Journal of Algebra 323 (2010) Contents lists available at ScienceDirect. Journal of Algebra.

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1 Journal of Algebra 323 (2010) Contents lists available at ScienceDirect Journal of Algebra Composition Diamond lemma for tensor product of free algebras L.A. Bokut a,b,1, Yuqun Chen a,, Yongshan Chen a a School of Mathematical Sciences, South China Normal University, Guangzhou , PR China b Sobolev Institute of Mathematics, Russian Academy of Sciences, Siberian Branch, Novosibirsk , Russia article info abstract Article history: Received 17 June 2008 Available online 11 March 2010 Communicated by Efim Zelmanov MSC: 16S10 16S15 13P10 In this paper, we establish a Composition Diamond lemma for the tensor product k k Y of two free algebras over a field. As an application, we construct a Gröbner Shirshov basis in k k Y by lifting a given Gröbner Shirshov basis in the tensor product k[] k Y in which k[] is the polynomial algebra Elsevier Inc. All rights reserved. Keywords: Gröbner Shirshov basis Gröbner basis Free algebra Polynomial algebra Tensor product 1. Introduction In [31], A.I. Shirshov established the theory of one-relator Lie algebras Lie( s = 0). This theory is in full analogy, by statements but not by method, of the celebrated theory of Magnus on one-relater groups [22,23] (see also [24,21]). In particular, A.I. Shirshov provided the algorithmic decidality of the word problem for any one-relator Lie algebra. In order to proceed his ideas, he first created the socalled Gröbner Shirshov bases theory for Lie algebras Lie( S) which are presented by generators and defining relations. The main notion of Shirshov s theory was a notion of composition ( f, g) w of two Supported by the NNSF of China (Nos , ) and the NSF of Guangdong Province (No ). * Corresponding author. addresses: bokut@math.nsc.ru (L.A. Bokut), yqchen@scnu.edu.cn (Y.Q. Chen), jackalshan@126.com (Y.S. Chen). 1 Supported by RFBR , LSS and SB RAS Integration grant No (Russia) /$ see front matter 2010 Elsevier Inc. All rights reserved. doi: /j.jalgebra

2 L.A. Bokut et al. / Journal of Algebra 323 (2010) Lie polynomials, f, g Lie() relative to some associative word w. Based on this notion, he defined an infinite algorithm of adding all non-trivial compositions to some set S until a set S c is obtained which is closed under compositions. In addition, S and S c generate the same ideal, i.e., Id(S) = Id(S c ). We now call S c the Gröbner Shirshov basis of Id(S). The following lemma was proved by Shirshov [31]. Let Lie() k be a free Lie algebra over a field k which is regarded as an algebra of the Lie polynomials in the free algebra k, and let S be a subset of Lie().If f Id(S),then f = u sv, where s S c,u, v, f, s are the leading associative words of the corresponding Lie polynomials f, s with respect to the deg-lex order on and is the free monoid generated by. The following corollary is an easy consequence of the above lemma. Irr(S c ) ={[u] u a sb, s S, a, b } is a linear basis of the algebra Lie( S) = Lie()/Id(S),whereu is an associative Lyndon Shirshov word in and [u] is the corresponding non-associative Lyndon Shirshov word under the Lie brackets [xy]=xy yx. In order to define the Lie composition ( f, g) w of two monic Lie polynomials, where f = ac, ḡ = cb, c 1, a, b, c are associative words and w = acb, A.I. Shirshov first defined the associative composition fb ag. Then by putting on fb and ag some special brackets [ fb], [ag] (see [29]), he obtained the Lie composition ( f, g) w =[fb] [ag]. Following [31], one can easily deduce the same lemma for a free associative algebra, that is, let S k and S c be defined as before. If f Id(S), then f = a sb for some s S c, a, b. This lemma was later formulated by L.A. Bokut [3] as an analogy of Shirshov s Lie composition lemma, and by G. Bergman [1] under the name Diamond lemma after the Newman s Diamond lemma for graphs [28]. Nowadays, Shirshov s lemma is named the Composition Diamond lemma for Lie and associative algebras. The formulation of this lemma will be given in the next section of this paper. This kind of ideas were also independently discovered by H. Hironaka [17] for power series algebras and by B. Buchberger [9,10] for polynomial algebras. The name Gröbner bases was suggested by B. Buchberger. The applications of Gröbner bases in mathematics are now well known and are well recognized, particularly in algebraic geometry, computer science and information science. At present, there are many Composition Diamond lemmas (CD-lemma for short) for different classes of non-commutative or non-associative algebras. We now list some of them below. A.I. Shirshov [30] proved himself the first CD-lemma for commutative (anti-commutative) nonassociative algebras, and he mentioned that this CD-lemma is also valid for non-associative algebras. It gave a solution of the word problem for these classes of algebras. For non-associative algebras, a version of CD-lemma was proved by A.I. Zhukov in [33]. A.A. Mikhalev [25] proved a CD-lemma for Lie superalgebras. T. Stokes [32] proved a CD-lemma for left ideals of an algebra k[] E k (Y ), the tensor product of a polynomial algebra and an exterior (Grassmann) algebra. A.A. Mikhalev and E.A. Vasilieva [26] proved a CD-lemma for free supercommutative polynomial algebras. A.A. Mikhalev and A.A. Zolotykh [27] proved a CD-lemma for k[] k Y, the tensor product of a polynomial algebra and a free algebra. L.A. Bokut, Y. Fong and W.F. Ke [7] proved a CD-lemma for associative conformal algebras. L. Hellström [16] proved a CD-lemma for a non-commutative power series algebra. S.-J. Kang and K.-H. Lee [18,19] and E.S. Chibrikov [13] proved a CD-lemma for a module over an algebra (see also [12]). D.R. Farkas, C.D. Feustel and E.L. Green [15] proved a CD-lemma for path algebras. L.A. Bokut and K.P. Shum [8] proved a CD-lemma for Γ -algebras. Y. Kobayashi [20] proved a CD-lemma for algebras based on well-ordered semigroups. L.A. Bokut, Yuqun Chen and Cihua Liu [5] proved a CD-lemma for dialgebras (see also [4]). Yuqun Chen, Yongshan Chen and Yu Li [11] proved a CD-lemma for differential algebras. L.A. Bokut, Yuqun Chen and Jianjun Qiu [6] proved a CD-lemma for associative algebras with multiple linear operators.

3 2522 L.A. Bokut et al. / Journal of Algebra 323 (2010) Let and Y be sets and k k Y be the tensor product of free algebras. In this paper, we will give a Composition Diamond lemma for the algebra k k Y (see Theorem 3.4). As a result, we prove a theorem on the pair of algebras (k[] k Y, k k Y ) following the spirit of Eisenbud Peeva Sturmfels theorem [14] on the pair (k[], k ). Namely, we construct a Gröbner Shirshov basis S in k k Y by lifting a given Gröbner Shirshov basis S in the tensor product k[] k Y (in the sense of [27]) such that (k k Y )/Id(S ) = (k[] k Y )/Id(S). Also, we give another proof of the Eisenbud Peeva Sturmfels theorem above. 2. Preliminaries We first cite some known concepts and results in the literature [31,2,3] concerning the Gröbner Shirshov bases theory of associative algebras. Let k be a field, k thefreeassociativealgebraoverk generated by and thefreemonoid generated by, where the empty word is the identity which is denoted by 1. For a word w,we denote the length of w by w. A well ordering > on is monomial if it is compatible with the multiplication of the words, that is, for u, v,wehave u > v w 1 uw 2 > w 1 vw 2, for all w 1, w 2. A standard example of monomial ordering on is the deg-lex ordering to compare two words first by their degrees and then lexicographically, where is a well-ordered set. Let f k with the leading word f. Then we call f monic if f has coefficient 1. If f and g are two monic polynomials in k and > awellorderingon, then there are two kinds of compositions: (i) If w is a word such that w = fb= aḡ for some a, b with f + ḡ > w, then the polynomial ( f, g) w = fb ag is called the intersection composition of f and g with respect to w. (ii) If w = f = aḡb for some a, b, then the polynomial ( f, g) w = f agb is called the inclusion composition of f and g with respect to w. Let S k with each s S monic. Then the composition ( f, g) w is called trivial modulo (S, w) if ( f, g) w = α i a i s i b i, where each α i k, a i, b i, s i S and a i s i b i < w. If this is the case, then we write ( f, g) w In general, for p, q k, wewritep qmod(s, w) which means that p q We now call the set S a Gröbner Shirshov basis in k with respect to the monomial ordering > if any composition of polynomials in S is trivial modulo S and corresponding w. Lemma 2.1 (Composition Diamond lemma for associative algebras). Let S k be a set of monic polynomials and > a monomial ordering on. Then the following statements are equivalent: (i) S is a Gröbner Shirshov basis in k. (ii) f Id(S) f = a sb for some s Sanda, b,whereid(s) is the ideal of k generated by S. (iii) Irr(S) ={u u a sb, s S, a, b } is a k-basis of the algebra A = k S. 3. Composition Diamond lemma for tensor product Let and Y be two well-ordered sets, T ={yx = xy x, y Y }. With the deg-lex ordering (y > x for any x, y Y )on( Y ), T is clearly a Gröbner Shirshov basis in k Y. Then, by Lemma 2.1,

4 L.A. Bokut et al. / Journal of Algebra 323 (2010) N = Y = Irr(T ) = { u = u u Y u and u Y Y } is the set of normal words of the tensor product k k Y =k Y T. Let kn be a k-space spanned by N. Foranyu = u u Y, v = v v Y N, wedefinethemultiplication of the normal words as follows uv = u v u Y v Y N. It is clear that kn is exactly the tensor product algebra k k Y, that is, kn = k Y T = k k Y. Now, we order the set N. Foranyu = u u Y, v = v v Y N, u > v u > v or ( u = v and ( u > v or ( u = v and u Y > v Y ))), where u = u + u Y is the length of u. Obviously,> is a monomial ordering on N. Suchanordering is also called the deg-lex ordering on N = Y. Throughout this paper, we will adopt this ordering unless otherwise stated. For any polynomial f k k Y, f has a unique presentation of the form f = α f f + αi u i, where f, u i N, f > ui, α f, α i k. The proof of the following lemma is straightforward and we hence omit the details. Lemma 3.1. Let f k k Y be a monic polynomial. Then uf v = u f vforanyu, v N. We give here the definition of compositions. Let f and g be two monic polynomials of k k Y and w = w w Y N. Then we have the following compositions. 1. Inclusion inclusion only Suppose that w = f = aḡ b for some a, b, and f Y, ḡ Y are disjoint. Then there are two compositions according to w Y = f Y cḡ Y and w Y = ḡ Y c f Y for c Y, respectively: and ( f, g) w1 = fcḡ Y f Y cagb, w 1 = f f Y cḡ Y ( f, g) w2 = ḡ Y cf agbc f Y, w 2 = f ḡ Y c f Y Y -inclusion only Suppose that w Y = f Y = cḡ Y d for c, d Y and f, ḡ are disjoint. Then there are two compositions according to w = f aḡ and w = ḡ a f for a, respectively: and ( f, g) w1 = faḡ f acgd, w 1 = f aḡ f Y ( f, g) w2 = ḡ af cgda f, w 2 = ḡ a f f Y.

5 2524 L.A. Bokut et al. / Journal of Algebra 323 (2010) , Y -inclusion Suppose that w = f = aḡ b for some a, b and w Y = f Y = cḡ Y d for some c, d Y.Then ( f, g) w = f acgbd. The transformation f ( f, g) w = f acgbd is said to be the elimination of leading word (ELW) of g in f. 1.4., Y -skew-inclusion Suppose that w = f = aḡ b for some a, b and w Y = ḡ Y = c f Y d for some c, d Y.Then ( f, g) w = cfd agb. 2. Intersection intersection only Suppose that w = f a = bḡ for some a, b with f + ḡ > w,and f Y, ḡ Y are disjoint. Then there are two compositions according to w Y = f Y cḡ Y and w Y = ḡ Y c f Y for c Y, respectively: and ( f, g) w1 = facḡ Y f Y cbg, w 1 = w f Y cḡ Y ( f, g) w2 = ḡ Y cfa bgc f Y, w 2 = w ḡ Y c f Y Y -intersection only Suppose that w Y = f Y c = dḡ for some c, d Y with f Y + ḡ Y > w Y, and f, ḡ are disjoint. Then there are two compositions according to w = f aḡ and w = ḡ a f for a, respectively: and ( f, g) w1 = fcaḡ f adg, w 1 = f aḡ w Y ( f, g) w2 = ḡ af c dga f, w 2 = ḡ a f w Y. 2.3., Y -intersection If w = f a = bḡ for some a, b and w Y = f Y c = dḡ Y for some c, d Y together with f + ḡ > w and f Y + ḡ Y > w Y,then ( f, g) w = fac bdg. 2.4., Y -skew-intersection If w = f a = bḡ for some a, b and w Y = c f Y = ḡ Y d for some c, d Y together with f + ḡ > w and f Y + ḡ Y > w Y,then ( f, g) w = cfa bgd. 3. Both inclusion and intersection inclusion and Y -intersection There are two subcases to consider. If w = f = aḡ b for some a, b and w Y = f Y c = dḡ Y for some c, d Y with f Y + ḡ Y > w Y,then ( f, g) w = fc adgb.

6 L.A. Bokut et al. / Journal of Algebra 323 (2010) If w = f = aḡ b for some a, b and w Y = c f Y = ḡ Y d for some c, d Y with f Y + ḡ Y > w Y,then ( f, g) w = cf agbd intersection and Y -inclusion There are two subcases to consider. If w = f a = bḡ for some a, b with f + ḡ > w and w Y = f Y = cḡ Y d for some c, d Y,then ( f, g) w = fa bcgd. If w = f a = bḡ for some a, b with f + ḡ > w and w Y = c f Y d = ḡ Y for some c, d Y,then ( f, g) w = cfad bg. From Lemma 3.1, it follows that for any case of compositions ( f, g) w < w. If Y =, then the compositions of f, g are the same in k as mentioned in Section 2. Let S be a monic subset of k k Y and f, g S. A composition ( f, g) w is said to be trivial modulo (S, w), denotedby ( f, g) w mod(s, w), if ( f, g) w = i α ia i s i b i, where a i, b i N, s i S, α i k and a i s i b i < w for any i. In general, for any p, q k k Y, wehavep qmod(s, w) if p q We call S a Gröbner Shirshov basis in k k Y if all compositions of elements in S are trivial modulo S and corresponding w. Lemma 3.2. Let S be a Gröbner Shirshov basis in k k Y and s 1, s 2 S. If w = a 1 s 1 b 1 = a 2 s 2 b 2 for some a i, b i N, i = 1, 2, thena 1 s 1 b 1 a 2 s 2 b 2 Proof. There are four cases to consider. Case 1. Inclusion inclusion only Suppose that w 1 = s 1 = a s 2 b, a, b and s Y 1, sy 2 are disjoint. Then a 2 = a 1 a and b 2 = bb 1. There are two subcases to consider: w Y 1 = sy 1 c sy 2 and w Y 1 = sy 2 c sy 1, where c Y. For w Y 1 = sy 1 c sy 2, we have w 1 = s 1 sy 1 c sy 2, ay 2 = ay 1 sy 1 c, by 1 = c sy 2 by 2, w = a 1 w 1 b 1 by 2 = a 1 s 1 ac s 2 b 2 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 s 1 b 1 c sy 2 by 2 a 1 aay 1 sy 1 cs 2bb 1 by 2 = a 1 ( s1 c s Y 2 sy 1 cas 2b ) b 1 by 2 = a 1 (s 1, s 2 ) w1 b 1 by 2 For w Y 1 = sy 2 c sy 1,wehavew 1 = s 1 sy 2 c sy 1, ay 1 = ay 2 sy 2 c, by 2 = c sy 1 by 1, w = a 1 ay 2 w 1b 1 and hence

7 2526 L.A. Bokut et al. / Journal of Algebra 323 (2010) a 1 s 1 b 1 a 2 s 2 b 2 = a 1 ay 2 sy 2 cs 1b 1 a 1 aay 2 s 2bb 1 c sy 1 by 1 ) = a 1 ay Y 2 ( s 2 cs 1 as 2 bc s Y 1 b1 = a 1 ay 2 (s 1, s 2 ) w1 b Y -inclusion only This case is similar to , Y -inclusion We may assume that s 2 is a subword of s 1, i.e., w 1 = s 1 = ac s 2 bd, a, b, c, d Y, a 2 = a 1 a, b 2 = bb 1, ay 2 = ay 1 c and by 2 = dby 1.Thus,a 2 = a 1 ac, b 2 = bdb 1, w = a 1 w 1 b 1 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 s 1 b 1 a 1 acs 2 bdb 1 = a 1 (s 1 acs 2 bd)b 1 = a 1 (s 1, s 2 ) w1 b , Y -skew-inclusion Assume that w 1 = s 1 = a s 2 b, a, b and w Y 1 = sy 2 = c sy 1 d, c, d Y.Thena 2 = a 1 a, b 2 = bb 1, a Y 1 = ay 2 c and by 1 = dby 2.Thus,w = a 1 ay 2 w 1b 1 by 2 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 ay 2 cs 1b 1 dby 2 a 1 aay 2 s 2bb 1 by 2 = a 1 ay 2 (cs 1d as 2 b)b 1 by 2 = a 1 ay 2 (s 1, s 2 ) w1 b 1 by 2 Case 2. Intersection intersection only We may assume that s 1 is at the left of s 2, i.e., w 1 = s 1 b = a s 2, a, b and s 1 + s 2 > w 1. Then a 2 = a 1 a and b 1 = bb 2. There are two subcases to consider: w Y 1 = sy 1 c sy 2 and w Y 1 = sy 2 c sy 1, c Y. For w Y 1 = sy 1 c sy 2, i.e., w 1 = s 1 bc s Y 2,wehaveaY 2 = ay 1 sy 1 c, by 1 = c sy 2 by 2, w = a 1 s 1 ac s 2 b 2 = a 1 w 1 b 2 and a 1 s 1 b 1 a 2 s 2 b 2 = a 1 s 1 bb 2 c sy 2 by 2 a 1 aay 1 sy 2 cs 2b 2 = a 1 ( s1 bc s Y 2 a sy 2 cs 2) b2 = a 1 (s 1, s 2 ) w1 b 2 For w Y 1 = sy 2 c sy 1, i.e., w 1 = s Y 2 c s 1b, wehavea Y 1 = ay 2 sy 2 c, by 2 = c sy 1 by 1, w = a 1 ay 2 w 1b 2 by 1 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 ay 2 sy 2 cs 1bb 2 by 1 a 1 aay 2 s 2b 2 c sy 1 by 1 ) = a 1 ay Y 2 ( s 2 cs 1b as 2 c s Y 1 b 2 by 1

8 L.A. Bokut et al. / Journal of Algebra 323 (2010) = a 1 ay 2 (s 1, s 2 ) w1 b 2 by Y -intersection only This case is similar to , Y -intersection Assume that w 1 = s 1 b = a s 2, w Y 1 = sy 1 d = c sy 2, a, b, c, d Y, s 1 + s 2 > w 1 and sy 1 + s Y 2 > w Y 1.Thena 2 = a 1 a, b 1 = bb 2, ay 2 = ay 1 c, by 1 = dby 2, w = a 1 w 1 b 2 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 s 1 bb 2 dby 2 a 1 aay 1 cs 2b 2 = a 1 (s 1 bd acs 2 )b 2 = a 1 (s 1, s 2 ) w1 b , Y -skew-intersection Assume that w 1 = s 1 b = a s 2, w Y 1 = c sy 1 = sy 2 d, s 1 + s 2 > w 1, sy 1 + sy 2 > w Y 1, a, b, c, d Y.Thena 2 = a 1 a, b 1 = bb 2, ay 1 = ay 2 c, by 2 = dby 1, w = a 1 ay 2 w 1b 2 by 1 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 ay 2 cs 1bb 2 by 1 a 1 aay 2 s 2b 2 dby 1 = a 1 ay 2 (cs 1b as 2 d)b 2 by 1 = a 1 ay 2 (s 1, s 2 ) w1 b 2 by 1 Case 3. Both inclusion and intersection inclusion and Y -intersection We may assume that w 1 = s 1 = a s 2 b, a, b.thena 2 = a 1 a and b 2 = bb 1.Theretwocasesto consider: w Y 1 = sy 1 d = c sy 2 and w Y 1 = c sy 1 = sy 2 d, where c, d Y, s Y 1 + sy 2 > w Y 1. For w Y 1 = sy 1 d = c sy 2,wehaveaY 2 = ay 1 c, by 1 = dby 2, w = a 1 w 1 b 1 by 2 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 s 1 b 1 dby 2 a 1 aay 1 cs 2bb 1 by 2 = a 1 (s 1 d acs 2 b)b 1 by 2 = a 1 (s 1, s 2 ) w1 b 2 For w Y 1 = c sy 1 = sy 2 d,wehaveay 1 = ay 2 c, by 2 = dby 1, w = a 1 ay 2 w 1b 2 dby 1 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 ay 2 cs 1b 1 a 1 aay 2 s 2bb 1 dby 1 = a 1 ay 2 (cs 1 as 2 bd)b 1 = a 1 ay 2 (s 1, s 2 ) w1 b 1

9 2528 L.A. Bokut et al. / Journal of Algebra 323 (2010) intersection and Y -inclusion Assume that w 1 = s 1 b = a s 2, a, b Y with s 1 + s 2 > w 1.Thena 2 = a 1 a, b 1 = bb 2.There are two subcases to consider: w Y 1 = sy 1 = c sy 2 d and sy 2 = c sy 1 d, where c, d Y. For w Y 1 = sy 1 = c sy 2 d,wehaveay 2 = ay 1 c, by 2 = dby 1, w = a 1 w 1 b 2 by 1 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 s 1 bb 2 by 1 a 1acs 2 b 2 dby 1 = a 1 (s 1 b acs 2 d)b 2 by 1 = a 1 (s 1, s 2 ) w1 b 2 by 1 For w Y 1 = sy 2 = c sy 1 d,wehaveay 1 = ay 2 c, by 1 = dby 2, w = a 1 ay 2 w 1b 2 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 ay 2 cs 1bb 2 dby 2 a 1 aay 2 s 2b 2 = a 1 ay 2 (cs 1bd as 2 )b 2 = a 1 ay 2 (s 1, s 2 ) w1 b 2 Case 4. s 1 and s 2 are disjoint For w = w w Y, by symmetry, there are two subcases to consider: w Y = a Y 1 sy 1 c sy 2 by 2 and w Y = a Y 2 sy 2 c sy 1 by 1, where w = a 1 s 1 a s 2 b 2, a, a 2 = a 1 s 1 a, b 1 = a s 2 b 2 and c Y. For w = a 1 s 1 a s 2 b 2 ay 1 sy 1 c sy 2 by 2 = a 1 s 1 ac s 2 b 2,wehavea 2 = a 1 s 1 ac, b 1 = ac s 2 b 2 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 s 1 ac s 2 b 2 a 1 s 1 acs 2 b 2 = a 1 (s 1 s 1 )acs 2 b 2 a 1 s 1 ac(s 2 s 2 )b 2 For w = a 1 s 1 a s 2 b 2 ay 2 sy 2 c sy 1 by 1,wehaveaY 1 = ay 2 sy 2 c, by 2 = c sy 1 by 1 and hence a 1 s 1 b 1 a 2 s 2 b 2 = a 1 ay 2 sy 2 cs 1a s 2 b 2 by 1 a 1 s 1 aay 2 s 2b 2 c sy 1 by 1 = a 1 ay Y 2 ( s 2 cs 1a s 2 ) s 1 as 2c s Y 1 b 2 by 1. Let s 1 = n i=1 α iu 1i uy 1i and s 2 = m j=1 β ju 2 j uy 2 j, where α 1 = β 1 = 1. Then n m s Y 2 cs 1a s 2 s 1 as 2c s Y 1 = α i u 1i a s 2cu Y 1i β i u Y 2 j c s 1au 2 j i=2 j=2 n m = α i u 1i a( s 2 s 2 )cu Y 1i + β j u Y 2 j c(s 1 s 1 )au 2 j i=2 j=2 n m + α i u 1i as 2cu Y 1i β j u Y 2 j cs 1au 2 j i=2 j=2

10 L.A. Bokut et al. / Journal of Algebra 323 (2010) n m m α i β j u 1i au 2 j uy 2 j cuy 1i n α i β j u Y 2 j cuy 1i u 1i au 2 j i=2 j=2 j=2 i=2 mod(s, w 1 ) where w 1 = s Y 2 c s 1a s 2 = s 1 a s 2c s Y 1.Sincew = a 1 ay 2 w 1b 2 by 1,wehave a 1 s 1 b 1 a 2 s 2 b 2 = a 1 ay 2 Y ( s 2 cs 1a s 2 ) s 1 as 2c s Y 1 b 2 by 1 This completes the proof. Lemma 3.3. Let S k k Y with each s S monic and Irr(S) ={w N w asb, a, b N, s S}. Then for any f k k Y, f = a i s i b i f where each α i,β j k, a i, b i N, s i Sandu j Irr(S). α i a i s i b i + β j u j u j f Proof. Let f = α i iu i k k Y, where 0 α i k and u 1 > u 2 >. If u 1 Irr(S), then let f 1 = f α 1 u 1.Ifu 1 / Irr(S), then there exist some s S and a 1, b 1 N such that f = a 1 s 1 b 1.Let f 1 = f α 1 a 1 s 1 b 1.Inbothcases,wehave f 1 < f. Then the result follows from induction on f. By summing up the above lemmas, we arrive at the following theorem. Theorem 3.4 (Composition Diamond lemma for tensor product k k Y ). Let S k k Y with each s S monic and < the ordering on N = Y as before. Then the following statements are equivalent: (i) S is a Gröbner Shirshov basis in k k Y. (ii) f Id(S) f = asb for some a, b N, s S. (iii) Irr(S) ={w N w asb, a, b N, s S} is a k-linear basis for the factor algebra k k Y /Id(S). Proof. (i) (ii). Suppose that 0 f Id(S). Then f = α i a i s i b i for some α i k, a i, b i N, s i S. Let w i = a i s i b i and w 1 = w 2 = = w l > w l+1. We will prove that f = a sb for some a, b N, s S, by using induction on l and w 1.Ifl = 1, then the result is clear. If l > 1, then w 1 = a 1 s 1 b 1 = a 2 s 2 b 2. Now, by (i) and Lemma 3.2, we see that a 1 s 1 b 1 a 2 s 2 b 2 mod(s, w 1 ).Thus, α 1 a 1 s 1 b 1 + α 2 a 2 s 2 b 2 = (α 1 + α 2 )a 1 s 1 b 1 + α 2 (a 2 s 2 b 2 a 1 s 1 b 1 ) (α 1 + α 2 )a 1 s 1 b 1 mod(s, w 1 ). By induction on l and w 1, we obtain the desired result. (ii) (iii). For any 0 f k k Y, by Lemma 3.3, we can express f as f = α i a i s i b i + β j u j, where α i,β j k, a i, b i N, s i S and u j Irr(S). ThenIrr(S) generates the factor algebra. Moreover, if 0 h = β j u j Id(S), u j Irr(S), u 1 > u 2 > and β 1 0, then u 1 = h = a sb for some a, b N, s S by (ii), a contradiction. This shows that Irr(S) is a linear basis of the factor algebra.

11 2530 L.A. Bokut et al. / Journal of Algebra 323 (2010) (iii) (i). For any f, g S, wehaveh = ( f, g) w Id(S). The result is now trivial if ( f, g) w = 0. Assume that ( f, g) w 0. Then, by Lemma 3.3 and (iii), we have h = a i s i b i h α i a i s i b i. Now, by noting that h = ( f, g) w < w, we see immediately that (i) holds. Remark. Theorem 3.4 is valid for any monomial ordering on Y. Remark. Theorem 3.4 is precisely the Composition Diamond lemma for associative algebras (Lemma 2.1) when Y =. 4. Applications Now, we give some applications of Theorem 3.4. Example 4.1. Suppose that for the deg-lex ordering, S 1 and S 2 are Gröbner Shirshov bases in k and k Y respectively. Then for the deg-lex ordering on Y as before, S 1 S 2 is a Gröbner Shirshov basis in k Y T =k k Y. It follows that k S 1 k Y S 2 =k Y T S 1 S 2. Proof. The possible compositions in S 1 S 2 are -including only, -intersection only, Y -including only and Y -intersection only. Suppose that f, g S 1 and ( f, g) w1 mod(s 1, w 1 ) in k. Then in k k Y, ( f, g) w = ( f, g) w1 c, where w = w 1 c for any c Y. From this it follows that each composition in S 1 S 2 is trivial modulo S 1 S 2. A special case of Example 4.1 is the following: Example 4.2. Let, Y be well-ordered sets, k[] the free commutative associative algebra generated by. ThenS ={x i x j = x j x i x i > x j, x i, x j } is a Gröbner Shirshov basis in k k Y with respect to the deg-lex ordering. Therefore, k[] k Y =k Y T S. In [14], a Gröbner Shirshov basis in k is constructed by lifting a commutative Gröbner basis and adding some commutators. Let ={x 1, x 2,...,x n },let[] be the free commutative monoid generated by and k[] the polynomial ring. Let S 1 ={h ij = x i x j x j x i i > j} k. Then, consider the natural map γ : k k[] which maps x i to x i and the lexicographic splitting of γ, which is defined as the k-linear map δ : k[] k, x i1 x i2 x ir x i1 x i2 x ir if i 1 i 2 i r. For any u [], wepresentu = x l 1 1 xl 2 2 xl n n, where l i 0. We use any monomial ordering on []. Following [14], we define an ordering on using the ordering x 1 < x 2 < < x n as follows: for any u, v, u > v γ (u)>γ (v) in [] or ( γ (u) = γ (v) and u >lex v ). It is easy to check that this ordering is monomial on and δ(s) = δ( s) for any s k[]. Moreover, for any v γ 1 (u), v δ(u).

12 L.A. Bokut et al. / Journal of Algebra 323 (2010) For any m = x i1 x i2 x ir [], i 1 i 2 i r, denote the set of all the monomials u [x i1 +1,...,x ir 1] by U(m). The proofs of the following lemmas are straightforward. Lemma 4.3. Let a, b,a= δ(γ (a)),b= δ(γ (b)) and s k[].ifw= aδ( s)b = δ(γ (ab) s),then,ink, aδ(s)b δ ( γ (ab)s ) mod(s 1, w). Proof. Suppose that s = s + s and h = aδ(s)b δ(γ (ab)s). Since aδ( s)b = δ(γ (ab) s), we have h = aδ(s )b δ(γ (ab)s ), and h < w. By noting that γ (aδ(s )b) = γ (δ(γ (ab)s )), h mod(s 1, w). Lemma 4.4. Let f, g k[], ḡ = x i1 x i2 x ir (i 1 i 2 i r ) and w = δ( f ḡ).then,ink, δ ( ( f f )g ) α i a i δ(u i g)b i mod(s 1, w) where α i k, a i [x x x i1 ],b i [x x x ir ],u i U(ḡ) and γ ( α i a i u i b i ) = f f. Theorem 4.5. (See [14].) Let the orderings on [] and be defined as above. If S is a minimal Gröbner basis in k[],thens ={δ(us) s S, u U( s)} S 1 is a Gröbner Shirshov basis in k. Proof. We will show that all the possible compositions of elements in S are trivial. Let f = δ(us 1 ), g = δ(vs 2 ) and h ij = x i x j x j x i S. (i) f g Case 1. f and g have a composition of including, i.e., w = δ(u s 1 ) = aδ(v s 2 )b for some a, b and a = δ(γ (a)), b = δ(γ (b)). If s 1 and s 2 have no composition in k[], i.e., lcm( s 1 s 2 ) = s 1 s 2,thenu = u s 2, γ (ab)v = u s 1 for some u []. By Lemmas 4.3 and 4.4, we have ( f, g) w = δ(us 1 ) aδ(vs 2 )b δ(us 1 ) δ ( ) γ (ab)vs 2 δ ( u s ) ( ) 2 s 1 δ u s 1 s 2 δ ( u ) ( (s 1 s 1 )s 2 δ u ) (s 2 s 2 )s 1 ( 0 mod S, ) w. Since, in k[], S is a minimal Gröbner basis, the possible compositions are only intersection. If s 1 and s 2 have composition of intersection in k[], i.e., (s 1, s 2 ) w = a s 1 b s 2, where a, b [], w = a s 1 = b s 2 and w < s 1 + s 2,thenw is a subword of γ (w). Hence,wededucethatw = δ(tw ) = δ(ta s 1 ) = δ(tb s 2 ) and u = ta, γ (ab)v = tb for some t []. Then ( f, g) w = δ(us 1 ) aδ(vs 2 )b δ(us 1 ) δ ( ) γ (ab)vs 2 δ ( ta ) ( ) s 1 δ tb s2 δ ( ( t a s 1 b )) s 2

13 2532 L.A. Bokut et al. / Journal of Algebra 323 (2010) δ ( ) t(s 1, s 2 ) w ( 0 mod S, ) w since t(s 1, s 2 ) w < tw = γ (w). Case 2. If f and g have a composition of intersection, we may assume that f is on the left of ḡ, i.e., w = δ(u s 1 )a = bδ(v s 2 ) for some a, b and a = δγ (a), b = δγ (b). Similarly to Case 1, we have to consider whether s 1 and s 2 have compositions in k[] or not. One can check that both cases are trivial mod(s, w) by Lemmas 4.3 and 4.4. (ii) f h ij By noting that h ij = x i x j cannot be a subword of f = δ(u s 1 ) since i > j, only possible compositions are intersection. Suppose that s 1 = x i1 x ir x i (i 1 i 2 i r i). Then f = δ(u s 1 ) = x i1 vx i for some v k, v = δγ (v) and w = δ(u s 1 )x j. If j i 1,then If j > i 1,thenux j U( s 1 ) and ( f, h ij ) w = δ(us 1 )x j x i1 v(x i x j x j x i ) = δ ( u(s 1 s 1 ) ) x j + x i1 vx j x i x j δ ( u(s 1 s 1 ) ) + x j x i1 vx i ( ( x j δ u(s1 s 1 ) ) + δ(u s 1 ) ) x j δ(us 1 ) mod ( S, w ). ( f, h ij ) w = δ(us 1 )x j x i1 v(x i x j x j x i ) = δ ( u(s 1 s 1 ) ) x j + x i1 vx j x i δ ( ux j (s 1 s 1 ) ) + δ(x i1 vx i x j ) δ ( ux j (s 1 s 1 ) ) + δ(ux j s 1 ) δ(ux j s 1 ) ( 0 mod S, ) w. Thus, the proof is completed. Now we extend γ and δ as follows: γ 1: k k Y k[] k Y, u u Y γ ( u ) u Y, δ 1: k[] k Y k k Y, u u Y δ ( u ) u Y. Any polynomial f k[] k Y has a presentation f = α i u i u Y i, where α i k, u i [] and u Y i Y. Let the orderings on [] and Y be any monomial orderings respectively. We order the set []Y = {u = u u Y u [], u Y Y } as follows. For any u, v []Y, u > v u Y > v Y or ( u Y = v Y and u > v ).

14 L.A. Bokut et al. / Journal of Algebra 323 (2010) Now, we order Y :foranyu, v Y, u > v γ ( u ) u Y > γ ( v ) v Y or ( γ ( u ) u Y = γ ( v ) v Y and u > lex v ). This ordering is clearly a monomial ordering on Y. The following definitions of compositions and the Gröbner Shirshov basis are taken from [27]. Let f, g be monic polynomials of k[] k Y, L the least common multiple of f and ḡ. 1. Inclusion Let ḡ Y be a subword of f Y,say, f Y = cḡ Y d for some c, d Y.If f Y = ḡ Y then f ḡ and if ḡ Y = 1thenwesetc = 1. Let w = L f Y = Lcḡ Y d. We define the composition C 1 ( f, g, c) w = L f f L ḡ cgd. 2. Overlap Let a non-empty beginning of ḡ Y be a non-empty ending of f Y,say, f Y = cc 0, ḡ Y = c 0 d, f Y d = cḡ Y for some c, d, c 0 Y and c 0 1. Let w = L f Y d = Lcḡ Y. We define the composition C 2 ( f, g, c 0 ) w = L f fd L ḡ cg. 3. External Let c 0 Y be any associative word (possibly empty). In the case that the greatest common divisor of f and ḡ is non-empty and f Y, ḡ Y are non-empty, we define the composition C 3 ( f, g, c 0 ) w = L f fc 0 ḡ Y L ḡ f Y c 0 g where w = L f Y c 0 ḡ Y. Let S be a monic subset of k[] k Y. ThenS is called a Gröbner Shirshov basis (standard basis) in k[] k Y if for any element f Id(S), f contains s as its subword for some s S. It is defined as usual that a composition is trivial modulo S and corresponding w. We also have that S is a Gröbner Shirshov basis in k[] k Y if and only if all the possible compositions of its elements are trivial. A Gröbner Shirshov basis in k[] k Y is called minimal if for any s S and all s i S \{s}, s i is not a subword of s. Similar to the proof of Theorem 4.5, we have the following theorem. Theorem 4.6. Let the orderings on []Y and Y be defined as before. If S is a minimal Gröbner Shirshov basis in k[] k Y,thenS ={δ 1(us) s S, u U( s )} S 1 is a Gröbner Shirshov basis in k k Y, where S 1 ={h ij = x i x j x j x i i > j}. Proof. We will show that all the possible compositions of elements in S are trivial. For s 1, s 2 S, let f = δ 1(us 1 ), g = δ 1(vs 2 ), h ij = x i x j x j x i S and L = lcm( s 1, s 2 ). 1. f g In this case, all the possible compositions are related to the ambiguities w s (in the following, a, b, c, d Y ) inclusion only w = δ ( us ) 1 = aδ ( ) v s 2 b, w Y = s Y 1 c sy 2 or w Y = s Y 2 c sy Y -inclusion only w = δ ( u s 1 ) ( ) aδ v s 2 or w = δ ( v s 2 ) ( ) aδ u s 1, w Y = s Y 1 = c sy 2 d.

15 2534 L.A. Bokut et al. / Journal of Algebra 323 (2010) , Y -inclusion w = δ 1(u s 1 ) = acδ 1(v s 2 )bd. 1.4., Y -skew-inclusion w = δ(u s 1 ) = aδ(v s 2 )b, w Y = s Y 2 = c sy 1 d intersection only w = δ ( u s 1 ) ( ) a = bδ v s 2, w Y = s Y 1 c sy 2 or w Y = s Y 2 c sy Y -intersection only w = δ ( u s 1 ) ( ) aδ v s 2 or w = δ(u s 2 )aδ(v s 1 ), w Y = s Y 1 c = d sy , Y -intersection 2.4., Y -skew-intersection w = δ(u s 1 )a = bδ(v s 2 ), w Y = s Y 1 c = d sy 2. w = δ ( u s inclusion and Y -intersection w = δ ( u s intersection and Y -inclusion w = δ ( u s 1 ) ( ) a = bδ v s 2, w Y = c s Y 1 = sy 2 d. ) ( ) = aδ v s 2 b, w Y = s Y 1 c = d sy 2 or w Y = c s Y 1 = sy 2 d. ) ( ) a = bδ v s 2, w Y = s Y 1 = c sy 2 d or w Y = s Y 2 = c sy 1 d. We only check the cases 1.1, 1.2 and 1.3. Other cases are similarly checked inclusion only Suppose that w = δ(u s 1 ) = aδ(v s 2 )b, a, b and s Y 1, sy 2 aredisjoint.therearetwocasesto consider: w Y = s Y 1 c sy 2 and w Y = s Y 2 c sy 1, where c Y. We will only prove the first case and the second one are similar. If s 1 and s 2 have no composition in k[] k Y, i.e., lcm( s 1, s 2 ) = s 1 s 2,thenu = u s 2, γ (ab)v = u s 1 for some u []. By the proof of Theorem 4.5, we have ( f, g) w = δ 1(us 1 )c s Y 2 sy 1 caδ 1(vs 2)b δ ( 1 us 1 γ ( )) ( c s Y 2 δ 1 γ Y ( s 1 c) ) γ (ab)vs 2 δ ( 1 u s ) ) 2 s 1c s Y 2 δ 1 Y ( s 1 cu s 1 s 2 δ ( 1 u ) ( ) s 1 c s 2 δ 1 u s 1 cs 2 δ ( 1 u ) ( (s 1 s 1 )cs 2 δ 1 u s1 c(s 2 s 2 ) ) mod ( S, w ).

16 L.A. Bokut et al. / Journal of Algebra 323 (2010) If s 1 and s 2 have composition of external (the elements of S have no composition of inclusion because S is minimal and s 1 and s 2 have no composition of overlap because s Y 1 and sy 2 are disjoint) in k[] k Y, i.e., C 3 (s 1, s 2, c) w = L s s 1 γ (c s Y 2 ) L γ ( s Y 1 s 1 c)s 2 = t 2 s 1 γ (c s Y 2 ) t 1γ ( s Y 1 c)s 2 where 2 gcd( s 1, s 2 ) = t 1, s 1 = tt 1, s 2 = tt 2 and L = tt 1 t 2, w = Lγ ( s Y 1 c sy 2 ), then w is a subword of γ (w). Therefore, we have w = δ 1(mw ) and u = mt 2, γ (ab)v = mt 1 since ut 1 = γ (ab)vt 2 and gcd(t 1, t 2 ) = 1. Then ( f, g) w = δ 1(us 1 )c s Y 2 sy 1 caδ 1(vs 2)b δ ( 1 us 1 γ ( )) ( c s Y 2 δ 1 γ Y ( s 1 c) ) γ (ab)vs 2 δ ( 1 mt 2 s 1 γ ( )) ( c s Y 2 δ 1 mt1 γ ( s Y 1 c) ) s 2 δ ( ) 1 mc 3 (s 1, s 2, c) w ( 0 mod S, ) w since mc 3 (s 1, s 2, c) w < mw = γ (w) Y -inclusion only Suppose that w Y = s Y 1 = c sy 2 d, c, d Y and δ(u s 1 ), δ(v s 2 ) are disjoint. Then there are two compositions according to w = δ(u s 1 )aδ(v s 2 ) and w = δ(v s 2 )aδ(u s 1 ) for a.weonlyprovethe first. ( f, g) w = δ 1(us 1 )aδ ( ) ( ) v s 2 δ u s 1 acδ 1(vs2 )d δ ( 1 us 1 γ (a)v s 2 u s γ (a)vγ 1 (c)s 2γ (d) ) δ ( 1 uγ ( (a)v s 1 s 2 s γ 1 (c)s 2γ (d) )) δ ( ( 1 uγ (a)vc 1 s1, s 2,γ (c) ) ) w ( 0 mod S, ) w where w = s 1 s 2 sy 1 = s 1 s 2 γ (c) sy 2 γ (d) and uγ (a)vc 1(s 1, s 2, γ (c)) w < uγ (a)vw = γ (w). 1.3., Y -inclusion We may assume that ḡ is a subword of f, i.e., w = δ 1(u s 1 ) = acδ 1(v s 2 )bd, a, b, c, d Y. Then u s 1 = γ (ab)v s 2 = ml for some m [], u sy 1 = γ (c) sy 2 γ (d) ( f, g) w = δ 1(us 1 ) acδ 1(vs 2 )bd δ ( 1 us 1 γ (ac)vs 2 γ (bd) ) ( δ 1 m L s s 1 m L ) γ 1 s (c)s 2 γ (d) 2 δ ( ( 1 mc 1 s1, s 2,γ (c) ) ) w ( 0 mod S, ) w where w = Lγ (c) s Y γ 2 (d) and mc 1(s 1, s 2, c) w < mw = γ (w). 2. f h ij Similar to the proof of Theorem 4.5, they only have compositions of -intersection. Suppose that s 1 = x i 1 x ir x i (i 1 i 2 i r i). Then f = δ 1(u s 1 ) = x i1 vx i s Y 1 for some v k, and v = δγ (v) and w = δ 1(u s 1 )x j s Y 1.

17 2536 L.A. Bokut et al. / Journal of Algebra 323 (2010) If j i 1,then If j > i 1,thenux j U( s 1 ) and ( f, h ij ) w = δ 1(us 1 )x j x i1 v s Y 1 (x ix j x j x i ) = δ 1 ( u(s 1 s 1 ) ) x j + x i1 vx j x i s Y 1 x j δ ( 1 u(s 1 s 1 ) ) + x j x i1 vx i s Y 1 ( ( x j δ 1 u(s1 s 1 ) ) + δ 1(u s 1 ) ) x j δ 1(us 1 ) mod ( S, w ). ( f, h ij ) w = δ 1(us 1 )x j x i1 v s Y 1 (x ix j x j x i ) = δ ( 1 u(s 1 s 1 ) ) x j + x i1 vx j x i s Y 1 δ ( 1 ux j (s 1 s 1 ) ) + δ ( ) 1 x i1 vx i x j s Y 1 δ ( 1 ux j (s 1 s 1 ) ) + δ 1(ux j s 1 ) δ 1(ux j s 1 ) ( 0 mod S, ) w. This completes the proof. As an application of the above result, we have now constructed a Gröbner Shirshov basis for the tensor product k k Y by lifting a given Gröbner Shirshov basis in the tensor product k[] k Y. References [1] G.M. Bergman, The Diamond lemma for ring theory, Adv. 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 Logika 15 (1976) [4] L.A. Bokut, Yuqun Chen, Gröbner Shirshov bases: Some new results, in: Proceedings of the Second International Congress in Algebra and Combinatorics, World Scientific, 2008, pp [5] L.A. Bokut, Yuqun Chen, Cihua Liu, Gröbner Shirshov bases for dialgebras, Internat. J. Algebra Comput., in press, ariv.org/abs/ [6] L.A. Bokut, Yuqun Chen, Jianjun Qiu, Gröbner Shirshov bases for associative algebras with multiple operators and free Rota Baxter algebras, J. Pure Appl. Algebra 214 (2010) [7] L.A. Bokut, Y. Fong, W.F. Ke, Composition Diamond lemma for associative conformal algebras, J. Algebra 272 (2004) [8] L.A. Bokut, K.P. Shum, Relative Gröbner Shirshov bases for algebras and groups, Algebra and Analysis 19 (6) (2007) 1 12 (in Russian). [9] B. Buchberger, An algorithm for finding a basis for the residue class ring of a zero-dimensional polynomial ideal, PhD thesis, University of Innsbruck, Austria, 1965 (in German). [10] B. Buchberger, An algorithmical criteria for the solvability of algebraic systems of equations, Aequationes Math. 4 (1970) (in German). [11] Yuqun Chen, Yongshan Chen, Yu Li, Groebner Shirshov bases for differential algebras, Arab. J. Sci. Eng. 34 (2A) (2009) [12] Yuqun Chen, Yongshan Chen, Chanyan Zhong, Composition Diamond lemma for modules, Czechoslovak Math. J. 60 (135) (2010) [13] E.S. Chibrikov, On free Lie conformal algebras, Vestnik Novosibirsk State Univ. 4 (1) (2004) [14] D. Eisenbud, I. Peeva, B. Sturmfels, Non-commutative Gröbner bases for commutative algebras, Proc. Amer. Math. Soc. 126 (3) (1998)

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