PI-algebras with slow codimension growth

Transcription

1 Journal of Algebra 284 (2005) PI-algebras with slow codimension growth A. Giambruno, D. La Mattina Dipartimento di Matematica e Applicazioni, Via Archirafi 34, Palermo, Italy Received 19 July 2004 Available online 2 November 2004 Communicated by Susan Montgomery Abstract Let c n (A), n = 1, 2,..., be the sequence of codimensions of an algebra A over a field F of characteristic zero. We classify the algebras A (up to PI-equivalence) in case this sequence is bounded by a linear function. We also show that this property is closely related to the following: if l n (A), n = 1, 2,..., denotes the sequence of colengths of A, counting the number of S n -irreducibles appearing in the nth cocharacter of A, then lim n l n (A) exists and is bounded by Elsevier Inc. All rights reserved. Keywords: Polynomial identity; T-ideal; Codimensions 1. Introduction Given an algebra A over a field F one can associate to A a numerical sequence c n (A), n = 1, 2,..., called the sequence of codimensions of A, giving a measure of the polynomial identities satisfied by A. In general c n (A) is bounded from above by n!, butin case A is a PI-algebra, i.e., satisfies a non-trivial polynomial identity, a celebrated theorem of Regev asserts that c n (A) is exponentially bounded [17]. When the field F is of characteristic zero, it turns out that the sequence of codimensions of any PI-algebra is either polynomially bounded or grows exponentially (see [11]). For general PI-algebras the exponential rate of growth was computed in [5] and [6] and it turns out to be a non-negative * Corresponding author. addresses: agiambr@unipa.it (A. Giambruno), daniela@math.unipa.it (D. La Mattina) /$ see front matter 2004 Elsevier Inc. All rights reserved. doi: /j.jalgebra

2 372 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) integer. In case the codimensions are polynomially bounded, Kemer in [12] gave the following characterization. Let G be the infinite dimensional Grassmann algebra and let UT 2 be the algebra of 2 2 upper triangular matrices. Then c n (A), n = 1, 2,..., is polynomially bounded if and only if G, UT 2 / var(a),wherevar(a) denotes the variety of algebras generated by A. The aim of this paper is to try to refine Kemer s result through the knowledge of the polynomial rate of growth of c n (A). In general, given an integer t 0 can one find a (finite) number of algebras M 1,...,M d depending on t, such that lim sup n log n c n (A) t if and only if M 1,...,M d / var(a)? This question seems out of reach in this generality but we shall give a complete answer for some values of t. We first need to formulate an apparently different problem about representations of the symmetric group S n. An equivalent formulation of Kemer s result can be given as follows. Let F X be the free algebra on a countable set X ={x 1,x 2,...} and let Id(A) be the T-ideal of polynomial identities of the algebra A. The permutation action of S n on the space V n of multilinear V polynomials in the first n variables induces a structure of S n -module on n V n Id(A) and let χ n (A) be its character. By complete reducibility we can write χ n (A) = λ n m λχ λ where χ λ is the irreducible S n -character associated to the partition λ of n and m λ 0isthe corresponding multiplicity. Then l n (A) = λ n m λ is the nth colength of A.NowKemer s result can be stated as follows [15]: c n (A) is polynomially bounded if and only if the sequence of colengths is bounded by a constant, i.e., l n (A) k,forsomek 0andforall n 1 (see [3]). In this setting one can ask if it is possible to give a finer classification depending on the value of the constant k. In particular, given k 0, can one find a finite number of algebras M 1,...,M d, depending on k, such that lim sup n l n (A) k if and only if M 1,...,M d / var(a)? In this paper we are able to answer this question in the positive in case k 2. We shall also show that this is strictly related to the codimensions of A being linearly bounded. As a consequence we are able to classify up to PI-equivalence the algebras A such that l n (A) 2orc n (A) kn. It turns out that for n large enough the only sequences of codimensions allowed are c n (A) = 0, 1,nand c n (A) = 2n Generalities Throughout this paper, we shall denote by F a field of characteristic zero and by A an associative algebra over F. We refer the reader to [4] and [18] for the basic definitions and properties of PI-algebras. Let F X be the free associative algebra on the countable set X ={x 1,x 2,...} and let Id(A) ={f F X f 0onA} be the set of polynomial identities of A. Clearly Id(A) is a T-ideal of F X, i.e., an ideal invariant under all endomorphisms of F X. It is well known that in characteristic zero Id(A) is completely determined by its multilinear polynomials and we denote by V n = span F {x σ(1) x σ(n) σ S n }

3 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) the space of multilinear polynomials in the indeterminates x 1,...,x n. The symmetric group S n acts on the left on V n :ifσ S n and f(x 1,...,x n ) V n,then σf (x 1,...,x n ) = f(x σ(1),...,x σ(n) ). Since the subspace V n Id(A) is invariant under this action, V n (A) = V n V n Id(A) has a structure of S n -module. The S n -character of V n (A), denoted χ n (A), is called the nth cocharacter of A and c n (A) = dim F V n (A) is the nth codimension of A. It is well known that in characteristic zero there is a one-to-one correspondence between irreducible S n -characters and partitions λ n. Ifχ λ denotes the irreducible S n -character correspondingto λ then, since char F = 0, by complete reducibility we can write χ n (A) = λ n m λ χ λ, (1) where m λ 0 is the multiplicity of χ λ in the given decomposition. Also l n (A) = λ n m λ is called the nth colength of A. The following remark lists some basic properties of the sequenceof cocharacters, codimensions and colengths. Remark 1. Let A and B be F -algebras and let A B be their direct sum. If χ n (A) = λ n m λχ λ, χ n (B) = λ n m λ χ λ,andχ n (A B) = λ n m λ χ λ are the nth cocharacters of A, B and A B, respectively, then m λ m λ + m λ, l n (A B) l n (A) + l n (B) and c n (A B) c n (A) + c n (B). In case B is a nilpotent algebra and B k = 0, then for all n k we have m λ = m λ. Hence l n (A B) = l n (A) and c n (A B) = c n (A) in this case. Proof. Consider the map of S n -modules α : V n V n V n Id(A) V n V n Id(B) such that α(f ) = (f + (V n Id(A)), f + (V n Id(B))). Sinceker(α) = V n Id(A) Id(B) = V n Id(A B), we have an embedding of S n -modules V n V n Id(A B) V n V n Id(A) V n V n Id(B).

4 374 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) Thus m λ m λ + m λ and l n(a B) l n (A) + l n (B), c n (A B) c n (A) + c n (B). If B k V = 0 then, for all n k, n V V n Id(B) is the zero module. It follows that for all n k. V n V n Id(A) n V n Id(A B) = GivenanalgebraA, letvar(a) denote the variety of algebras generated by A. Another fact that we shall use throughout the paper is that if A and B are F -algebras and B var(a) then l n (B) l n (A) and c n (B) c n (A). This is clear since in this case, Id(A) Id(B) and so, V n (B) can be embedded in V n (A). In the next sections we prefer to work with the representation theory of the general linear group which is well related to that of the symmetric group. To this end we need to introduce the space of homogeneous polynomials in a given set of variables. Let F m X =F x 1,...,x m denote the free associative algebra in m variables and let U = span F {x 1,...,x m }. The group GL(U) = GL m acts naturally on the left on the space U and we can extend this action diagonally to get an action on F m X. The space F m X Id(A) is invariant under this action, hence F m (A) = F m X F m X Id(A) inherits a structure of left GL m -module. Let Fm n be the space of homogeneous polynomials of degree n in the variables x 1,...,x m.then F n m (A) = F n m F n m Id(A) is a GL m -submodule of F m (A) and we denote its character by ψ n (A). Write ψ n (A) = λ n m λ ψ λ where ψ λ is the irreducible GL m -character associated to the partition λ and m λ is the corresponding multiplicity. It was proved in [1] and [2] that if the nth cocharacter of A has the decomposition given in (1) then m λ = m λ,forallλ n whose corresponding diagram has height at most m. It is also well known (see for instance [4, Theorem ]) that any irreducible submodule of F n m (A) corresponding to λ is generated by a non-zero polynomial f λ, called highest weight vector, of the form λ 1 f λ = St hi (λ)(x 1,...,x hi (λ)) α σ σ, (2) σ S n i=1

5 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) where α σ F, the right action of S n on F n m (A) is defined by place permutation, h i(λ) is the height of the ith column of the diagram of λ and St r (x 1,...,x r ) = τ S r (sgn τ)x τ(1) x τ(r) is the standard polynomial of degree r. Recall that f λ is unique up to a multiplicative constant. For a Young tableau T λ, denote by f Tλ the highest weight vector obtained from (2) by considering the only permutation σ S n such that the integers σ(1),..., σ (h 1 (λ)), inthis order, fill in from top to bottom the first column of T λ, σ(h 1 (λ) + 1),..., σ (h 1 (λ) + h 2 (λ)) the second column of T λ,etc. We also have the following (see for instance [4, Proposition ]). Remark 2. If ψ n (A) = λ n m λ ψ λ is the GL m -character of F n m (A),then m λ 0 if and only if there exists a tableau T λ such that the corresponding highest weight vector f Tλ is not a polynomial identity for A. Moreover m λ is equal to the maximal number of linearly independent highest weight vectors f Tλ in F n m (A). 3. Computing the identities of some PI-algebras The purpose of this section is to compute the cocharacters, the codimensions, the T-ideals, etc. of some PI-algebras that will play a basic role in the next section. Given polynomials f 1,...,f n F X let us denote by f 1,...,f n T the T-ideal generated by f 1,...,f n. We shall also denote by y,z,t,w the variables of X. Also, given an algebra A let J(A)denote its Jacobson radical. In order to shorten the proof of next lemma and of Lemma 6 we use a result of Guterman and Regev [9] even though a direct proof can be easily found. Lemma 3. Let M 1 = ( ) 0 F 0 F and M2 = ( ) FF 0 0. Then for all n>1, 1. χ n (M 1 ) = χ n (M 2 ) = χ (n) + χ (n 1,1). 2. l n (M 1 ) = l n (M 2 ) = {x i1 x in,i 2 < <i n } is a basis of V n (mod V n Id(M 1 )) and {x i1 x in,i 1 < <i n 1 } is a basis of V n (mod V n Id(M 2 )). 4. c n (M 1 ) = c n (M 2 ) = n. 5. Id(M 1 ) = z[x,y] T and Id(M 2 ) = [x,y]z T.

6 376 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) Proof. The conclusion clearly holds for n = 2, hence we may assume that n 3. By [9, Theorem 1], Id(M 1 ) = z[x,y] T,Id(M 2 ) = [x,y]z T and c n (M 1 ) = c n (M 2 ) = n. Hence the elements x i1 x in, i 2 < <i n span V n (mod V n Id(M 1 )). Since their number equals n = c n (M 1 ) = dim V n /(V n Id(M 1 )) we have that {x i1 x in i 2 < <i n } is a basis of V n (mod V n Id(M 1 )). We now determine the decomposition of the nth cocharacter of M 1.Lete ij be the usual matrix units. Since c n (M 1 ) = n is polynomially bounded and J(M 1 ) = Fe 12 satisfies J(M 1 ) 2 = 0, by [7, Theorem 3] we have that χ n (M 1 ) = m λ χ λ = m (n) χ (n) + m (n 1,1) χ (n 1,1). λ n λ λ 1 <2 If λ = (n), the corresponding highest weight vector f λ = x n is not an identity of M 1 since f λ (e 22 ) = e Then, being x n the only highest weight vector corresponding to λ = (n), it follows that m (n) = 1. Since c n (M 1 ) = m (n) degχ (n) + m (n 1) degχ (n 1) and by the hook formula [10], degχ (n) = 1, degχ (n 1,1) = n 1, it follows that n = 1 + m (n 1) (n 1) and, so, m (n 1) = 1. Therefore and χ n (M 1 ) = χ (n) + χ (n 1,1) l n (M 1 ) = λ n m λ = 2. A similar proof gives the desired results about M 2. In what follows we use the left normed notation for Lie commutators. Hence we write [...[[x 1,x 2 ],x 3 ],...,x n ]=[x 1,x 2,...,x n ]. Lemma 4. Let M 3 = {( abc 0 ad 00a ) } a,b,c,d F. Then for all n>3, 1. χ n (M 3 ) = χ (n) + χ (n 1,1) + χ (n 2,1,1). 2. l n (M 3 ) = {x 1 x n,x i1 x in 2 [x i,x j ], i 1 < <i n 2, i > j} is a basis of V n (mod V n Id(M 3 )). 4. c n (M 3 ) = n(n 1) Id(M 3 ) = [x,y,z], [x,y][z, w] T. Proof. Let Q = [x,y,z], [x,y][z, w] T.Since[M 3,M 3 ] span{e 13 } it is clear that [x,y,z] and [x,y][z, w] are identities of M 3 and Q Id(M 3 ). Before proving the opposite inclusion, we find a generating set of V n modulo V n Q.

7 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) It is well known (see for instance [4, Theorem 5.2.1]) that any multilinear polynomial of degree n can be written, modulo [x,y][z, w] T, as a linear combination of polynomials of the type x i1 x im [x k,x j1,...,x jn m 1 ] where i 1 < <i m, j 1 < <j n m 1, k>j 1, m n 1. Thus, because of the identity [x,y,z], the elements x 1 x n,x i1 x in 2 [x i,x j ], i 1 < <i n 2,i>j (3) span V n modulo V n Q. We next prove that these elements are linearly independent modulo Id(M 3 ). Suppose that f = n i,j=1 i>j α ij x i1 x in 2 [x i,x j ]+βx 1 x n 0 ( mod V n Id(M 3 ) ). By making the evaluation x k = e 11 + e 22 + e 33,forallk = 1,...,n,wegetβ = 0. Also, for fixed i and j, the evaluation x i = e 12, x j = e 23 and x k = e 11 + e 22 + e 33 for k/ {i, j} gives α ij = 0. Thus the elements in (3) are linearly independent modulo V n Id(M 3 ).Since V n Q V n Id(M 3 ), this proves that Id(M 3 ) = Q and the elements in (3) are a basis of V n modulo V n Id(M 3 ). By counting we obtain c n (M 3 ) = dim V n V n Id(M 3 ) n(n 1) + 2 =. 2 Since deg χ (n) + degχ (n 1,1) + degχ (n 2,1,1) = 1 + (n 1) + (n 1)(n 2) 2 = n(n 1)+2 2,if we find for each λ {(n), (n 1, 1), (n 2, 1, 1)} a highest weight vector which is not an identity of M 3, we may conclude that χ n (M 3 ) = χ (n) + χ (n 1,1) + χ (n 2,1,1) and the nth cocharacter has the wished decomposition. Clearly x n, the highest weight vector corresponding to λ = (n), is not an identity of M 3. Let f (n 1,1) =[x 1,x 2 ]x n 2 1 be a highest weight vector corresponding to λ = (n 1, 1). Taking a 1 = (e 11 + e 22 + e 33 ) + e 12 and a 2 = e 23, we get that f (n 1,1) (a 1,a 2 ) = e Thus f (n 1,1) is not an identity of M 3. Finally f λ = St 3 (x 1,x 2,x 3 )x1 n 3 is a highest weight vector corresponding to λ = (n 2, 1, 1),whereSt 3 is the standard polynomial of degree 3. By choosing a 1 = e 11 + e 22 + e 33, a 2 = e 12 and a 3 = e 23, we obtain f λ (a 1,a 2,a 3 ) = e 13 0.

8 378 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) Thus, since these polynomials are not identities of M 3, it follows that and χ n (M 3 ) = χ (n) + χ (n 1,1) + χ (n 2,1,1) l n (M 3 ) = 3. The proof of the lemma is complete. Lemma 5. Let B = M 1 M 2. Then for all n 3, 1. χ n (B) = χ (n) + 2χ (n 1,1). 2. l n (B) = {[x 1,x i ]x j1 xˆ i x jn 2,x j1 xˆ i x jn 2 [x 1,x i ],x 1 x n,j 1 < <j n 2 } is a basis of V n (mod V n Id(B)), where the symbol ˆx i means that the variable x i is omitted. 4. c n (B) = 2n Id(B) = St 3 (x,y,z),z[x,y]w,[x,y][z, w] T. Proof. Let Q = St 3 (x,y,z),z[x,y]w,[x,y][z, w] T. It is easy to check that Q Id(B). Next we claim that the set of polynomials { } [x1,x i ]x j1 ˆx i x jn 2,x j1 ˆx i x jn 2 [x 1,x i ],x 1 x n,j 1 < <j n 2 (4) span V n modulo V n Q. Since for n = 3 this is clear, we assume that n>3. Now, as in the proof of the previous lemma, any multilinear polynomial of degree n can be written, modulo [x,y][z, w] T,as a linear combination of polynomials of the type x i1 x im [x k,x j1,...,x jn m 1 ] where i 1 < <i m, j 1 < j n m 1, k>j 1, m n 1. Since [x k,x j1,...,x jn m 1 ] [x k,x j1 ]x j2 x jn m 1 ± x jn m 1 x j2 [x k,x j1 ] ( mod z[x,y]w T ) and x i1 x im [x k,x j1 ]x j2 x jn m 1 0 ( mod z[x,y]w T), we have that for m 1, x i1 x im [x k,x j1,...,x jn m 1 ] x k1 x kn 2 [x k,x j1 ] (mod Q).

9 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) Also, since [x,y][z, w] Q, we may assume that k 1 < <k n 2.Now,sincen 4, if j 1 1wehavethat x k1 x kn 2 [x k,x j1 ] wx 1 x k x j1 wx 1 x j1 x k wx k x 1 x j1 wx j1 x 1 x k wx k [x 1,x j1 ] wx j1 [x 1,x k ] (mod Q) where w = x k2 x kn 2. Thus x k1 x kn 2 [x k,x j1 ] x h1 x hn 2 [x 1,x j1 ] x l1 x ln 2 [x 1,x k ] (mod Q) where h 1 < <h n 2 and l 1 < <l n 2. Similarly [x k,x j1 ]x k1 x kn 2 [x 1,x j1 ]x h1 x hn 2 [x 1,x k ]x l1 x ln 2 (mod Q). It follows that the polynomials [x 1,x i ]x j1 xˆ i x jn 2,x j1 ˆx i x jn 2 [x 1,x i ], x 1 x n,j 1 < <j n 2 generate V n (mod V n Q) for n 3. We next show that the elements in (4) are linearly independent modulo Id(B). Let f Id(B) be a linear combination of the elements in (4): f = n α i [x 1,x i ]x xˆ j1 i x jn 2 i=1 j 1 < <j n 2 + n j=1 j 1 < <j n 2 β j x j1 xˆ i x jn 2 [x 1,x j ]+γx 1 x n. By making the evaluation x i = (e 22, 0), foralli = 1,...,n,wegetγ = 0. Also for a fixed k, the evaluation x k = (e 12, 0) and x i = (e 22, 0) for i k, givesα k = 0. Similarly x k = (0,e 12 ) and x i = (0,e 11 ) for i k, givesβ k = 0. Therefore the elements in (4) are linearly independent modulo V n Id(B). Since V n Id(B) V n Q it follows that Id(B) = Q and the elements in (4) are a basis of V n modulo V n Id(B) for n 3. Thus V n c n (B) = dim = 2n 1, for n 3. V n Id(B)

10 380 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) We now determine the decomposition of the nth cocharacter of this algebra. For an algebra A write χ n (A) = λ n m λ(a)χ λ. Then, since B = M 1 M 2, by Remark 1, m λ (B) m λ (M 1 ) + m λ (M 2 ) for all λ n. Hence by Lemma 3, it follows that χ n (B) = m (n) χ (n) + m (n 1,1) χ (n 1,1) and m (n),m (n 1,1) 2. Since deg χ (n) = 1andx n is not an identity of B, clearly m (n) = 1. Moreover, since c n (B) = 2n 1 and degχ (n 1,1) = n 1, it follows that m (n 1,1) = 2. Thus and Lemma 6. Let M 4 = ( ) FFF 0 0 F, M 5 = χ n (B) = χ (n) + 2χ (n 1,1) l n (B) = λ n m λ = 3. ( ) 0 FF 00F and M 6 = 00F ( ) 0 FF 0 FF. Then for all n>3, χ n (M 4 ) = χ n (M 5 ) = χ n (M 6 ) = χ (n) + 2χ (n 1,1) + χ (n 2,2) + χ (n 2,1,1). 2. l n (M 4 ) = l n (M 5 ) = l n (M 6 ) = {x i1 x in 2 x i x j, i 1 < <i n 2 } is a basis of V n (mod V n Id(M 4 )), {x i x j x i1 x in 2,i 1 < <i n 2 } is a basis of V n (mod V n Id(M 5 )) and {x i x i1 x in 2 x j, i 1 < <i n 2 } is a basis of V n (mod V n Id(M 6 )). 4. c n (M 4 ) = c n (M 5 ) = c n (M 6 ) = n(n 1). 5. Id(M 4 ) = [x,y]zw T, Id(M 5 ) = zw[x,y] T and Id(M 6 ) = z[x,y]w T. Proof. By [9] we have that Id(M 4 ) = [x,y]zw T,Id(M 5 ) = zw[x,y] T,Id(M 6 ) = z[x,y]w T and c n (M 4 ) = c n (M 5 ) = c n (M 6 ) = n(n 1). As in the proof of the previous lemmas, it is easy to see that the elements x i1 x in 2 x i x j, i 1 < <i n 2 form a basis of V n (mod V n Id(M 4 )). A similar remark holds for M 5 and M 6. We next determine the decomposition of the nth cocharacter of M 4. A similar proof will give the decomposition of the nth cocharacter of M 5 and M 6.SinceJ(M 4 ) = Fe 12 + Fe 13 + Fe 23,andJ(M 4 ) 3 = 0, by [7, Theorem 3] we have that χ n (M 4 ) = m (n) χ (n) + m (n 1,1) χ (n 1,1) + m (n 2,2) χ (n 2,2) + m (n 2,1,1) χ (n 2,1,1). Since M 4 is not nilpotent, clearly m (n) = 1. Let λ = (n 1, 1) and denote by T (i) λ the standard tableau containing the integer i = 2,...,nin the only box of the second row. Then f T (i) λ = x 1 x1 i 2 x 2 x1 n i x 2 x1 i 2 x 1 x1 n i

11 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) is the highest weight vector corresponding to T (i) λ. By making the evaluation x 1 = e 11 +e 23 and x 2 = e 12, we get that f T (n) λ = e 12 and f T (n 1) λ = e 13. This says that f (n) T and f (n 1) λ T are not identities of M 4. Moreover these polynomials are λ linearly independent (mod Id(M 4 )). In fact if f = α 1 f T (n) λ + α 2 f T (n 1) λ 0 ( mod Id(M4 ) ), make the evaluation x 1 = e 11 and x 2 = e 12 to get α 1 = 0. Similarly, choose x 1 = e 11 + e 23 and x 2 = e 12 to obtain α 2 = 0. Thus m (n 1,1) 2. Since c n (M 4 ) = n(n 1) and degχ (n) + 2degχ (n 1,1) + deg χ (n 2,2) + degχ (n 2,1,1) = 1 + 2(n 1) + n(n 3) 2 + (n 2)(n 1) 2 = n(n 1), in order to prove the given decomposition of the nth cocharacter, it is enough to find for each λ {(n 2, 2), (n 2, 1, 1)} a highest weight vector which is not an identity of M 4. For λ = (n 2, 2), consider the following standard tableau 1 2 n 1 n and the corresponding highest weight vector f Tλ = (sgn στ)x σ(1) x τ(1) x1 n 4 x σ(2) x τ(2). σ,τ S 2 Evaluating x 1 = e 11 and x 2 = e 12 + e 23,wegetf Tλ = e 13 0andf Tλ is not an identity of M 4. Finally consider f Tλ = σ S 3 (sgn σ)x σ(1) x1 n 3 x σ(2) x σ(3), the highest weight vector corresponding to the standard tableau Since 1 n 1 n. f Tλ (e 11,e 12,e 23 ) = e 13 0 we have that f Tλ is not an identity of M 4. Hence χ n (M 4 ) = χ (n) + 2χ (n 1,1) + χ (n 2,2) + χ (n 2,1,1) and l n (M 4 ) = 5.

12 382 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) When codimensions are bounded by a constant A main tool in this and the next section is the following result on the decomposition of the Jacobson radical of a finite dimensional algebra. Lemma 7 [8, Lemma 2]. LetA be a finite dimensional algebra over F and suppose that A = B + J,whereB is a semisimple subalgebra and J = J(A) is its Jacobson radical. Then J can be decomposed into the direct sum of B-bimodules J = J 00 J 01 J 10 J 11, where, for i {0, 1}, J ik is a left faithful module or a 0-left module according as i = 1 or i = 0, respectively. Similarly, J ik is a right faithful module or a 0-right module according as k = 1 or k = 0, respectively. Moreover, for i, k, l, m {0, 1}, J ik J lm δ kl J im where δ kl is the Kronecker delta and J 11 = BN for some nilpotent subalgebra N of A commuting with B. An obvious remark is that if A = B + J is an algebra satisfying the hypothesis of the previous lemma and 1 B is the unit element of B,wehavethat 1 B a1 B = a, 1 B b = b and c1 B = c for all a J 11,b J 10,c J 01. In the sequel we shall also use the following notation. Definition 8. Let A and B be algebras. We say that A is PI-equivalent to B and we write A PI B, if Id(A) = Id(B). In the following lemmas we shall assume that A is a finite dimensional algebra of the type A = F +J where J = J(A)is the Jacobson radical of A. Also we shall tacitly assume that J has the decomposition given in Lemma 7. We start with the following. Lemma 9. Let A = F + J.If[J 11,J 11 ] 0,thenM 3 var(a). Proof. Notice that A = F + J 11 is a subalgebra of A. We shall prove that M 3 var(a ) var(a). Hence without lost of generality we may assume that A = F + J and J = J 11. Since by hypothesis J is not commutative, J 2 0andletk be the least integer such that J k = 0andJ k 1 0. We proceed by induction on k. If k = 3 we shall prove that A itself is PI-equivalent to M 3. In fact, in this case, since 1 F commutes with J, [A,A] J 2 ; hence [A,A,A]=0and[A,A][A,A]=0. By Lemma 4 this says that Id(M 3 ) Id(A). Conversely, let f Id(A) be a multilinear polynomial of degree n. By Lemma 4, one can write f as f = αx 1 x n + α ij x x i1 in 2 [x i,x j ]+g i>j i 1 < <i n 2

13 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) where g Id(M 3 ). Choosing x i = 1 F for all i = 1,...,n we get α = 0. Since [J,J] 0, there exist two elements a,b J such that [a,b] 0. Let i 0,j 0 {1,...,n} be such that α i0 j 0 0. Take c 1,...,c n A, with c i0 = a, c j0 = b and c k = 1 F for all k/ {i 0,j 0 }.Itis easy to check that f(c 1,...,c n ) = α i0 j 0 [a,b] 0, a contradiction. Hence f = g Id(M 3 ) and Id(A) = Id(M 3 ). This settles the case k = 3. Let now k>3 and suppose first that [J k 2,J] 0. Let a J k 2 and b J be such that [a,b] 0andletB be the subalgebra of A generated by 1 F,a,b.ThenB is the linear span of the elements {1 F,a,b,ab,ba,b 2,...,b k 1 }, and an easy computation shows that [B,B] J k 1 ; hence [B,B,B] =0and[B,B][B,B]=0. By Lemma 4 this says that Id(M 3 ) Id(B). The other inclusion is proved as above. Thus B PI M 3, M 3 var(a) and we are done in this case too. Therefore we may assume that [J k 2,J]=0. Suppose first that [J,J] J k 2. Then, since J k = 0wehavethat [J,J,J]=0, [J,J][J,J]=0 (5) and since A = F + J, J = J 11 we also have that [x,y,z], [x,y][z, w] are identities of A. Then by Lemma 4, Id(M 3 ) Id(A). Notice that, by hypothesis, there exist a,b J such that [a,b] 0. Let f = αx 1 x n + i>j i 1 < <i n 2 α i>jx i1 x in 2 [x i,x j ]+g be an identity of A with g Id(M 3 ). The same procedure as above proves that f = g Id(M 3 ),andid(a) = Id(M 3 ) and we are done. In order to finish the proof of the lemma we have only to study the case [J,J] J k 2. This says that if Ā = A/J k 2, Ā is not commutative. Then, from Lemma 7, J(Ā) = J 11 (Ā) = J/J k 2 is still non-commutative and J(Ā) k 2 = 0 with k>4. By the inductive hypothesis M 3 var(ā) and since Ā var(a), wehavethatm 3 var(ā) var(a). The proof of the lemma is now complete. Lemma 10. Let A = F + J with J 01 0(respectively J 10 0). ThenB = F + J 01 or B = F + J 01 + J 11,incaseJ 11 is commutative, is a subalgebra PI-equivalent to M 1 (respectively B = F + J 10 or B = F + J 10 + J 11 is PI-equivalent to M 2 ). Proof. It is clear that B = F + J 01 or B = F + J 01 + J 11 is a subalgebra of A. In any case by the hypotheses we have that [B,B] J 01 and so, B[B,B]=0. This says, by Lemma 3,

14 384 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) that Id(M 1 ) Id(B).Letf Id(B) be a multilinear polynomial of degree n. By Lemma 3, f can be written as f = n α x i1 i1 x in + g i 1 =1 i 2 < <i n where g Id(M 1 ). Suppose that there exists t {1,...,n} such that α t 0. Choose c i = 1 F,fori t and c t = b J 01, b 0. Since g is an identity of B, wehave f(c 1,...,c n ) = α t b1 F = α t b 0, a contradiction. Hence f = g Id(M 1 ) and Id(B) = Id(M 1 ) follows. Thus B is a subalgebra of A which is PI-equivalent to M 1. Similarly it is proved that B = F + J 10 and B = F + J 10 + J 11, in case J 11 is commutative, are PI-equivalent to M 2. Let G denotes the Grassmann algebra of countable rank over F. Hence G is generated by the set {e 1,e 2,...} over F satisfying the relations e i e j = e j e i, i, j = 1, 2,...Let also UT 2 denote the algebra of 2 2 upper triangular matrices over F.Wehave Remark 11. M 3 var(ut 2 ) var(g). Proof. By [14], Id(UT 2 ) = [x,y][z, w] T andby[13],id(g) = [x,y,z] T. Hence, since [x,y][z, w], [x,y,z] Id(M 3 ) it follows that Id(UT 2 ) Id(G) Id(M 3 ) and, so, M 3 var(ut 2 ) var(g). We next state a result that will be used throughout the paper. Theorem 12 [7, Theorem 2].LetA be an F -algebra whose codimensions are polynomially bounded. Then there exists an algebra B such that A PI B and B = B 1 B m,where B 1,...,B m are finite dimensional algebras such that dim B i /J (B i ) 1 with J(B i ) the Jacobson radical of B i, 1 i m. We are now in a position to prove the main result of this section. Theorem 13. Let A be an F -algebra. Then M 1,M 2,M 3 / var(a) if and only if lim n l n (A) exists and is bounded by 1. Proof. Suppose first that lim n l n (A) 1. Then if M i var(a) for some i {1, 2, 3}, it would follow that l n (M i ) l n (A). But by Lemmas 3 and 4 we have that for all n>3, l n (M i )>1 and this contradicts the assumption that lim n l n (A) 1. Conversely, suppose that M 1,M 2,M 3 / var(a). By Remark 11, UT 2, G/ var(a), hence by [12] the codimensions of A are polynomially bounded, i.e., c n (A) dn t,for some constants d,t. Since the codimensions and the cocharacters of an algebra do not

15 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) change upon extension of the base field (see for instance [5]) we assume, as we may, that F is algebraically closed. Also by Theorem 12 we may assume that A = A 1 A m where A 1,...,A m are finite dimensional algebras such that, dim A i /J (A i ) 1, 1 i m. Notice that this says that either A i = F + J(Ai ) or A i = J(A i ) is a nilpotent algebra. If A i = J(A i ) is nilpotent for all i,thena is a nilpotent algebra and for n large enough l n (A) = 0. Hence we are done in this case. Therefore we may assume that there exists i {1,...,m} such that A i = F + J(A i ) and let J(A i ) = J 00 + J 11 + J 10 + J 01 where J 00,J 11,J 10,J 01 are the bimodules defined in Lemma 7. If J 11 is non-commutative, by Lemma 9, M 3 var(a i ) var(a), a contradiction. Therefore J 11 must be commutative. Since M 1,M 2 / var(a), by Lemma 10 we have that J 10 = J 01 = 0. Thus A i = (F + J 11 ) J 00 is a direct sum of algebras and F + J 11 is a commutative subalgebra of A i. We have proved that if for some i, A i is not nilpotent, then A i is the direct sum of a commutative algebra and a nilpotent algebra. Recalling that A = A 1 A m and putting together all pieces, it turns out that we can write A = C N where C is a commutative non-nilpotent algebra and N is a nilpotent algebra. Since for all n, l n (C) = 1andfornlarge enough l n (N) = 0, by Remark 1 we obtain l n (A) = l n (C) = 1forn large enough, and the conclusion of the theorem follows. Notice that the proof of the previous theorem actually shows that M 1,M 2, M 3 / var(a) if and only if there exists n 0 > 3 such that l n0 (A) 1. Also in this case A is PI-equivalent to either a nilpotent algebra or to the direct sum of a nilpotent algebra and a commutative algebra. Thus l n (A) has constant value (equal to 0 or 1) for n large. Reading these conclusions in terms of codimensions we get the following corollary. Corollary 14. For an F -algebra A, the following conditions are equivalent. 1. M 1,M 2,M 3 / var(a). 2. There exists n 0 > 3 such that l n0 (A) lim n l n (A) exists and is bounded by Either A PI N or A PI C N where N is a nilpotent algebra and C is a commutative algebra. 5. c n (A) k for some constant k 0, for all n 1. We remark that as a consequence, if the codimensions of an algebra A are bounded, then they are eventually bounded by 1. This result and essentially property 4 of the above corollary were proved by Olsson and Regev in [16].

16 386 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) Characterizing varieties of colength 2 In this section we shall deal with the case l n (A) 2. We start with the following. Lemma 15. Let A = F + J be an F -algebra. 1. If J 10 J 00 0,thenM 4 var(a). 2. If J 00 J 01 0,thenM 5 var(a). Proof. Suppose that J 10 J 00 0andletk 1 be the largest integer such that J 10 J00 k 0. Then there exist a J 10 and b J00 k such that ab 0. Let B be the subalgebra of A generated by 1 F,a,b.Sinceab 2 = ba = a 2 = a1 F = 1 F b = b1 F = 0, it is easily seen that [x,y]zw Id(B). Hence by Lemma 6, Id(M 4 ) Id(B). Letnowf Id(B) be a multilinear polynomial of degree n. By Lemma 6 we can write f n i,j=1 i 1 < <i n 2 α ij x i1 x in 2 x i x j ( mod Id(M4 ) ) and suppose that there exist i 0,j 0 {1,...,n} such that α i0 j 0 0. Take c 1,...,c n B, such that c i0 = a, c j0 = b and c k = 1 F,forallk/ {i 0,j 0 }. It is easy to check that f(c 1,...,c n ) = α i0 j 0 ab 0, a contradiction. Thus f Id(M 4 ) and Id(B) = Id(M 4 ) implies M 4 var(a). Property 2 is proved similarly. Lemma 16. Let A = F + J be such that J 10 0,J 01 0 and J 10 J 01 = J 01 J 10 = 0. IfJ 11 is commutative, B = F + J 10 + J 01 + J 11 is a subalgebra PI-equivalent to M 1 M 2. Proof. Under the hypotheses of the lemma, B is a subalgebra of A. Since[B,B] J 01 + J 10, it is immediate that z[x,y]w, [x,y][z, w] are identities of B. Also it can be checked that St 3 vanishes in B. By Lemma 5 this implies that Id(M 1 M 2 ) Id(B).Now,Id(B) Id(F + J 10 + J 11 ) Id(F + J 01 + J 11 ). Since by Lemma 10, Id(F + J 10 + J 11 ) = Id(M 2 ) and Id(F + J 01 + J 11 ) = Id(M 1 ) we obtain Id(B) Id(M 1 ) Id(M 2 ) = Id(M 1 M 2 ). Thus Id(B) = Id(M 1 M 2 ) and we are done. Lemma 17. Let A = A 1 A 2,whereA 1 = F + J(A 1 ) with J(A 1 ) 10 0 and A 2 = F + J(A 2 ) with J(A 2 ) 01 0.ThenM 1 M 2 var(a). Proof. It is clear that B = (F + J(A 1 ) 10 ) (F + J(A 2 ) 01 ) is a subalgebra of A. Hence Id(B) = Id ( ) ( ) F + J(A 1 ) 10 Id F + J(A2 ) 01 andbylemma10,id(b) = Id(M 1 ) Id(M 2 ) = Id(M 1 M 2 ) Thus M 1 M 2 var(a).

17 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) Theorem 18. Let A be an F -algebra. Then M 1 M 2,M 3,M 4,M 5 / var(a) if and only if lim n l n (A) exists and is bounded by 2. Proof. Suppose first that there exists an integer n 0 > 3 such that l n0 (A) 2. Then, since for n 4, M 1 M 2,M 3,M 4,M 5 have colength sequence bounded from below by 3, it is clear that M 1 M 2,M i / var(a),foreveryi = 3, 4, 5. Conversely, suppose that M 1 M 2,M i / var(a), fori = 3, 4, 5. Since, by Remark 11, UT 2,G / var(a), as in the proof of Theorem 13, we may assume that F is algebraically closed and that A = A 1 A m with A i either a nilpotent algebra or A i isomorphic to F + J(A i ), i = 1,...,m. By eventually reordering the algebras A i, we may write A = A 1 A k A k+1 A m, where A i = F + J(A i ) for i = 1,...,k,andA k+1,...,a m are nilpotent algebras. Since A k+1 A m is still a nilpotent algebra, by Remark 1 l n (A) = l n (A 1 A k ) for n large enough. Hence, without lost of generality we may assume that A = A 1 A m and for each i = 1,...,m, A i = F + J(A i ). Since M 1 M 2 / var(a), by Lemma 17, A can be only of one of the following types: (1) for every i = 1,...,m, A i = F + J(A i ) with J(A i ) 01 = 0; (2) for every i = 1,...,m, A i = F + J(A i ) with J(A i ) 10 = 0; (3) there exists i such that A i = F + J(A i ) with J(A i ) 10 0, J(A i ) 01 0 and for every j i, A j = F + J(A j ) with J(A j ) 10 = J(A j ) 01 = 0. Notice that since M 3 / var(a i ) var(a), by Lemma 9, we have that J(A i ) 11 is commutative, for all i = 1,...,m. We start by considering the first case, i.e., J(A i ) 01 = 0, for all i. Write A i = F + J with J = J(A i ).IfJ 10 J 00 0 then, by Lemma 15 we would get M 4 var(a i ) var(a), a contradiction. Thus J 10 J 00 = 0 and this says that A i = (F + J 10 + J 11 ) J 00, a direct sum of algebras. But then by Lemma 10 we have that either A i PI M 2 J 00 or A i PI C J 00, for some commutative algebra C, according as J 10 0orJ 10 = 0, respectively. Summing up over all algebras A i, recalling that C var(m 2 ), it turns out that either A PI M 2 M 2 C N PI M 2 C N PI M 2 N

18 388 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) or or A PI M 2 M 2 N PI M 2 N A PI C N where N is a nilpotent algebra and C is a commutative non-nilpotent algebra. Thus for n large, the sequence of colengths of A is constant and l n (A) equals either 1 or 2 as wished. It is clear that in case the second possibility occurs, the same proof with the due changes shows that either or A PI M 1 N A PI C N and l n (A) takes constant value equal to 1 or 2, for n large. Next we show that the third possibility cannot occur. Since M 4,M 5 / var(a i ),by Lemma 15, J 10 J 00 = J 00 J 01 = 0. This implies that J 10 J 01 + J 01 J 10 is a two-sided ideal of A i.letā i be the algebra Ā i = A i /(J 10 J 01 + J 01 J 10 ). It is easy to check that Ā i = F + J(Ā i ) and the Jacobson radical J(Ā i ) satisfies J ( Ā i )01 J ( ) A i 10 = 0 and J ( ) Ā i 10 J ( Ā i )01 = 0. Obviously J 10, J 01 are not contained in J 10 J 01 + J 01 J 10, therefore J(Ā i ) 10, J(Ā i ) Hence by Lemma 16, M 1 M 2 var(ā i ) var(a i ) var(a), a contradiction. This completes the proof of the theorem. From the proof of Theorem 18 it follows that M 1 M 2,M 3,M 4,M 5 / var(a) if and only if there exists n 0 > 3 such that l n0 (A) 2. Also in this case A is PI-equivalent to either one of the algebras of Corollary 14 or to M 1 N or to M 2 N with N a nilpotent algebra. Thus l n (A) has constant value (equal to 0 or 1 or 2) for n large. Reading also these conclusions in terms of codimensions we have. Corollary 19. For an F -algebra A, the following conditions are equivalent. 1. M 1 M 2,M 3,M 4,M 5 / var(a). 2. There exists n 0 > 3 such that l n0 (A) lim n l n (A) exists and is bounded by 2.

19 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) A is PI-equivalent to one of the algebras N, C N, M 1 N or M 2 N, wheren is a nilpotent algebra and C is a commutative algebra. 5. c n (A) n for n large enough. In conclusion we have the following classification where N denotes a nilpotent algebra and C is a commutative non-nilpotent algebra: for any algebra A and n large enough, 1. l n (A) = 0 if and only if A PI N. 2. l n (A) = 1 if and only if A PI C N. 3. l n (A) = 2 if and only if either A PI M 1 N or A PI M 2 N. 6. Algebras with linear codimension growth In this section we introduce a new algebra denoted M 7 defined as follows: M 7 = {( ) } a b c 0 0 d a,b,c,d F. 0 0 a We claim that the sequence of codimensions of M 7 is bounded from below by n 2.Infact,in order to show this, it is enough to find an S n -character appearing with non-zero multiplicity in χ n (M 7 ), whose degree is n 2. Consider the partition λ = (n 2, 1, 1) and let f Tλ be the highest weight vector corresponding to the standard Young tableau containing the integers 1, 2, 3 into the first column. Choosing a 1 = e 11 + e 33, a 2 = e 12 and a 3 = e 23 we get that f Tλ = 2e This says that χ (n 2,1,1) appears with non-zero multiplicity in χ n (M 7 ). Therefore, since by the hook formula, degχ (n 2,1,1) = (n 1)(n 2)/2, c n (M 7 ) grows asymptotically at least as n 2 /2. Lemma 20. Let A = F + J be an F -algebra. If J 10 J 00 = J 00 J 01 = 0 and J 01 J 10 0,then M 6 var(a). Proof. From the hypotheses it follows that J 10 J 01 is a two-sided ideal of A. Then the quotient algebra Ā = J A 10 J 01 still satisfies the hypotheses and moreover J ( Ā ) 01 J ( Ā ) 10 0 and J ( Ā ) 10 J ( Ā ) 01 = 0. We shall prove that M 6 var(ā), hence we may assume that A itself satisfies J 01 J 10 0 and J 10 J 01 = 0. Let a J 01,b J 10 be such that ab 0andletB be the algebra generated by 1 F,a and b. Since ba = aba = 0 and 1 F a = b1 F = 0,a1 F = a, 1 F b = b, then B = span{1 F,a,b,ab} and [B,B] span{a,b,ab}. Thus B[B,B] span{b,ab} and B[B,B]B = 0. By Lemma 6, Id(M 6 ) Id(B).

20 390 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) Conversely, let f Id(B) be a multilinear polynomial of degree n. By Lemma 6, one can write n ( f α ij x i x i1 x in 2 x j mod Id(M6 ) ). i,j=1 i 1 < <i n 2 Suppose that there exist i 0,j 0 {1,...,n} such that α i0 j 0 0. Take c 1,...,c n B, with c i0 = a, c j0 = b and c k = 1 F for all k/ {i 0,j 0 } it is easy to check that f(c 1,...,c n ) = α i0 j 0 ab 0, a contradiction. Therefore f Id(M 6 ) and Id(B) = Id(M 6 ). Hence M 6 var(a). Lemma 21. If A = F + J with J 10 J 01 0 and J 01 J 10 = 0.ThenM 7 var(a). Proof. Let a J 10,b J 01 be such that ab 0. We claim that the subalgebra B generated by 1 F,a and b over F is isomorphic to M 7. This is easily seen since J 2 01 = J 2 10 = J 01 J 10 = 0 implies that a 2 = b 2 = ba = 0. Hence B = span{1 F,a,b,ab} is isomorphic to M 7 through the map ϕ such that ϕ(1 F ) = e 11 + e 33, ϕ(a)= e 12, ϕ(b)= e 23, ϕ(ab)= e 13. Theorem 22. Let A be an F -algebra. Then the following conditions are equivalent. 1. c n (A) kn for all n 1, for some constant k. 2. M 3,M 4,M 5,M 6,M 7 / var(a). 3. A is PI-equivalent to either N or C N or M 1 N or M 2 N or M 1 M 2 N where N is a nilpotent algebra and C is a commutative algebra. Proof. First suppose that the sequence of codimensions of A is linearly bounded. By Lemmas 4 and 6 and by the claim proved at the beginning of the section, M 3,M 4,M 5,M 6,M 7 / var(a). Suppose now that property 2 holds. Then, since M 3 / var(a), by Remark 11, UT 2,G / var(a) and, as in the proof of the previous theorems, we may assume that F is algebraically closed and A = A 1 A m where for every i = 1,...,m, either A i = F + J(Ai ) or A i = J(A i ) is a nilpotent algebra. Suppose that for some i, A i is not a nilpotent algebra. Since M 3 / var(a), by Lemma 9, J 11 is commutative. Also, since M 4,M 5,M 6,M 7 / var(a), by Lemmas 15, 20 and 21 we have that J 10 J 00 = J 00 J 01 = J 01 J 10 = J 10 J 01 = 0. Under these conditions J 00 is a two-sided nilpotent ideal of A i and A i = F + J 01 + J 10 + J 11 J 00. By Lemmas 10 and 16 we obtain that A i PI B where B is either M 1 N, or

21 A. Giambruno, D. La Mattina / Journal of Algebra 284 (2005) M 2 N or M 1 M 2 N or C N. The four cases appear according as J 01 0,J 10 = 0 or J 01 = 0,J 10 0orJ 01 0,J 10 0orJ 01 = J 10 = 0, J 11 0 respectively. Since A = A 1 A m we obtain that property 3 holds. It is clear that since each of the algebras N, C N, M 1 N, M 2 N, M 1 M 2 N has codimensions linearly bounded then property 3 implies property 1 and we are done. It is worth noticing that the previous theorem allows us to classify all possible linearly bounded codimension sequences. Corollary 23. Let A be an F -algebra such that c n (A) kn for all n 0. Then there exists n 0 such that for all n>n 0 we must have either c n (A) = 0 or c n (A) = 1 or c n (A) = n or c n (A) = 2n 1. Acknowledgment Research partially supported by MIUR of Italy. References [1] A. Berele, Homogeneous polynomial identities, Israel J. Math. 42 (1982) [2] V. Drensky, Representation of the symmetric group and varieties of linear algebras, Mat. Sb. 115 (1981) (in Russian). English translation in: Math. USSR Sb. 43 (1981) [3] V. Drensky, Relations for the cocharacter sequence of T-ideals, Contemp. Math. Part (1992) [4] V. Drensky, Free Algebras and PI-Algebras, Grad. Course in Algebra, Springer-Verlag, Singapore, [5] A. Giambruno, M. Zaicev, On codimension growth of finitely generated associative algebras, Adv. Math. 140 (1998) [6] A. Giambruno, M. Zaicev, Exponential codimension growth of PI algebras: an exact estimate, Adv. Math. 142 (1999) [7] A. Giambruno, M. Zaicev, A characterization of algebras with polynomial growth of the codimensions, Proc. Amer. Math. Soc. 129 (2000) [8] A. Giambruno, M. Zaicev, Asymptotics for the standard and the Capelli identities, Israel J. Math. 135 (2003) [9] A. Guterman, A. Regev, On the growth of identities, in: Algebra, Moscow, 1998, de Gruyter, Berlin, 2000, pp [10] G. James, A. Kerber, The representation theory of the symmetric group, Encyclopedia Math. Appl., vol. 16, Addison Wesley, London, [11] A.R. Kemer, T-ideals with power growth of the codimensions are Specht, Sibirsk. Mat. Zh. 19 (1978) (in Russian). English translation in: Siberian Math. J. 19 (1978) [12] A.R. Kemer, Varieties of finite rank, in: Proc. 15th All the Union Algebraic Conf., Krasnoyarsk, 1979, p. 73, vol. 2 (in Russian). [13] D. Krakowski, A. Regev, The polynomial identities of the Grassmann algebra, Trans. Amer. Math. Soc. 181 (1973) [14] Yu.N. Maltsev, A basis for the identities of the algebra of upper triangular matrices, Algebra i Logika 10 (1971) , (Russian). [15] S. Mishchenko, A. Regev, M. Zaicev, A characterization of PI algebras with bounded multiplicities of the cocharacters, J. Algebra 219 (1999) [16] J.B. Olsson, A. Regev, Applications of representation theory to PI-algebras, Proc. Amer. Math. Soc. 55 (1976) ; Proc. Amer. Math. Soc. 38 (1976) [17] A. Regev, Existence of identities in A B, Israel J. Math. 11 (1972) [18] L.H. Rowen, Ring Theory, Academic Press, New York, 1988.