Linear Algebra and its Applications

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1 Linear Algebra and its Applications 432 (2010) Contents lists available at ScienceDirect Linear Algebra and its Applications journal homepage: Graded identities for tensor products of matrix (super)algebras over the Grassmann algebra Onofrio Mario Di Vincenzo a,1, Plamen Koshluov b,,2, Ednei Aparecido Santulo Jr. c,3 a Dipartimento di Matematica e Informatica, Università degli Studi della Basilicata, Potenza, Italy b IMECC, UNICAMP, Cx. P. 6065, Campinas, SP, Brazil c Departamento de Matemática, UEM, Av. Colombo, 5790, Maringá , PR, Brazil A R T I C L E I N F O A B S T R A C T Article history: Received 10 September 2009 Accepted 13 September 2009 Available online 28 October 2009 Submitted by H. Schneider AMS classification: Primary 16R10 Secondary: 16R50, 16W50, 15A75 Keywords: Graded identities PI equivalence Matrices over Grassmann algebras Relatively free algebras In this paper we study the graded identities satisfied by the superalgebras M a,b over the Grassmann algebra and by their tensor products. These algebras play a crucial role in the theory developed by A. Kemer that led to the solution of the long standing Specht problem. It is well nown that over a field of characteristic 0, the algebras M pr+qs,ps+qr and M p,q M r,s satisfy the same ordinary polynomial identities. By means of describing the corresponding graded identities we prove that the T-ideal of the former algebra is contained in the T-ideal of the latter. Furthermore the inclusion is proper at least in case (r, s) = (1, 1). Finally we deal with the graded identities satisfied by algebras of type M 2 n 1,2 n 1 and relate these graded identities to the ones of tensor powers of the Grassmann algebra. Our proofs are combinatorial and rely on the relationship between graded and ordinary identities as well as on appropriate models for the corresponding relatively free graded algebras Elsevier Inc. All rights reserved. 1. Preliminaries Let K be an infinite field of characteristic different from 2. All algebras and vector spaces, as well as their tensor products considered in this paper will be over K.If G is an additive abelian group then the Corresponding author. addresses: onofrio.divincenzo@unibas.it (O.M. Di Vincenzo), plamen@ime.unicamp.br (P. Koshluov), easjunior@ uem.br (E.A. Santulo). 1 Partially supported by MIUR and Università della Basilicata. 2 Partially supported by CNPq grant /2008-0, and by FAPESP grant 2005/ Supported by FAPESP grant 03/ /$ - see front matter 2009 Elsevier Inc. All rights reserved. doi: /j.laa

2 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) algebra A is G-graded when A = g G A g is a direct sum of vector subspaces A g such that A g A h A g+h for all g, h G. Similarly one adapts this definition for multiplicative groups G. An important example of graded algebra that will be used frequently in the paper is given by the Grassmann algebra E. More precisely, let e 1, e 2, be a basis of the infinite dimensional vector space V. The Grassmann algebra E = E(V) is the vector space with basis {e i1 e i2 e i 0} where i 1 < i 2 < < i ;if = 0we denote the corresponding element by 1. The multiplication in E(V) is induced by e i e j = e j e i for all i and j. SetE i the span of all basis elements with i (mod 2). Then E 0 is the centre of E and the elements of E 1 anticommute, and E = E 0 E 1 is Z 2 -graded. Kemer s celebrated theory (see for an account [14]) was one of the most important achievements in the theory of graded algebras and their (graded) identities. Kemer s theory led to the classification of the T-ideals in characteristic 0, and to the positive solution of the famous Specht problem. The T-prime algebras are fundamental in describing the T-ideals in characteristic 0. Recall that an algebra A is T-prime if its T-ideal is prime inside the class of the T-ideals in the free associative algebra K X. Such T-ideals are called T-prime as well. According to Kemer s theory, the non-trivial T-prime T-ideals in characteristic 0 are T(M n (K)), T(M n (E)), and T(M,l ). Here M n (K) and M n (E) are the algebra of n n matrices over K and over E, respectively. Moreover M,l = M,l (E) is a subalgebra of M +l (E); ( ) u v it consists of the matrices such that u M w t (E 0 ), t M l (E 0 ), v M l (E 1 ), w M l (E 1 ). All these algebras are graded in a natural way; we define these gradings later on. As a by-product of his structure theory Kemer proved that the tensor product of any two T-prime algebras is PI equivalent to a T-prime algebra. More precisely let us say that the algebras A and B are PI equivalent if they satisfy the same polynomial identities, that is T(A) = T(B). In this case we write A B. Kemer s Tensor product theorem is the following. Theorem 1. Let char K = 0, then M a,b E M a+b (E); M a,b M c,d M ac+bd,ad+bc ; M 1,1 E E. Notice that the remaining tensor products of T-prime algebras from the list above yield isomorphisms. This theorem admits proofs that do not depend on the structure theory developed by Kemer. The first such proof was given by Regev in [17]. By using appropriate gradings and graded identities, the third statement of Theorem 1 wasprovedin[9,15], the second in [10,11], and the first in [12,3,4]. Also Berele [8] proved parts of Kemer s theorem. Recall that all of this research was conducted under the assumption char K = 0. If K is infinite and char K = p > 2 it is nown that Kemer s theorem holds only at multilinear level, that is the multilinear elements of the corresponding T-ideals are the same [3]. Furthermore it was shown that in general the theorem fails in positive characteristic, see [3,4,1]. Let us recall the main notions concerning graded polynomial identities. Let K X be the free associative algebra freely generated over K by the countable set of variables X, and let G be a finite abelian group. Assume X = g G X g where X g are disjoint and countable. Then the algebra K X is G-graded in a natural manner, by letting K X g to be the span of all monomials m = x i1 x i2 x in such that x it X gt and g 1 + g 2 + +g n = g. The polynomial f (x 1,..., x n ) is a G-graded polynomial identity for the G-graded algebra A if f (a 1,..., a n ) = 0 for every a i where the G-degree of a i is the same as that of x i. The set of all G-graded identities for A is the T G -ideal T G (A); it is closed under the G-graded endomorphisms of K X, sometimes we call it the graded T-ideal of the identities of A. Graded identities became an object of independent interest soon after Kemer s wor. The interested reader can find more information and recent results in graded PI theory in [6]. Let A be a G-graded algebra. If v A is a G-homogeneous element of A, that is v A g for some g G, we shall write G (v) = g. We recall that a G-grading on A is fine if dim A g 1 for all g G. On the other extreme are the so-called elementary gradings. Let M n (K) be the full matrix algebra, a grading on it is elementary if all matrix units e ij are G-homogeneous. It was proved in [5] that, when K is an algebraically closed field, every grading on M n (K) can be obtained as a tensor product of an elementary and a fine grading. Here we give two examples of gradings on A = M n (K), the first being the canonical elementary grading while the second is fine. Let G = Z n, and set A g = span e ij j i g (mod n). Then it is immediate to prove that this defines a grading on A. The identities for this grading were described in [18] when char K = 0, and in [2] when char K = p > 2.

3 782 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) Now suppose G = Z n Z n, and let ω K be a primitive n-th root of 1. Put u = ω n 1 e 11 + ω n 2 e ωe n 1,n 1 + e n,n and v = e 12 + e e n 1,n + e n1. Then u n = v n = 1, uv = ωvu, and the spans of u i v j,1 i, j n, determine a fine G-grading. If A is G-graded and B is H-graded where G and H are groups then A B is G H-graded: the (g, h)- homogeneous component equals A g B h, g G, h H. In this way one defines a Z n Z 2 -grading on M n (E). But M a,b is a subalgebra of M a+b (E), and it is easy to see that it is a homogeneous subalgebra. Therefore the Z n Z 2 -grading on M n (E) induces, by restriction, one on M a,b when a + b = n. In this paper we study the graded identities of the algebras M p,q M r,s and M pr+qs,ps+qr. More precisely put a = pr + qs,b = ps + qr, m = p + q and n = r + s, we consider a subalgebra M p,q,r,s (E) of M mn (E) which is isomorphic to M a,b. In Theorem 13 we describe a set of generators for the graded identities of M p,q,r,s (E) in the case when the field K is infinite and char K = p /= 2. We prove that these polynomials are graded identities for the algebra M p,q M r,s (see Corollary 14) and, as a consequence (Corollary 15), we obtain the inclusion T(M pr+qs,ps+qr ) T(M p,q M r,s ) for the corresponding ordinary polynomial identities. Furthermore we consider the algebras of the type M 2 n 1,2 n 1. Starting from a fine grading on M 2 n(k), we define a Z 2n 2 -grading on the former algebra, and a Z2n+1 2 -grading on M 2 n(e). In Sections 3.2 and 3.3 we describe bases of the graded identities for these algebras. One considers also E with the natural Z 2 -grading on it. Let G = Z 2, we prove that when = 2n is even, the G-graded T-ideal of the latter algebra contains that of M 2 n 1,2n 1. When = 2n + 1 is odd we prove that T G (M 2 n(e)) T G (E ) (see Theorem 34). In this way we obtain an alternative proof of one case of Kemer s theorem in characteristic 0 (see Theorem 1), as well as the corresponding results in characteristic p > 2, see [3,4,1]. This represents a generalisation of Regev s results in [16]. In Section 3.4 we describe a chain of G-graded T-ideals between T G (M 2 n 1,2 n 1) and T G(E ) when = 2n is even (Theorem 36). In theorem 38 we obtain a similar description for a chain of graded ideals between T G (M 2 n(e)) and T G (E 2n+1 ). 2. Bases of graded identities Here we construct appropriate models for the relatively free (or universal) graded algebras for M pq M rs and for M pr+qs,ps+qr. Furthermore we exhibit generating sets for the ideals of graded identities these algebras satisfy, and draw several corollaries of independent interest Models for the relatively free graded algebras We fix the notation p + q = m, r + s = n and set G = Z mn Z 2. In the full matrix algebra M mn (K) we consider the canonical elementaryz mn -grading, moreover if R is anyz 2 -graded algebra we consider in M mn (R) = M mn (K) R the G-grading defined as above for the tensor product of graded algebras. For h N we denote I h ={1, 2,..., h} N, we assume that p q and r s and we define the function γ p,q : I p+q Z 2 by γ p,q (i) = 0 when 1 i p and γ p,q (i) = 1 otherwise. Similarly, we consider the function γ r,s : I r+s Z 2. Consider the following sets of variables: Y ={y ij 1 i, j m}, Z ={z ij 1 i, j n}, U ={u ij 1 i, j mn}, where = 1, 2,... Form the free algebra K Y Z and define a Z 2 -grading on it by putting Z2 (y ) = γ ij p,q(i) + γ p,q (j) and Z2 (z ) = γ ij r,s(i) + γ r,s (j). Then let P 1 be its ideal determined by the relations: [y 1 i 1 j 1, z 2 i 2 j 2 ]=0, [y 1 i 1 j 1, y 2 i 2 j 2 ]=0, if γ p,q (i 1 ) + γ p,q (j 1 ) = 0, [z 1 i 1 j 1, z 2 i 2 j 2 ]=0, if γ r,s (i 1 ) + γ r,s (j 1 ) = 0,

4 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) y 1 i 1 j 1 y 2 i 2 j 2 = 0, if γ p,q (i 1 ) + γ p,q (j 1 ) = γ p,q (i 2 ) + γ p,q (j 2 ) = 1, z 1 i 1 j 1 z 2 i 2 j 2 = 0, if γ r,s (i 1 ) + γ r,s (j 1 ) = γ r,s (i 2 ) + γ r,s (j 2 ) = 1, for all 1, 2, i 1, i 2, j 1, j 2. Here [a, b] =ab ba is the commutator of a and b, and a b = ab + ba is the Jordan product of a and b. Define R 1 = K Y Z /P 1. We shall use the same letters y ij and z ij for the images of y ij and z ij under the projection K Y Z R 1. It follows from the above relations that R 1 is a Z 2 -graded algebra, moreover the set Y generates a free supercommutative algebra (see for example [7] for the definition) as well as the set Z does, and the elements of Y commute with the elements of Z. Let (t, a) G = Z mn Z 2 and define the following matrices in M mn (R 1 ): A (t,a) = n(j i)+w v=t δ a,(γp,q (i)+γ r,s (v)+γ p,q (j)+γ r,s (w))y ij z vw e n(i 1)+v,n(j 1)+w. Here δ v,w is the Kronecer symbol and, as above, e vw stands for the matrix unit, that is the matrix whose only nonzero entry is at position (v, w) and equals 1. Clearly A (t,a) is an homogeneous element in the G-graded algebra M mn (R 1 ). Put G (t,a) to be the set of all matrices A (t,a), 1, and G = (t,a) G G (t,a). Finally define the algebra F p,q,r,s as the one generated by the set G. Then the algebra F p,q,r,s is a G-graded subalgebra of M mn (R 1 ). Here we give as an example the construction of the algebra F 1,1,1,1. It is generated by the matrices A (0,0) = y 11 z 11 e 11 + y 11 z 22 e 22 + y 22 z 11 e 33 + y 22 z 22 e 44, A (1,0) = y 12 z 21 e 23 + y 21 z 21 e 41, A (1,1) = y 11 z 12 e 12 + y 22 z 12 e 34, A (2,1) = y 12 z 11 e 13 + y 12 z 22 e 24 + y 21 z 11 e 31 + y 21 z 22 e 42, A (3,0) = y 12 z 12 e 14 + y 21 z 12 e 32, A (3,1) = y 11 z 21 e 21 + y 22 z 21 e 43, 1, and the components of degrees (0,1) and (2,0) are zero. Now we construct the algebras L p,q,r,s as follows. For w I mn we write w = n(w 1 1) + w 2, w 1 I m, w 2 I n and we denote ε(w) = γ p,q (w 1 ) + γ r,s (w 2 ). Observe that ε(w) is well defined since w 1 and w 2 are determined uniquely by w.setp 2 the ideal in the free associative algebra K U determined by the relations [u 1 i 1 j 1, u 2 i 2 j 2 ]=0, if ε(i 1 ) + ε(j 1 ) = 0, u 1 i 1 j 1 u 2 i 2 j 2 = 0, if ε(i 1 ) + ε(j 1 ) = ε(i 2 ) + ε(j 2 ) = 1. Let R 2 = K U /P 2, then it is easy to see that R 2 is a Z 2 -graded algebra which is free supercommutative; its even variables are all u ij such that ε(i) + ε(j) = 0. (Here as above we use the same letters u ij for the generators of K U and for their images in R 2.) Denote a = pr + qs b = ps + qr and let (t, c) G. Define H (t,c) to be the set of all matrices B (t,c) M mn (R 2 ) where B (t,c) = δ c,ε(j)+ε(i) u ij e ij. i,j j i=t Now put H = (t,c) G H (t,c).letl p,q,r,s be the algebra generated by the set H. It is immediate that L p,q,r,s is a G-graded subalgebra of M mn (R 2 ) in a natural way. As an example we give below the generators of the algebra L 1,1,1,1.

5 784 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) B (0,0) = u 11 e 11 + u 22 e 22 + u 33 e 33 + u 44 e 44, B (1,0) = u 23 e 23 + u 41 e 41, B (1,1) = u 12 e 12 + u 34 e 34, B (2,1) = u 13 e 13 + u 24 e 24 + u 31 e 31 + u 42 e 42, B (3,0) = u 14 e 14 + u 32 e 32, B (3,1) = u 21 e 21 + u 43 e 43, and the components of degrees (0, 1) and (2, 0) equal zero. Remar 2. Let (t, c) G = Z mn Z 2, and fix p, q, r, s. Due to the gradings of F p,q,r,s and L p,q,r,s the positions of the nonzero entries of the matrix B (t,c) are the same as those of A (t,c) l. Recall that G = Z mn Z 2. The algebra M p,q M r,s is G-graded as follows. Let b ij e ij M p,q and b vw e vw M r,s. Then the element C ij,vw = b ij e ij b vw e vw M p,q M r,s is homogeneous when b ij, b vw E 0 E 1 are such that Z2 (b ij ) = γ p,q (i) + γ p,q (j) and Z2 (b vw ) = γ r,s (v) + γ r,s (w). The G-degree of the matrix C ij,uv M p,q M r,s is defined as follows: G (C ij,vw ) = (n(j i) + w v, Z2 (b ij ) + Z2 (b vw )). This determines a G-grading on the tensor product M p,q M r,s. One readily verifies that this is indeed a G-grading, see [12, Section 2]. Lemma 3. The algebra F p,q,r,s is relatively free in the variety of G-graded algebras determined by M p,q M r,s. Proof. Let K X =K X g1,..., X g2mn be the free G-graded algebra where the set X gi consists of the variables {x (g i) t t 1} of G-degree g i, and X = X gi. The homomorphism of G-graded algebras ϕ: K X F p,q,r,s defined by x (g i) l A (g i) l is onto. In order to prove the Lemma it suffices to see that er ϕ = T G (M p,q M r,s ). 1. The inclusion er ϕ T G (M p,q M r,s ) is established by repeating the argument from the proof of [13, Lemma 1.4.2, pp. 10, 11]. 2. For the opposite inclusion we use a modification of the argument from the proof of [13, Theorem 1.4.4, p. 11]. Assume f (x (g 1) 1,..., x (g ) ) er ϕ and let a 1,..., a M p,q M r,s be homogeneous elements such that g (a i ) = g i,1 i. We draw the reader s attention that we wor with the grading on M p,q M r,s defined in the remars preceding this lemma. The homogeneous element C ij,vw defined there is of degree (t, a) and if nonzero it corresponds to the summand y ij z vw e n(i 1)+v,n(j 1)+w. The sets Y and Z generate free supercommutative algebras in R 1. The free supercommutative algebra is a generic (that is relatively free) algebra for the Grassmann algebra therefore one can obtain a i as a specialization of A (g i) t. (That is we do not need linear combinations of the A (g i) t in order to obtain a i.) These specializations extend to a Z 2 -graded homomorphism R 1 E, and consequently to a G-graded homomorphism M mn M p,q M r,s such that A (g i) t a i for all i. But f er ϕ thus f (A (g 1) 1,..., A (g ) ) = 0, therefore f (a 1,..., a ) = 0. Recall a = pr + qs and b = ps + qr. It was constructed in [12] a subalgebra M p,q,r,s (E) M mn (E) that is isomorphic to M a,b. The algebra M p,q,r,s (E) is defined in the following manner. The element

6 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) A = i,j a ij e ij M p,q,r,s (E) if and only if a ij E ε(j)+ε(i). Furthermore it was shown in [12, Section 2] that M p,q,r,s (E) = M a,b, a graded isomorphism. Lemma 4. The algebra L p,q,r,s is relatively free in the variety of G-graded algebras generated by M p,q,r,s (E). Proof. The proof is quite similar to that of Lemma 3 bearing in mind the G-graded isomorphism M p,q,r,s (E) = M a,b. The G-graded homomorphism K X L p,q,r,s defined by x (g i) l B (g i) l is onto. Its ernel is the ideal of the G-graded identities of M p,q,r,s (E). Thus we have constructed convenient models of the relatively free G-graded algebras for M p,q M r,s and for M p,q,r,s (E). In fact these models are the G-graded analogs of the generic matrix algebras Graded identities of M p,q,r,s (E) and of L p,q,r,s In this section in order to simplify the notation, given the G-graded variables x 1,..., x d K X of degree (t i, a i ) G, we shall write the generators B (t i,a i ) i of L p,q,r,s simply as B i. Moreover, for any G-homogeneous element v of any G-graded algebra A, we shall write G (v) = (v) = (α(v), β(v)) where α(v) is the Z mn -degree of v and β(v) is the Z 2 -degree of v. In other words we represent g G as g = (α, β)for α Z mn, β Z 2. Finally, if B is a matrix we denote by B ij its (ij)-th entry. We observe that the matrix A (t,a) contains at most one nonzero entry in each row and in each column, and the same for B (t,c). Then we have the following couple of lemmas. Lemma 5. Let M(x 1,..., x d ) and N(x 1,..., x d ) K X be two graded monomials such that M(B 1,..., B d ) ij =±N(B 1,..., B d ) ij /= 0 for some i and j. Then the monomials M and N are of the same N d -multidegree as ordinary monomials. Furthermore if M = m x h m then N = n x h n for suitable n and n such that (m ) = (n ) and (m ) = (n ). Here some of the monomials m, m, n and n might be void. Proof. Let M = x 1 x 2 x l, where { 1,..., l }={1,..., d} and (x i ) = (t i, a i ) = (B i ) for all i = 1,..., d. Then we write the (i, j)-th entry M ij of M(B 1,..., B d ) as M ij = u 1 ij 1 u 2 j 1 j 2 u l j l 1 j. We repeat the same procedure for the monomial N. Now, the statement about the multidegrees is straightforward since the entries of B i are variables that appear only in the respective matrix. In order to prove the second statement, if x h = x v then u h j v 1 j v appears in M ij and (m ) = (j v 1 i, ɛ(i) + ɛ(j v 1 )). Clearly u h j v 1 j v appears in N ij in some position and using the fact that this entry is nonzero, we are done. If B L p,q,r,s is a G-homogeneous element we denote by supp (B) the number of nonzero entries of B and call it the support of B. The next lemma is immediate. Lemma 6. The following inequality holds: supp (B (t 1,c 1 ) 1 B (t 2,c 2 ) 2 ) min{ supp (B (t 1,c 1 ) 1 ), supp (B (t 2,c 2 ) 2 ) }. If M = M(x 1,..., x d ) is a graded monomial, we define the density of M in L p,q,r,s as the number of nonzero entries of the matrix M( B 1,..., B d ). Here B stands for the matrix of the same size as B and obtained from B by substituting all nonzero entries of B by 1 K, and preserving the zero entries. Definition 7. The graded monomial M is said to be sparse in L p,q,r,s if its density in L p,q,r,s equals 0.

7 786 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) Here we draw the reader s attention to the condition M(B 1,..., B d ) i,j /= 0 imposed in Lemma 5. Without imposing that condition it might happen that M(B 1,..., B d ) i,j = 0 as a result of some cancellations among the entries of the B i s. On the other hand the matrices B i have entries 0 and 1, so such cancellations cannot happen with them. That is the reason we define density and sparsity. Remar 8. It is immediate to see that if a monomial has a submonomial of G-degree (0,1) then it is sparse in L p,q,r,s. Furthermore when p = q then the existence of a submonomial of G-degree (mn/2, 0) also implies that the given monomial is sparse. Analogously to the case of the algebra L p,q,r,s we define supp (A) for a homogeneous element A F p,q,r,s. Thus we have the notion of sparse monomials in F p,q,r,s. It follows immediately from Remar 2 that Lemma 9. A monomial is sparse in F p,q,r,s if and only if it is sparse in L p,q,r,s. Let J K X be the T G -ideal of G-graded identities generated by the polynomials [x (0,0) 1, x (0,0) 2 ], x (t,0) 1 x ( t,0) 2 x (t,0) 3 x (t,0) 3 x ( t,0) 2 x (t,0) 1, x (t,1) 1 x ( t,1) 2 x (t,1) 3 + x (t,1) 3 x ( t,1) 2 x (t,1) 1, where t Z mn, and by all monomials that are sparse in L p,q,r,s. Here we recall that the G-grading on M p,q M r,s was defined just before Lemma 3; its relations to the G-grading on M p,q,r,s were outlined in the proof of Lemma 3. Lemma 10. Let M(x 1,..., x d ) K X be a graded monomial. Suppose that M(B 1,..., B d ) = 0. Then M(x 1,..., x m ) 0(mod J). Proof. Since M(B 1,..., B d ) = 0 then either M is sparse or some odd variable u ij appears at least twice among the entries of the monomial M. In the first case, we are done. In the second one, we have M = m 1 x (t,1) m 2 x (t,1) m 3 with α(m 1 ) = α(m 1 x m 2 ) and α(x m 2 ) = 0. But if (x m 2 ) = (0, 1) then M J.If (x m 2 ) = (0, 0) then (m 2 ) = ( t,1) and hence M = m 1 x (t,1) m ( t,1) 2 x (t,1) m 3 m 1 x (t,1) m ( t,1) 2 x (t,1) m 3 = M(mod J). Thus 2M J and since char K /= 2 we obtain that M J. Lemma 11. Let M and N be two graded monomials in K X. Suppose that for some (i, j) we have M(B 1,..., B d ) ij =±N(B 1,..., B d ) ij /= 0. Then M ±N(mod J). Proof. According to Lemma 5 the monomials M and N have the same multidegree. Suppose that x 1 is the leftmost variable that appears in M, that is M = x 1 m. Once again by Lemma 5 we can write N = n 1 x 1 n 2 where (n 1 ) = (0, 0). Letl be the length (total degree) of M and of N. We induct on l, the base of the induction being l = 1 which is obvious. Suppose l>1. If n 1 is the empty monomial then we are done since we can apply the induction to m and n 2. Hence we shall assume that n 1 is nonempty. We consider the following three cases. Case 1. Let M = x 1 m 1 x 1 m 2 with (x 1 m 1 ) = (0, 0). Then according to Lemma 5, we have N = n 1 x 1 n 2 x 1 n 3 with (n 1 ) = (0, 0) and (x 1 n 2 ) = (0, 0). Then applying the first generator of J, we have that N = n 1 x 1 n 2 x 1 n 3 x 1 n 2 n 1 x 1 n 3 (mod J). Putting n = n 2 n 1 x 1 n 3 we have N x 1 n (mod J) and we may apply the induction to m and n.

8 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) Case 2. LetM = x 1 m 1 x a x b m 2 and N = n 1 x a n 2 x 1 n 3 x b n 4 be such that (n 1 x a ) = (x 1 m 1 x a ) = (n 1 x a n 2 x 1 n 3 ), (n 1 x a n 2 ) = (0, 0). If n 2 is the empty monomial then (n 1 x a ) = (x 1 n 3 ) = (0, 0). If on the other hand n 2 is nonempty then (n 1 x a ) = (x 1 n 3 ) = (t, c) and (n 2 ) = ( t, c). In both situations we conclude N ±x 1 n 3 n 2 n 1 x a x b n 4 and we finish this case in the same manner as Case 1. Case 3. Let none of the above cases hold. Suppose N starts with the letter x j and write M = x 1 m 1 m 2 x j m 3 with (x 1 m 1 m 2 )=(0, 0). Assume that N = x j n 1 x 1 n 2, thus we have (x j n 1 ) = (0, 0). But (M) = (x j m 3 ) = (x 1 n 2 ) = (N). It follows from (x 1 n 2 ) = (M) that some initial submonomial of M mustbeofg-degree equal to that of M, and furthermore x j cannot appear in that initial submonomial. Therefore we can choose m 1 and m 2 such that (x 1 m 1 ) = (x 1 n 2 ) = (M). Then we have (m 2 x j m 3 ) = (x j n 1 ) = (0, 0). As in Case 2 we distinguish two subcases. If m 2 is the empty monomial then (x 1 m 1 ) = (x j m 3 ) = (0, 0). If, on the other hand, m 2 is nonempty we get (x j m 3 ) = (x 1 m 1 ) = (t, c) and (m 2 ) = ( t, c). In each one of them we conclude that M ±x j m 3 x 1 m 1 and we are done by the induction argument, this time starting with n. Proposition 12. The ideal of G-graded identities of L p,q,r,s coincides with the ideal J. In other words T G (L p,q,r,s ) = J. Proof. Recall the definition of the ideal of graded identities J given just after Lemma 9.Firstweprove that its generators are G-graded identities for L p,q,r,s. Clearly all sparse monomials in L p,q,r,s are graded identities for L p,q,r,s. We observe that all remaining generators of J are multilinear polynomials thus we may substitute the variables by the matrices B (t,c) only. Once again due to the multilinearity we may restrict the substitutions to the elements δ c,ε(i)+ε(j) u ij e ij only. Recall that here c Z 2 and j i = t. 1. Let (t, c) = (0, 0). Then j = i, ε(i) + ε(j) = 0. In this case we have u 1 ii e iiu 2 μ,μ e μμ = u 2 μ,μ e μμu 1 ii e ii since u 1 ii u2 μ,μ = u2 μ,μ u1 ii. (If i /= μ both the above products are 0.) Hence [x(0,0) 1, x (0,0) 2 ] is a graded identity for L p,q,r,s. 2. We prove that x (t,0) 1 x ( t,0) 2 x (t,0) 3 x (t,0) 3 x ( t,0) 2 x (t,0) 1 vanishes on L p,q,r,s.ift = 0 this follows from (1) above. Therefore substitute x (t,0) 1, x ( t,0) 2, x (t,0) 3 by δ 0,ε(i)+ε(i+t) u 1 i,i+t e i,i+t, δ 0,ε(i+t)+ε(i) u 2 i+t,i e i+t,i, δ 0,ε(i)+ε(i+t) u 3 i,i+t e i,i+t where ε(i) + ε(i + t) = 0 and the sums in the lower indices of the u and e are modulo mn. It suffices to consider only substitutions of this type since the rules of the matrix multiplication yield 0 for all other possibilities for x ( t,0) 2 and x (t,0) 3. But in this situation u 1 i,i+t u2 i+t,i u3 = i,i+t u3 i,i+t u2 i+t,i u1 i,i+t and we are done. 3. The computation for x (t,1) 1 x ( t,1) 2 x (t,1) 3 + x (t,1) 3 x ( t,1) 2 x (t,1) 1 is quite similar to the above and we omit it. In this way we obtain the inclusion J T G (L p,q,r,s ). Below we prove the opposite inclusion. Suppose that the multihomogeneous polynomial f is a G-graded identity for L p,q,r,s but f / J. Choose the least positive integer such that f can be written, modulo J, as a linear combination of monomials. That is f i=1 a im i (mod J) where a i K are all nonzero, and m i K X are graded monomials. Since f / J we have 1. By Lemma 10 we may suppose that, up to reordering the monomials m 1,..., m,wehavem 1 (B 1,..., B d ) /= 0. We get a 1 m 1 (B 1,..., B d ) = i=2 a im i (B 1,..., B d ). Hence for some h, 2 h, wehavem 1 (B 1,..., B d ) ij =±m h (B 1,..., B d ) ij /= 0. But according to Lemma 11 it holds m 1 ±m h (mod J). Therefore we can represent f as h 1 f a i m i (a 1 ± a h )m 1 + a i m i + a i m i (mod J). i=1 i=2 i=h+1 But this contradicts the minimality of, therefore f J and we are done.

9 788 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) This proposition and lemma 4 imply the following theorem. Theorem 13. Let K be an infinite field, char K M p,q,r,s (E) coincides with J. /= 2. Then the ideal of the G-graded identities of the algebra Corollary 14. T G (M p,q,r,s (E)) T G (M p,q M r,s ). Proof. Since J = T G (M p,q,r,s (E)) it suffices to verify that the generators of J are G-graded identities for M p,q M r,s. It is clear that the sparse monomials vanish on M p,q M r,s. The remaining generators of J are graded identities for M p,q M r,s as it was established in [12, Lemma 10]. It is well nown that if the G-graded algebras A and B satisfy T G (A) T G (B) then T(A) T(B). Since M p,q,r,s (E) is isomorphic to M pr+qs,ps+qr ([12, Section 2]), we obtain: Corollary 15. T(M pr+qs,ps+qr ) T(M p,q M r,s ). In [1] it was proved that the above two T-ideals fail to be equal when char K > 2. Namely there was constructed a polynomial that is an (ordinary) identity for M p,q M r,s but not for M pr+qs,ps+qr, when (r, s) = (1, 1). Thus we have the following theorem. Theorem 16. If K is infinite, char K > 2 then one has the proper inclusion T(M p+q,p+q ) T(M p,q M 1,1 ). In this way we are able to give a more precise answer to the respective problem posed in [4]. 3. Matrices over supercommutative algebras and their gradings In this section we extend results of Regev s [16] to algebras over infinite fields of characteristic different from 2, and add more flavour to Regev s results. Here we consider gradings with groups of the type G = Z 2 for suitable positive integers. In such case, if a A is a homogeneous element of the G-graded algebra A, we shall denote its homogeneous degree by a. Further if α = (α 1,..., α ) and β = (β 1,..., β ) Z 2 we shall denote by α, β = α 1 β 1 + +α β, s α = α 1 + +α, s α,i = s α α i, where the operations are in Z 2. If the number is clear from the context we shall use α, β instead of α, β. Definition 17. The Z 2-graded algebra A is -supercommutative if for every two homogeneous a, b A we have ab = ( 1) a, b ba. These algebras are colour commutative algebras; when = 1 we have just a supercommutative algebra, see for example [7]. If A is Z 2 -graded and B is Zl 2-graded then A B is Z+l 2 -graded in a natural way. Its homogeneous component of degree (α 1,..., α +l ) is the span of all a b where a A and b B are homogeneous elements and a =(α 1,..., α ), b =(α +1,..., α +l ). Proposition 18. Let A and B be -supercommutative and l-supercommutative algebras, respectively. Then A Bis( + l)-supercommutative with respect to the induced grading. Proof. We show that A B satisfies all graded identities xy ( 1) x, y +l yx. But these are multilinear. Thus it suffices to substitute x and y by homogeneous elements of the basis of A B. Let

10 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) x =(α 1,..., α +l ), y =(β 1,..., β +l ). Tae a 1, a 2 A and b 1, b 2 B with a 1 =(α 1,..., α ), a 2 =(β 1,..., β ), b 1 =(α +1,..., α +l ), and b 2 =(β +1,..., β +l ). Then a 1 a 2 b 1 b 2 = ( 1) a 1, a 2 a 2 a 1 b 1 b 2 = ( 1) a 1, a 2 + b 1, b 2 l a 2 a 1 b 2 b 1. Since a i b i =( a i, b i ) and a 1, a 2 + b 1, b 2 l = ( a 1, b 1 ), ( a 2, b 2 ) +l, we are done. Let n N, denote c n (x, y) = xy ( 1) x, y yx, where x and y are variables in the free Z n 2-graded algebra K X, and x and y are their homogeneous degrees, respectively. Clearly, a Z n 2-graded algebra is n-supercommutative if and only if the polynomials c n (x, y) are graded polynomial identities of A for all x, y X. Let us denote by J n the Z n 2 -graded T-ideal of K X generated by the polynomials c n(x, y), when x and y run independently over Z n 2. We recall that the (graded) T-ideals of K X are generated by their multihomogeneous elements, since K is an infinite field. So we shall wor with multihomogeneous (graded) polynomials. We have: Lemma 19. Let f (x 1,..., x ) be a multihomogeneous Z n 2 -graded polynomial of multidegree (t 1,..., t ), then, for some λ K, f (x 1,..., x ) λx t x t (mod J n). Moreover, if s xi = 1 and t i 2 then f J n. Proof. We wor modulo J n, therefore first statement is immediate because the variables of f either commute or anticommute. Furthermore, if x i =α, then s α = 1 implies x 2 i = x 2 i (mod J n ) and the result follows from char K /= 2. An easy example of 1-supercommutative Z 2 -graded algebra is given by the Grassmann algebra E with respect to its naturalz 2 -grading. Clearly, by Proposition 18 the algebra E n is n-supercommutative. In this case the grading is given by setting E n (α 1,...,α n ) to be span K a 1 a n Z2 (a j ) = α j. In the next subsection we will define gradings for the algebras M 2 n 1,2 n 1 and M 2 n(e) The gradings First, we recall the definition of the matrix groups G n ( ) ( ) GL 2 n(k) givenin[16]. The group G 1 is generated by the matrices A 11 =, B =. Inductively, we define 1 0 A i,n+1 = ( ) 0 Ain, B A in 0 i,n+1 = ( ) 0 Bin, 1 i n. B in 0 Denote C n = A 1n A 2n A nn B 1n...B nn and set further ( ) ( ) 0 Cn 0 I2 n A n+1,n+1 =, B C n 0 n+1,n+1 =, I 2 n 0 where I 2 n GL 2 n(k) is the identity matrix. Then the group G n+1 is generated by A 1,n+1,..., A n+1,n+1, B 1,n+1,..., B n+1,n+1. Thus the group G n is generated by the matrices A 1n,,A nn and B 1n,,B nn GL 2 n(k).

11 790 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) The proof of the following results is given in [16]: Lemma 20. The following relations hold in G n : A 2 in = B2 = jn C2 = n I 2 n; A inc n = C n A in ; B in C n = C n B in ; A in A jn = A jn A in, i /= j; B in B jn = B jn B in, i /= j; A in B jn = B jn A in. It follows that every element of G n can be written uniquely in the form ±A α 1 1n Aα 2 A α n nn Bα n+1 1n B α 2 2n Bα 2n nn, (α 1,..., α 2n ) Z 2n 2. 2n Let H n ={A α 1 1n Aα 2 2n Aα n nn Bα n+1 1n B α 2 2n Bα 2n nn (α 1,..., α 2n ) Z 2n 2 }. Denote H i,n the subset of H n consisting of all elements such that 2n j=1 α j = i in Z 2, i = 0, 1. Lemma 21 ([16, Lemma 3]). The set H n is a basis, consisting of invertible matrices, of the vector space M 2 n. Its subsets H 0,n and H 1,n are bases of the subspaces ( ) ( ) M2 n 1(K) 0 0 M M 2 n(k) 0 =, M 0 M 2 n 1(K) 2 n(k) 1 = 2 n 1(K) M 2 n 1(K) 0 respectively. Now we define a Z 2n 2 -grading on M 2 n 1,2 = n 1 M 2 n 1,2n 1(E). Thus the homogeneous component M 2 n 1,2 n 1(E) (α 1,...,α 2n ) of degree (α 1,..., α 2n ) is the subspace H (α1,...,α 2n ) ={ra α 1 1n Aα n nn Bα n+1 1n B α 2n nn r E α 1 + +α 2n }. According to Lemma 21, M 2 n 1,2 n 1 α Z 2n 2. Lemma 22. The graded algebra M 2 n 1,2n 1 is 2n-supercommutative. = α Z 2n 2 H α while Lemma 20 yields H α H α H α+α for all α, Proof. Let M, N be homogeneous elements of M 2 n 1,2 n 1 of Z2n 2 -degrees α M and α N, respectively. Then M = ra α 1 1n Bα 2n nn, N = r A α 1 1n Bα 2n nn where r E s α and r E s α. Hence α M = (α 1,..., α 2n ) and α N = (α 1,..., α 2n ). One computes MN = rr A α 1 1n Bα 2n nn Aα 1 1n Bα 2n nn = ( 1) s α,1α 1rr A α 1 1n Aα 1 1n Bα 2n nn Aα 2 2n Bα 2n nn = ( 1) s α,1α 1 + +s α,2nα 2nrr A α 1 1n Bα 2n nn A α 1 1n Bα 2n nn = ( 1) s α,1α 1 + +s α,2nα 2n +s αs α r ra α 1 1n Bα 2n nn A α 1 1n Bα 2n nn which equals ( 1) α,α NM. In order to define a Z 2n+1 2 -grading on M 2 n(e) we need the following lemma. Lemma 23. Let r EbeZ 2 -homogeneous, and let M H n. Then the element rm M 2 n(e) can be written uniquely as r A α 1 1n Bα 2n nn Cα 2n+1 n where r is a Z 2 -homogeneous element of E and Z2 (r) = s α. Proof. We have rm = ra α 1 11 Bα 2n nn where α = (α 1,..., α ) 2n Z2n 2 and this expression for rm is unique. If Z2 (r) = s α, then we set α = α if 2n, α 2n+1 = 0 and r = r. If Z2 (r) /= s α we put

12 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) α 2n+1 = 1 and α = α + 1if 2n. NowC n = A 11 B nn, and we obtain M =±A α 1 11 Bα 2n nn Cα 2n+1 n since the matrices A in, B jn, C n anticommute. In this case we put r =±rand we are done. Now assume r 1 A β 1 11 Bβ 2n nn C β 2n+1 n = rm = r 2 A γ 1 11 Bγ 2n nn C γ 2n+1 n, where s β = Z2 (r 1 ) = Z2 (r) = Z2 (r 2 ) = s γ. Then, as above, we have β + β 2n+1 = α = γ + γ 2n+1 for all = 1,...,2n. Hence 2n β i=1 i = 2n γ i=1 i and we obtain β 2n+1 = γ 2n+1 from s β = s γ. Therefore β = γ for all. Thus we have the uniqueness. Corollary 24. The algebra M 2 n(e) isz 2n+1 2 -graded. Its homogeneous component of degree(α 1,..., α 2n+1 ) is the subspace W (α1,...,α 2n+1 ) ={ra α 1 11 Bα 2n nn Cα 2n+1 n r E α1 + +α 2n+1 }. Proof. It follows immediately from the previous lemma and from Lemma 20. The next assertion (together with its proof) is analogous to Lemma 22. Lemma 25. The graded algebra M 2 n(e) is 2n + 1-supercommutative The graded identities in characteristic 0 In this subsection we give new proofs of the main results of [16]. Recall that Theorem 10 of [16]was proved there by means of constructing appropriate ring homomorphisms; here we employ graded identities. In this subsection we fix char K = 0, then the(graded) identities of an algebra are determined by the multilinear ones. In the next proposition we describe bases of the graded identities for M 2 n 1,2 n 1(E) and for M 2 n(e). Proposition The Z 2n 2 -graded identities of M 2 n 1,2 n 1 follow from the polynomials c 2n(x, y) when x and y run independently over Z 2n The Z 2n+1 2 -graded identities for M 2 n(e) follow from the polynomials c 2n+1 (x, y). Proof. 1. Obviously all c 2n are graded identities for M 2 n 1,2 n 1. Denote by J = J 2n the graded ideal of K X generated by the polynomials c 2n.Letf(x 1,..., x ) be a graded multilinear identity for M 2 n 1,2n 1. Then, by Lemma 19, f (x 1,..., x ) λx 1 x (mod J), where λ K. Ifλ = 0 then f is a consequence of the polynomials in J. Supposing λ /= 0, we shall show that the monomial x 1 x cannot be a graded identity for M 2 n 1,2 n 1. Let x i = (α 1,..., α 2n ).Ifs xi = 0 we put x i = A α 1 1n Bα 2n nn ;ifs x i = 1 we put x i = e i A α 1 1n Bα 2n nn. Then x 1 x evaluates to ±e i1 e ir D where i 1 < < i r correspond to the variables x i with s xi = 1, and D is an invertible matrix. Thus f is not a graded identity for M 2 n 1,2n 1, a contradiction. 2. The second assertion of the proposition is proved in the same way as above. Now we describe the generators of the graded T-ideal of E n. Proposition 27. The polynomials c n (x, y), when x and y run independently over Z n 2, generate the graded T-ideal of E n. Proof. As we noticed above, E n is n-supercommutative. This means that the polynomials c n (x, y) are graded identities for E n.letj = J n be the graded ideal generated by the polynomials c n. Suppose f (x 1,..., x ) is a multilinear graded identity for E n and f / J, then by Lemma 19, f λx 1 x (mod J) for some 0 /= λ K. But, if x i =(α 1,..., α n ) we substitute x i by a i = t 1 t n where t j = 1 if α j = 0, and t j = e (i 1)n+j when α j = 1. Thus f cannot be a graded identity for E n.

13 792 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) Theorem 28. Let char K = 0, the algebra E m is (graded) PI equivalent to M 2 n 1,2n 1 when m = 2n is even, and to M 2 n(e) when m = 2n + 1 is odd. Proof. The graded version of the theorem follows from Propositions 26 and 27. This of course implies the ungraded version. As an immediate consequence we obtain a new proof of one of the assertions of Kemer s theorem. Corollary 29. If char K = 0 the algebras M 1,1 and E E are PI equivalent The graded identities in characteristic p > 2 Here we fix the infinite field K of positive characteristic p /= 2. We now that the algebras M 1,1 and E E share the same T-ideal in characteristic 0. But if char K = p > 2 they satisfy different 2-graded identities ([15]), and also ordinary identities: we have T(M 1,1 ) T(E E), a proper inclusion, see [3]. We start by studying the graded identities of M 2 n 1,2n 1. First we prove that part (1) of Proposition 26 holds in the case of positive characteristic. Corollary 30. If char K = p > 2 then the graded identities of M 2 n 1,2 n 1 c 2n (x, y). follow from the polynomials Proof. If f (x 1,..., x ) is a graded multihomogeneous identity for M 2 n 1,2n 1 that does not follow from the polynomials c 2n (x, y) then modulo these polynomials we can write f as f = λx t 1 1 x t,0 /= λ K. Letx i be of homogeneous degree x =(α 1,..., α 2n ) in the Z 2n 2 -grading. By Lemma 19 if t i > 1 we must have s xi = 0. Then we can use the same substitution as in Proposition 26 in order to get a nonzero value of f on M 2 n 1,2n 1, a contradiction. Analogously we have the following corollary: Corollary 31. Let char K = p > 2. The ideal of graded identities for M 2 n is generated by the polynomials c 2n+1 (x, y). Now we study the Z n 2 -graded identities of E n in positive characteristic. Lemma 32. Let char K = p > 2 and let x be a variable in the free Z n 2-graded algebra K X of homogeneous degree α = (α 1,..., α n ). Suppose further s α = 0 and α i = 1 for some i, 1 i n. Then x p is a graded identity for E n. Proof. Let a = j=1 a j1 a jn E n be a Z n 2-homogeneous element of the same homogeneous degree as x. Since the summands of a have the same degree α and s α = 0 they commute. Hence a p = j=1 (a j1 a jn ) p = j=1 (a j1) p (a jn ) p. But a ji E 1, hence a 2 ji = 0 and we have a p = 0. Denote by P the graded T-ideal generated by all polynomials c n (x, y) and by all polynomials x p where x is a variable whose Z n 2 -degree α = (α 1,..., α n ) satisfies s α = 0 and α i = 1 for some i, 1 i n. Proposition 33. The ideal of Z n 2 -graded identities of E n equals P. Proof. Suppose first that f (x 1,..., x ) is a multihomogeneous Z n 2-graded polynomial, f / P. Then, by Lemma 19, one concludes f λx t 1 1 x t where 0 /= λ K and t i = 1 whenever s xi = 1. Moreover, by our assumptions, if s xi = 0 and x i /= (0,...,0) then t i < p. We shall show that x t 1 1 x t is not a graded identity for E n.

14 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) In characteristic p, the variable xt is equivalent to its complete linearization g t (y 1,..., y t ) = σ S t y σ(1) y σ(t) for any t < p. Moreover, if s x = 0 then, modulo P, the variables y j commute, so g t t!y 1 y t (mod P). With other words, we may assume that in x t 1 1 x t we have t i = 1 for all i such that x i /= (0,...,0). Now, as in the proof of Proposition 27,if x i =(α 1,..., α n ) we substitute x i by a i = b 1 b n where b j = 1ifα j = 0, and b j = e (i 1)n+j when α j = 1. Thus x t 1 1 x t cannot be a graded identity for E n and we are done. Theorem The Z 2n 2 -graded T-ideal of E 2n contains properly that of M 2 n 1,2 n The Z 2n+1 2 -graded T-ideal of E 2n+1 contains properly that of M 2 n. Proof. 1. It suffices to show that x p, x =(1, 1, 0...,0) is not an identity of M 2 n 1,2 n 1. But(A 1nA 2n ) p /= 0 as a direct computation shows. The assertion 2. is proved analogously A lattice of graded T-ideals Denote by L m,n the algebra M 2 m 1,2 m 1 E n. It is equipped with a Z 2m+n 2 -grading that is induced by the Z 2m 2 -grading on M 2 m 1,2 m 1 and by the Zn 2 -grading on E n. Proposition 35. The Z 2m+n 2 -graded T-ideal of L m,n is generated by the identities c 2m+n (x, y) and by the elements x p where x =α = (α 1,..., α 2m+n ) Z 2m+n 2 are such that s α = 0 and α j = 1 for some j, 2m + 1 j 2m + n. Proof. First we notice that M 2 m 1,2 m 1 is 2m-supercommutative and E n is n-supercommutative. Hence L m,n is 2m + n-supercommutative and this says that L m,n satisfy the graded identities c 2m+n (x, y). Moreover, the validity of the graded identities of the type x p is verified as in Lemma 32. Finally if f is a multihomogeneous polynomial, we can write f, modulo the identities of the statement, as a single monomial multiplied by a scalar, that is f λx u 1 1 xu where 0 /= λ K. If s xi is odd then, by Lemma 25, we can assume that u i = 1. Furthermore if s xi = 0, x i =α as in the statement and α j = 1 for some j {2m + 1, 2m + n} then u i < p. We shall prove that this monomial does not vanish on L m,n. As in the proof of Proposition 33, in order to find a non zero substitution, we shall assume that u i = 1 for each x i which satisfies the condition of the statement. Hence, if u i > 1 then x i =(α 1,..., α 2m+n ) where s xi = 0 and α j = 0 for all j = 2m + 1,...,2m + n. Now, for any i = 1,...,,lety i, z i be graded variables of homogeneous Z 2m 2 and Z n 2 degree y i and z i respectively, such that x i =( y i, z i ) Z 2m+n 2. Let us consider the monomials g = y u 1 1 yu and h = z u 1 1 zu. We remar that, by our assumptions, if u i > 1 then z i =(α 2m+1,..., α 2m+n ) = 0 Z n 2 and so s y i = s xi + s zi = s xi = 0. By Proposition 33 it follows that h does not vanish on E n. Therefore, there exist homogeneous elements b 1,..., b E n such that b i = z i and b u 1 1 bu /= 0. Similarly, by Proposition 26 and Corollary 30, it follows that g is not a graded polynomial identity of M 2 m 1,2 m 1 and so au 1 1 au /= 0 for some homogeneous elements a 1,..., a M 2 m 1,2m 1 such that a i = y i. Therefore f (a 1 b 1,..., a b ) = λa u 1 1 au bu 1 1 bu /= 0 and we are done. As a consequence we have the following theorem. Theorem 36. Let G = Z 2n 2, then T G(L n,0 ) T G (L n 1,2 ) T G (L 0,2n ). The algebra L m,n = M 2 m(e) E n is Z 2m+n+1 2 -graded. As in Proposition 35 one has Lemma 37. The Z 2m+n+1 2 -graded T-ideal of L m,n is generated by the polynomials c 2m+n+1 (x, y) and by all x p, x =α = (α 1,..., α 2m+n+1 ) Z 2m+n+1 2. Here x is such that s α = 0 and α j = 1 for some j {2m + 2,...,2m + n + 1}.

15 794 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) Now we have all ingredients of the following theorem. Theorem 38. Let K be an infinite field, char K /= 2 and let G = Z 2n. 2 Then T G(L n,0 ) T G (L n 1,1 ) T G (L n 1,2 ) T G (L n 2,3 ) T G (L 1,2n 2 ) T G (L 0,2n 1 ). Denote G = Z 2n+1 2, then T G (L n,0 ) T G (L n,1 ) T G (L n 1,2 ) T G (L n 1,3 ) T G (L 1,2n 2 ) T G (L 1,2n 1 ) T G (L 0,2n ). If char K = 0 all inclusions become equalities; when char K = p > 2 all inclusions are proper ones. Proof. When char K = 0 the theorem follows from Theorem 28. Suppose char K = p > 2. Then Proposition 35 together with Lemma 37 yield the proper inclusions. Corollary 39. Let char K char K = 0. /= 2. Then T G (M 2 n 1,2 n 1 E) T G(M 2 n(e)). The equality holds only when Proof. It follows from the above theorem since M 2 n 1,2 n 1 E = L n 1,1 and M 2 n(e) = L n,0. 4. More cases of the Tensor product theorem In this section we compare the graded T-ideals of the algebras M 2 n 1,2 n 1 and M 2 1,2 1 M 2 l 1,2 l 1 where + l = n, over an infinite field K of positive characteristic p /= 2. First we consider the Z 2n 2 - grading on M 2 1,2 1 M 2 l 1,2l 1 induced by the Z2 2 -grading on M 2 1,2 1 and the Z2l 2 -grading on M 2 l 1,2 l 1. Thus we have a Z 2n = 2 Z2 2 Z2l 2 -grading on this algebra. Moreover, by Propositions 18 and 22 we have that the algebra M 2 1,2 1 M 2 l 1,2l 1 is 2n-supercommutative. Lemma 40. Let G = Z 2n 2 and let x be a variable. Assume x =α = (α 1,..., α 2n ), s α = 0 and α 1 + +α 2 = α α 2n = 1. Then x p is a graded identity for M 2 1,2 1 M 2 l 1,2 l 1. Proof. Let a M be homogeneous of G-degree α. Then a = a 1 + +a t where a i = r i A α 1 1 Bα 2 r i Aα 2+1 1l B α 2n ll, and r, r E 1 are odd elements of the Grassmann algebra for all i. Since s α = 0 the homogeneous elements a i commute, hence a p = a p 1 + +ap t. Since r i E 1 we have a p i = 0as desired. Now let J be the graded T-ideal generated by the polynomials c n (x, y) and by all x p where the variable x satisfies the condition of Lemma 40. Theorem 41. The graded T-ideal of M 2 1,2 1 M 2 l 1,2l 1 equals J. Proof. Let f (x 1,..., x q )/ J be a multihomogeneous G-graded polynomial of multidegree(u 1,..., u q ). By Lemma 19 we write f λx u 1 1 xu q q (mod J ) where 0 /= λ K and u i = 1ifs xi = 1. Furthermore if the G-degree of some x i satisfies the condition of Lemma 40 then u i < p. Therefore, as in the proof of Proposition 33, in order to find a non zero substitution, we shall assume that u i = 1 for each x i which satisfies the condition of Lemma 40. Hence, if u i > 1 then x i =(α 1,..., α 2n ) where 2 α i=1 i = 0 = 2l α i=1 2+i. We describe a substitution that does not vanish f on M 2 1,2 1 M 2 l 1,2 l 1. Tae a variable x i of f and set α = x i.ifs α = 1 then we put x i r i A α 1 1, Bα 2, r i Aα 2+1 1,l B α 2n l,l. Here r i equals e ip when α 1 + +α 2 is even and 1 if α 1 + +α 2 is odd, and analogously for r i. In case both α 1 + +α 2 and α α 2n are odd then x p i vanishes. So we put x i a 1 + +a p. Here a j = r j A α 1 1, Bα 2n l,l r j A α 1 1, Bα 2n l,l (the tensor square) and r j = e (i 1)p+j. Otherwise put x i A α 1 1, Bα 2n l,l A α 1 1, Bα 2n l,l.

16 O.M. Di Vincenzo et al. / Linear Algebra and its Applications 432 (2010) Corollary 42. Let K be infinite, char K /= 2. Assume + l = n, then T G (M 2 n 1,2 n 1) T G(M 2 1,2 1 M 2 l 1,2l 1). These two graded ideals coincide only in characteristic 0. Corollary 43. Let 1 + l 1 = 2 + l 2, 1 l 1 and 2 l 2, and assume 1 /= 2. Assume K infinite and char K /= 2. Then T Z 2n (M ,2 1 1 M 2 l 1 1,2 l 1) is equal to T 1 Z 2n(M ,2 2 1 M 2 l 2 1,2 l 2 1) only when char K = 0. If char K = p > 2 neither of these two ideals is contained in the other. Acnowledgments We than the Referees whose valuable comments and positive criticism improved much the exposition of the paper. References [1] S.M. Alves, P. Koshluov, Polynomial identities of algebras in positive characteristic, J. Algebra 305 (2) (2006) [2] S.S. Azevedo, Graded identities for the matrix algebra of order n over an infinite field, Comm. Algebra 30 (12) (2002) [3] S.S. Azevedo, M. Fidelis, P. Koshluov, Tensor product theorems in positive characteristic, J. Algebra 276 (2) (2004) [4] S.S. Azevedo, M. Fidelis, P. Koshluov, Graded identities and PI equivalence of algebras in positive characteristic, Comm. Algebra 33 (4) (2005) [5] Yu. Bahturin, V. Drensy, Graded polynomial identities of matrices, Linear Algebra Appl. 357 (2002) [6] Yu. Bahturin, M. Zaicev, Graded algebras and graded identities, Polynomial Identities and Combinatorial Methods, Lect. Notes Pure Appl. Math. 235, M. Deer, 2003, pp [7] A. Berele, Generic verbally prime algebras and their GK-dimensions, Comm. Algebra 21 (5) (1993) [8] A. Berele, Supertraces and matrices over Grassmann algebras, Adv. Math. 108 (1) (1994) [9] O.M. Di Vincenzo, On the graded identities of M 1,1 (E), Israel J. Math. 80 (3) (1992) [10] O.M. Di Vincenzo, V. Nardozza, Graded polynomial identities for tensor products by the Grassmann algebra, Comm. Algebra 31 (3) (2003) [11] O.M. Di Vincenzo, V. Nardozza, Z +l Z 2 -graded polynomial identities for M l (E) E, Rend. Sem. Mat. Univ. Padova 108 (2002) [12] O.M. Di Vincenzo, V. Nardozza, Graded polynomial identities of verbally prime algebras, J. Algebra Appl. 6 (3) (2007) [13] A. Giambruno, M. Zaicev, Polynomial identities and asymptotic methods, Math. Surveys Monographs 122, Amer. Math. Soc., [14] A. Kemer, Ideal of identities of associative algebras, Amer. Math. Soc. Transl. Ser. 87, [15] P. Koshluov, S.S. Azevedo, Graded identities for T-prime algebras over fields of positive characteristic, Israel J. Math. 128 (2002) [16] A. Regev, Homomorphisms for tensor products of Grassmann algebras, Israel Math. Conf. Proc. 1 (1989) [17] A. Regev, Tensor products of matrix algebras over the Grassmann algebra, J. Algebra 133 (2) (1990) [18] S.Yu. Vasilovsy, Z n -graded polynomial identities of the full matrix algebra of order n, Proc. Amer. Math. Soc. 127 (12) (1999)