Gröbner-Shirshov Bases for Free Product of Algebras and beyond

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1 Southeast Asian Bulletin of Mathematics (2006) 30: Southeast Asian Bulletin of Mathematics c SEAMS Gröbner-Shirshov Bases for Free Product of Algebras and beyond Yuqun Chen School of Mathematical Sciences, South China Normal University, Guangzhou , China yqchen@scnu.edu.cn Jianjun Qiu School of Mathematical Sciences, South China Normal University, Guangzhou , China jianjunqiu@126.com AMS Mathematics Subject Classification (2000): 16S15, 13P10 Abstract. Let K be a commutative ring with identity 1. In this paper, the Gröbner- Shirshov bases for the free product and tensor product of associative K-algebras (not necessarily with identity) were given, and the Gröbner-Shirshov basis for the free product of Lie K-algebras is also obtained. As an application, we prove the Normal Form Theorem for free product of groups. Keywords: Algebras; Gröbner-Shirshov bases; Tensor products; Free products; Normal form. 1. Preliminaries Let k be a field, k X the free associative algebra over k generated by X, the free monoid generated by X, where the empty word is the identity which is denoted by 1. For a word w X, we denote the length of w by deg(w). Let X be a well ordered set. If f k X having the leading word f, then we say that f is monic when f has coefficient 1. Definition 1.1. ([13], see also [2], [3]) Let f and g be two monic polynomials in k X. Then, there are two kinds of compositions: (1) If w is a word such that w = fb = aḡ for some a, b X with deg( f)+deg(ḡ) >deg(w), then the polynomial (f, g) w = fb ag is called the The research is supported by the National Natural Science Foundation of China (Grant No ) and the Natural Science Foundation of Guangdong Province (Grant No ; ).

2 812 Y.Q. Chen and J.J. Qiu intersection composition of f and g with respect to w. (2) If w = f = aḡb for some a, b X, then the polynomial (f, g) w = f agb is called the inclusion composition of f and g with respect to w. Definition 1.2. ([2], [3], cf. [13]) Let S k X with each s S monic. Then the composition (f, g) w is called trivial modulo S if (f, g) w = α i a i s i b i, where each α i k, a i, b i X and a i s i b i < w. If this is the case, then we write (f, g) w 0 mod(s, w). In general, for p, q k X, we write p q mod(s, w) which means that p q = α i a i s i b i, where each α i k, a i, b i X and a i s i b i < w. Definition 1.3. ([2], [3], cf. [13]) We call the set S endowed with the well order < a Gröbner-Shirshov set (basis) in k X if any composition of polynomials in S is trivial relative to S. A well order < on X is monomial if it is compatible with the multiplication of words, that is, for u, v X, we have u > v w 1 uw 2 > w 1 vw 2, for all w 1, w 2 X. The following lemma was proved by Shirshov [13] for free Lie algebras (with deg-lex ordering) in 1962 (see also Bokut [2]). In 1976, Bokut [3] specialized the approach of Shirshov to associative algebras (see also Bergman [1]). For commutative polynomials, this lemma is known as the Buchberger s Theorem (see [7] and [8]). Lemma 1.4. (Composition-Diamond Lemma) Let k be a field, A = k X S = k X /Id(S) and < a monomial order on X, where Id(S) is the ideal of k X generated by S. Then the following statements are equivalent: (1) S is a Gröbner-Shirshov basis. (2) f Id(S) f = a sb for some s S and a, b X. (3) Red(S) = {u X u a sb, s S, a, b X } is a basis of the algebra A = k X S.

3 Gröbner-Shirshov Bases for Free Product of Algebras and beyond 813 In this paper, otherwise stated, K is a commutative ring with identity 1. Except for 6, all algebras are associative algebras over a commutative ring K. It is easy to check that if we replace field k by commutative ring K, then all concepts and results in the above statements are true, in particular, we have the Composition-Diamond Lemma of the algebra K X over the commutative ring K. This paper consists of 6 sections. In 2, 4 ( 3, 5), by using the Gröbner- Shirshov basis, we give some characterizations of the free (tensor) product of K-algebras (with identity and not necessarily with identity, respectively) which are free K-modules. In 6, we give some characterizations of the free product of Lie K-algebras which are free K-modules by using Gröbner-Shirshov basis. Lemma 1.5. Let A be a K-algebra with identity 1. Suppose A is a free K-module with a linearly ordered basis {a 0 } {a i i I}, where a 0 = 1. For i, j I, we let a i a j = n I {0} αn ij a n, α n ij K and denote {a ia j } = n I {0} αn ij a n. Then S = {a i a j {a i a j } i, j I} is a Gröbner-Shirshov basis of A with generators {a i i I}. Proof. Since Red(S) = {1} {a i i I}, by the Composition-Diamond Lemma, S is a Gröbner-Shirshov basis. Remark 1.6. If K is a field and A a K-algebra with identity 1, then A is a K-free module and there exists a K-basis L of A such that 1 L. Most part of this paper is a mathematical folklore (see [9], [4]). 2. Free Product of Associative Algebras with Identity In this section and the next section, each K-algebra is with identity 1 and each K-algebra homomorphism preserves 1. Definition 2.1. Let A, A 1 and A 2 be K-algebras and e i : A i A, the K-algebra homomorphism, i = 1, 2. Then (A, (e 1, e 2 )) is called a free product of A 1 and A 2 if for any K-algebra B and K-algebra homomorphism f i : A i B, i = 1, 2, there exists a unique K-algebra homomorphism ϕ : A B such that the following diagram commutes, i.e., ϕe i = f i, i = 1, 2. f i B ϕ e i A A i

4 814 Y.Q. Chen and J.J. Qiu Theorem 2.2. The free product of A 1 and A 2, if it exists, is uniquely determined up to isomorphism. In this case, we denote the free product of A 1 and A 2 by A 1 A 2. The following theorem shows that, for any K-algebras A 1 and A 2, the free product A 1 A 2 exists. Theorem 2.3. Let A i = K X i S i, i = 1, 2. Then A 1 A 2 = K X 1 X 2 S 1 S 2. Proof. Let A = K X 1 X 2 S 1 S 2. Define e i : A i A, e i (x i ) = x i, x i X i, i = 1, 2. Then e i can be uniquely extended to a K-algebra homomorphism, i = 1, 2. For any K-algebra B and K-algebra homomorphism f i : A i A, i = 1, 2, define ϕ : A B, ϕ(x) = f i (x) if x X i. It is easy to see that ϕ is a K-algebra homomorphism, f i = ϕe i, i = 1, 2 and such ϕ is unique. Theorem 2.4. Let A i be associative K-algebra with identity 1 and L i = T i {1} be linearly ordered K-basis for A i, i=1,2, where T 1 = {a i i I 1 }, a 0 = 1 and T 2 = {b h h I 2 }, b 0 = 1. For any a i, a j T 1, b h, b k T 2, let a i a j = n I 1 {0} α n ija n, b h b k = m I 2 {0} γ m hkb m where α n ij, γm hk K, and let M 1 = { a i a j α n ij a n i, j I 1 } and M 2 = { b h b k γ m hk b m h, k I 2 }. Now define an order on T 1 T 2 by a i < b h for any i I 1, h I 2 and the words in (T 1 T 2 ) in this alphabet are given in the deg-lex way. Then the following statements hold. (a) M 1 M 2 is a Gröbner-Shirshov basis in the algebra K T 1 T 2. (b) A 1 A 2 = K T 1 T 2 M 1 M 2. (c) Let Ω be the set which consists of u = c 1 c 2 c t, t 0 such that (1) if t = 0, then u = 1; (2) if t = 1, then u T 1 T 2 ; (3) if t > 1, then c i T 1 T 2, and successive c i and c i+1 are not in the same set T 1 or T 2. Then A 1 A 2 is a free K-module with basis Ω. Proof. (a) Let a ij = a i a j αij n a n and b hk = b h b k γhk m b m. The only types of compositions are (a ij, a jk ) aia ja k and (b hk, b kq ) bh b k b q. Since (a i a j )a p = n α n ij(a n a p ) = n α n ij( s αjpa s s ) = αijα n jpa s s, n,s

5 Gröbner-Shirshov Bases for Free Product of Algebras and beyond 815 a i (a j a p ) = n α n jp(a i a n ) = n α n jp( s α s ina s ) = n,s α n jpα s ina s, and (a i a j )a p = a i (a j a p ), we have αijα n np s = n n α n jpα s in for every s I 1. Thus (a ij, a jk ) aia ja k = a ij a k a i a jk = (a i a j n α n ija n )a k a i (a j a k n α n jka n ) = n α n ij(a n a k ) + n α n jk(a i a n ) s = 0 ( n α n ijα s nk n α n jkα s in)a s mod(m 1 M 2, a i a j a k ) and similarly, (b hk, b kq ) bh b k b q 0 mod(m 1 M 2, b h b k b q ). Hence M 1 M 2 is a Gröbner-Shirshov basis. This shows (a). (b) By Lemma 1.4 and Theorem 2.4. (c) This part follows from the Composition-Diamond Lemma. Now we turn to the free product of groups. Definition 2.5. ([10]) Let A, A 1 and A 2 be groups and e i : A i A group homomorphism, i = 1, 2. Then (A, (e 1, e 2 )) is called a free product of A 1 and A 2 if, for any group B and group homomorphism f i : A i B, i = 1, 2, there exists a unique group homomorphism ϕ : A B such that the following diagram commutes, i.e., ϕe i = f i, i = 1, 2. f i B ϕ e i A A i Theorem 2.6. ([10]) Let A i = gp X i S i, i = 1, 2. Then the free product of the groups A 1 and A 2 is A 1 A 2 = gp X 1 X 2 S 1 S 2.

6 816 Y.Q. Chen and J.J. Qiu Definition 2.7. ([10]) Let A 1 and A 2 be groups and A 1 A 2 the free product of A 1 and A 2. Then a reduced sequence (or normal form) is a sequence g 1, g 2,, g n, n 0, of elements of A 1 A 2 such that each g i 1, g i A 1 A 2 and successive g i, g i+1 are not in the same factor. (We allow n = 0 for the empty sequence.) The following lemma follows from Composition-Diamond Lemma. Lemma 2.8. Let K be a commutative ring, M = mon X S the momoid generated by X with relation S and < a monomial order on X. If S K X is a Gröbner- Shirshov basis, then Red(S) = {u X u a sb, s S, a, b X } is a normal form for M. Let A i be group with identity 1 and A i = T i {1}, i = 1, 2. Then T i {1} is a K-basis for the algebra K T i M i where M i = {a r a t {a r a t } a r, a t T i }. M 1 M 2 is obviously a Gröbner-Shirshov basis of the algebra K T 1 T 2. By using Theorem 2.3 and Lemma 2.8, we can easily obtain the following corollary. Corollary 2.9. (The Normal Form Theorem for Free Products, [10] Theorem 4.1.2) Let A 1, A 2 be groups. Then, each element w of A 1 A 2, the free product of groups A 1 and A 2, can be uniquely expressed as w = g 1 g 2 g n, where n 0 and g 1, g 2,, g n is a reduced sequence. 3. Tensor Product of Associative Algebras with Identity Definition 3.1. Let A, A 1 and A 2 be K-algebras and e i : A i A K-algebra homomorphism with e 1 (a 1 )e 2 (a 2 ) = e 2 (a 2 )e 1 (a 1 ), a 1 A 1, a 2 A 2. Then (A, (e 1, e 2 )) is called a tensor product of A 1 and A 2 if, for any K-algebra B and K-algebra homomorphism f i : A i B with f 1 (a 1 )f 2 (a 2 ) = f 2 (a 2 )f 1 (a 1 ), a i A i, i = 1, 2, there exists a unique K-algebra homomorphism ϕ : A B such that the following diagram commutes, i.e., ϕe i = f i, i = 1, 2. f i B ϕ e i A A i

7 Gröbner-Shirshov Bases for Free Product of Algebras and beyond 817 Theorem 3.2. The tensor product of A 1 and A 2, if it exists, is uniquely determined up to isomorphism. In this case, we denote the tensor product of A 1 and A 2 by A 1 A 2. The following theorem shows that the tensor product A 1 A 2 exists for any K-algebras A 1 and A 2, and gives a relationship between free product and tensor product. Theorem 3.3. Let A i = K X i S i, i = 1, 2 and S 3 = { x 1 x 2 x 2 x 1 x i X i, i = 1, 2}. Then A 1 A 2 = K X 1 X 2 S 1 S 2 S 3. It follows that A 1 A 2 = (A 1 A 2 )/Id(S 3 ). Proof. Let A = K X 1 X 2 S 1 S 2 S 3. Define e i : A i A, e i (x i ) = x i, x i X i, i = 1, 2. For any K-algebra B and K-algebra homomorphism f i : A i A, i = 1, 2 with f 1 (x 1 )f 2 (x 2 ) = f 2 (x 2 )f 1 (x 1 ), x i X i, i = 1, 2, define ϕ : A B, ϕ(x) = f i (x), if x X i. Since ϕ(x 1 x 2 x 2 x 1 ) = f 1 (x 1 )f 2 (x 2 ) f 2 (x 2 )f 1 (x 1 ) = 0, we can see that ϕ is a K-algebra homomorphism. Clearly, f i = ϕe i, i = 1, 2 and ϕ is unique. Theorem 3.4. Let A i be K-algebra with identity 1 and free K-module with linearly ordered K-basis L i = T i {1}, i = 1, 2, where T 1 = {a i i I 1 }, and T 2 = {b j j I 2 }. Let M 1 = { a i a j {a i a j } i, j I 1 }, M 2 = { b h b k {b h b k } h, k I 2 } and M 3 = {b h a i a i b h i I 1, h I 2 }. We order T 1 T 2 by a i < b h for any i I 1, h I 2 and the words in (T 1 T 2 ) in this alphabet are in the deg-lex way. Then the following statements hold. (a) M 1 M 2 M 3 is a Gröbner-Shirshov basis in the algebra K T 1 T 2. (b) A 1 A 2 = K T 1 T 2 M 1 M 2 M 3. (c) Let T 3 = {a i b h i I 1, h I 2 }. Then A 1 A 2 is a free K-module with basis {1} T 1 T 2 T 3. Proof. (a) Let l hi = b h a i a i b h. Then the possible kinds of compositions are (a ij, a jk ) aia ja k, (b hk, b kq ) bh b k b q, (b hk, l ki ) bh b k a i and (l hi, a ij ) bh a ia j. Similar to Theorem 2.4, we have (a ij, a jk ) aia ja k 0 and (b hk, b kq ) bh b k b q 0. Since (b hk, l ki ) bh b k a i = (b h b k {b h b k })a i b h (b k a i a i b k ) = {b h b k }a i + b h a i b k {b h b k }a i a i b h b k {b h b k }a i a i {b h b k } {b h b k }a i {b h b k }a i = 0

8 818 Y.Q. Chen and J.J. Qiu and similarly (l hi, a ij ) bh a ia j 0, we see that M 1 M 2 M 3 is a Gröbner-Shirshov basis. (b) By Lemma 1.4 and Theorem 3.3. (c) This part follows the Composition-Diamond Lemma. 4. Free Product of Associative Algebras (Not Necessarily with 1) The K-span of all nonempty words in X is a free associative algebra without identity and denoted by K X +. It is clear that, for every algebra A (not necessarily with 1), we have A = K X + /I = K X + S for some X and S, where I = Id(S) is the ideal of K X + generated by S. Let X = {x i i I} be a linearly ordered set and > a monomial order on X +, the semigroup of nonempty words on X. Let f, g K X + be monic polynomials. If w = f = aḡb for some a, b X, then we define the inclusion composition (f, g) w = f agb and if w = fa = bḡ and deg( f) + deg(ḡ) > deg(w) for some a, b X +, we define the intersection composition (f, g) w = fa bg as in 1. Definition 4.1. ([2], [3], cf. [13]) A monic subset S K X + is called a Gröbner-Shirshov basis if all compositions (f, g) w are trivial modulo S, where f, g S. The proof of the following lemma is essentially the same as in [3]. Lemma 4.2. ([2], [3], cf. [13], Composition-Diamond Lemma) Let S be monic subset of K X + and > a monomial order on X +. Then the following statements are equivalent: (1) S is a Gröbner-Shirshov basis. (2) f Id(S) f = a sb for some s S and a, b X. (3) Red(S) = {u X + u a sb, s S, a, b X } is a K-basis of the algebra A = K X + S. Lemma 4.3. Let A be K-algebra (not necessarily with identity) and A a free K-module with a linearly ordered basis {a i i I}. For i, j I, we let a i a j = α k ij a k, α ij K and {a i a j } = α k ij a k. Then S = {a i a j {a i a j } i, j I} is a Gröbner-Shirshov basis of A with generators {a i i I}.

9 Gröbner-Shirshov Bases for Free Product of Algebras and beyond 819 Proof. The proof is similar to the proof of Theorem 2.4. Definition 4.4. Let A, A 1 and A 2 be K-algebras and e i : A i A K-algebra homomorphism, i = 1, 2. Then (A, (e 1, e 2 )) is called a -free product of A 1 and A 2 if, for any K-algebra B and K-algebra homomorphism f i : A i B, i = 1, 2, there exists a unique K-algebra homomorphism ϕ : A B such that the following diagram commutes, i.e., ϕe i = f i, i = 1, 2. f i B ϕ e i A A i Theorem 4.5. The -free product of A 1 and A 2, if it exists, is uniquely determined up to isomorphism. If this is the case, then we denote by A 1ˆ A 2 the -free product of A 1 and A 2. The following theorem shows that the -free product A 1ˆ A 2 exists for any K-algebras A 1 and A 2. The proof is similar to the in Theorem 2.3. Theorem 4.6. Let A i = K X i+ S i, i = 1, 2. Then A 1ˆ A 2 = K (X 1 X 2 ) + S 1 S 2. Theorem 4.7. Let A i be K-algebra and free K-module with linearly ordered basis T i, i = 1, 2, where T 1 = {a i i I 1 } and T 2 = {b h h I 2 }. Let M 1 = { a i a j α n ij a n i, j I 1 } and M 2 = { b h b k γ m hk b m h, k I 2 }. Now we order T 1 T 2 by a i < b h for any i I 1, h I 2 and the words in (T 1 T 2 ) + in this alphabet are in the deg-lex way. Then the following statements hold. (a) M 1 M 2 is a Gröbner-Shirshov basis in the algebra K (T 1 T 2 ) +. (b) A 1ˆ A 2 = K (T 1 T 2 ) + M 1 M 2. (c) Let Ω be the set which consists of u = c 1 c 2 c t, t 1 such that (1) if t = 1, then u T 1 T 2 ; (2) if t > 1, then c i T 1 T 2, and successive c i and c i+1 are not in the same set T 1 or T 2. Then A 1ˆ A 2 is a free K-module with basis Ω.

10 820 Y.Q. Chen and J.J. Qiu Proof. Similar to the Theorem 2.4. Remark 4.8. Assume that A 1 and A 2 are K-algebras with identities 1 1 and 1 2, respectively. Then A 1 A 2 and A 1ˆ A 2 are different. The next theorem displays a relationship between them. Theorem 4.9. Let K be a field, A 1 and A 2 the K-algebras with identities 1 1 and 1 2, respectively. Then A 1 A 2 = (A1ˆ A 2 )/Id( ). Proof. Let T 1 = {a i i I 1 } and T 2 = {b h h I 2 } be linearly ordered bases of A 1 and A 2, respectively, such that 1 i T i, i = 1, 2. We order T 1 T 2 by a i < b h for any i I 1, h I 2 and (T 1 T 2 ) + by the deg-lex way in this alphabet. Let M 1 = {a i a j {a i a j } i, j I 1 }, M 2 = {b h b l {b h b l } h, l I 2 } and M 3 = { , b h 1 1 b h, 1 1 b h b h h I 2 }. Then, we can easily to check that M 1 M 2 M 3 is a Gröbner-Shirshov basis of the algebra K (T 1 T 2 ) +. Clearly, K (T 1 T 2 ) + M 1 M 2, = K (T 1 T 2 ) + M 1 M 2 M 3. Now, by the Composition-Diamond Lemma, Theorem 2.4 and Theorem 4.7, we have A 1 A 2 = (A1ˆ A 2 )/Id( ). 5. Tensor Product of Associative Algebras (Not Necessarily with 1) Definition 5.1. Let A, A 1 and A 2 be K-algebras (not necessarily with 1 ) and e i : A i A, i = 1, 2, the K-algebra homomorphism with e 1 (a 1 )e 2 (a 2 ) = e 2 (a 2 )e 1 (a 1 ), a 1 A 1, a 2 A 2. Then (A, (e 1, e 2 )) is called a -tensor product of A 1 and A 2 if, for any K-algebra B and K-algebra homomorphism f i : A i B with f 1 (a 1 )f 2 (a 2 ) = f 2 (a 2 )f 1 (a 1 ), a i A i, i = 1, 2, there exists a unique K- algebra homomorphism ϕ : A B such that the following diagram commutes, i.e., ϕe i = f i, i = 1, 2. f i B ϕ e i A A i

11 Gröbner-Shirshov Bases for Free Product of Algebras and beyond 821 Theorem 5.2. The -tensor product of A 1 and A 2, if it exists, is uniquely determined up to isomorphism. In this case, we denote the -tensor product of A 1 and A 2 by A 1 A 2. The proofs of the following theorems are similar to those in the case of the K-algebra with identity (see Theorem 3.3 and Theorem 3.4). We hence omit the details. Theorem 5.3. Let A i = K X i+ S i, i = 1, 2 and S 3 = { x 1 x 2 x 2 x 1 x i X i, i = 1, 2}. Then A 1 A 2 = K (X 1 X 2 ) + S 1 S 2 S 3.Hence, it follows that A 1 A 2 = (A 1ˆ A 2 )/Id(S 3 ). Theorem 5.4. Let A i be K-algebra and free K-module with linearly ordered basis T i, i = 1, 2, where T 1 = {a i i I 1 } and T 2 = {b h h I 2 }. Let M 1 = { a i a j α n ij a n i, j I 1 }, M 2 = { b h b k γ m hk b m h, k I 2 } and M 3 = {b h a i a i b h i I 1, h I 2 }. We order T 1 T 2 by a i < b h for any i I 1, h I 2 and words in (T 1 T 2 ) + in this alphabet in the deg-lex way. Then the following statements hold. (a) M 1 M 2 M 3 is a Gröbner-Shirshov basis in the algebra K (T 1 T 2 ) +. (b) A 1 A 2 = K (T 1 T 2 ) + M 1 M 2 M 3. (c) If T 3 = {a i b h i I 1, h I 2 }, then A 1 A 2 is a free K-module with basis T 1 T 2 T 3. Definition 5.5. Let A i = K X i+ S i, i = 1, 2. The subalgebra of A 1 A 2 generated by x 1 x 2, x i X i, i = 1, 2, is called the -tensor product of A 1 and A 2 and is denoted by A 1 A 2. Theorem 5.6. Let A i be K-algebra and free K-module with linearly ordered K- basis T i, i = 1, 2, where T 1 = {a i i I 1 } and T 2 = {b h h I 2 }. Then A 1 A 2 is a free K-module with basis {a i b h i I 1, h I 2 }. Proof. It is easy to check. Theorem 5.7. Let A i be K-algebra and free K-module with linearly ordered basis T i, i = 1, 2, where T 1 = {a i i I 1 } and T 2 = {b h h I 2 }. Let K (T 1 T 2 ) +

12 822 Y.Q. Chen and J.J. Qiu be free algebra and S = { (a i, b h )(a j, b k ) n,m α n ijγ m hk(a n, b m ) a i, a j T 1, b h, b k T 2 }, where a i a j = α n ij a n, i, j I 1, b h b k = γ m hk b m, h, k I 2. Then S is a Gröbner-Shirshov basis in K (T 1 T 2 ) + and K (T 1 T 2 ) + S is a -tensor product of A 1 and A 2, that is, K (T 1 T 2 ) + S = A 1 A 2. Proof. Let f ihjk = (a i, b h )(a j, b k ) n,m αn ij γm hk (a n, b m ). Then there is only one type of compositions (f ihjk, f jkpq ) w with w = (a i, b h )(a j, b k )(a p, b q ). Just like the proof of the Theorem, for every s I 1 and t I 2, we have αijα n np s = n n α n jpα s in Then for any s I 1, t I 2, since we have and n,m α n ijγ m hkα s npγ t mq = n,m γhkγ m mq t = m m α n jpγ m kqα s inγ t hm, γ m kqγ t hm (f ihjk, f jkpq ) w = ((a i, b h )(a j, b k ) n,m α n ijγ m hk(a n, b m ))(a p, b q ) (a i, b h )((a j, b k )(a p, b q ) n,m α n jpγ m kq(a n, b m )) = αijγ n hk(b m n, b m )(a p, b q ) + αjpγ n kq(a m i, b h )(a n, b m ) n,m n,m ( αijγ n hk( m n,m s,t α s npγ t mq(a s, b t )) n,m α n jpγ m kq( s,t α s inγ t hm(a s, b t )) mod(s, w) = ( αijγ n hkα m npγ s mq t αjpγ n kqα m inγ s hm)(a t s, b t ) s,t n,m n,m = 0. Now, all the compositions are trivial modulo S. Hence, S is a Gröbner-Shirshov basis. By the Composition-Diamond Lemma, T 1 T 2 is a K-basis of K (T 1 T 2 ) + S. It is easy to see that K (T 1 T 2 ) + S is a -tensor product of A 1 and A 2. Remark 5.8. Let K be a field and A 1 and A 2 K-algebras with identities 1 1 and 1 2, respectively. Then in the preceding statements, we can assume that 1 1 T 1 and 1 2 T 2. It is clear that (1 1, 1 2 ) is the identity of A 1 A 2.

13 Gröbner-Shirshov Bases for Free Product of Algebras and beyond 823 Theorem 5.9. Let K be a field, A 1 and A 2 the K-algebras with identities 1 1 and 1 2, respectively. Then A 1 A 2 = (A1 A 2 )/Id( ). Proof. The proof follows from the Theorem 3.3, Theorem 5.3 and Theorem Free Product of Lie Algebras A nonassociative algebra L over a commutative ring K is called Lie algebra if the following two identities are satisfied in L : (xy)z + (yz)x + (zx)y = 0 and x 2 = 0. Let Lie(X) be the free Lie algebra generated by X over the commutative ring K. Then for every Lie algebra L, we have L = Lie(X)/Id(S) = Lie(X S). for some X and S. Such an algebra is called the Lie algebra with generator X and defining relation S. The definitons of the associative Lyndon-Shirshov word and nonassociative Lyndon-Shirshov word in X are the same as in [6]. As we known, the nonassociative Lyndon-Shirshov words in X form a bases of the Lie algebra Lie(X) (see[12]). Following [13], for monic f, g Lie(X), if w = f = aḡb for some words a and b, we define the inclusion composition (f, g) w = f [agb]ḡ and if w = fa = bḡ, deg( f) + deg(ḡ) > deg(w), we define the intersection composition (f, g) w = [fa] f [bg]ḡ. Here [agb]ḡ, [fa] f, [bg]ḡ are special Shirshov s bracketing of agb, f a, bg, respectively [12] (see, for example [6]). Definition 6.1.([13]) A monic subset S Lie(X) is called a Gröbner-Shirshov basis if all Lie compositions (f, g) w is trivial modulo S, where f, g S. The following theorem is from the paper of Shirshov in 1962 [13] (see also [2]). Lemma 6.2. (Composition-Diamond Lemma) Let Lie(X) be the free Lie algebra generated by X over the commutative ring K. If S is monic subset of Lie(X), then the following statements are equivalent: (1) S is a Gröbner-Shirshov basis. (2) f Id(S) f = a sb for some s S and a, b X. (3) The set Red(S) = {[u] u a sb, s S} of S-reduced Lyndon Shirshov words in X is a K-basis of the Lie algebra L = Lie(X S).

14 824 Y.Q. Chen and J.J. Qiu The proof of the following lemma is straightforward. Lemma 6.3. Let L be Lie algebra over K and free K-module with a linear ordered basis {a i i I}. For i, j I, we let [a i a j ] = αij k a k, αij k K and {a i a j } = αij k a k. Then S = {[a i a j ] {a i a j } i > j, i, j I} is a Gröbner-Shirshov basis of L with generators {a i i I}. Definition 6.4. Let L, L 1 and L 2 be Lie algebras over K and e i : L i L, Lie algebra homomorphism, i = 1, 2. Then (L, (e 1, e 2 )) is called a free product of L 1 and L 2 if, for any Lie algebra over B and Lie algebra homomorphism f i : A i B, i = 1, 2, there existsb a unique Lie algebra homomorphism ϕ : L B such that the following diagram commutes, i.e., ϕe i = f i, i = 1, 2. f i B ϕ e i L L i Theorem 6.5. The free product of L 1 and L 2, if it exists, is uniquely determined up to isomorphism. In this case, we denote the free product of L 1 and L 2 by L 1 L 2. The following theorem shows that the free product exists for any two Lie K-algebras. Theorem 6.6. Let L i = Lie(X i S i ), i = 1, 2. Then L 1 L 2 = Lie(X 1 X 2 S 1 S 2 ). Proof. Similar to Theorem 2.3. The following theorem who established by A.I.Shirshov in 1962 [14] in case K is a field. Theorem 6.7. Let L i be Lie K-algebra and free K-module with linearly ordered basis T i, i = 1, 2, where T 1 = {a i i I 1 } and T 2 = {b h h I 2 }. Let M 1 = {[a i a j ] {a i a j } i > j, i, j I 1 } and M 2 = {[b h b k ] {b h b k } h > k, h, k I 2 }.

15 Gröbner-Shirshov Bases for Free Product of Algebras and beyond 825 We order T 1 T 2 by a i < b h for any i I 1, h I 2 and words in (T 1 T 2 ) in this alphabet in the deg-lex way. Then the following statements hold. (a) M 1 M 2 is a Gröbner-Shirshov basis in Lie(T 1 T 2 ). (b) L 1 L 2 = Lie(T 1 T 2 M 1 M 2 ). (c) Let Ω be the set which consists of [u], where u is Lyndon Shirshov word and u does not contain subwords a i a j, i > j and b h b k, h > k. Then L 1 L 2 is a free K-module with basis Ω. Proof. (a) Let f ij = [a i a j ] {a i a j } and g hk = [b h b k ] {b h b k }. Then by Lemma 6.2, M 1 and M 2 are Gröbner-Shirshov bases of Lie(T 1 M 1 ) and Lie(T 2 M 2 ), respectively. Hence (f ij, f jk ) aia ja k 0 and (g ij, g jk ) bib jb k 0 and thereby, M 1 M 2 is Gröbner-Shirshov basis of Lie(T 1 T 2 ). (b) This part follows from Theorem 6.6. (c) From the Composition-Diamond Lemma and (a), the result follows. Theorem 6.8. Let L 1 and L 2 be Lie algebras as above. Then U(L 1 L 2 ) = U(L 1 ) U(L 2 ), where U(L ) presents the enveloping of the Lie algebra L. Proof. By the definition of enveloping, we have U(L 1 ) = K T 1 M ( ) 1, where M ( ) 1 = {a i a j a j a i {a i a j } i > j, i, j I 1 }. Similarly, U(L 2 ) = K T 2 M ( ) 2. By the Theorem 6.7, L 1 L 2 = Lie(T 1 T 2 M 1 M 2 ). Then by the Theorem 2.4. (L 1 L 2 ) = K T 1 T 2 M ( ) 1 M ( ) 2 = K T 1 M ( ) 1 K T 2 M ( ) 2 = U(L 1 ) U(L 2 ) Theorem 6.9. (PBW-Theorem [11]) Let L 1 be Lie algebra as above. Then L 1 can be embedded into Lie algebra U(L 1 ), where [x, y] is defined by xy yx. Proof. It is easy to prove that M ( ) 1 is a is Gröbner-Shirshov basis (see, for example, [5] ). By the Composition-Diamond Lemma, the set {a i1 a i2 a it i 1 i 2 i t, t = 0, 1, }

16 826 Y.Q. Chen and J.J. Qiu is a basis of the algebra U(L 1 ). It follows that L 1 can be embedded into Lie algebra U(L 1 ). Acknowledgement: The authors would like to express their deepest gratitude to Professor L. A. Bokut for his kind guidance, useful discussions and enthusiastic encouragement during his visit to the South China Normal University. References [1] G. M. Bergman: The diamond lemma for ring theory, Adv. in Math. 29, (1978). [2] L.A. Bokut: Unsolvability of the word problem, and subalgebras of finitely presented Lie algebras, Izv. Akad. Nauk. SSSR Ser. Mat. 36, (1972). [3] L.A. Bokut: Imbeddings into simple associative algebras, Algebra i Logika 15, (1976). [4] L.A. Bokut, G.P Kukin: Algorithmic and combinatorial algebra, Mathematics and its Appilication 225, Kluwer Academic Publishers Group, Dordrecht,1994. [5] L.A.Bokut, Y.Fong, W.F.Ke: Gröbner Shirshov bases and composition lemma for associative conformal algebras: an example, Contemp. Math. 264, (2000). [6] L.A. Bokut, K.P.Shum: Gröbner and Gröbner-Shirshov bases in algebra: an elmentary approach, Southeast Asian Bulletin of Mathematics 29, (2005). [7] B. Buchberger: An Algorithm for Finding a Basis for the Residue Class Ring of a Zero-dimensional Polynomial Ideal, Ph.D. thesis, University of Innsbruck, Austria, (1965). (in German) [8] B. Buchberger: An algorithmical criteria for the solvability of algebraic systems of equations, Aequationes Math. 4, (1970). (in German) [9] P.M. Cohn: Free Rings and Their Relations, Academic Press, 2nd edition, [10] R.C. Lyndon, P.E. Schupp: Combinatorial Group Theory, Springer-Verlag, [11] C.Reutenauer: Free Lie Algebras, Oxford University Press, [12] A.I. Shirshov: On free Lie rings, Sibirsk. Mat. Sb. 45(87), (1958). (in Russian) [13] A.I. Shirshov: Some algorithmic problem for Lie algebras, Sibirsk. Mat. Z. 3, (1962) (in Russian); English translation in SIGSAM Bull. 33(2), 3-6 (1999). [14] A.I. Shirshov: On one hypothesis of the theory of Lie aglebras, Sibirsk. Mat. Z. 3, (1962). (in Russian)