Z n -GRADED POLYNOMIAL IDENTITIES OF THE FULL MATRIX ALGEBRA OF ORDER n

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1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 127, Number 12, Pages S (99) Article electronically published on May 13, 1999 Z n -GRADED POLYNOMIAL IDENTITIES OF THE FULL MATRIX ALGEBRA OF ORDER n SERGEI YU VASILOVSKY (Communicated by Ken Goodearl) Abstract The algebra M n(f )ofalln nmatrices over a field F has a natural Z n-grading M n(f )= (α) α Z n M n In this paper graded identities of the Z n-graded algebra M n(f ) over a field of characteristic zero are studied It is shown that all the Z n-graded polynomial identities of M n(f ) follow from the following: x 1 x 2 x 2 x 1 =0, α(x 1 )=α(x 2 )=0; x 1 xx 2 x 2 xx 1 =0, α(x 1 )=α(x 2 )= α(x) Introduction Let F be a field of characteristic zero, let M n (F ) be the algebra of all square matrices of order n over F The ideal T (M n (F )) of all polynomial identities of the algebra M n (F ) plays an important role in the theory of varieties of associative algebras As follows from Kemer s result [7] about the finite basis property of identities of associative algebras over a field of characteristic zero, all polynomial identities of M n (F ) follow from finitely many basic identities in T (M n (F )) for any n However, the problem of finding an explicit finite basis for the ideal T (M n (F )) has so far been resolved only for n = 2 (cf [5], [9]), there is still no solution in sight even for n =3 One of the general methods in the modern theory of varieties of algebras which has proved to be quite effective consists in exping the signature of the algebra under consideration with an additional structure such as a trace or a grading that is inherent to the algebra studying generalized identities of the arising algebraic system rather than the ordinary polynomial identities of the original algebra (see, for instance, [2], [6], [8], [10], [11]) In particular, YuP Razmyslov C Procesi have independently established that, for all n, all the trace identities of M n (F ) are obtained from the Cayley-Hamilton theorem (cf [10] [8]) Let Z n denote the additive group of integers modulo n In [3] OM Di Vincenzo has described a finite basis for the Z 2 -graded polynomial identities of M 2 (F ) Subsequently, IP Shestakov has posed the problem of finding an explicit finite basis for the graded polynomial identities of the algebra M n (F ) equipped with its natural Z n -grading for any n In the present paper this problem is completely resolved Received by the editors April 10, 1997, in revised form, February 26, Mathematics Subject Classification Primary 16R40 Key words phrases Graded polynomial identities, full matrix algebra 3517 c 1999 American Mathematical Society

2 3518 SERGEI YU VASILOVSKY Theorem All graded polynomial identities of the Z n -graded algebra M n (F ) follow from (01) x 1 x 2 x 2 x 1 =0, α(x 1 )=α(x 2 )=0, (02) x 1 xx 2 x 2 xx 1 =0, α(x 1 )=α(x 2 )= α(x) 1 Definitions preliminary results Denote by E i,j,1 i, j n, then nmatrix whose only nonzero entry is 1 in the ith row jth column The matrices E i,j s, known as matrix units, form a basis of M n (F ) as a vector space For t Z, lettdenote the residue class in Z n that contains t For α Z n,letm n (α) be the subspace of M n (F ) spanned by all matrix units E i,j such that j i = α, sothatm (0) n consists of the matrices of the form a 1,1 a 2,2 0 0, a 1,1,a 2,2,,a n,n F, an,n, for 0 <t n 1, M (t) n consists of the matrices of the form 0 0 a 1,t+1 0 a 2,t a n t,n, a n t+1, a n,t 0 0 where a 1,t+1,a 2,t+2,,a n t,n,a n t+1,1,a n t+2,2,,a n,t F ThenM n (F)isa direct sum of the subspaces M n (α) s: M n (F )= M (α) (11) n α Z n Recall (cf, for instance, [4]) that a family A (α), α G} of subspaces of an algebra A, wheregis an additive group, defines a G-grading of the algebra A if A = α G A(α), for α, β G, A (α) A (β) A (α+β) Since Ei,s, if j = r, E i,j E r,s = 0, if j r, it follows that, for p, q 0,1,,n 1},wehaveM (p) M (q) M (p+q),so that the decomposition (11) defines a Z n -grading of the algebra M n (F ) We will denote the Z n -graded algebra M n (F )bym n Let } X (α),α Z n be a family of disjoint countable sets X (α) Put X = α Z n X (α) denote by A[X] the free associative algebra freely generated by the set X The monomials x i1 x i2 x ik : k =1,2, ; x i1,x i2,,x ik X}

3 Z n-graded POLYNOMIAL IDENTITIES OF THE FULL MATRIX ALGEBRA 3519 form a basis of A[X] as a vector space An indeterminate x X is said to be of homogeneous degree α, written α(x) =α,ifx X (α) The homogeneous degree of a monomial n = x i1 x i2 x ik is defined to be α(n) =α(x i1 )+α(x i2 )+ +α(x ik ) For α Z n,denotebya[x] (α) the subspace of A[X] spanned by all the monomials having homogeneous degree α Notice that A[X] (α) A[X] (β) A[X] (α+β) for all α, β Z n Thus A[X]= A[X] (α) α Z n proves to be a Z n -graded algebra The elements of the Z n -graded algebra A[X] are referred to as Z n -graded polynomials or, simply, graded polynomials An ideal I of A[X] issaidtobeat n -ideal if it is invariant under all F -endomorphisms γ : A[X] A[X] such that γ(a[x] (α) ) A[X] (α) for all α Z n Let A = α Z n A (α) be a Z n -graded associative algebra A Z n -graded polynomial f(x 1,x 2,,x k ), or the expression f(x 1,x 2,,x k ) = 0, is said to be a graded polynomial identity of the Z n -graded algebra A if f(a 1,a 2,,a k )=0for all a 1,a 2,,a k α Z n A (α ) such that a s A (α(xs)), s =1,2,,k The set T n (A) of all graded identities of a Z n -graded algebra A is a T n -ideal of A[X] A Z n -graded polynomial identity f = 0 is said to follow from a family of Z n -graded polynomial identities g λ =0,λ Λ, if the graded polynomial f lies in the smallest T n -ideal containing all the graded polynomials g λ, λ Λ Lemma 1 The graded algebra M n satisfies (01) (02) Proof Since any two diagonal matrices commute, it follows that M n satisfies graded identity (01) The identities (02) being multilinear, it is only required to show that this identity holds when x 1 = E i1,j 1,x 2 =E i2,j 2, x = E r,s for E i1,j 1,E i2,j 2 M n (t),e r,s M n (n t),where0<t n 1; so that i1 + t, if i j 1 = 1 + t n, i 1 + t n, if i 1 + t>n; j2 t, if j i 2 = 2 t 1, j 2 t + n, if j 2 t<1; s + t, if s + t n, r = s + t n, if s + t>n Note that E i1,j 1 E r,s E i2,j 2 0onlyifj 1 =r s = i 2 We claim that in this case i 1 = s = i 2 j 1 = r = j 2 Observe that, if j 1 = i 1 +t r = s + t n, then, from j 1 = r, it follows that n = s i 1, which is impossible So the equalities j 1 = i 1 + t r = s + t n cannot hold simultaneously The same applies to the equalities s = r t i 2 = j 2 t + n Thus,whenj 1 =i 1 +t,wehaver=s+t i 2 = j 2 t, sothat i 2 =s=r t=j 1 t=i 1 r = j 1 = i 1 + t = i 2 + t = j 2

4 3520 SERGEI YU VASILOVSKY Similarly, when j 1 = i 1 + t n, wehaver=s+t n i 2 = j 2 t + n, whence i 2 = s = r t + n = j 1 t + n = i 1 r = j 1 = i 1 + t n = i 2 + t n = j 2, proving the above assertion This enables us to conclude that E i1,j 1 E r,s E i2,j 2 0 if only if i 1 = s = i 2 j 1 = r = j 2, if only if E i2,j 2 E r,s E i1,j 1 0 In this case E i1,j 1 E r,s E i2,j 2 = E i1,j 2 = E i2,j 1 = E i2,j 2 E r,s E i1,j 1 Otherwise, E i1,j 1 E r,s E i2,j 2 =0=E i2,j 2 E r,s E i1,j 1, proving that (02) does hold in M n From now on let I n be the T n -ideal generated by the graded identities (01) (02) For a positive integer k, denotebys k the set of all permutations of the set 1, 2,,k}Forx 1,x 2,,x k X S k,let m =m (x 1,x 2,,x k )=x (1) x (2) x (k) The multilinear monomial in x 1,x 2,,x k corresponding to the identity permutation will be denoted by m = m(x 1,x 2,,x k )=x 1 x 2 x k Obviously, α(m) =α(m )=α(x 1 )+α(x 2 )+ +α(x k ) It is clear that every multilinear graded polynomial f(x 1,x 2,,x k ) can be expressed as f = a m, S k where a F By a stard substitution we will underst a substitution S of the form (12) x 1 = E i1,j 1, x 2 = E i2,j 2,,x k =E ik,j k, where (13) j s i s = α(x s ) so that E is,j s M (α(xs)) n, s =1,2,,k For a graded polynomial f(x 1,x 2,,x k ) a substitution S, wedenotebyf S the value of f corresponding to the substitution S It is easy to see that, if a multilinear graded polynomial f(x 1,x 2,,x k ) is such that f S = 0 for every stard substitution S, thenf= 0 is a graded identity of M n Observe that, when a substitution (12) is made, the value of a monomial m (x 1,x 2,,x k ) differs from zero only if (14) j (1) = i (2), j (2) = i (3),, j (k 1) = i (k), in which case m (12) = E i(1),j (k) For a monomial m (x 1,x 2,,x k ), S k, any two integers 1 p q k, denote by m [p,q] the subword obtained from m by discarding the first p 1 the last k q factors: m [p,q] = x (p) x (p+1) x (q)

5 Z n-graded POLYNOMIAL IDENTITIES OF THE FULL MATRIX ALGEBRA 3521 Lemma 2 For any S k, the exists a stard substitution S such that m (x 1,x 2,,x k ) S 0 Proof We apply induction on k The case k = 1 is trivial Let k>1 let α(x k )=tfor some integer 0 t<n If, when some stard substitution (15) x (1) = E i1,j 1,x (2) = E i2,j 2,,x (k 1) = E ik 1,j k 1 is made, m [k 1] = E i1,j k 1 0, then, setting, in addition to (15), x (k) = E jk 1,j k, where jk 1 + t, if j j k = k 1 + t n, j k 1 + t n, if j k 1 + t>n, we get m (x 1,x 2,,x k )=E i1,j k 1 E jk 1,j k = E i1,j k 0, completing the proof Lemma 3 If m (12) 0, then, for any 1 p q k, α(m [p,q] )=j (q) i (p) Proof In view of (13) (14), we have α(m [p,q] )=α(x (q) )+α(x (q 1) )+ +α(x (p) ) =(j (q) i (q) )+(j (q 1) i (q 1) )+ +(j (p) i (p) )=j (q) i (p), completing the proof Lemma 4 If, for a permutation S k, there exists a stard substitution (12) such that m (x 1,x 2,,x k ) (12) = m(x 1,x 2,,x k ) (12) 0, then m (x 1,x 2,,x k ) x 1 n(x 2,x 3,,x k )(mod I n ) for some monomial n (x 2,x 3,,x k )=x l2 x l3 x lk Proof Suppose (1) 1 Then1= 1 ((1)) < 1 (1) Let t be the least positive integer such that 1 (t +1)< 1 (1) Obviously, 1 1 (t +1)< 1 (1) 1 (t) Since E i1,j 1 E i2,j 2 E ik,j k = E i(1),j (1) E i(2),j (2) E i(k),j (k) 0, it follows that i 1 = i (1), j t = i t+1, for s>1, j (s 1) = i (s) So, setting p = 1 (t +1), q = 1 (1) r = 1 (t), we have 1 p < q r with j (q 1) = i (q) = i (1), j (r) = i (p), when p > 1, j (p 1) = i (p) First consider the case p>1 From the equalities j (r) = i (p) = j (p 1) j (q 1) = i (q) = i (1), we infer that j (p 1) i (1) = i (p) j (q 1) = j (r) i (q) = t 0

6 3522 SERGEI YU VASILOVSKY for some t 0 Z By Lemma 3, we have α(m [1,p 1] α(m [p,q 1] Hence, using (02), we get m = m [1,p 1] m [q,r] m [p,q 1] m [1,p 1] )=j (p 1) i (1) = t 0 ; )=j (q 1) i (p) = t 0 ; α(m [q,r] )=j (r) i (q) =t 0 m [p,q 1] m [q,r] m [r+1,k] m [r+1,k] = x (q) x l2 x lk =x 1 x l2 x lk (mod I n ) Now let p =1 Inthiscasej (q 1) = i (q) = i (1) = j (r), which, in view of Lemma 3, gives α(m [1,q 1] )=j (q 1) i (1) = 0; α(m [q,r] )=j (r) i (q) =0 Then it follows from (01) that m [q,r] m [1,q 1] This completes the proof m = m [1,q 1] m [q,r] m [r+1,k] m [r+1,k] = x (q) x l2 x lk =x 1 x l2 x lk (mod I n ) Lemma 5 If, for a permutation S k, there exists a stard substitution (12) such that m (x 1,x 2,,x k ) (12) = m(x 1,x 2,,x k ) (12) 0, then m (x 1,x 2,,x k ) m(x 1,x 2,,x k )(mod I n ) Proof Let r be the greatest positive integer such that m (x 1,x 2,,x k ) x 1 x 2 x r n(x r+1,,x k )(modi n ) for some multilinear monomial n = n(x r+1,,x k ) We need only show that r = k Suppose, on the contrary, r<k Then, obviously, r k 2 We have x 1 x 2 x r n (12) = m (12) = m (12) 0 Combining this with x 1 x 2 x r n (12) = E i1,j 1 E ir,j r n (12) = E i1,j r n (12) m (12) = E i1,j 1 E ir,j r x r+1 x r+2 x k } (12) = E i1,j r x r+1 x r+2 x k } (12), we get n(x r+1,x r+2,,x k ) (12) = x r+1 x r+2 x k (12) 0 By Lemma 4, there exists a multilinear monomial n (x r+2,x r+3,,x k ) such that n(x r+1,x r+2,,x k ) x r+1 n (x r+2,,x k )(modi n ) so that m x 1 x r n(x r+1,x r+2,,x k ) x 1 x r x r+1 n (x r+2,,x k )(modi n ), which contradicts our choice of the number r This completes the proof

7 Z n-graded POLYNOMIAL IDENTITIES OF THE FULL MATRIX ALGEBRA 3523 From Lemma 5, we immediately obtain the following Corollary 6 If, for two permutations, τ S k, there exists a stard substitution S such that m (x 1,x 2,,x k ) S =m τ (x 1,x 2,,x k ) S 0, then m (x 1,x 2,,x k ) m τ (x 1,x 2,x k )(mod I n ) 2 Proof of the theorem Since the characteristic of the basic field is zero, we need only prove that an arbitrary multilinear graded polynomial identity f(x 1,x 2,,x k )=0ofM n lies in I n (cf, for instance, [1, Theorem 423]) Let r be the least non-negative integer such that the polynomial f can be expressed, modulo I n, as a linear combination of r multilinear monomials: r f a q m q (mod I n ), q=1 where 0 a q F, 1, 2,, r S k Show that r = 0 Suppose, on the contrary, r>0 In view of Lemma 2, we can find a stard substitution S such that m 1 S 0 Since m q S E i,j : i, j =1,2,,n} 0},q=1,2,,r, r a 1 m 1 S = ( a q )m q S, q=2 it follows that there is at least one integer p 2,3,,r}such that m p S = m 1 S Then, by Corollary 6, m p m 1 (mod I n ), so that f r q=1 p 1 a q m q (a 1 + a p )m 1 + a q m q + q=2 r q=p+1 a q m q (mod I n ), that is, f can be expressed, modulo I n, as a linear combination of no more than r 1 multilinear monomials, which contradicts our choice of the number r Thus f 0(modI n ) This completes the proof of the theorem Acknowledgment I want to thank Professor IP Shestakov for introducing me to this problem References 1 YuA Bahturin, Identities in Lie algebras, Nauka, Moskva, 1985 [Russian] Translation: VNU Sci Press, Utrecht, 1987 MR 86k: A Berele, Trace identities Z/2Z-graded invariants, Trans AMS, 309(2) (1988), MR 89m: OM Di Vincenzo, On the graded identities of M 1,1 (E), Israel J Math 80(3) (1992), MR 94f: OM Di Vincenzo, Cocharacters of G-graded algebras, Comm Algebra 24(10) (1996), MR 97f:16041

8 3524 SERGEI YU VASILOVSKY 5 VS Drenski, A minimal basis for the identities of a second order matrix algebra over a field of characteristic 0, Algebra i Logika, 20 (1981), [Russian] Translation: Algebra Logic, 20 (1981), MR 83g: AR Kemer, Varieties Z 2 -graded algebras, Izv Akad Nauk SSSR Ser Mat, 48(5) (1984), [Russian] Translation: Math USSR Izv, 25 (1985), MR 86f: AR Kemer, Finite basis property of identities of associative algebras, Algebra i Logika, 26(5) (1987), [Russian] Translation: Algebra Logic, 26(5) (1987), MR 90b: C Procesi, The invariant theory of n n matrices, Adv in Math, 19 (1976), MR 54: YuP Razmyslov, Finite basing of the identities of a matrix algebra of second order over a field of characteristic zero, Algebra i Logika, 12 (1973), [Russian] Translation: Algebra Logic, 12 (1974), MR 49: YuP Razmyslov, Trace identities of full matrix algebra over a field of characteristic zero, Izv Akad Nauk SSSR Ser Mat 38(4) (1974), [Russian] Translation: Math USSR Izv, 8(4) (1974), MR 58: YuP Razmyslov, Trace identities central polynomials in the matrix superalgebras M n,k, Mat Sb, 128(2) (1985) [Russian] Translation: Math USSR Sb, 56(1) (1987), MR 87f:16014 Department of Mathematics, University of Swazil, Private Bag 4, Kwaluseni, Swazil, Southern Africa Institute of Mathematics, Novosibirsk , Russia address: vasilovs@realnetcosz