NILPOTENCY INDEX OF NIL-ALGEBRA OF NIL-INDEX 3

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J. Appl. Math. & Computing Vol. 20(2006), No. 1-2, pp. 569-573 NILPOTENCY INDEX OF NIL-ALGEBRA OF NIL-INDEX 3 WOO LEE Abstract. Nagata and Higman proved that any nil-algebra of finite nilindex is nilpotent of finite index.the Nagata-Higman Theorem can be formulated in terms of T -ideals. TheT -ideal generated by a n for all a A is also generated by the symmetric polynomials. The symmetric polynomials play an importmant role in analyzing nil-algebra. We construct the incidence matrix with the symmetric polynomials. Using this incidence matrix, we determine the nilpotency index of nil-algebra of nil-index 3. AMS Mathematics Subject Classification : 15A72 Key words and phrases : Nil-algebra, nilpotency index, nil-index. 1. Introduction. Let F be a field of characteristic 0 and A be a F -algebra. We do not assume that A has units. Suppose that there is a positive integer n N such that a n =0 for all a A, then A is called a nil-algebra and the natural number n is called the nil-index of A. If A m = 0, but A m 1 0, then A is said to be nilpotent of index m or A has nilpotency m. In other words, A is nilpotent of index m if any product of m elements of A is 0. Theorem 1.1. [1, 4] (Nagata-Higman Theorem) If there exists a positive integer n N such that a n =0for all a A, then there is a natural number N (n) such that a 1 a 2 a N =0for all a 1, a 2,..., a N A. Let s review the brief history of this theorem and the following consequence. Nagata in 1953 first proved the above theorem. Higman provided a simpler proof in 1956 and showed that n 2 e 2 N(n) 2n 1. Received June 3, 2005. c 2006 Korean Society for Computational & Applied Mathematics and Korean SIGCAM. 569

570 Woo Lee It was Razmyslov [6] in 1974 who improved Higman s upper bound to n 2, that is, N (n) n 2. It was Kuzmin [2] in 1975 who heightened the lower bound to n(n+1) 2, i.e., n(n +1) N(n). 2 It is known that N (2) = 3. In this work, we show that N (3) = 6. Determining N (n) for n 4 is still an open question[3]. Let C(n) be the commutative ring (with unit) generated by the traces, T (σ) where σ is a monomial in the generic n n matrices X 1,X 2,... and R(n) be the ring generated by C(n) and the generic n n matrices. Let s denote by R + (n) the two-sided ideal of R(n) generated by X 1,X 2,... Let P(n) be the least integer such that R + (n) is generated as a C(n)-module by monomials in the generic matrices X 1,X 2,... of degree P(n). And Q(n) is the least integer such that C(n) is generated as a F -algebra by elements of degree Q(n). Theorem 1.2. [5] The natural numbers N (n), P(n)+1 and Q(n) are equal 2. Symmetric Polynomials and Incidence matrices. Let s begin with the definition of symmetric polynomial. Definition 2.1. The symmetric polynomial of degree n is the polynomial S n (a 1,a 2,...,a n )= a σ(1) a σ(2) a σ(n), σ Sym(n) where Sym(n) is the symmetric group on n letters. Let T n be the T -ideal generated by a n for a A. For example, when n =2, a 2 1, a2 2 T 2, also (a 1 + a 2 ) 2 T 2. Thus a 2 1 + a 1a 2 + a 2 a 1 + a 2 2 T 2 implies S 2 (a 1, a 2 )=a 1 a 2 + a 2 a 1 T 2. By multilinearizing, we can show that T n is generated by the symmetric polynomial S n (a 1,a 2,...,a n ) of degree n for a 1,a 2,...,a n A. Given a partition P = (p 1,...,p n )ofm with n parts, where p i p i+1, 1 i n 1, the incidence matrix denoted by n, m P is constructed as the following. We label the columns by the (multilinear) monomials of m variables lexicographically. In other words, the first column is labeled by a 1 a 2 a m 2 a m 1 a m, the second by a 1 a 2 a m 2 a m a m 1 and so on. Thus the last column is indexed by a m a m 1 a m 2 a 2 a 1. We use 1, 2,..., for a 1, a 2,..., if there is no risk of confusion. Suppose that j-th column is indexed by

Nilpotency index of nil-algebra 571 Table 1. The 2, 3 (1,2) -incidence matrix. 123 132 213 231 312 321 S 2 (1, 23) 1 0 0 1 0 0 S 2 (1, 32) 0 1 0 0 0 1 S 2 (2, 13) 0 1 1 0 0 0 S 2 (2, 31) 0 0 0 1 1 0 S 2 (3, 12) 1 0 0 0 1 0 S 2 (3, 21) 0 0 1 0 0 1 i 1 i m. Then j-th row of the incidence matrix corresponding to the partition P =(p 1,...,p n ) is labeled by ). In the j-th row, one places 1 for the columns labeled by the monomials that appear in that row index, and 0 elsewhere. In other words, if ) = i 1 i p1 i p1+1 i p1+p 2 i p1+ +p n 1+1 i p1+ +p n + + i p1+ +p n 1+1 i p1+ +p n i p1+1 i p1+p 2 i 1 i p1, then put 1 for the columns labeled by i 1 i p1 i p1+1 i p1+p 2 i p1+ +p n 1+1 i p1+ +p n,..., i p1+ +p n 1+1 i p1+ +p n i p1+1 i p1+p 2 i 1 i p1. For instance, the 2, 3 (1,2) -incidence matrix is in Table 1. This matrix has determinant 4 and this fact implies that the nil-algebra of nil-index 2 is nilpotent of index 3. 3. Properties of Incidence matrices. Let (1 r1 2 r2 k r k ) be a partition of m with n parts where the superscript r i is the multiplicity of part of size i. Hence 1 r 1 +2 r 2 + + k r k = m and r 1 + r 2 + + r k = n. Proposition 3.1. The n, m P -incidence matrix is a {0, 1}-matrix, with constant row and column sums for both.

572 Woo Lee Proof. The n, m P -incidence matrix has m! columns. In each row, we assign 1 for all the monomials of ), which contains monomials. Therefore the row sum of n, m P is. By the symmetry of n, m P, we only need to determine the sum of the first column sum, which is labeled by 12 m = a 1 a 2 a m.the number of ways dividing the string 12 m into n sets is. But in each division, one get n, m P is distinct monomials. So the column sum of =. Even though we have the column sum of n, m P as, some of the columns can be identical. The number of the distinct rows in the n, m P is. Now we are ready to prove our main theorem. Theorem 3.2. The nil-algebra A of nil-index 3 is nilpotent of index 6. Proof. The partition of 6 with 3 parts are (1, 1, 4), (1, 2, 3) and (2, 2, 2). By the Proposition 3.1, the 3, 6 -incidence matrix is of dimension 6! = 720. Each of 3, 6 (1,1,4), 3, 6 (1,2,3) and 3, 6 (2,2,2) has the row and column sums 3! = 6. Furthermore 3, 6 (1,1,4) has 6! 2! = 360 distinct rows and 3, 6 (2,2,2) has 6! = 120 distinct rows. This means that 3, 6 (1,1,4) and 3, 6 (2,2,2) have ranks less than 720, 3! which implies none of them are invertible. If we consider the sum of 3, 6 (1,1,4), 3, 6 (1,2,3) and 3, 6 (2,2,2), the resulting matrix has nonzero determinant 2 208 3 291 5 10 7 16 11 10 13 10 17 14 23 9 29 9 67 5, which implies that the sum of 3 incidence matrices is invertible. Thus the monomial a 1 a 2 a 6 can be expressed by the sum of S 3 (w 1,w 2,w 3 ) where w i are words in a 1,...,a 6 and 3 length(w i )=6

Nilpotency index of nil-algebra 573 References 1. G. Higman. On a conjecture of Nagata, Proc. Cambridge Philos. Soc 52 (1956), 1 4. 2. E. N. Kuzmin. On the Nagata-Higman theorem, in Mathematical Structures- Computational Mathematics-Mathematical Modeling, Proceedings dedicated to the sixtieth birthday of academican L. Iliev, Sofia, 1975. (Russian) 3. W. Lee. The Poincare series of generic 2 by 2 matrices, J. Appl. Math. & Computing 18 (2005), No. 1-2, 585 589. 4. M. Nagata. On the nilpotency of nil-algebras, J. Math. Soc. Japan 4 (1953), 296 301. 5. C. Procesi. The invariant theory of n n matrices, Adv. in Math. 19 (1976), 306 381. 6. Y. P. Razmyslov. Trace identities of full matrix algebras over a field of characteristic zero, Izv. Akad. Nauk SSSR 38 (1974), 723 756; English transl., Math. USSR-Izv. 8 (1974), 727 760. Woo Lee received his BA from Sogang Univ and PH.D at Pennsylvania State Univ. under the supervision of Edward Formanek. Since 1998, he has been at Kwangju University. His research interests include noncommutative rings, braid groups and representations. Department of Information and Telecommunication, Kwangju University, Kwangju 503-703 South Korea. email : woolee@gwangju.ac.kr