) If u is a non-zero associator, then u is not a zero divisor. (5) If n ∈ N is non-zero, then n is not a zero divisor.Proof. (1) This follows from Lemma 2.4(5). Indeed ( w, n )( w, n ) u = 0 , forany associator u. Hence ( w, n ) = 0 . (2) This follows the fact (Facts 2.2(2)) that [ v , R, R ] v = 0 . (3) Let u = ( x, y, z ) , and suppose that u = 0 . By hypothesis x, y, z can-notcommute pairwise. Without loss of generality assume v = ( x, y ) is not zero.Now [K2, Lemma 3, p. 130] implies that u and v anticommute, which is tosay uv + vu = 0 . If also u and v were to commute we would get 2 uv = 0 , which contradicts our hypotheses.Thus ( u, v ) is a non-zero commutator, so 0 = a := ( u, v ) ∈ C, by (2).Since u and v anti-commute, We have(3.1) a = ( u, v ) = uvuv − uvvu − vuuv + vuvu = − uuvv. This shows that u = 0 . Let c := − v . Then c ∈ C, and cu = a, withnon-zero center elements a, c. For arbitrary elements r and s in R we get from equation 2.1, c ( u , r, s ) =( cu , r, s ) = 0 . Then c ( u , r, s ) = 0 . By hypothesis ( u , r, s ) = 0 , so u ∈ N. Similarly, 0 = ( cu , r ) = c ( u , r ) , so ( u , r ) = 0 , and u is a non-zero elementin the center.(4) Let u = ( x, y, z ) be a non-zero commutator. Suppose ut = 0 , for some0 = t ∈ R. As in the proof of (3) above, we may assume that v = ( x, y ) = 0 . As above, ( u, v ) = 0 . By equation (3.1), and since u ∈ C, we have ( u, v ) t = − v u t = 0 . This contradicts the hypothesis that ( u, v ) is not a zero divisor.(5) Let n ∈ N be non-zero. By (1), n ∈ C . Since we are assuming thatcommutative subrings of R are associative, and that R is not associative,we know that R is not commutative. Let ( v, w ) be a non-zero commutatorin R. Then 0 = ( v, w ) n = ( v, wn ) . Now if nx = 0 , for some x ∈ R, then( v, wn ) x = ( v, w ) nx = 0 , hence x = 0 . (cid:3) Proposition 3.2. R has no zero divisors.Proof. Let 0 = x ∈ R. If x ∈ N, then x is not a zero divisor by Lemma 3.1(5).So assume x / ∈ N. Then there are y, z ∈ R such that u := ( x, y, z ) = 0 . Recallthat by [K2, equation (4), p. 130], ux = ( x, y, z ) x = ( x, y, xz ) and hence u + ux = ( x, y, z + xz ) . KLEINFELD AND SEGEV
We have ( u + ux ) = u + u x + uxu + ( ux ) = ⇒ u x + uxu = ( u + ux ) − u − ( ux ) . Note that 0 / ∈ { ux, ( ux ) , u } by Lemma 3.1(4&5). Note also that u , ( ux ) , ( u + ux ) ∈ N = C, by Lemma 3.1(4).Set a := u and b := ( u + ux ) − u − ( ux ) . Then a, b ∈ C, and ax + uxu = b. Multiplying this last equality by x on the right we see that ax + c = bx, where 0 = c := ( ux ) ∈ C. Suppose now that xt = 0 , for some t ∈ R . Then0 = bxt = ax t + ct. But x t = 0 , so ct = 0 , contradicting Lemma 3.1(5). (cid:3) This completes the proof of the Theorems 1.1, 1.3 and 1.5.
Proof of Theorem 1.6.
So here, instead of assuming that N has no zero divisors, we assume that R has an identity, and that N is a simple ring. We prove that N has no zerodivisors.First we show that N is a field. Recall from Lemma 2.4 that nuclearelements commute with associators. Set I := { n ∈ N | n ( x, y, z ) = 0 , for all x, y, z ∈ R } . Then ( n ′ n )( x, y, z ) = n ′ ( n ( x, y, z ) = 0 , and ( nn ′ )( x, y, z ) = ( x, y, z )( nn ′ ) =(( x, y, z ) n ) n ′ = 0 , for n ∈ I, n ′ ∈ N, x, y, z ∈ R. So I is an ideal of N. Hence I = 0 , since / ∈ I. However by Lemma 2.4(3), any commutator of elementsin N is in I, so N must be commutative. This plus simple is enough to make N into a field.Now let 0 = n ∈ N, such that nt = 0 , for some t ∈ R. Then t = t =( n − n ) t = n − ( nt ) = 0 . Note that the identity of R is the identity of N. Hence by Theorem 1.5, and by [BK, Corollary 1], Theorem 1.6 holds. (cid:3)
Acknowledgement.
We would like to thank Professor Holger Peterssonfor remarks that improved the results of this paper.
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Yoav Segev, Department of Mathematics, Ben-Gurion University, Beer-Sheva 84105, Israel
Email address : [email protected] Erwin Kleinfeld, 1555 N. Sierra St. Apt 120, Reno, NV 89503-1719, USA
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