MatrixType of Some Algebras over a Field of Characteristic p

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1 Journal of Algebra 251, (2002 doi: /jabr MatrixType of Some Algebras over a Field of Characteristic p Alexander Kemer Department of Mathematics, Ulyanovsk University, Ulyanovsk, , Russia Communicated by Efim Zelmanov Received June 15, 2001 Key Words: polynomial identity; standard identity; variety of algebras. 1. INTRODUCTION It was proven in 1 that for every associative PI-algebra A over a field F of characteristic p there exists a number m such that the algebra A satisfies all multilinear polynomial identities of the matrixalgebra M m F. A minimal m with such a property is said to be the matrixtype of the algebra A. Denote by t n the matrixtype of the algebra M n G, where G is the Grassmann algebra of infinite rank over a field F of characteristic p 2. It is quite easy to prove that the algebra M n G satisfies all multilinear polynomial identities of the algebra M np F (Lemma 7. It is also easy to see that the matrixtype of the Grassmann algebra is greater than p 1 2 because this algebra does not satisfy the standard identity of degree p 1. Thus we have the trivial estimations for t n p 1 n<t 2 n pn The main result of the paper (Theorem 2 gives a precise formula for t n. It follows from Theorem 1, which gives a good estimation for the minimal degree of the standard identity of the algebra M n G. Theorem 1. If k>1 then d k > 2k 3 p 1. Theorem 2. t n = pn /02 $ Elsevier Science (USA All rights reserved.

2 850 alexander kemer Let X be a countable set and let F X be the free associative algebra over F generated by the set X. We denote by T A the ideal of identities of the algebra A. This ideal is a T -ideal of the algebra F X. Theorem 2 is equivalent to the statement: There exists a multilinear polynomial f T M np 1 F such that f T M n G. The problem of calculating such polynomials is very difficult even for n = 1. This problem is similar to the problem which has arisen in 2. In 3 some calculations of such types were made with the help of a computer. 2. TECHNICAL STATEMENTS In this section we assume that char F = p>2. Let F X be the free associative algebra with a unit generated by the countable set X. We denote by S n the symmetric group of permutations acting on the set 1 n. The polynomials S n x 1 x n = 1 σ x σ 1 x σ n S + n x 1 x n = σ S n σ S n x σ 1 x σ n are called the standard polynomial and the symmetric standard polynomial of degree n, respectively. By the well-known theorem of Kaplansky the matrixalgebra M k F satisfies the standard identity of degree 2k. The first statement is well known. Lemma 1. The algebra M k F satisfies identity S + pk = 0. Proof. Let C be the commutative algebra generated by X satisfying relations x 2 = 0 for every x X. Evidently the algebra C satisfies identity x p = 0. Take an arbitrary element x M k C. By the Cayley Hamilton theorem we have the equality in M k C x k = k 1 m=1 for some c m C. This equality implies x pk = k 1 m=1 c m x m c p m xpm = 0 because cm p = 0. Thus the algebra M k C satisfies identity x pk = 0. Linearizing this identity we obtain the identity S + pk = 0. It follows from this

3 matrix algebra 851 that the algebra M k F satisfies identity S + pk = 0 because the algebra C is commutative and nonnilpotent. The lemma is proved. The algebra M k F does not satisfy the identity S + p k 1 = 0 and even more, the following statement is valid. Lemma 2. Let k>2 α i F i = If either α 1 α 2 or α 2 α 3 then the algebra M k F does not satisfy identity 3 α i i=1 σ S 2k 4 p x σ 1 x σ k i p vx σ k i p+1 x σ 2k 4 p = 0 (1 To prove this lemma, which is the main statement of this section, we need prove a few technical lemmas. Later we fixthe number k>1. Denote by e i j the matrixunits of the algebra M k F. Lete = 1 e k k. Lemma 3. Let m t s l be the nonnegative integers, m t + s + l; x i em k F e y i = e k k z i = e k 1 k u i = e k k 1 w i = e k 1 k 1. Let b = es + m x 1 x m t s l y 1 y t z 1 z s u 1 u l e 1. If either s l or s = l = 0 t>0, then b = 0; 2. If s = l = t = 0 then 3. If s = l>0 then b = S + m x 1 x m b = s s + t 1! S + m s t x 1 x m 2s t w 1 w s (2 Proof. The first and the second equalities are trivial, so we will prove the third equality. Let R be the set of all integer rows r = r 1 r s of length s satisfying the properties r i = t r i 0. The number of the elements of the set R is equal to Ct s+t 1. Since z i e = eu i = 0 the left side of equality (2 is equal to the sum of all elements of the form S + m s t x 1 x m 2s t z σ 1 y τ 1 y τ r1 u 1 z σ s y τ t rs +1 y τ t u s where σ S s τ S t r 1 r s R. Evidently these elements are equal to S + m s t x 1 x m 2s t w 1 w s and the number of them is equal to s! t! Ct s+t 1 is proved. = s s + t 1!. The lemma

4 852 alexander kemer We denote by F X M k F the free product with amalgamated 1 of the free algebra with unit F X and M k F. The elements of this algebra are called the generalized polynomials on the variables from the set X. Let f = f x 1 x n be a generalized polynomial. We say that the algebra M k F satisfies a generalized identity f = 0 iff for every a i M k F the equality f a 1 a n =0 holds in the algebra M k F. Let µ = p+1. Take the arbitrary integers s m n satisfying the properties 2 0 s µ m s, and n µ s. Let f s m n = f s m n( x1 x m+n µ v F X M k F be the sum of all generalized polynomials of the form S + m( xi1 x im s y 1 y s vs + n ( xj1 x jn µ s y s+1 y µ where x i v X y l = e k k i 1 < <i m s j 1 < <j n µ s, { i1 i m s } { j1 j n µ s } = { 1 m+ n µ } Lemma 4. Let m i = pq i + r i 0 r i <p i= 1 2. Ifr 1 + r 2 = p 1 then f s m 1 m 2 xm1 +m 2 µ=x m1 +m 2 =x m1 +m 2 p+2=1 = f s pq 1 pq 2 (3 Proof. Let a i x X a i x. Evidently S + m a 1 a m 1 x x=1 = ms + m 1 a 1 a m 1 S + m a 1 a m x=1 = S + m a 1 a m Using these formulas and the definition of the polynomials fm s n we obtain that the left side of formula (3 is equal to where p 1 r=0 α r f s m 1 r m 2 p 1 r α r = C r p 1 m 1 m 1 1 m 1 r + 1 m 2 m 2 1 m 2 p + r + 2 It is easy to see that α r = 0ifr r 1, in F. The lemma is proved. α r1 = C r p 1 r 1! r 2! = p 1! = 1

5 matrix algebra 853 Remark. F X Later we will often use the following well-known identities in σ S m v xσ 1 x σ m = S + p v x 1 x p 1 = σ S m σ S p 1 v S + p x 1 x p = σ S P m 1 t Cm t x σ 1 x σ t vx σ t+1 x σ n v xσ 1 x σ p 1 v xσ 1 x σ p Lemma If α β then the algebra M 2 F does not satisfy a generalized identity α e2 2 x σ 1 x σ p 1 xσ p x σ 2p 1 σ S 2p 1 + β σ S 2p 1 x σ 1 x σ p e2 2 x σ p+1 x σ 2p 1 = 0 (4 2. If l 1 r p l + m<p l+ r<2p then the algebra M 2 F does not satisfy a nontrivial generalized identity of the form α x σ 1 x σ l e2 2 x σ l+1 x σ l+m xσ l+m+r σ S l+m+r +β σ S l+m+r x σ 1 x σ r e2 2 x σ r+1 x σ r+m xσ l+m+r =0 (5 Proof. Assume that the algebra M 2 F satisfies identity (4. Applying Lemma 1 and the remark this identity can be written in the form p 1 γ i=0 σ S 2p 1 x 1 x i e 2 2 x σ i+1 x σ 2p 1 = 0 where γ = α β 0. Substitute into the left side of this identity x i = e i p 1 x i = e 2 1 p i 2p 3, and x 2p 2 = x 2p 1 = e 1 1 e 2 2 and denote by d the result of the substitution. To calculate the element d we need a few relations. First, the elements x i x j = x i x j + x j x i are central for every i j, because Tr x i =0. It can be verified that x i x 2p 1 = x i x 2p 2 = 0 for i 2p 3; x 2p 1 x 2p 2 = 2. Using these relations and the evident equality σ S i x σ 1 x σ i = 1 2 σ S i x σ 1 x σ i 2 ( xσ i 1 x σ i

6 854 alexander kemer we have the equalities in M 2 F ( p + 1 d = γ e x σ 1 + p 1 x 2 σ 1 e 2 2 x σ 2 x σ 2p 1 σ S 2p 1 = γ 2 p σ S 2p 1 = γ 2 p p 1 e2 2 x σ 1 ( xσ 2 x σ 3 ( xσ 2p 2 x σ 2p 1 σ S 2p 3 ( x2p 1 x 2p 2 e2 2 x σ 1 ( x σ 2 x σ 3 ( xσ 2p 4 x σ 2p 3 = γ 2 p 1 p 1! p 2! e 2 2 e 1 2 ( e1 2 e 2 1 p 1 = γe 1 2 We obtain the contradiction because γ Assume that the algebra M 2 F satisfies identity (5. This identity can be written in the form l i=0 σ S l+m+r + l α i e2 2 x σ 1 x σ m+i xσ m+i+1 x σ l+m+r i=0 σ S l+m+r β i x σ m+i+1 x σ l+m+r e2 2 x σ 1 x σ m+i = 0 where α i = 1 i C i l α β i = C i l β. Since l 1 then α i β i for some i. Let s be a minimal number with the property α s β s. Substituting into this identity x i = 1 m+ s + p<i l + m + r we obtain the identity e2 2 x σ 1 x σ m+s xσ m+s+p α s σ S m+s+p + β s σ S m+s+p x σ m+s+1 x σ m+s+p e2 2 x σ 1 x σ m+s This identity implies the identity (we recall that m + s<p α s e2 2 x σ 1 x σ p 1 xσ p x σ 2p 1 σ S 2p 1 + β s σ S 2p 1 x σ p x σ 2p 1 e2 2 x σ 1 x σ p 1 = 0 This contradicts what has been proven. The lemma is proved.

7 Lemma 6. matrix algebra 855 Let k>3 α 0 α µ β 0 β µ F. If µ 1 t α t 0 (6 then the algebra M k F does not satisfy a generalized identity µ µ α t ef t p k 1 p k 2 e + β t ef t p k 2 p k 1 e = 0 (7 Proof. Assume the contrary. Let k>3 be a minimal number such that the algebra M k F satisfies identity (7 whose coefficients satisfy property (6. Substituting into (7 x i em k e i = 1 p 2k 5 ; v em k F e x j = e k 1 k p 2k 5 < j p 2k 5 +µ; x q = e k k 1 p 2k 5 +µ < q p 2k 5 +2µ = p 2k 4 +1; x r = e p 2k 4 +1 <r p 2k 3 and applying Lemmas 3 and 4, we obtain the generalized identity of the algebra M k 1 F α 0 f 0 p k 1 p k 4 + α µ f µ p k 3 p k 2 + β 0 f 0 p k 2 p k 3 + β µ f µ p k 4 p k 1 where + s=1 α s f s p k 2 p k 3 + β s f s p k 3 p k 2 = 0 α 0 = α 0µ 2µ 1! = 0 α µ = α µµ 2µ 1! = 0 β 0 = β 0µ 2µ 1! = 0 β µ = β µµ 2µ 1! = 0 α s = ( Cµ s 2s µ µ s s + t 1! 2µ s t 1! α t β s = ( Cµ s 2s µ µ s s + t 1! 2µ s t 1! β t if s 0 µ.leta = µ 1 t α t.fors 0 µ we have the equalities in F α s = 1 µ µ µ 1 Cs µ Cs 1 µ 2 s + t 1! p s t! α t = 4Cµ s Cs 1 µ 2 µ 1 s+t p 1! α t = 4 1 s Cµ s Cs 1 µ 2 a In the same way β s = 4 1 s C s µ Cs 1 µ 2 b, where b = µ 1 t β t. We will prove the inequality 1 s α s 0 s=1

8 856 alexander kemer in F. It is sufficient to prove s=1 C s µ Cs 1 µ 2 0 modulo p. Consider the rational function φ x = 1 + x µ 1 + x 1 µ 2. This function can be written in the form µ φ x = γ r x r r= µ+2 for some integer γ r. Calculate the coefficient γ 1. Since then On the other hand φ x = γ 1 = µ µ 2 Cµ t Cs µ 2 xt s s=0 s=1 C s µ Cs 1 µ x p 1 φ x = x µ 2 and γ 1 = C p 1 0 (mod p. Thus we have proved that the algebra M k 1 F satisfies some general identity of form (7 k = k 1 whose coefficients satisfy property (6 α t = α t. It follows from this by the minimality of k that k 1 = 3 and the algebra M 3 F satisfies some nontrivial generalized identity of the form s=1 1 s C s µ Cs 1 µ 2( αef s 2p p e + βef s p 2p e = 0 where e = e e 2 2 e k k = e 3 3. Substituting into this identity v = e 3 3 x 3p µ = e 2 3 x 3p = e 3 2 x i em 3 F e, i = 1 3p µ 2, we obtain the generalized identity of the algebra M 2 F 0 = σ S 3p µ 2 s=1 x σ 2p s x σ 3p µ 2 + αs! µ s! C s µ 1 s C s 1 µ 2 x σ 1 x σ 2p s 1 e 2 2 σ S 3p µ 2 s=1 x σ 1 x σ p s 1 e 2 2 x σ 2p s x σ 3p µ 2 βs! µ s! C s µ 1 s C s 1 µ 2

9 = σ S 3p µ 2 αµ! x σ 1 x σ 2p µ matrix algebra 857 ( µ 2 1 t+1 Cµ 2 t x σ 2p µ+1 x σ 2p t 2 e 2 2 x σ 2p t 1 x σ 2p 2 x σ 2p 1 x σ 3p µ 2 + σ S 3p µ 2 βµ! x σ 1 x σ p µ ( µ 2 1 t+1 Cµ 2 t x σ p µ+1 x σ p t 2 e 2 2 x σ p t 1 x σ p 2 x σ p 1 x σ 3p µ 2 = αµ! 1 x σ 1 x σ 2p µ e2 2 x σ 2p µ+1 x σ 2p 2 σ S 3p µ 2 x σ 2p 1 x σ 3p µ 2 + σ S 3p µ 2 βµ! 1 x σ 1 x σ p µ e 2 2 x σ p µ+1 x σ p 2 xσ p 1 x σ 3p µ 2 This contradicts Lemma 5. The lemma is proved. Proof of Lemma 2. Assume that the algebra M k F satisfies identity (1. Since the algebra M k F is antiisomorphic to itself then this algebra also satisfies identity 3 γ i i=1 σ S 2k 4 p x σ 1 x σ k i p vx σ k i p+1 x σ 2k 4 p = 0 where γ 1 = α 3, γ 2 = α 2, γ 3 = α 1. Therefore we can assume α 1 α 2. Let k>4. Identity (1 implies the generalized identity 3 α i i=1 σ S 2k 4 p+ ex σ 1 x σ k i p+ v x σ 2k 4 p+ e = 0 where µ = p+1 2. Substituting into this identity v x i em k e,1 i 2k 5 p µ; x j = e, 2k 5 p µ + 1 j 2k 4 p µ 1; x r = ek 1, k, 2k 4 p µ j 2k 4 p 1; x q = e k k 1 2k 4 p q 2k 4 p + µ 1 and applying Lemmas 3 and 4, we obtain the generalized identity of the algebra M k 1 F µ g + h + α t f t p k 2 p k 3 = 0 where α t = ( C t µ 2t! µ t! α2 = µ! C t µ

10 858 alexander kemer if t µ, α µ = µ! α 1; the polynomial h is a linear combination of the polynomials f s p k 3 p k 2, and the polynomial g is a linear combination of the polynomials fi s j satisfying the property either i = k 1 p or j = k 1 p. By Lemma 1 let g = 0 be an identity of the algebra M k 1 F. Thus we have proven that the algebra M k 1 F satisfies some identity of form (7 k = k 1 > 3 α i = α i. This contradicts Lemma 6 because ( µ 1 t α t = 1 µ µ! α t Cµ t α 2 = 1 µ µ! α 1 α 2 0 Let k = 4. Substituting into (1 k = 4 v = e 4 3, x 4p = e 3 4, x i = e, i> 3p; x j em 4 F e, j 3 p and using Lemma 1 we obtain the generalized identity of the algebra M 3 F 2 α i i=1 σ S 3p x σ 1 x σ 3 i p e 3 3 x σ 3 i p+1 x σ 3p = 0 Substituting into this identity x 3p = e 2 3, x 3p 1 = e 3 2, x i em 3 F e, i 3p 2 e = 1 e 3 3 we obtain the generalized identity of the algebra M 2 F α 1 x σ 1 x σ 2p 1 e 2 2 x σ 2p x σ 3p 2 σ S 3p 2 + α 2 σ S 3p 2 x σ 1 x σ p 1 e 2 2 x σ p x σ 3p 2 = 0 By Lemma 1 this identity can be written in the form x σ 1 x σ 2p 1 e2 2 x σ 2p x σ 3p 2 α 1 σ S 3p 2 + α 2 σ S 3p 2 e2 2 x σ 1 x σ p 1 xσ p x σ 3p 2 = 0 Substituting into the last identity x i = 1, p + 1 i 2p 1 we obtain the identity of form (4. This contradicts Lemma 5, because α 1 α 2. Consider the last case: k = 3. Substituting into (1 k = 3 v = e 3 2, x 2p = e 2 3, x i = e, p<i<2p; x j em 4 F e, j p, we obtain the generalized identity of the algebra M 2 F α 1 x σ 1 x σ p e α 2 e 2 2 x σ 1 x σ p = 0 σ S p σ S p This identity is a nontrivial identity of form (4. We again obtain the contradiction. The lemma is proved.

11 matrix algebra THE MAIN RESULTS Denote by t k the matrixtype of the algebra M k G, where G is the Grassmann algebra of infinite rank. Lemma 7. t k pk. Proof. Let A be the algebra M k G with a unit adjoined. It is well known that the algebras A and M k G have the same multilinear identities. Define the trace in A trivially: Tr a =0 for every a A. Then by Lemma 1 the algebra A satisfies every multilinear trace identity of degree kp whose nontrace part is equal to S + kp. In particular the algebra A satisfies the multilinear Cayley Hamilton identity of degree kp. Hence by the main theorem of 1 the algebra A satisfies every multilinear identity of the algebra M kp. The lemma is proved. Let d k be a minimal number m such that the algebra M k G satisfies the standard identity of degree m. By Lemma 7 and the theorem of Kaplansky we have the inequality: d k 2kp. The following theorem gives a low estimation for d k. Theorem 1. If k>1 then d k > 2k 3 p 1. Proof. Denote by g 1 g 2 the generators of the algebra G satisfying the relations g i g j = g j g i. In the case of k = 2 the conclusion of the theorem is trivial because the algebra G does not satisfy the standard identity of degree <p+ 1. Let k>2. If d k 2k 3 p 1 then the algebra M k G satisfies the standard identity S 2k 3 p 1 y 1 y 2k 3 p 1 =0 Substituting into this identity y i = x i g i, i 2k 3 p 2 y 2k 3 p 1 = vg 2k 3 p 1 g 2k 3 p, where x i v M k F, we obtain hg 1 g 2k 3 p = 0, where h = σ S 2k 3 p 2 2k 3 p 2 i=0 1 i x σ 1 x σ i v x σ 2k 3 p 2 It follows from this that the algebra M k F satisfies identity h = 0. This identity and identity S + 2k 3 p 1 = 0, which holds in M k G by Lemma 1, imply the identity x σ 1 x σ 2m v x σ 2k 3 p 2 = 0 σ S 2k 3 p 2 0 2m 2k 3 p 2

12 860 alexander kemer The last identity can be written in the form x σ 1 x σ 2sp h σ where h σ = σ S 2k 3 p 2 0 2sp 2k 5 p 1 p 1 m=0 x σ 2 s+1 p 1 x σ 2k 3 p 2 = 0 x σ 2sp+1 x σ 2sp+2m vx σ 2sp+2m+1 x σ 2 s+1 p 2 Then by the remark we have the identity x σ 1 x σ 2sp σ S 2k 3 p 2 0 2sp 2k 5 p 1 v x σ 2sp+1 x σ 2sp+2 x σ 2 s+1 p 3 x σ 2 s+1 p 2 x σ 2 s+1 p 1 x σ 2k 3 p 2 = 0 Substituting into this identity x i = 1, 2k 4 p 2k 3 p 2 we obtain x σ 1 x σ 2sp v xσ 2sp+1 x σ 2s+1 p 1 σ S 2k 4 p 1 0 2sp 2k 5 p 1 x σ 2s+1 p x σ 2k 4 p 1 = 0 This identity implies σ S 2k 4 p 0 2sp 2k 5 p 1 x σ 2s+1 p+1 x σ 2k 4 p = 0 x σ 1 x σ 2sp v xσ 2sp+1 x σ 2s+1 p By the remark and Lemma 1 the last identity can be written in the form x σ 1 x σ 2sp vx σ 2sp+1 x σ 2k 4 p σ S 2k 4 p k 3 2s k 1 σ S 2k 4 p k 3 2s k 1 x σ 1 x σ 2s+1 p v x σ 2k 4 p = 0 It is easy to see that this identity is an identity of form (1, where α 1 = 1, α 2 = 1, and α 3 = 1 if the number k is odd and α 1 = 1, α 2 = 1, and α 3 = 1 if the number k is even. Since α 1 α 2, by Lemma 2 we obtain the contradiction. The theorem is proved. The problem of calculating the number d k is very interesting. We formulate the following conjecture. Conjecture. d k = 2k 1 p + 1.

13 matrix algebra 861 Now we can prove the main theorem. Theorem 2. t n = pn. Proof. If t n pn for some n 1 then the algebra M n G satisfies every multilinear identity of the matrixalgebra M pn 1 F. It follows from this that for every s the algebra M sn G =M s F M n G satisfies every multilinear identity of the algebra M s pn 1 F. In particular, by the theorem of Kaplansky the algebra M sn G must satisfy the standard identity of degree 2s pn 1. On the other hand, by Theorem 1 this algebra does not satisfy the standard identity of degree 2sn 3 p 1. Hence we have the inequality 2s pn 1 > 2sn 3 p 1, which is not true for s 3p+1. The 2 theorem is proved. Let Z be the ring of integers, Q be the field of rational numbers, Q p be the subring of Q consisting of all fractions whose denominators are not divisible by p, and F p be the simple field of characteristic p. Denote by φ the natural homomorphism φ Q p X F p X The following statement looks like the lemma in 3. Theorem 3. For n 3p + 1 p 1 there exists a multilinear polynomial f Q p X such that 1 m! S n = f (8 modulo the T -ideal T M p 1 T G, where m = n if n be even, and m = n 1 if n be odd. Proof. Let s = 3p+1. Consider the algebra M 2 s Q p X. Take the elements s a k = x k i j e i j M s Z X i j=1 where x k i j X, k = 1 2s p 1 = 3p + 1 p 1. We have the equality ( s S 2s p 1 a1 a 2s p 1 = h i j e i j i j=1 for some h i j Z X. Since the algebra M s p 1 Q satisfies the standard identity of degree 2s p 1 then the algebra M p 1 Q satisfies identities h i j = 0 for all i j. By Theorem 1 for some i j the Grassmann algebra G over field F p does not satisfy identity φ h i j =0 because 2s 3 p 1 = 2s p 1. It follows from this that there exists a multilinear polynomial h = h x 1 x l Z X, l 2s p 1, satisfying the properties: (1 The

14 862 alexander kemer algebra M p 1 Q satisfies identity h = 0 but the Grassmann algebra over F p does not satisfy identity φ h =0; (2 The polynomial h is unitary; i.e., the polynomial h is a linear combination of the products of long commutators. It is well known that if the polynomial h is unitary and h / T G then l = 2k and h = α x 1 x 2 x 2k 1 x 2k +g for some integer α and polynomial g T G with integer coefficients. Since the Grassmann algebra over F p does not satisfy identity φ h =0 then α 0 modulo p. It follows from this that for every q s p 1 the algebra M p 1 Q satisfies identity of the form x 1 x 2 x 2q 1 x 2q +g q = 0 for some g q T G Q p X. Denote the left side of this identity by h q. Let n 3p + 1 p 1. Letf = 2 q h q if n = 2q, and 2q+1 f = 2 q i=1 x i h q x 1 x i 1 x i+1 x 2q+1 if n = 2q + 1. We will prove identity (8. Evidently the algebra M p 1 Q satisfies this identity. It remains to show that the algebra G satisfies identity (8. Indeed, using the identity x y z t = x t z y which holds in G, we have the identities modulo T G 1 2q! S 1 2q = 2 q 2q! σ S 2q 1 σ x σ 1 x σ 2 xσ 2q 1 x σ 2q = 2 q x 1 x 2 x 2q 1 x 2q =2 q h q = f 1 2q! S 2q+1 x 1 x 2q+1 = 1 2q! 2q+1 1 i S 2q x 1 x i 1 x i+1 x 2q+1 i=1 2q+1 = 2 q 1 i h q x 1 x i 1 x i+1 x 2q+1 =f i=1 The theorem is proved.

15 matrix algebra 863 REFERENCES 1. A. Kemer, Multilinear identities of the algebras over a field of characteristic p, Internat. J. Algebra Comput. 2 (1995, A. Kemer, Multilinear components of the prime subvarieties of the variety Var M 2 F, Algebras Represent. 1 (2001, A. Kemer, T. Antipova, and A. Antipov, On the traditional way of description of the prime varieties in characteristic p, Analele stiintifice Univ. Ovidius. Constanta 9 (2001.