On commutative nilalgebras of low dimension

We prove that every commutative non-associative nilalgebra of dimension ≤7, over a field of characteristic zero or sufficiently large is solvable.

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Date:	2010
Main Author:	Gutierrez Fernandez, J.C.
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Language:	English
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Cite this:	On commutative nilalgebras of low dimension / J.C. Gutierrez Fernandez // Algebra and Discrete Mathematics. — 2010. — Vol. 9, № 1. — С. 16–30. — Бібліогр.: 13 назв. — англ.

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spelling	irk-123456789-1544972019-06-16T01:31:39Z On commutative nilalgebras of low dimension Gutierrez Fernandez, J.C. We prove that every commutative non-associative nilalgebra of dimension ≤7, over a field of characteristic zero or sufficiently large is solvable. 2010 Article On commutative nilalgebras of low dimension / J.C. Gutierrez Fernandez // Algebra and Discrete Mathematics. — 2010. — Vol. 9, № 1. — С. 16–30. — Бібліогр.: 13 назв. — англ. 1726-3255 2000 Mathematics Subject Classification:17A05,17A30. http://dspace.nbuv.gov.ua/handle/123456789/154497 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection	DSpace DC
language	English
description	We prove that every commutative non-associative nilalgebra of dimension ≤7, over a field of characteristic zero or sufficiently large is solvable.
format	Article
author	Gutierrez Fernandez, J.C.
spellingShingle	Gutierrez Fernandez, J.C. On commutative nilalgebras of low dimension Algebra and Discrete Mathematics
author_facet	Gutierrez Fernandez, J.C.
author_sort	Gutierrez Fernandez, J.C.
title	On commutative nilalgebras of low dimension
title_short	On commutative nilalgebras of low dimension
title_full	On commutative nilalgebras of low dimension
title_fullStr	On commutative nilalgebras of low dimension
title_full_unstemmed	On commutative nilalgebras of low dimension
title_sort	on commutative nilalgebras of low dimension
publisher	Інститут прикладної математики і механіки НАН України
publishDate	2010
url	http://dspace.nbuv.gov.ua/handle/123456789/154497
citation_txt	On commutative nilalgebras of low dimension / J.C. Gutierrez Fernandez // Algebra and Discrete Mathematics. — 2010. — Vol. 9, № 1. — С. 16–30. — Бібліогр.: 13 назв. — англ.
series	Algebra and Discrete Mathematics
work_keys_str_mv	AT gutierrezfernandezjc oncommutativenilalgebrasoflowdimension
first_indexed	2025-07-14T06:35:18Z
last_indexed	2025-07-14T06:35:18Z
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fulltext	Jo u rn al A lg eb ra D is cr et e M at h . Algebra and Discrete Mathematics RESEARCH ARTICLE Volume 9 (2010). Number 1. pp. 16 – 30 c© Journal “Algebra and Discrete Mathematics” On commutative nilalgebras of low dimension Juan C. Gutierrez Fernandez Communicated by Efin Zelmanov Abstract. We prove that every commutative non-associative nilalgebra of dimension ≤ 7, over a field of characteristic zero or sufficiently large is solvable. Introduction Throughout this paper the term algebra is understood to be a commu- tative not necessarily associative algebra. We will use the notations and terminology of [6] and [7]. Let A be an (commutative nonassociative) al- gebra over a field F . We define inductively the following powers, A1 = A and A s = ∑ i+j=sA i A j for all positive integers s ≥ 2. We shall say that A is nilpotent if there is a positive integer s such that A s = (0). The least such number is called the index of nilpotency of the algebra A. The algebra A is called nilalgebra if given a ∈ A we have that alg(a), the subalgebra of A generated by a, is nilpontent. The (principal) powers of an element a in A are defined recursively by a1 = a and ai+1 = aai for all integers i ≥ 1. The algebra A is called left-nilalgebra if for every a in A there exists an integer k = k(a) such that ak = 0. The smallest positive integer k which this property is the index. Obviously, every nilalgebra is left-nilalgebra. For any element a in A, the linear mapping La of A defined by x → ax is called multiplication operator of A. An Engel al- gebra is an algebra in which every multiplication operator is nilpotent in the sense that for every a ∈ A there exists a positive integer j such that Lj a = 0. Partially Supported by FAPESP 05/01790-9 and FAPESP 2005/60337-2, Brazil 2000 Mathematics Subject Classification: 17A05,17A30. Key words and phrases: solvable, commutative, nilalgebra. Jo u rn al A lg eb ra D is cr et e M at h .J. C. G. Fernandez 17 An important question is that of the existence of simple nilalgebras in the class of finite-dimensional algebras. We have the following Shestakov’s Conjecture: there exists an example of commutative finite-dimensional simple nilalgebras. In [6] we proved that every nilagebra A of dimension ≤ 6 over a field of characteristic 6= 2, 3, 5 is solvable and hence A 2 & A. For power-associative nilalgebras of dimension ≤ 8 over a field of characteristic 6= 2, 3, 5, we have shown in [8] that they are solvable, and hence there is no simple algebra in this subclass. See also [4] and [6] for power-associative nilalgebras of dimension ≤ 7. We show now the process of linearization of identities, which is an important tool in the theory of varieties of algebras. See [9], [12] and [13] for more information. Let P be the free commutative nonassociative polynomial ring in two generators x and y over a field F . For every α1, . . . , αr ∈ P , the operator linearization δ[α1, . . . , αr] can be defined as follows: if p(x, y) is a monomial in P , then δ[α1, . . . , αr]p(x, y) is obtained by making all the possible replacements of r of the k identical arguments x by α1, . . . , αr and summing the resulting terms if x−degree of p(x, y) is ≥ r, and is equal to zero in other cases. Some examples of this operator are δ[y](x2(xy)) = 2(xy)2+x2y2, δ[x2, y](x2) = 2x2y and δ[y, xy2, x](x2) = 0. For simplicity, δ[α : r] will denote δ[α1, . . . , αr], where α1 = · · · = αr = α. We observe that if p(x) is a polynomial in P , then p(x + y) = p(x) + ∑ ∞ j=1 δ[y : j]p(x), where δ[y : j]p(x) is the sum of all the terms of p(x+y) which have degree j with respect to y. The following known results are a basic tool in our investigation. See [2], [3] and [7]. Lemma 1. Let A be a commutative left-nilalgebra of index ≤ 4 over a field F of characteristic different form 2 or 3. Then A satisfies the identities x2x3 = −x(x2x2), x3x3 = (x2)3 = x(x(x2x2)), (1) x3y = −x(x2y)− 2x(x(xy)), (2) A is a nilalgebra of index ≤ 7 and every monomial in P of x-degree ≥ 10 and y-degree 1 is an identity in A. Furthermore, for every a ∈ A the associative algebra Aa generated by all Lc with c ∈ alg(a) is in fact generated by La and La2. For simplicity, we will denote by L and U the multiplication operators, Lx and Lx2 respectively, where x is an element in A. Lemma 2. [7] Let A be a commutative algebra over a field of character- istic 6= 2 or 3 satisfying the identities x4 = 0 and x(x2x2) = 0. Then A Jo u rn al A lg eb ra D is cr et e M at h .18 On Commutative nilalgebras satisfies the following multiplication identities: Lx2x2 = −4LUL, UU = −2ULL+ 2LUL+ 4L4, (3) Table i, Multiplication identities of degree 5, ULU LUL 2 L 3 U L 5 UUL 0 2 0 0 LUU 0 −2 −2 −4 L 2 UL 0 0 −1 −4 UL 3 0 0 0 2 Table ii, Multiplication identities of degree 6, ULLU L 4 U L 6 UUU −2 4 8 UULL 0 0 4 ULUL −1 2 4 LUUL 0 2 0 LULU 0 0 4 ULLU L 4 U L 6 LLUU 0 −4 −4 UL 4 0 0 2 LUL 3 0 0 2 L 2 UL 2 0 1 0 L 3 UL 0 −1 −4 Furthermore, every monomial in P of x-degree ≥ 7 and y-degree 1 is an identity in A and the algebra generated by Lx and Lx2 is spanned, as vector space, by L,U, L2, UL, LU,L3, UL2, LUL,L2U,L4, ULU,LUL2, L3U , L5, UL2U , L4U , L6. Lemma 3. [7] Let A be a commutative algebra over a field of charac- teristic 6= 2, 3 or 5, satisfying the identities x4 = 0 and x(x(x2x2)) = 0. Then A satisfies the following multiplication identities: LUU = −2LUL2 − 2L3U − 4L5, (4) LUL3 = − 1 2 ( L2UL2 + L3UL ) , (5) L4UL = −3L5U − 16L7, (6) L2ULU = −L3UL2 + 5L5U + 28L7, (7) UL4U = − 1 2 L2UL2U + 24L6U + 62L8, (8) L2UL2U = 48L6U + 156L8, (9) L6U = −2L8. (10) Furthermore, every monomial in P of x-degree ≥ 9 and y-degree 1 is an identity in A. We now study (commutative nonassociative) nilalgebras of dimension ≤ 7, over a field F of characteristic zero or sufficiently large. We will show that nilalgebras over F with dimension ≤ 7, are solvable. An algebra A is called solvable if there exists a positive integer t such that A [t] = (0), Jo u rn al A lg eb ra D is cr et e M at h .J. C. G. Fernandez 19 where we define inductively A [1] = A and A [j+1] = A [j] A [j] for all positive integers j. Let A be a finite-dimensional nilgalgebra over F . We will denote by deg(A), the degree of A, the smallest number m such that for every a ∈ A, the subalgebra alg(a) of A generated by a has dim(alg(a)) ≤ m. If deg(A) ≤ 2, then A satisfies the identity x3 = 0 and hence this algebra is Jordan. It is well-known that any finite-dimensional Jordan nilalgebra is nilpotent. Therefore A is nilpotent if deg(A) ≤ 2. Because any nilpotent algebra is solvable, we have that A is solvable if deg(A) ≤ 2. The following lemma, proved in [6], is an immediate consequence of a result of [10] and [11] for linear spaces of nilpotent matrices. Lemma 4. Let A be a nilalgebra over the field F . Then A 2 A 2 ⊂ B for every subalgebra B of codimension ≤ 2. By above lemma, if deg(A) ≥ dim(A)− 2, then A 2 A 2 is nilpotent and hence A is solvable. Summarizing, A is solvable in the following cases: (i) dim(A) ≤ 5; (ii) dim(A) = 6 and deg(A) 6= 3; (iii) dim(A) = 7 and deg(A) 6= 3 or 4. Thus, for dim(A) ≤ 7, it remains to be shows that A is solvable if deg(A) = 3 or 4. The following lemma is clear from Lemma 1. For any subset S of A we denote by 〈S〉 the vector space spanned by S. Lemma 5. Let A be an algebra over F satisfying the identity x4 = 0. Consider an element a in A. (i) If a(a(a2a2)) 6= 0, then dim(alg(a)) = 6; (ii) If a(a(a2a2)) = 0 and a(a2a2) 6= 0, then dim(alg(a)) = 5. Proof. By Lemma 1 we observe that A is a nilalgebra of nilindex ≤ 7 and alg(a) = 〈a, a2, a3, a2a2, a(a2a2), a(a(a2a2))〉. Assume a(a(a2a2)) 6= 0. We will prove that a, a2, a3, a2a2, a(a2a2), a(a(a2a2)) are linearly inde- pendent. Let λ1a+ λ2a 2 + λ3a 3 + λ4a 2a2 + λ5a(a 2a2) + λ6a(a(a 2a2)) = 0. Then 0 = L2 aLa2La(0) = L2 aLa2La(λ1a + λ2a 2 + λ3a 3 + λ4a 2a2 + λ5a(a 2a2) + λ6a(a(a 2a2))) = λ1a(a(a 2a2)) and hence λ1 = 0. Anal- ogously, 0 = L2 aLa2(0) = L2 aLa2(λ2a 2 + λ3a 3 + λ4a 2a2 + λ5a(a 2a2) + λ6a(a(a 2a2))) = λ2a(a(a 2a2)) so that λ2 = 0. Next, 0 = LaLa2(0) = LaLa2(λ3a 3 + λ4a 2a2 + λ5a(a 2a2) + λ6a(a(a 2a2))) = −λ3a(a(a 2a2)) so that λ3 = 0. And analogously we can prove that λ4 = λ5 = λ6 = 0. The case (ii) is similar. Corollary 1. Let A be an algebra over F satisfying the identity x4 = 0. Assume deg(A) = 3 or 4 and let a be an element in A. Then alg(a) = 〈a, a2, a3, a2a2〉 and 〈a3, a2a2〉 · alg(a) = 0. Jo u rn al A lg eb ra D is cr et e M at h .20 On Commutative nilalgebras 1. The case degree(A)=3 Now we will study nilalgebras of degree 3. In this section A will be a nilalgebra of degree 3 and dimension ≤ 7 over the field F . Consider a an element in A. Because A is nilalgebra, there exists a positive integer t such that at = 0. We can assume that at = 0 and at−1 6= 0. Clearly, the elements a, a2, . . . , at−1 are linearly independent, and hence t ≤ 4, since deg(A) = 3. Consequently, the algebra A satisfies the identity x4 = 0. By Corollary 1, the sequence a3, a2a2 is linearly dependent and A satisfies the identities x(x2x2) = 0, x2x3 = 0. Consequently, A satisfies multiplication identities (3), Tables i and ii and Lemma 2. Lemma 6. Let A be a nilalgebra over the field F with dimension ≤ 7 and degree 3. Then L6 = 0 is a multiplication identity in A. Proof. Assume that there exist a, b ∈ A such that L6 a(b) 6= 0. Then the sequence Ψ = {Li a(b) : i = 0, 1, . . . , 6} is a basis of A. On the other hand, we note that from Table ii and (3) we have L6 a(b) = 1 2 a(a2(a(a(ab)))) = − 1 8 (a2a2)(a(ab)), so that a2a2 6= 0. Because Ψ is a basis and a(a2a2) = 0, we get that a2a2 = λL6 a(b), for any 0 6= λ ∈ F . Combining above relations we get that a2a2 = (a2a2)[(−λ/8)a(ab)], but this is impossible because A is an Engel algebra. Therefore L6 a = 0 for all a ∈ A. We may use (2) combined with (3) to yield Lx2x2L− 4Lx3L2 = 8L5. (11) We shall use this formula now. Lemma 7. Let A be a nilalgebra over the field F with dimension ≤ 7 and degree 3. Then L5 = 0 is a multiplication identity in A. Proof. Assume that there exist a, b ∈ A such that L5 a(b) 6= 0. By identity (11) we have that either a3 6= 0 or a2a2 6= 0. The proof now splits into two cases. Case 1. If a2a2 6= 0, then a3 = βa2a2 for any β ∈ F and using (11) we obtain 8L5 a = La2a2La − 4βLa2a2L 2 a. Multiplying this relation Jo u rn al A lg eb ra D is cr et e M at h .J. C. G. Fernandez 21 from the right side with La yields La2a2L 2 a = 0, so that La3L 2 a = 0 and La2a2La = 8L5 a. Now, it is easy to prove that Ψ = {a2a2, Li a(b) : i = 0, 1, . . . , 5} is linearly independent and hence a basis of A. Let a2 = λa2a2 + ∑5 i=0 µiL i a(b). Multiplying by a, 2 times, we get 0 = µ0L 2 a(b) + µ1L 3 a(b) + µ2L 4 a(b) + µ3L 5 a(b), so that µ0 = µ1 = µ2 = µ3 = 0. Now, multiplying with a2, we get a2a2 = λ(a2)3+µ4La2L 4 a(b)+µ5La2L 5 a(b) = 0, but this is impossible. Case 2. If a2a2 = 0, then La3L 2 a = −2L5 a. Now, it is easy to prove that Φ = {a3, Li a(b) : i = 0, 1, . . . , 5} is linearly independent and hence a basis of A. Let a = λa3 + ∑5 i=0 µiL i a(b). Multiplying by a, 3 times, we get 0 = µ0L 3 a(b) + µ1L 4 a(b) + µ2L 5 a(b), so that µ0 = µ1 = µ2 = 0. Next, multiplying by a two time, we have a3 = µ3L 5 a(b), but this is impossible because Φ is a basis. Lemma 8. Let A be a nilalgebra over the field F with dimension ≤ 7 and degree 3. Then every monomial in P of x-degree ≥ 6 and y-degree 1 is an identity in A. Proof. By Lemma 2 and Lemma 6 we only need to prove that L4U = 0 and UL2U = 0 are multiplication identities in A. Using identity (3), Table ii and relation 0 = δ[x2]{x(x(x(x(xy))))} we have that 0 = UL4 + LUL3+L2UL2+L3UL+L4U = L2UL2+L3UL+L4U = L4U . Now, from Lemma 2, multiplication identities L6 = 0 and L4U = 0, and identity (11), we see that ULLU = −ULUL = 1 4 ULx2x2 = ULx3L. Let a ∈ A. If a2a2 = 0, if follows immediately that La2L 2 aLa2 = 0. If a2a2 6= 0, then there exists λ ∈ F such that a3 = λa2a2. Therefore, La2L 2 aLa2 = La2La3La = λLa2La2a2La = 0. This proves the lemma. Using identity (2), Lemma 2 and Lemma 7, we can prove easily the following multiplication identities L3U = −L2UL = −L2Lx3 = LLx3L = 1 4 LLx2x2 , LUL2 = −Lx3L2 = − 1 4 Lx2x2L, for nilalgebras of dimension ≤ 7 and degree 3 over the field F . We shall use these formulas now. Lemma 9. Let A be a nilalgebra over the field F with dimension ≤ 7 and degree 3. Then L3U = 0 and LUL2 = 0 are multiplications identities in A. Jo u rn al A lg eb ra D is cr et e M at h .22 On Commutative nilalgebras Proof. Let a be an element in A. If a2a2 = 0 then, from above identities we obtain immediately that L3 aLa2 = (1/4)LaLa2a2 = 0 and LaLa2L 2 a = −(1/4)La2a2La = 0. If a2a2 6= 0 then there exists λ ∈ F such that a3 = λa2a2. This means that La3 = λLa2a2 . Then we have L3 aLa2 = L2 aLa3 = λL2 aLa2a2 = 0 and LaLa2L 2 a = −La3L 2 a = λLa2a2L 2 a = 0 by Lemma 8. This proves the lemma. Lemma 10. Let A be a nilalgebra over the field F with dimension ≤ 7 and degree 3. Then LUL = 0 is a multiplication identity in A. Proof. We will assume the contrary, there exist two elements a, b ∈ A such that a(a2(ab)) 6= 0. We know by (3) that a(a2(ab)) = −(1/4)(a2a2)b. Therefore, a2a2 6= 0 and also the sequence {a2a2, a(a2(ab))} is linearly independent, because Lb is nilpotent. For any λ ∈ F we have that a3 = λa2a2. Obviously, this forces La3 = λLa2a2 . From identity (2) and above lemma, we have immediately that LaLa2La = −La3La − 2L4 a = −λLa2a2La − 2L4 a = 4λLaLa2L 2 a − 2L4 a = −2L4 a, that is LaLa2La = −2L4 a. (12) We will now prove that Ψ = {b, ab, a2(ab), a(a2(ab)), a, a2, a2a2} is a basis of A. Let λ1b+λ2ab+λ3a 2(ab)+λ4a(a 2(ab))+µ1a+µ2a 2+µ3a 2a2 = 0, with λi, µj ∈ F . Multiplying with a, a2 and a successively, we get λ1 = 0. Multiplying with a2 and a successively, we have λ2 = 0. Multiplying with a and a2 successively, we obtain µ1 = 0, so that λ3a 2(ab) + λ4a(a 2(ab)) + µ2a 2 + µ3a 2a2 = 0. Multiplying with a it follows that λ3 = 0 since a2a2, a(a2(ab)) are linearly independent and a3 ∈ 〈a2a2〉. Multiplying with a2 we have µ2 = 0. Now, relation λ4a(a 2(ab))+µ3a 2a2 = 0 forces λ4 = µ3 = 0. Therefore, we have proved that the sequence Ψ is linearly independent. Since dim(A) ≤ 7, it follows that Ψ is a basis of A. On the other hand, because Ψ is a basis of A, we have a representation a(ab) = α1b+α2ab+α3a 2(ab)+α4a(a 2(ab))+α5a+α6a 2+α7a 2a2, with αi ∈ F . Using the operators LaLa2La, LaLa2 , La2La and LaLa, we prove that α1 = 0, α2 = 0, α5 = 0 and a(a(a(ab))) = 0 respectively, but this is impossible since by identity (12) we have that 2a(a(a(ab))) = −a(a2(ab)) and by hypothesis this element is diferent form zero. This proves the lemma. Jo u rn al A lg eb ra D is cr et e M at h .J. C. G. Fernandez 23 It was proved in [8] the following result for power-associative nilalge- bras. Lemma 11. Every commutative power-associative nilalgebra of dimen- sion ≤ 8 over a field of characteristic 6= 2, 3 or 5 is solvable. Theorem 1. Let A be a nilalgebra over the field F with dimension ≤ 7 and degree 3. Then, the algebra A is solvable. Proof. By Lemma 10 we have that a2a2 belong to the annihilator of the algebra A, for every a ∈ A. This means that the linear subspace J = 〈a2a2 : a ∈ A〉 is an ideal of A and AJ = 0. Thus, A/J is a commutative power-associative nilalgebra of dimension ≤ 7, and hence solvable. This implies that A is solvable. 2. The case degree(A) = 4 and x(x(xx))=0 For any subalgebra B of an algebra A, the set st(B) = {x ∈ A : xB ⊂ B} is called stabilizer of B in A. For every element a ∈ st(B), we can define a linear transformation La on the quotient vector space A = A/B as follows, La(x+B) = ax+B, for all x ∈ A. We will now denote by MB the linear space {La : a ∈ st(B)} and by NB the linear subspace {Lb : b ∈ B}. Evidently, we have that NB ⊂ MB. The following result will be useful. Items (iv), (v), (vi) and (vii) follow immediately from (i)-(iii) proved in [6]. Lemma 12. Let V be a vector space of dimension 3 over a field F of characteristic 6= 2 and let M be a vector space of nilpotent linear endo- morphisms in EndF (V ). Then dimM ≤ 3 and either M 3 = 0 or: (i) dimM = 2; (ii) for every nonzero f ∈ M we have that rank(f) = 2; (iii) if M = 〈f1, f2〉, then there exists a basis φ of V and 0 6= λ ∈ F such that the matrices (using columns) of f1 and λf2 with respect to φ are   0 0 0 1 0 0 0 −1 0   ,   0 1 0 0 0 1 0 0 0   respectively; (iv) if f, g and fg are all in M, then f = 0 or g = 0; (v) if f, g, h and f(g + fh) are all in M and f 6= 0, then g = 0 and h ∈ 〈f〉; (vi) if f, g, h ∈ M and f(f + gh) = 0, then f = 0; (vii) if f, g ∈ M and f2g2 = 0, then the sequence {f, g} is linearly dependent. Jo u rn al A lg eb ra D is cr et e M at h .24 On Commutative nilalgebras Let A be a nilalgebra over the field F with degree 4 and dimension ≤ 7 satisfying the identity x4 = 0. From Lemma 4, A is solvable if dim(A) ≤ 6, so that throughout this section we will assume that A has dimension 7. By Corollary 1, the algebra A satisfies the identities x(x2x2) = 0 and x2x3 = 0. Now we may take an element b in A such that B, the subalgebra of A generated by b, has dimension 4. By Corollary 1, we have B = 〈b, b2, b3, b2b2〉, and 〈b3, b2b2〉B = (0). (13) If dimNB = 0, then B is an ideal of A and hence A is solvable because A/B is solvable. If MB is nilpotent, then there exists a ∈ A but not in B such that f(a+B) = 0 +B, (14) for all f ∈ MB. There exists a smallest integer m, 1 ≤ m ≤ 3, such that Mm B = (0). If m = 1 take a ∈ A but not in B; if m > 1, take 0 6= g ∈ Mm−1 B and a+B in g(A/B) with a+B 6= 0+B. Then (14) is satisfied. Since a ∈ st(B) we have that La ∈ MB. Then relation (14) implies that 0 +B = La(a +B) and hence a2 ⊂ B. Let B ′ = 〈b, b2, b3, b2b2, a〉. We have that B ′ is a subalgebra of A with codimension 2. Using Lemma 4 we get that A 2 A 2 ⊂ B ′ so that A is solvable. We now consider the case NB 6= (0) and M3 B 6= (0). Then MB satisfies (i)-(vii) of Lemma 12. By Lemma 1 we have that NB is nilpotent, so that Lemma 12 forces dim(NB) = 1 since NB ⊂ MB and MB is not nilpotent. Let 0 6= h ∈ NB. Then Lbi = αih for any αi ∈ F and for i = 1, 2, 3. From identities (3) and (2) we have Lb2b2 = −4α2 1α2h 3 = 0 and α3h = Lb3 = −α1α2h 2 − 2α3 1h 3 = −α1α2h 2 so that Lb3 = 0 since h3 = 0. Next (3) forces Lb2 2 = −2α2 1α2h 3 + 2α2 1α2h 3 + 4α4 1h 4 = 0. Therefore Lb2 = 0 since Lb2 ∈ MB and from Lemma 12 every nonzero element in MB is nilpotent of index 3. Thus, we have proved that B 2 A = 〈b2, b3, b2b2〉A ⊂ B. This yields NB = 〈Lb〉. Because NB MB and dim(MB) = 2, we can take a ∈ st(B), but not in B such that MB = 〈Lb, La〉. By Lemma 12 there exists a basis Φ = {v1 +B, v2 +B, v3 +B} of A/B such that the matrices of Lb and La with respect to Φ are respectively   0 0 0 1 0 0 0 −1 0   and 1 λ   0 1 0 0 0 1 0 0 0   Jo u rn al A lg eb ra D is cr et e M at h .J. C. G. Fernandez 25 for any 0 6= λ ∈ F . This means that v3+B = αa+B, v2+B = αa2+B and v1 +B = αa3 +B for any α ∈ F , α 6= 0. We can assume, without lost of generality, that λ = 1 and α = 1. Now, by equation (3) and (13) we have (b2b2)a = −4b(b2(ba)) ⊂ b(b2B) = (0), so that (b2b2)a = 0. On the other hand, ab can be expressed as a linear combination of b, b2, b3, b2b2. Let ab = µ1b+µ2b 2+µ3b 3+µ4b 4b2. Then cb = µ1b+µ4b 2b2, where c = a − µ2b − µ3b 2. Therefore c(cb) = µ1cb + µ4c(b 2b2) = µ1cb. Since A is an Engel algebra, Lc is nilpotent and hence either µ1 = 0 or cb = 0. This implies ab ∈ B 2. Using relation (3) we have that b2(b2a) = −2b2(b(ba)) + 2b(b2(ba)) + 4b(b(b(ba))) = 0. This forces b2a ∈ B 3. Finally, using (2) and (3) we have (b2b2)a2 = −4b(b2(ba2)) ∈ 〈b(b2(−a+B))〉 = 〈b(b2a)〉 = 0, (b2b2)a3 = −a(a2(b2b2))− 2a(a(a(b2b2))) = 0, and hence b2b2 ∈ ann(A). Let J = 〈b2b2 : b ∈ A, dim(alg(b)) = 4〉. Then A/J is a commutative nilalgebra of dimension ≤ 6 and degree ≤ 3, so that solvable. This implies that A is solvable. 3. The case degree(A) = 4 Let A be a nilalgebra with degree 4. If a ∈ A, then there exists an integer t such that at 6= 0 and at+1 = 0 so that the elements a, a2, . . . , at are linearly independent. Since deg(A) = 4, we have that t ≤ 4 and hence A satisfies the identity x5 = 0. Now we will see that A is a nilalgebra of index ≤ 9. Let B be a subalgebra of A generated by a single element and let k1 be the index of B as left- nilalgebra, that is k1 is the smallest integer such that xk1 = 0 for all x ∈ B. Evidently, dimB ≤ 4 and k1 ≤ 5. If k1 ≤ 3, then B is a Jordan algebra and hence nilpotent with B k1 = 0. If k1 = 4, then by Lemma 1 and Lemma 5, we have that p(x) = 0 is an identity in B for every monomial p(x) of degree ≥ 5. Finally, if k1 = 5, then there exists b ∈ B such that B = 〈b, b2, b3, b4〉. Now, because B is nilpotent, we have that b2b2 ∈ 〈b3, b4〉, b2b3, b3b3 ∈ 〈b4〉, b4B = (0). Jo u rn al A lg eb ra D is cr et e M at h .26 On Commutative nilalgebras Thus, B3,B4 ⊂ 〈b3, b4〉, B5 ⊂ 〈b4〉 and B t = 0 for all t ≥ 9. It has the following consequences. Lemma 13. The algebra A satisfies the identities xi(xj(xtx2)) = 0, i, j, t ≥ 1, and p(x) = 0 for every monomial p(x) ∈ P of degree ≥ 9. Linearizing the above identities we have the following multiplication identities (in order to simplify, we will write L3 instead of Lx3 and L4 instead of Lx4): L4 + LL3 + L2U + 2L4 = 0, (15) Lx2x3 + 2LL3L+ LUU + 2LUL2 = 0, (16) 2L4L 2 + L4U + L3L3 + L3LU + 2L3L 3 = 0, (17) Lx3x3 + 2LL3U + 4LL3L 2 = 0, (18) U3 + 2UUL2 + 2UL3L+ 2Lx2x3L = 0, (19) L4(L3 + LU + 2L3) = 0. (20) Lemma 14. [9] Every nilalgebra of bounded index over F is an Engel algebra. We have proved that the index of a nilalgebra of degree 4 is ≤ 9. We then apply Lemma 14 to obtain Corollary 2. Every nilalgebra of degree 4 over F is an Engel algebra. Theorem 2. Let A be a nilalgebra over the field F . If dim(A) ≤ 7, then A is solvable. Proof. We already prove that A is solvable if either deg(A) 6= 4 or x4 = 0 is an identity. Thus, it remains to prove that A is solvable if dim(A) = 7, deg(A) = 4 and x4 = 0 is not an identity in A. Let A be a nilalgebra of dimension 7 and degree 4 such that there exists b ∈ A with b4 6= 0. Let B be the subalgebra of A generated by b. Because A has degree 4, we have B = 〈b, b2, b3, b4〉 and b5 = 0. As in Section 3, if MB is nilpotent, then the algebra A is solvable. Also, the algebra is solvable if NB = 0. Thus, we can assume that MB is not nilpotent and dimNB ≥ 1. By Lemma 12, we have that dim(MB) = 2. From (15) we have Lb4 = −LbLb3 − Lb 2 Lb2 = −Lb(Lb3 + LbLb2) ∈ NB. Jo u rn al A lg eb ra D is cr et e M at h .J. C. G. Fernandez 27 Combining above relation and (v) of Lemma 12 we get that Lb4 = 0. Now (17) implies Lb3(Lb3 + LbLb2) = 0 and by (vi) of Lemma 12 we get that Lb3 = 0. This means that B 3 A ⊂ B, (21) and NB = 〈Lb, Lb2〉. Now relation (19) for x = b forces 0 = Lb2 3 + 2Lb2 2 Lb 2 + 2Lb2Lb3Lb + 2Lb2b3Lb = 2Lb2 2 Lb 2 and hence using (vii) of Lemma 12 we have that dim(NB) = 1. (22) We can assume, without loss of generality, that Lb 6= 0, (23) since if Lb = 0, then Lb2 6= 0 and we can take 0 6= λ ∈ F such that (b+ λb2)4 = b4 + λ[b(b2b2) + b2b3] + λ2(b2)3 6= 0. Because dim(NB) = 1, there exists α ∈ F such that Lb2 = αLb. As in Section 3 there exists a ∈ A such that MB = 〈Lb, La〉, Φ = {a3 +B, a2 +B, a+B} is a basis of A/B and the matrices of Lb and La with respect to Φ are respectively   0 0 0 1 0 0 0 −1 0   and   0 1 0 0 0 1 0 0 0   . This means that ba3 − a2, b2a3 − αa2, ba2 + a, b2a2 + αa, ba, b2a, a4 ∈ B. (24) By (24) we have that ba ∈ B so that ba = λ1b + λ2b 2 + λ3b 3 + λ4b 4, with λi ∈ F . Therefore [a− λ2b− λ3b 3 − λ4b 3]b = λ1b. This implies that λ1 = 0 and hence ba ∈ B 2, (25) since every multiplication operator on A is nilpotent. Let j be a positive integer. From (15) and (21) we get b4aj = −b(b3aj)− b(b(b2aj))− 2b(b(b(baj))) ∈ B 2, (26) so that b4A ⊂ B 2. (27) By (20) we see that 0 = b4(b3a3 + b(b2a3) + 2b(b(ba3))) = b4(b(b2a3)) = −αb4a and now using identity (17) we have 2b4(b(ba3)) + b4(b2a3) + b3(b3a3) + b3(b(b2a3)) + 2b3(b(b(ba3))) = 0 and hence −2b4a + α(b4a2 − b3a) = −b3(b3a3) ∈ 〈b4〉. Therefore b4a = 0, α(b4a2 − b3a) ∈ 〈b4〉, (28) since A is an Engel algebra and αb4a = 0. The proof now splits into two cases: Jo u rn al A lg eb ra D is cr et e M at h .28 On Commutative nilalgebras Case 1. The relation x3x3 = 0 is not an identity in A. In this case, we can assume without loss of generality that b4 6= 0, and b3b3 6= 0. (29) Let b1 be an element in A such that b31b 3 1 6= 0 and B the subalgebra of A generated by b1. Then {b1, b 2 1, b 3 1, b 3 1b 3 1} is a basis of B and products satisfy the following properties, b21b 2 1 ∈ 〈b31, b 3 1b 3 1〉, b 3 1B ⊂ 〈b31b 3 1〉 and (b31b 3 1)B = (0), because B is nilpotent. By Corollary 1 we have that x4 = 0 is not an identity in B. Thus, there exists an element b in B of the form λ1b1 + λ2b 2 1 + λ3b 3 1 + λ4b 3 1b 3 1 such that b4 6= 0 and also we can assume that Lb 6= 0. Evidently, we have λ1 6= 0. Now b2 ∈ λ2 1b 2 1 + 〈b31, b 3 1b 3 1〉, b 3 ∈ λ3 1b 3 1+ 〈b31b 3 1〉 and b3b3 = λ6 1b 3 1b 3 1 6= 0. Then (29) is satisfied. Evidently, we have b3b3 = γb4 for any 0 6= γ ∈ F . Combining (18) and (28) it follows 0 = (b3b3)a2 + 2b(b3(b2a2)) + 4b(b3(b(ba2))) = (b3b3)a2 − 2αb(b3a) = γb4a2 − 2αb(b4a2). Thus b4b2 = (2α/γ)b(b4a2). Since A is an Engel algebra it follows that b4a2 = 0. Combining this identity with (28) we have that αb3a ∈ 〈b4〉. (30) Now, relation (16) with x = b for the element a2 implies 2b(b3a) + αb(b2a) ∈ 〈b4〉. Combining this relation with (30) we see that b(b3a) ∈ 〈b4〉, so that b3a ∈ 〈b3, b4〉. Therefore b3a ∈ 〈b4〉, since A is an Engel algebra. Next, we put x = b in (15) to obtain 0 = b4a+ bb3a+ b(b(b2a)) + 2b(b(b(ba))) = b(b(b2a)), and hence b2a ∈ 〈b3, b4〉. Now, by Lemma 13 we know that x(x2x3) = 0 is an identity in A and hence 0 = (1/48)δ[b : 4, a : 2]{x(x2x3)} = b(b2(ba2)) + 2b(b2(a(ba))) + 4b((ba)(b(ba))) + 2a(b2(b(ba))) + 2b((ba)(b2a)) + a(b2(b2a)) + b(a2b3) + 2a((ba)b3), forces b(b3a2) ∈ 〈b4〉 so that b3a2 ∈ 〈b4〉. Finally, (18) implies 0 = (b3b3)a3 + 2b(b3(b2a3)) + 4b(b3(b(ba3))) = (b3b3)a3+2b(b3(αa2−2a)) = (b3b3)a3. Therefore we must have b4a3 = 0. Consequently, we have proved, in this case, that b4 ∈ ann(A). Let J = Jo u rn al A lg eb ra D is cr et e M at h .J. C. G. Fernandez 29 〈c4 : c3c3 6= 0, c ∈ A〉. Then A = A/J is a nilalgebra of dimension ≤ 6 and hence solvable. This forces the solvability of A. Case 2. The relation x3x3 = 0 is an identity in A. Linearizing this identity we have that A satisfies the identity x3(x2y) + 2x3(x(xy)) = 0. Taking x = b and y = a2 it follows immediately that αb3a ∈ 〈b4〉 and for x = b and y = a3 this identity forces αb3a2 − 2b3a ∈ 〈b4〉. Therefore b3a ∈ 〈b4〉. Next, (15) forces 0 = b4a+ b(b3a) + b(b(b2a)) + 2b(b(b(ba))) = b(b(b2a)), so that b2a ∈ 〈b3, b4〉. Now, taking the identity δ[b : 4, a : 3]{x4x3} = 0 we have −b4a3 = [ b3a+ b(b2a) + 2b(b(ba)) ] · [ ba2 + 2a(ab) ] + [ b(ba2) + 2b(a(ba)) + 2a(b(ba)) + a(ab2) ] · [ b2a+ 2b(ba) ] + [ ba3 + a(a2b) + 2a(a(ab)) ] · [ b3 ] ∈ 〈b4〉 · 〈 a, b, b2, b3, b4〉+B · 〈b3, b4〉+B · 〈b3〉 ⊂ 〈b4〉 since by (24) we have that ba3 + a(a2b) ∈ B. This means that b4a3 = 0, because La3 is nilpotent. From δ[b : 3, a : 2]x5 = 0 we get −b(b(ba2)) = a(ab3)+a(b(ab2))+2a(b(b(ba)))+b(a(ab2))+2b(a(b(ba)))+2b(b(a(ba))) = 0. This means that b(ba2) ∈ 〈b4〉. Finally, from δ[b : 4, a : 2]{x4x2} = 0 it follows that −b4a2 = [ b3a+ b(b2a) + 2b(b(ba)) ] · [ 2ba ] + [ b(ba2) + 2b(a(ba)) + 2a(b(ba)) + a(ab2) ] · [ b2 ] ∈ 〈b4〉 · 〈b2, b3, b4〉+ 〈b4〉 · 〈b2〉 = (0). Therefore b4 ∈ ann(A) and as in the case 1, this implies the solvability of A. References [1] A.A. Albert, Power-associative rings, Trans. Amer. Math. Soc., N.64, 1948, pp.552-593. Jo u rn al A lg eb ra D is cr et e M at h .30 On Commutative nilalgebras [2] I. Correa, I.R. Hentzel, A. Labra, On the nilpotence of the multiplication operator in commutative right nil algebras, Comm. in Algebra, N.30, 2002, pp.3473-3488. [3] A. Elduque, A. Labra, On the Classification of commutative right-nilalgebras of dimension at most four, Comm. in Algebra, N.35, 2007, pp.577-588. [4] L. Elgueta, A. Suazo, (2004). Solvability of commutative power-associative nilal- gebras of nilindex 4 and dimension ≤ 8, Proyecciones, N.23, 2004, pp.123-129. [5] L. Elgueta, J.C.G. Fernandez, A. Suazo, Nilpotence of a class of commutative power-associative nilalgebras, Journal of Algebra, N.291, 2005, pp.492-504. [6] J.C.G. Fernandez, On commutative power-associative nilalgebras, Comm. in Al- gebra, N.32, 2004, pp.2243-2250. [7] J.C.G. Fernandez, On commutative left-nilalgebras of index 4, Proyecciones, N.27(1), 2008, pp. 103-112. [8] J.C.G. Fernandez, A. Suazo, Commutative power-associative nilalgebras of nilin- dex 5 Result. Math., N.47, 2005, pp.296-304. [9] M. Gerstenhaber, On nilalgebras and linear varieties of nilpotent matrices II, Duke Math. J., N.27, 1960, pp.21-31. [10] M. Gerstenhaber, On nilalgebras and linear varieties of nilpotent matrices III, Ann. of Math., N.70, 1959, pp.167-205. [11] B. Mathes, M. Omladic, H. Radjavi, Linear spaces of nilpotent matrices, Lin. Alg. Appl., N.149, 1991, pp.215-225. [12] J.M. Osborn, Varieties of algebras, Adv. in Math., N.8, 1972, pp.163-396. [13] K.A. Zhevlakov, A.M. Slinko, I.P. Shestakov, A.I. Shirshov, A.I. (1982). Rings that are nearly associative, Academic Press, 1982. Contact information J. C. G. Fernandez Departamento de Matemática-IME, Univer- sidade de São Paulo Caixa Postal 66281, CEP 05315-970, São Paulo, SP, Brazil E-Mail: jcgf@ime.usp.br Received by the editors: 19.09.2008 and in final form 19.09.2008.

On commutative nilalgebras of low dimension

Institution

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