On the centre of the generic algebra of M 1,1

Transcription

1 On the centre of the generic algebra of M 1,1 Thiago Castilho de Mello University of Campinas PhD grant from CNPq, Brazil

2 F is a field of characteristic zero; F X = F x 1, x 2,... is the free associative algebra, freely generated by the set X = {x 1, x 2,... }; E = E 0 E 1 is the Grassmann algebra of an infinite countable dimensional vector space over F. If A is a PI-algebra, we denote by T (A), its ideal of polynomial identities and we use the following notations for its relatively free algebras F (A) = F x 1, x 2,... T (A) F k (A) = F x 1,..., x k T (A) F x 1,..., x k

3 A T-ideal P of F X is T-prime if for any T-ideals I and J of F X such that IJ P, we have I P or J P (or both inclusions). A PI-algebra A is called a T-prime algebra if T (A) is T-prime. Kemer has proved that if P is a nontrivial T-prime T-ideal then P is the ideal of identities of one of the following algebras: (i) M n (F ) (ii) M n (E) (iii) M a,b where M a,b = M a,b (E) and for any 2-graded algebra A = A 0 A 1, M a,b (A) is defined by M a,b (A) = ( Ma (A 0 ) M a b (A 1 ) M b a (A 1 ) M b (A 0 ) )

4 Berele [Generic verbally prime algebras and their GK dimensions, Commun. Algebra 21, (1993)] has constructed models for the relatively free algebras of the T-prime algebras, using matrices over the free supercommutative algebra. To construct the free supercommutative algebra, we consider the free associative algebra F X Y and we induce on it a Z 2 -grading by setting deg x = 0 and deg y = 1 if x X and y Y. Finally, we ensure supercommutativity by modding out by all ab ( 1) deg a deg b ba, for all a, b X Y. The resulting algebra is the free supercommutative algebra, denoted by F [X ; Y ].

5 Now we consider X = {X (r) ij ; i, j {1,..., n}, r N} and Y = {Y (r) ij ; i, j {1,..., n}, r N}. If a + b = n and r N, we construct the generic matrices over F [X ; Y ]: A r := (X (r) ij ) B r := (X (r) ij C r := ( (X (r) + Y (r) ij ) ij ) a a (Y (r) ij ij ) b a (X (r) ij (Y (r) ) a b ) b b )

6 And we have Theorem (Berele) For any k 2, the following isomorphisms hold: F k (M n (F )) = F [A1,..., A k ] F k (M n (E)) = F [B1,..., B k ] F k (M a,b ) = F [C1,..., C k ] In the same paper, Berele proved that the centre of F k (M a,b ) is the sum of the field and a nilpotent ideal of the centre. He also asks whether or not such centre contains non-scalar elements.

7 In this talk, we work on the generic algebra of rank 2 of M 1,1, F [C 1, C 2 ] = F 2 (M 1,1 ). For this algebra we will: Answer Berele s question; Give a description of the centre of such algebra; Find a basis for its ideal of polynomial identities.

8 Popov [Identities of the tensor square of a Grassmann algebra. Algebra and Logic 21 (4) (1982)] has proved that the T-ideal of polynomial identities of M 1,1 is generated by the polynomials: [[x 1, x 2 ] 2, x 1 ] = 0 and [[[x 1, x 2 ], [x 3, x 4 ]], x 5 ] = 0 Since F 2 (M 1,1 ) is the free algebra of rank 2 in the variety of the identities of M 1,1 it also satisfies the identities above. In particular, the element [C 1, C 2 ] 2 is central in F [C 1, C 2 ]. In order to simplify the notations, we consider ( ) ( X1 Y C 1 = 1 X2 Y Y 1 X 1 C 2 = 2 Y 2 X 2 )

9 Since where ( [C 1, C 2 ] 2 2h1 h = 4 2h 2 2h 3 2h 1 + h 4 ) h 1 = Y 1 Y 2 Y 1 Y 2 h 2 = Y 1 Y 2 (Y 1 (X 2 X 2) Y 2 (X 1 X 1)) h 3 = Y 1 Y 2 (Y 1(X 2 X 2) Y 2 (X 1 X 1)) h 4 = (Y 1 (X 2 X 2) Y 2 (X 1 X 1))(Y 1 (X 2 X 2) Y 2 (X 1 X 1)), we have that [C 1, C 2 ] 2 is a non-scalar central element of F [C 1, C 2 ], answering positively Berele s question for F 2 (M 1,1 ).

10 Proposition ( ) a b Let A = F [C c d 1, C 2 ]. If A is central in F [C 1, C 2 ] there exist f 1 and f 2 F [X ] such that b = f 2 h 2 c = f 2 h 3 d a = f 1 h 1 + f 2 h 4 As a consequence, there exist f 1 and f 2 F [X ] such that ( ) ( ) h2 A = ai 2 + f 1 + f 0 h 2 1 h 3 h 4

11 It is well known that any element of F x 1, x 2 can be written as a linear combination of monomials of the type x1 n x2 m u k ukr r where u i L(x 1, x 2 ) are left-normed commutators, that is commutators of the form [x i1,..., x ir ] = [[x i1,..., x ir 1 ], x ir ] Lemma Let f (x 1, x 2 ) L(x 1, x 2 ) be a left-normed commutator. If f 0 then f (C 1, C 2 ) is not central in F [C 1, C 2 ].

12 We say that an element A of F [C 1, C 2 ] is central ideal if A is central in F [C 1, C 2 ] and the same holds for AB, for any B F [C 1, C 2 ]. Lemma Let f 1, f 2, f 3 L(x 1, x 2 ) be left-normed commutators, g = f 1 f 2 and h = f 1 f 2 f 3. Then g(c 1, C 2 ) is a nonzero central ideal element and h(c 1, C 2 ) = 0. Proposition Let f (C 1, C 2 ) = C1 nc 2 muk ukr r F [C 1, C 2 ], with u i L[C 1, C 2 ] left-normed commutators. Then, f is central if and only if k k r 2.

13 Lemma Let u 1,...,u r L(x 1, x 2 ) be left-normed commutators with the same degrees in x 1 and in x 2, and let f (C 1, C 2 ) = α i u i (C 1, C 2 ) F [C 1, C 2 ].Then f (C 1, C 2 ) is central i if and only if it is central ideal, and if and only if α i = 0 i

14 Proposition Let f (C 1, C 2 ) F [C 1, C 2 ]. We know that ( f (C 1, C 2 ) = αi 2 + j C n j 1 C m j 2 ( i α (j) i u (j) i ) ) + g(c 1, C 2 ) with g(c 1, C 2 ) = C n 1 C m 2 u 1... u k, k 2 and u i left normed commutators. Then, f (C 1, C 2 ) is central in F [C 1, C 2 ] if and only if, for all j, = 0, where Λ is the set of i such that the i Λ corresponding u i have the same degrees in x 1 and in x 2. Corollary α (j) i Z(F [C 1, C 2 ]) = F I Where I is a nilpotent ideal of F [C 1, C 2 ].

15 With the previous results, we are able to determine the polynomial identities of F [C 1, C 2 ]. Proposition Let f (C 1, C 2 ) = C1 nc 2 muk ukr r F [C 1, C 2 ], with u i L[C 1, C 2 ] left-normed commutators. Then, f (C 1, C 2 ) = 0 if and only if k k r 3. Theorem The algebra F [C 1, C 2 ] satisfies the identities [[x 1, x 2 ][x 3, x 4 ], x 5 ] and [x 1, x 2 ][x 3, x 4 ][x 5, x 6 ]

16 As charf = 0, we know that all polynomial identities of F [C 1, C 2 ] follows from its proper multilinear polynomial identities. The vector space of all proper polynomials of degree n is an S n -module, denoted by Γ n. It induces an S n -module structure on Γ n (M 1,1 ) = Γ n T (M 1,1 ) Γ n In the same paper mentioned before, Popov describes the S n -module structure of Γ n (M 1,1 ).

17 He showed that where M (s) p Γ n (M 1,1 ) = ( p+s=n M (s) p ) p+q+s=n p,q 2 M p,q (s) and M p,q (s) are the irreducible S n -submodules of (x 1,..., x p ) and Γ n (M 1,1 ) generated, respectively by ϕ (s) p ϕ (s) p,q(x 1,..., x p ), given by:

18 ϕ (s) p = ϕ (s) p,q= ±[x σ(1), x σ(2) ] [x σ(p), x (s) 1 ], p 1 (2) σ S p ±[x σ(1), x σ(2) ] [x σ(p 1), x σ(p), x (s) 1 ], p 0 (2) σ S p ±[x τ(1), x τ(2) ] [x τ(q 1), x τ(q) ]ϕ (s) p, q 0 (2) τ S q ±[x σ(1), x σ(2) ] [x σ(p 1), x σ(p) ]ϕ (s) q, q 1, p 0 (2) σ S p ±[x τ(1), x τ(2) ] [x σ(1), x σ(2) ] [x σ(p), x (s) 1, x τ(q)], q p 1 (2) τ S q σ S p

19 Now, we observe that for any s: ϕ (s) p is a consequence of [x 1, x 2 ][x 3, x 4 ][x 5, x 6 ], if p 5; ϕ (s) p,q is a consequence of [x 1, x 2 ][x 3, x 4 ][x 5, x 6 ], if p + q 5; ϕ (s) 4 is a consequence of ϕ (0) 4 ; ϕ (0) 4 = 2s 4. But we have: Theorem The polynomial s 4 is an identity for F [C 1, C 2 ]

20 Since F [C 1, C 2 ] satisfies all identities of M 1,1, we obtain that Γ n (F [C 1, C 2 ]) is a factor module of Γ n (M 1,1 ). Then, we remove from the decomposition of Γ n (M 1,1 ) the submodules generated by identities of F [C 1, C 2 ], and we obtain: and we have Lemma Γ n (F [C 1, C 2 ]) = M (n 2) 2 M (n 3) 3 M (n 4) 2,2 For every n, the polynomials ϕ (n 2) 2, ϕ (n 3) 3 and ϕ (n 4) 2,2 are not polynomial identities for F [C 1, C 2 ].

21 Finally, Corollary The T-ideal of polynomial identities of F 2 (M 1,1 ) is generated by the identities s 4, [[x 1, x 2 ][x 3, x 4 ], x 5 ],and [x 1, x 2 ][x 3, x 4 ][x 5, x 6 ].

22 Thank you!