On the solvability of the commutative power-associative nilalgebras of dimension 6

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1 Linear Algebra and its Applications 369 (2003) On the solvability of the commutative power-associative nilalgebras of dimension 6 Ivan Correa a,1, Irvin Roy Hentzel b, Luiz Antonio Peresi c,,2 a Departamento de Matemática, Universidad de La Serena, Benavente 980, La Serena, Chile b Department of Mathematics, Iowa State University, Ames, IA , USA c Departamento de Matemática, Universidade de São Paulo, Caixa Postal 66281, São Paulo, SP , Brazil Received 16 September 2002; accepted 3 December 2002 Submitted by H. Schneider Abstract We prove that commutative power-associative nilalgebras of dimension 6 over a field of characteristic /= 2, 3, 5 are solvable Elsevier Science Inc. All rights reserved. AMS classification: 17A05; 17A30 Keywords: Commutative; Power-associative; Nilalgebra; Albert s problem; Solvable; Nilpotent 1. Introduction Let A be any nonassociative algebra. We define inductively the following powers of A: A 1 = A, A n = A n 1 A + A n 2 A 2 + +AA n 1 ; A (0) = A, A (n) = (A (n 1) ) 2. We say that A is nilpotent (respectively, solvable) whena k = 0 (respectively, A (k) = 0) for some k. WhenA is nilpotent the smallest k such that A k = 0is Corresponding author. address: peresi@ime.usp.br (L.A. Peresi). 1 Supported by FONDECYT This paper was written while the third named author held a grant from CNPq of Brazil. Part of the research was done when this author was visiting Universidad de La Serena on a grant from FONDECYT /03/$ - see front matter 2003 Elsevier Science Inc. All rights reserved. doi: /s (02)

2 186 I. Correa et al. / Linear Algebra and its Applications 369 (2003) called the index of nilpotency of A. Analogously, we define the index of solvability of A. Clearly, if A is nilpotent then A is solvable. The algebra A is power-associative in case the subalgebra generated by each element of A is associative. For any algebra the (right) powers of an element x in A are defined by x 1 = x,x n+1 = x n x.ifa is power-associative then x i x j = x i+j. An element x in a power-associative algebra A is called nilpotent if there exists a k such that x k = 0. The index of nilpotency for such an element x is the smallest k such that x k = 0. A power-associative algebra is called a nilalgebra if each element is nilpotent. When there is a bound on the indices of nilpotency, the nilindex of the algebra is the smallest k such that x k = 0forallx in A. IfA is a power-associative nilalgebra of dimension n, then the nilindex of A is n + 1. In 1948, Albert [1] conjectured that any finite-dimensional commutative powerassociative nilalgebra is nilpotent. Suttles [8] gave in 1972 the following counterexample: let A be the commutative algebra with basis {e 1,...,e 5 } and nonzero products given by e 1 e 2 = e 2 e 4 = e 1 e 5 = e 3, e 1 e 3 = e 4, e 2 e 3 = e 5 ; A is a powerassociative nilalgebra of nilindex 4 that is solvable but is not nilpotent (since A 3 = A 2 and A 2 A 2 = 0). The following is an open problem. Albert s Problem. Is every finite-dimensional commutative power-associative nilalgebra solvable? Gerstenhaber and Myung [6] proved that every commutative power-associative nilalgebra of dimension 4 over a field of characteristic /= 2 is nilpotent. They also determined the isomorphism classes of all such algebras. A generalization of this result was obtained by Correa and Suazo [4]. They showed that commutative power-associative nilalgebras of nilindex n and dimension n are nilpotent of index n. They found necessary and sufficient conditions for such an algebra to be a Jordan algebra. When the algebra is a Jordan algebra, the corresponding isomorphism classes were given. The results require characteristic /= 2, 3. The noncommutative case was considered by Correa and Hentzel [2]. Let A be a noncommutative power-associative nilalgebra of dimension n and nilindex n over a field of characteristic /= 2, 3. They proved that A is solvable and A 2 is nilpotent. For any n>2, they presented two examples of noncommutative power-associative nilalgebras of dimension n. In the first example, the algebra has nilindex n and is not nilpotent. In the second example, the algebra has nilindex n 1 and is not solvable. Correa and Peresi [3] obtained that a commutative power-associative nilalgebra of dimension 5 is solvable of index 3. Moreover, they proved that a commutative power-associative nilalgebra of nilindex 3 and dimension 5 is nilpotent of index 4. Instead of what was stated in [3] the results require characteristic /= 2, 3. The purpose of this paper is to prove the following result. Theorem 1. Let A be a commutative power-associative nilalgebra of dimension 6 over a field of characteristic /= 2, 3, 5. Then A is solvable.

3 I. Correa et al. / Linear Algebra and its Applications 369 (2003) Basic results and notations A commutative algebra J is a Jordan algebra if it satisfies the Jordan identity (x 2 y)x = x 2 (yx). Any Jordan algebra is power-associative. The nilpotency of Jordan nilalgebras is given by the following result due to Albert. Proposition 2 [9, p. 92]. Any finite-dimensional Jordan nilalgebra over a field of characteristic /= 2 is nilpotent. If a commutative algebra A satisfies x 3 = 0forallx A then A is a Jordan algebra. Therefore we have the following consequence of Proposition 2. Corollary 3. Any finite-dimensional commutative nilalgebra of nilindex 3 is nilpotent (thus it is solvable). Let A be a commutative power-associative nilalgebra of dimension n and nilindex k. As we already noticed, k n + 1. If k = 1thenA = A 1 = 0. If k = 2thenfor any x,y A we have xy = (1/2)((x + y) 2 x 2 y 2 ) = 0. Therefore A 2 = 0. If k = 3thenA is nilpotent by Corollary 3. If k = n then A is nilpotent by Theorem 1 of Correa and Suazo [4]. If k = n + 1andx is an element of A such that x n+1 = 0 but x n /= 0, then {x 1,...,x n } is a basis of A. Using this basis it is easy to see that A is nilpotent. We collect these results in the following proposition. Proposition 4. Let A be a commutative power-associative nilalgebra of dimension n and nilindex k. If k = 1, 2, 3,n,n+ 1 then A is nilpotent (thus it is solvable). The following result will be useful to prove that an algebra is solvable. Proposition 5 [7, Proposition 2.2, p. 18]. If an algebra A contains a solvable ideal I, and if A/I is solvable, then A is solvable. If A is a commutative algebra we indicate by L x the left multiplication by x, i.e., yl x = xy = yx, wherex and y are elements of A. Wedenoteby a 1,a 2,...,a s the subspace of A generated by the elements a 1,a 2,...,a s. 3. Dimension 6 Throughout this section A is a commutative power-associative nilalgebra of dimension 6 over a field K. Unless otherwise stated, we assume that K has characteristic /= 2, 3, 5. By Proposition 4 A is solvable when the nilindex of A is 1, 2, 3, 6 and 7. We know that the nilindex of A is 7. Therefore, to obtain the result stated in the Theorem 1, it remains to prove that A is solvable when the nilindex of A is4or5.

4 188 I. Correa et al. / Linear Algebra and its Applications 369 (2003) Nilindex 4 Assume that A has nilindex 4. By Lemma 2 of [3] L 5 z = 0forallz A. Lemma 6. Any commutative power-associative nilalgebra of nilindex 4 over a field of characteristic /= 2 satisfies the following identities: 2((yx)x)x + (x 2 y)x + x 3 y = 0, (1) (yx)x 2 = 0, (2) 2(yx)(zx) + (yz)x 2 = 0, (3) 2(((yx)x)x)x + x 3 (yx) = 0, (4) (yx 2 )x 3 = 0, (5) (yx 3 )x 2 = 0. (6) Proof. Linearizing x 4 = 0 we obtain (1). Linearizing x 2 x 2 = 0 we obtain (2) and (3). Replacing y by yx in (1) and using (2) we obtain (4). Replacing x by x 2 and z by x in (3) we obtain (5). Replacing z by x 3 in (3) we obtain (6). Lemma 7. Let A be a commutative power-associative nilalgebra of dimension 6 and nilindex 4 over a field of characteristic /= 2. Then A satisfies the following identities: (yx)x 3 = 0, (7) x 3 (yz) = (zx 2 )(yx) 2((zx)x)(yx), (8) (yx 2 )x 2 = 0, (9) A 2 x 2 = (Ax) 2, (10) A 3 x 3 ((A 2 x)x)(ax) + (Ax) 3. (11) Proof. Identity (7) holds if x 3 = 0. Let x A with x 3 /= 0andX = x,x 2,x 3. Let y be an arbitrary element of A. Let A/X be the quotient vector space of A by X. SinceXL x X the linear map L x : A/X A/X, givenby(y + X)L x = yl x + X, is well-defined. Since L 5 x = 0wehave(L x) 5 = 0. Since A/X has dimension 3 we have (L x ) 3 = 0. This implies that yl 3 x X. Thus yl3 x = αx + βx2 + δx 3 (α, β, δ K). Since L 5 x = 0 we obtain αx3 = yl 5 x = 0andthenα = 0. It follows that yl 3 x X2 = x 2,x 3 and so yl 4 x X3 = x 3.Since(yx)x 3 = 2yL 4 x by (4) we obtain that (yx)x 3 = λx 3 (λ K). Since L 5 yx = 0wegetλ5 x 3 = x 3 L 5 yx = 0. Therefore λ = 0 and we obtain (yx)x 3 = 0. Therefore (7) is an identity of A. Linearizing (7) we obtain (8). Replacing z by x 2 in (3) and using (7) we get (9). From (3) we get (10). Letting z A 2 and y A in (8) and using (10) we get (11).

5 I. Correa et al. / Linear Algebra and its Applications 369 (2003) Throughout the rest of this subsection x represents an element of A such that x 3 /= 0andX = x,x 2,x 3. Thus L x /= 0andL 2 x /= 0. Using (4) and (7) we obtain L 4 x = 0. Therefore the minimal polynomial of L x is t 4 or t 3. Let J l denote the l l elementary Jordan matrix associated to the eigenvalue 0. Then the possible Jordan canonical forms of L x are: J [ ] J (a) J4 0, (b), (c) J , J [ ] (d) 0 J 2 0 J3 0, (e). 0 J The basis of A corresponding to each one of these matrices are: (a) {y,yx,(yx)x,((yx)x)x,a,b} with (((yx)x)x)x = 0, ax = 0, bx = 0, (b) {y,yx,(yx)x,((yx)x)x,a,ax} with (((yx)x)x)x = 0, (ax)x = 0, (c) {x,x 2,x 3,a,b,c} with ax = 0, bx = 0, cx = 0, (d) {x,x 2,x 3,z,zx,a} with (zx)x = 0, ax = 0, (e) {x,x 2,x 3,z,zx,(zx)x} with ((zx)x)x = 0. Assume that A has a basis of type (a): Writing x = α 1 y + α 2 yx + α 3 (yx)x + α 4 ((yx)x)x + α 5 a + α 6 b (α i K) and applying L 3 x we obtain α 1 = 0. Thus ((yx)x)x = 1 α 2 x 3. (12) It follows that x 2 = α 2 (yx)x + (α 3 /α 2 )x 3 and then (yx)x = 1 x 2 α 3 α 2 α2 2 x 3. (13) Therefore A = y,yx,x 2,x 3,a,b and we obtain that Ax = yx,x 2,x 3 and (Ax)x = x 2,x 3. Therefore by (2) and (7) we get ((Ax)x)(Ax) = 0. Then from (11) we obtain A 3 x 3 (Ax) 3. Now, (Ax) 3 = (yx) 3,(yx) 2 x 2,(yx) 2 x 3 by (2) and (7). Replacing y and z by yx in (3) we obtain (yx) 2 x 2 = 2((yx)x) 2 = 0 by (13). Replacing y and z by yx in (8) we obtain (yx) 2 x 3 = ((yx)x 2 )((yx)x) 2(((yx)x)x)((yx)x) = 0 by (2), (12) and (13). Replacing x by yx and z by x in (3) we obtain (yx) 3 = 2((yx)y)((yx)x). By (13) we have

6 190 I. Correa et al. / Linear Algebra and its Applications 369 (2003) ((yx)y)((yx)x) = 1 ((yx)y)x 2 α 3 α 2 α2 2 ((yx)y)x 3. Replacing y by yx and z by y in (3) we obtain ((yx)y)x 2 = 2((yx)x)(yx).Therefore ((yx)y)x 2 = 2 ( x 2 α ) 3 x 3 (yx) = 0 α 2 α 2 by (13), (2) and (7). Replacing z by yx in (8) we obtain ((yx)y)x 3 = ((yx)x 2 )(yx) 2(((yx)x)x)(yx) = 0 by (2), (12) and (7). Therefore (yx) 3 = 0. This proves that (Ax) 3 = 0. Therefore A 3 x 3 = 0. Finally, since X 3 = x 3 and A 3 x 3 = 0wehavethatX 3 is an ideal of A 3.The ideal X 3 is clearly solvable. Since A 3 /X 3 is a nilalgebra of dimension 5, it is solvable. Therefore A 3 is solvable by Proposition 5. It follows that A is solvable. Assume that A has a basis of type (b): As in the previous case, we have x = α 2 yx + α 3 (yx)x + α 4 ((yx)x)x + α 5 a + α 6 ax (α i K). Applying L x 2 to x and using (2) we get x 3 = α 5 ax 2 and then (ax 2 )x = 0. Therefore by (1) we have x 3 a = 2((ax)x)x (ax 2 )x = 0. Therefore a Ker(L x 3). If yx 3 = 0thenAx 3 = 0 by (7). Therefore X 3 is a solvable ideal of A. Since A/X 3 is a nilalgebra of dimension 5, it is solvable. Therefore A is solvable by Proposition 5. Assume now that yx 3 /= 0. Then Ker(L x 3) = yx,(yx)x,((yx)x)x,a,ax by (7). We claim that Ker(L x 3) is an ideal of A. We have to prove that tz Ker(L x 3) for any z {yx,(yx)x,((yx)x)x,a,ax} and t A. Replacing y by t in (8) we obtain (tz)x 3 = (zx 2 )(tx) 2((zx)x)(tx). For z = yx,(yx)x,((yx)x)x,ax we get (zx 2 )(tx) = 0 by (2). If z = a then (zx 2 )(tx) = (ax 2 )(tx) = (1/α 5 )x 3 (tx) = 0 by (7). Applying L 2 x to x we get x3 = α 2 ((yx)x)x. Thus if z = yx we obtain ((zx)x)(tx) = (((yx)x)x)(tx) = (1/α 2 )x 3 (tx) = 0 by (7). For z = (yx)x,((yx)x)x we obtain ((zx)x)(tx) = 0 since (((yx)x)x)x = 0. For z = a, ax we obtain ((zx)x)(tx) = 0since(ax)x = 0. Therefore (tz)x 3 = 0, i.e., tz Ker(L x 3). Since Ker(L x 3) and A/Ker(L x 3) are nilalgebras of dimension 5, they are solvable. Therefore A is solvable by Proposition 5. Assume that A has a basis of type (c): We have Ax = x 2,x 3 and then (Ax) 2 = 0. Since A 2 x 2 = (Ax) 2 by (10) it follows that A 2 x 2 = 0. Linearizing (3) we obtain A 2 x 3 (Ax 2 )(Ax). Since(Ax 2 )x 2 = 0by(9)and(Ax 2 )x 3 = 0by(5)we get (Ax 2 )(Ax) = 0. It follows that A 2 x 3 = 0. Since X 2 = x 2,x 3 it follows that A 2 X 2 = 0. Therefore X 2 is a solvable ideal of A 2.AlsoA 2 /X 2 is solvable since it is a nilalgebra of dimension 5. Therefore A 2 is solvable by Proposition 5. It follows that A is solvable.

7 I. Correa et al. / Linear Algebra and its Applications 369 (2003) Assume that A has a basis of type (d): We claim that A 3 x 3 = 0. Considering z A 2 and y A in (8) we obtain A 3 x 3 (A 2 x 2 )(Ax) + ((A 2 x)x)(ax). We will prove that (A 2 x 2 )(Ax) = 0. By (10) A 2 x 2 = (Ax) 2.WehavethatAx = x 2,x 3,zx and then (Ax) 2 = (zx) 2 by (2) and (7). Replacing y and z by zx in (3) we get (zx) 2 x 2 = 2((zx)x)((zx)x) = 0. Replacing x by zx, y by x 2 and z by x in (3) we get (zx) 2 x 3 = 2((zx)x 2 )((zx)x) = 0. Replacing x by zx and y by x in (3) we get (zx) 2 (zx) = 2((zx)x)((zx)z) = 0. Therefore (Ax) 2 (Ax) = 0 and then (A 2 x 2 )(Ax) = 0. We will prove that ((A 2 x)x)(ax) = 0. Since (Ax)x = x 3 and x 3 (Ax) = 0 by (7) we obtain ((Ax)x)(Ax) = 0. Since ((A 2 x)x)(ax) ((Ax)x)(Ax) it follows that ((A 2 x)x)(ax) = 0. Therefore A 3 x 3 = 0. Since X 3 = x 3 and A 3 x 3 = 0wehavethatX 3 is a solvable ideal of A 3.Since A 3 /X 3 is a nilalgebra and has dimension 5 it is solvable. Then A 3 is solvable by Proposition 5. Therefore A is solvable. AssumethatAhasabasisoftype(e):By(7)wehave(zx)x 3 = 0and((zx)x)x 3 = 0. Assume that zx 3 = 0. Then Ax 3 = 0 and this implies that X 3 = x 3 is a solvable ideal of A.AlsoA/X 3 is solvable since it is a nilalgebra of dimension 5. Therefore A is solvable by Proposition 5. Finally, assume that zx 3 /= 0. We have that Ker(L x 3) = x,x 2,x 3,zx,(zx)x. We can show that Ker(L x 3) is an ideal of A. The proof uses (8) with z in Ker(L x 3) and y any element of A.ThenKer(L x 3) and A/Ker(L x 3) are both solvable and so A is solvable by Proposition Nilindex 5 Assume that A has nilindex 5. Assuming that the characteristic is 0 or sufficiently large Gerstenhaber proved that L 7 z = 0forallz A (see [5, Theorem 1]). We independently verified that the result is true for characteristic /= 2, 3, 5. Linearizing x 2 x 2 = x 3 x we obtain 4(yx)x 2 = 2((yx)x)x + (x 2 y)x + x 3 y.replacing y by yx in this last identity we get 4((yx)x)x 2 + 2(((yx)x)x)x + ((yx)x 2 )x + (yx)x 3 = 0. (14) Linearizing x 3 x 2 = 0 we obtain 2((yx)x)x 2 + (yx 2 )x 2 + 2x 3 (yx) = 0. (15) Linearizing (x 2 x 2 )x = 0 we obtain 4((yx)x 2 )x + x 4 y = 0. (16) Throughout the rest of this subsection, x represents an element of A such that x 4 /= 0 and X = x,x 2,x 3,x 4. Let y be an arbitrary element of A.

8 192 I. Correa et al. / Linear Algebra and its Applications 369 (2003) The quotient vector space A/X has dimension 2. By the same argument we used in the beginning of the proof of Lemma 7, we have that yl 2 x and yl2 are x 2 in X. Writing (yx)x = α 1 x + α 2 x 2 + α 3 x 3 + α 4 x 4 (α i K) we obtain xl k z = αk 1 x for any k 1, where z = yx α 2 x α 3 x 2 α 4 x 3.SinceL 7 z = 0wehaveα 1 = 0. Hence (yx)x X 2. Therefore ((yx)x)x 2 and (((yx)x)x)x are in X 4. Our aim is to prove that yx 4 = 0. We will prove first that yx 4 X 4.By(16)we have yx 4 = 4((yx)x 2 )x. We will prove that ((yx)x 2 )x X 4. By (14) we have ((yx)x 2 )x = 4((yx)x)x 2 2(((yx)x)x)x (yx)x 3. But we already know that ((yx)x)x 2 and (((yx)x)x)x are in X 4. Therefore it remains to prove that (yx)x 3 X 4.Wehavethat(yx)x 3 = αx 4 ((yx)x 2 )x for some α K. Using (15) we obtain that (yx)x 3 = ((yx)x)x 2 (1/2)(yx 2 )x 2 X. Then, if (yx)x 3 = β 1 x + β 2 x 2 + β 3 x 3 + β 4 x 4 (β i K), we obtain xl k w = βk 1x for any k 1, where w = (yx)x 2 β 2 x β 3 x 2 + (α β 4 )x 3.SinceL 7 w = 0wehaveβ 1 = 0. Hence (yx)x 3 = β 2 x 2 + β 3 x 3 + β 4 x 4. Notice again that (yx)x 3 = ((yx)x)x 2 (1/2)(yx 2 )x 2. Then x 2 L k u = βk 2 x2 for any k 1, where u = β 3 x β 4 x 2 (yx)x (1/2)yx 2.SinceL 7 u = 0wehaveβ 2 = 0. Hence (yx)x 3 = β 3 x 3 + β 4 x 4. If v = yx β 4 x then x 3 L k v = βk 3 x3 for all k 1. Since L 7 v = 0 we obtain β 3 = 0. Therefore (yx)x 3 = β 4 x 4 X 4.Thisprovesthat((yx)x 2 )x X 4. Therefore yx 4 X 4.Ifyx 4 = βx 4 then x 4 L k y = βk x 4 for all k 1. Since L 7 y = 0 we obtain β = 0 and therefore yx 4 = 0. Since X 4 = x 4 and yx 4 = 0wehavethatX 4 is a solvable ideal of A. Also A/X 4 is solvable since it is a nilalgebra of dimension 5. Therefore A is solvable by Proposition 5. References [1] A.A. Albert, Power-associative rings, Trans. Am. Math. Soc. 64 (1948) [2] I. Correa, I.R. Hentzel, On solvability of noncommutative power-associative nilalgebras, J. Algebra 240 (2001) [3] I. Correa, L.A. Peresi, On the solvability of the five dimensional commutative power-associative nilalgebras, Results Math. 39 (2001) [4] I. Correa, A. Suazo, On a class of commutative power-associative nilalgebras, J. Algebra 215 (1999) [5] M. Gerstenhaber, On nilalgebras and linear varieties of nilpotent matrices. II, Duke Math. J. 27 (1960) [6] M. Gerstenhaber, H.C. Myung, On commutative power-associative nilalgebras of low dimension, Proc. Am. Math. Soc. 48 (1975) [7] R.D. Schafer, An Introduction to Nonassociative Algebras, Academic Press, New York, [8] D. Suttles, A counterexample to a conjecture of Albert, Notices Am. Math. Soc. 19 (1972) A-566. [9] K.A. Zhevlakov, A.M. Slin ko, I.P. Shestakov, A.I. Shirshov, Rings That Are Nearly Associative, Academic Press, New York, 1982.