The Standard Polynomial in Verbally Prime Algebras

Transcription

1 International Journal of Algebra, Vol 11, 017, no 4, HIKARI Ltd, wwwm-hikaricom The Standard Polynomial in Verbally Prime Algebras Geraldo de Assis Junior Departamento de Ciências Exatas e Tecnológicas Universidade Estadual de Santa Cruz Ilhéus, BA, Brasil Sérgio M Alves Departamento de Ciências Exatas e Tecnológicas Universidade Estadual de Santa Cruz Ilhéus, BA, Brasil Copyright c 017 Sérgio M Alves and Geraldo de Assis Junior This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Abstract In this paper we discuss the minimality of the degree of the standard polynomial as a polynomial identity for verbally prime algebras and its tensor products Mathematics Subject Classification: 16R10, 16R0, 16R40, 15A75 Keywords: PI Theory, Standard polynomial, Verbally prime algebra 1 Introduction Verbally prime algebras play a prominent role in the PI theory Recall that an algebra A is verbally prime if its T-ideal is prime in the class of all T-ideals in the free associative algebra Let V be an vector space over K of countable infinite dimension with basis {e 1, e, } The Grassmann algebra E of V is the associative algebra with K-basis consisting of 1 and all products of the form e i1 e i e im with i 1 < i < i m, m 1 and with multiplication induced

2 150 Sérgio M Alves and Geraldo de Assis Junior by e i = 0 and e i e j + e j e i = 0 When char K = p =, then obviously E is commutative and hence are not very interesting from the PI point of view Therefore, we restrict our attention the case p > We denote by M n (E) the algebra of n n matrices over E The algebra M( a,b (E) is ) the subalgebra of M a+b (E) that consists of the block matrices, where A M A B C D a (E 0 ), B M a b (E 1 ), C M b a (E 1 ) and D M b (E 0 ) If two algebras A and B satisfy the same polynomial identities we say that A is PI-equivalent to B and we denote by A B An important consequence of the Kemer s structure theory is the Tensor Product Theorem (TPT) Theorem 1 (TPT) Let chark = 0 Then we have M 1,1 (E) E E, M a,b (E) E M a+b (E) and M a,b (E) M c,d (E) M ac+bd,ad+bc (E) In [4] it has been proved that Kemer s tensor product theorem can not be transposed to fields of positive characteristic However we have the following multilinear version Let P (I) be the set of multilinear polynomials in the T -ideal I: Theorem ([6], theorem 5) Let chark = p Then we have P (T (M 1,1 (E))) = P (T (E E)), P (T (M a,b (E) E)) = P (T (M a+b (E))) and P (T (M a,b (E) M c,d (E))) = P (T (M ac+bd,ad+bc (E))) Next we will define the algebras the type S 1, S and S 3 Algebras of the type S 1, S and S 3 We denote by s k (x 1,, x k ) = σ S k ɛ(σ)x σ(1) x σ(k) the polynomial standard of degree k where ɛ(σ) is the sign of σ Note that if A satisfies s k with k minimal, then k must be even For if k were odd, then s k (x 1,, x k, 1) = s k 1 (x 1,, x k 1 ) and thus k would not be minimal The same remark holds for powers of standard identities We write µ(a) = n if A satisfies s n and no s k with k < n If A satisfies no standard identity we write µ(a) = We write in addition µ (A) = n if A satisfies a power of s n, and no power of s k with k < n Let A be an algebra Definition 1 We say that A is a S 1 n-algebra if A M n (K) and a S 1 - algebra if A is a S 1 n-algebra for some n

3 On the power of standard polynomial to M a,b (E) 151 Definition We say that A is a S n-algebra if A M n (E) and a S - algebra if A is a S n-algebra for some n Definition 3 We say that A is a S 3 ab-algebra if A M a,b (E) and a S 3 - algebra if A is a S 3 ab-algebra for some a and b Thus, acording to Kemer s Theory, all verbally prime PI-algebra over a field of characteristic zero are of the type S 1, S or S 3 The purpose of this paper is to study this classification for algebras over a field of characteristic positive and its tensor products 3 The S 1, S and S 3 algebras in char K = 0 The Amitsur-Levitzki theorem says that µ(m n (K)) = n Now if M n (K) satisfy a standard polynomial then satisfy a power itself Therefore µ(m n (K)) n Now let s show that if k < n then M n (K) does not satisfy s k Consider Q = s k (E 1, E 3,, E k,k+1, E k+1,k,, E 3, E 1 ) where the E ij are the matrices units Then the non-zero terms Q of the form E ii, and for i = 1 we will have only one non-zero term E 11 It follows that Q is an diagonal matrix with 1 in the upper left corner Then Q is not nilpotent, and therefore no power of s k satisfies M n (K) Thus we have the following µ(m n (K)) = µ (M n (K) = n Theorem 3 Let A be a S 1 -algebra, Then µ(a) = µ (A) and if A M n (K) we have µ(a) = n Proof Since A is a S 1 -algebra exists one positive integer n such that A M n (K) Therefore µ(a) = µ(m n (K)) and µ (A) = µ (M n (K)) From what has been stated above µ(a) = µ (A) = n Lemma 31 Let char K = 0 The Grassmann algebra E does not satisfy any standard polynomial, that is, µ(e) = Proof Let e 1,, e n we have that e i e j = e j e i Therefore e σ(1) e σ(n) = ɛ(σ)e 1 e n Follow that s n (e 1,, e n ) = ɛ(σ)e σ(1) e σ(n) = ɛ(σ)ɛ(σ)e 1 e n = n!e 1 e n σ S n σ S n The lemma below was proved in [1] for zero characteristic fields and extended in [5] for any characteristic In this article we will use the following isomorphisms M k (A) M k (K) A and M k (M 1,1 (E)) M k,k (E) ([4], lemma 1)

4 15 Sérgio M Alves and Geraldo de Assis Junior Lemma 3 [[5], lemma 1] On a field K of arbitrary characteristic we have: 1 M n (E) satisfies s k n for some k > 1, but does not satisfy s n nor s k n 1 for any k If a b Then M a,b (E) satisfies s k a for some k > 1 but does not satisfy s a or s k a 1 for any k Theorem 4 Let A be a S -algebra Then µ(a) = and if A M n (E) we have µ (A) = n Proof Since A is a S -algebra there is one positive integer n such that A M n (E) Therefore, µ(a) = µ(m n (E)) and µ (A) = µ (M n (E)) According to the lemma 3 we have µ (M n (E)) = n and since T (M n (E)) T (E) follows from the lemma 31 that µ(m n (E)) = Theorem 5 Let A be a S 3 -algebra Then µ(a) = and if A M a,b (E) we have µ (A) = max{a, b} Proof Since A is a S 3 -algebra there is positive integers a and b such that, A M a,b (E) However µ(a) = µ(m a,b (E)) e µ (A) = µ (M a,b (E)) According to the Lemma 3 we have µ (M a,b (E)) = max{a, b} And if d = min{a, b} we have P (T (M a,b (E))) P (T (M d,d (E))) = P (T (M d (E) E)) P (T (M d (E))) P (T (E)), These inclusions follow by natural immersions and equality follows from M d,d (E) M d (M 1,1 (E)) M d (K) M 1,1 (E) [M d (K) E] E M d (E) E The following lemma will be very useful in what follows Lemma 33 If A B and C D then A C B D Proof ([7],teorema 1) Theorem 6 If A and B are S 1 -algebras then A B is a S 1 -algebra Them µ(a B) = µ(a)µ(b), µ (A B) = µ (A)µ (B) and µ(a B) = µ (A B)

5 On the power of standard polynomial to M a,b (E) 153 Proof There are integers a and b such that A M a (K) and B M b (K) respectively Thus, by lemma 33 A B M a (K) M b (K) M ab (K) Therefore A B is a S 1 -algebra Them, µ(a) = µ (A) = a, µ(b) = µ (B) = b e µ(a B) = ab Theorem 7 If A is a S 1 -algebra and B is a S -algebra then A B is a S - algebra Them µ(a B) = and µ (A B) = µ (A)µ (B) = µ(a)µ (B) Proof There are integers a e b such that A M a (K) and B M b (E), respectively Thus, by lemma 33 we have A B M a (K) M b (E) On the other hand, M a (K) M b (E) M a (K) [M b (K) E] [M a (K) M b (K)] E M ab (K) E M ab (E) Therefore A B is a S -algebra Them µ(b) = µ(a B) =, µ(a) = µ (A) = a, µ (B) = b and µ (A B) = ab Theorem 8 If A is a S 1 -algebras and B is a S 3 -algebra then A B is a S 3 -algebra Them µ(a B) = and µ (A B) = µ (A)µ (B) = µ(a)µ (B) Proof We have A M a (K) e B M b,c (E) Thus, by lemma 33 A B M a (K) M b,c (E) On the other hand, M a (K) M b,c (E) M a (M b,c (E)) M ab,ac (E) Therefore A B is a S 3 -algebra Thus µ(b) = µ(a B) =, µ(a) = µ (A) = a, µ (B) = max{b, c} and µ (A B) = max{ab, ac} = a max{b, c} = µ (A)µ (B) Theorem 9 If A and B are S -algebras then A B is a S 3 -algebra Them µ(a B) = and µ (A B) = µ (A)µ (B) Proof Let A M a (E) and B M b (E) Thus by lemma 33 we have A B M a (E) M b (E) On the other hand, M a (E) M b (E) [M a (K) E] [M b (K) E] M ab (K) (E E) M ab (K) M 1,1 (E) M ab (M 1,1 (E)) M ab,ab (E) because E E M 1,1 (E) Then A B is a S 3 -algebra Thus µ(a B) =, µ (A) = a, µ (B) = b and µ (A B) = ab = µ (A)µ (B) Theorem 10 If A is a S -algebras and B is a S 3 -algebra then A B is a S -algebra Them µ(a B) = and if B M b,c (E) then µ (A B) = µ (A)(b + c)

6 154 Sérgio M Alves and Geraldo de Assis Junior Proof Let A M a (E) and B M b,c (E) Then by lemma 33 we have A B M a (E) M b,c (E) On the other hand, M a (E) M b,c (E) [M a (K) E] M b,c (E) M a (K) M b+c (E) M a (M b+c (E)) M a(b+c) (E) Therefore A B is a S -algebra Moreover µ(a) = µ(a B) =, µ (A) = a, and µ (A B) = a(b + c) = µ (A)(b + c) Theorem 11 If A and B are S 3 -algebras then A B is a S 3 -algebra Them µ(a B) = and µ (A B) > µ (A)µ (B) Proof If A and B are S 3 -algebras then there are integers a, b, c and d such that A M a,b (E) e B M c,d (E), respectively Thus by Kemer s Theorem A B M a,b (E) M c,d (E) M ad+bc,ac+db (E) Now A B is a S 3 -algebra Thus µ(a B) =, µ(a) = max{a, b}, µ (B) = max{c, d} e µ (A B) = max{ad + bc, ac + bd} Therefore, µ (A B) > µ (A)µ (B) 4 The S 1, S and S 3 algebras in char K = p > In this section we will transpose the results of the previous section into algebras over a infinite field characteristic field p 41 On the valour of µ (A) If A is a S 1 -algebra then there is n such that A M n (K) and therefore µ (A) = n because the theorem 3 is free of characteristic Besides that, if A is a S n or S 3 ab-algebra then µ (A) = n or µ (A) = max{a, b}, because the lemma 3 is free of characteristic For the tensor products we have the following Theorem 1 If A is a S 1 -algebra and B is a S -algebra, then A B is a S -algebra and µ (A B) = µ (A)µ (B) Proof Idem to the proof of the theorem 7 Theorem 13 If A is a S 1 -algebra and B is a S 3 -algebra then A B is a S 3 -algebra and µ (A B) = µ (A)µ (B) Proof Idem to the proof of the theorem 8 Theorem 14 Let A be a S -algebra and let B be a S 3 -algebra Then µ (A B) = µ (A)µ (B)

7 On the power of standard polynomial to M a,b (E) 155 Proof Since A is a S -algebra and B is a S 3 -algebra there are a, b and c such that A M a (E) and B M b,c (E), thus µ (A) = a and µ (B) = max{b, c} Now note that M a (E) M b,c (E) [M a (K) E] M b,c (E) [M a (K) M b,c (E)] E M ab,ac (E) E Let d = max{ab, ac}, according to ([5], Lemma 7), there exists integer t > 1 such that s t d satisfies M ab,ac (E) E, thus µ (M ab,ac (E) E) d On the other hand, since M ab,ac (E) E has an isomorphic copy of M d (K) follow that µ (M ab,ac (E) E) d Therefore µ (A B) = µ (M ab,ac (E) E) = max{ab, ac} = a max{b, c} = µ (A)µ (B) Theorem 15 If A and B are S -algebras then µ (A B) = µ (A)µ (B) Proof Since A and B are S -algebras there are a and b such that A M a (E) and B M b (E), thus µ (A) = a and µ (B) = b Now note that M a (E) M b (E) [M a (K) E] [M b (K) E] [M ab (K) E] E M ab (E) E Now, using ([5], Lemma 6) we obtain that there exists an integer t > 1 with t = t(a, b, p) such that s t ab satisfies M ab (E) E Thus µ (M ab (E) E) ab On the other hand, since M ab (E) E has an isomorphic copy of M ab (K), follows that µ (M ab (E) E) ab Therefore, µ (A B) = µ (M ab (E) E) = ab = µ (A)µ (B) Theorem 16 If A and B are S 3 -algebras then µ (A B) = µ (A)µ (B) Proof Since A and B are S 3 -algebras there are integers a, b, c and d such that A M a,b (E) and B M c,d (E) Without loss of generality, let us suppose that a b and c d Thus µ (A) = a and µ (B) = c According to ([4], lemma 1) s t ac satisfies M a,b (E) M c,c (E) for some t = t(a, b, p) Since T (M a,b (E) M c,c (E)) T (M a,b (E) M c,d (E)) then s t ac also satisfies M a,b (E) M c,d (E) On the other hand, since M a,b (E) M c,d (E) has an isomorphic copy of M a (K) M c (K) M ac (K) so it does not satisfy s t k for all t is k < ac Therefore, µ (A B) = µ (M a,b (E) M c,d (E)) = ac = µ (A)µ (B)

8 156 Sérgio M Alves and Geraldo de Assis Junior 4 On the valour of µ(a) In this section we study the valour of the µ(a) when A is an S 1, S and S 3 -algebra or tensor product it First, we observe that Theorem 3 is free of characteristic In [[], Lemma 5] the authors proved the following result Lemma 41 The standard polynomial s m is a polynomial identity of the E if and only if m p + 1 In [[3], Theorem 55] the authors proved the following result Theorem 17 If the algebra A satisfies polynomial identity s n, then the algebra M k (A) satisfies the polynomial identity s nk k +1 In [[7], Theorem 15] the authors proved the following result Theorem 18 If the algebra M k (A) satisfies the polynomial identity s n, then the algebra M k 1 (A) satisfies the polynomial identity s n Moreover, the algebra A satisfies the polynomial identity s n (k 1) Now we can prove the following: Theorem 19 If A is a S n-algebra then n + p 1 µ(a) pn + 1 Proof Since A is S -algebra, we see that A M n (E) We have that, µ(a) = µ(m n (E)) If the algebra M n (E) satisfies s m, by theorem 18, we obtain that the algebra E satisfies s m (n 1) By lemma 41, we have that m (n 1) p + 1 m n + p 1 Thus, we obtain that µ(a) (n+p 1) Now, since E satisfies s m with m = p + 1, by theorem 17, we have that M n (E) satisfies s mn n +1 We observe that Therefore µ(a) pn + 1 mn n + 1 = (p + 1)n n + 1 = pn + 1 This result will be the basis for the next theorems Theorem 0 Let A be a S 3 ab-algebra and d = min{a, b} Then d + p 1 µ(a) p(a + b) + 1

9 On the power of standard polynomial to M a,b (E) 157 Proof Since A M a,b (E) and µ(a) = µ(m a,b (E)), we obtain that P (T (M a,b (E))) P (T (M d,d (E))) P (T (M d (E) E)) P (T (M d (E))) Now, if s m T (M a,b (E)) then s m T (M d (E)), and by theorem 19, we obtain that µ(a) d+p 1 Them by theorem 17 s p(a+b) +1 satisfies M a,b (E), thus µ(a) p(a + b) + 1 Theorem 1 If A is a S a-algebra and B is a S 3 bc then (ab + ac) + p 1 µ(a B) p(ab + ac) + 1 Proof We know that M a (E) M b,c (E) M ab,ac (E) E By theorem P (T (M ab,ac (E) E)) = P (T (M ab+ac (E))) However µ(a B) = µ(m ab+ac (E)) Theorem If A is a S a-algebra and B is a S b -algebra then ab + p 1 µ(a) (ab) p Proof Since A M a (E) and B M b (E) then A B M a (E) M b (E) M ab (K) [E E] Since T (M 1,1 (E)) T (E E) then: T (M a (E) M b (E)) T (M ab (K) M 1,1 (E)) = T (M ab (M 1,1 (E))) = T (M ab,ab (E)) follows that M a (E) M b (E) satisfies s p(ab+ab) +1 Since p(ab + ab) + 1 is odd then M a (E) M b (E) satisfies s 4(ab) and therefore µ(a B) (ab) On the other hand, M a (E) M b (E) M ab (E) E contains an isomorphic copy of M ab (E) and therefore µ(a B) ab+p 1 Theorem 3 If A is a S 3 ab-algebra, B is a S 3 cd-algebra and d = min{ac + bd, ad + bc} then d + p 1 µ(a B) p(ac + bd + ad + bc) + 1 Proof By theorem we have P (T (M a,b (E) M c,d (E))) = P (T (M ac+bd,ad+bc (E))), thus µ(a B) = µ(m ac+bd,ad+bc (E)), and by theorem 0 follows the result Acknowledgements Sérgio M Alves is partially supported by PROPP/UESC

10 158 Sérgio M Alves and Geraldo de Assis Junior References [1] A Berele, Classification theorems for verbally semiprime algebras, Commun Algebra, 1 (1993), no 5, [] A Giambruno, P Koshlukov, On the identities of the Grassmann algebras in characteristic p > 0, Israel Journal of Mathematics, 1 (001), [3] M Domokos, Eulerian polynomial identities and algebras satisfying a standard identity, J Algebra, 169 (1994), [4] SM Alves, PI (non) equivalence and Gelfand-Kirillov dimension in positive characteristic, Rendiconti del Circolo Matematico di Palermo, 58 (009), no 1, [5] SM Alves, P Koshlukov, Polynomial identities of algebras in positive characteristic, J Algebra, 305 (006), no, [6] SS Azevedo, M Fidelis, P Koshlukov, Tensor product theorems in positive characteristic, J Algebra, 76 (004), no, [7] U Leron, A Vapne, Polynomial identities of related rings, Israel Journal of Mathematics, 8 (1970), Received: February 1, 017; Published: May 1, 017