Two-generated graded algebras

Two-generated graded algebras

The paper is devoted to classification of twogenerated graded algebras. We show that under some general assumptions there exist two classes of these algebras, namely quantum polynomials and Jordanian plane. We study prime spectrum, the semigroup of endomorphisms and the Lie algebra of derivations...

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Дата:2005
Автор: Shirikov, E.N.
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Опубліковано: Інститут прикладної математики і механіки НАН України 2005
Назва видання:Algebra and Discrete Mathematics
Онлайн доступ:http://dspace.nbuv.gov.ua/handle/123456789/157199
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Цитувати:Two-generated graded algebras / E.N. Shirikov // Algebra and Discrete Mathematics. — 2005. — Vol. 4, № 3. — С. 60–84. — Бібліогр.: 8 назв. — англ.

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spelling irk-123456789-1571992019-06-20T01:29:38Z Two-generated graded algebras Shirikov, E.N. The paper is devoted to classification of twogenerated graded algebras. We show that under some general assumptions there exist two classes of these algebras, namely quantum polynomials and Jordanian plane. We study prime spectrum, the semigroup of endomorphisms and the Lie algebra of derivations of Jordanian plane. 2005 Article Two-generated graded algebras / E.N. Shirikov // Algebra and Discrete Mathematics. — 2005. — Vol. 4, № 3. — С. 60–84. — Бібліогр.: 8 назв. — англ. 1726-3255 http://dspace.nbuv.gov.ua/handle/123456789/157199 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection DSpace DC
language English
description The paper is devoted to classification of twogenerated graded algebras. We show that under some general assumptions there exist two classes of these algebras, namely quantum polynomials and Jordanian plane. We study prime spectrum, the semigroup of endomorphisms and the Lie algebra of derivations of Jordanian plane.
format Article
author Shirikov, E.N.
spellingShingle Shirikov, E.N.
Two-generated graded algebras
Algebra and Discrete Mathematics
author_facet Shirikov, E.N.
author_sort Shirikov, E.N.
title Two-generated graded algebras
title_short Two-generated graded algebras
title_full Two-generated graded algebras
title_fullStr Two-generated graded algebras
title_full_unstemmed Two-generated graded algebras
title_sort two-generated graded algebras
publisher Інститут прикладної математики і механіки НАН України
publishDate 2005
url http://dspace.nbuv.gov.ua/handle/123456789/157199
citation_txt Two-generated graded algebras / E.N. Shirikov // Algebra and Discrete Mathematics. — 2005. — Vol. 4, № 3. — С. 60–84. — Бібліогр.: 8 назв. — англ.
series Algebra and Discrete Mathematics
work_keys_str_mv AT shirikoven twogeneratedgradedalgebras
first_indexed 2025-07-14T09:35:27Z
last_indexed 2025-07-14T09:35:27Z
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fulltext Jo u rn al A lg eb ra D is cr et e M at h . Algebra and Discrete Mathematics RESEARCH ARTICLE Number 3. (2005). pp. 60 – 84 c© Journal “Algebra and Discrete Mathematics” Two-generated graded algebras Evgenij N. Shirikov Communicated by V. A. Artamonov To my parents Abstract. The paper is devoted to classification of two- generated graded algebras. We show that under some general as- sumptions there exist two classes of these algebras, namely quan- tum polynomials and Jordanian plane. We study prime spectrum, the semigroup of endomorphisms and the Lie algebra of derivations of Jordanian plane. Introduction Let A0 = K be a field and A = ∞ ⊕ n=0 An the associative graded algebra generated over K by elements X,Y ∈ A1. Suppose that dimA2 = 3. In the paper we find a criterion for A to be a domain when dimAn+1 = n+1 (see Corollary 5.5). We also show that if K has no quadratic extensions, A is a domain and either dimAn+1 = n+1 or A is a central algebra then A is either the algebra of quantum polynomials in two variables Λ1 (K, λ) = K 〈X,Y 〉/(Y X − λXY ), λ ∈ K∗, or Jordanian plane Λ2 (K) = K 〈X,Y 〉 / ( Y X −XY − Y 2 ) (see Theorems 5.3, 5.4). The other sections of this paper are devoted to Jordanian plane. We describe its center (see Theorem 2.2), deriva- tions (see Theorem 4.2) and the Lie algebra of outer derivations for an This research was partially supported by RFBR 03-01-00167 grant. Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 61 arbitrary field K (see Theorems 4.6, 4.10, 4.16, 4.20, 4.23). In the case charK = 0 we describe prime spectrum (see Theorem 2.4), the group of automorphisms (see Theorem 3.1), the endomorphisms with non-trivial kernels (see Proposition 3.2). Similar problems for quantum polynomials have been considered by V. A. Artamonov [1]. Some properties of quan- tum polynomials are also considered in details in [4]. Note that a study of non-commutative graded algebras is motivated by non-commutative algebraic geometry [6]. The author is grateful to professor V. A. Artamonov for constant attention to this work and useful discussions. 1. Definitions Definition 1.1. The single parameter algebra of quantum polyno- mials in two variables over K is the K−algebra Λ1 (K, λ), λ ∈ K∗, given by generators X and Y and defining relation Y X = λXY , i.e. Λ1 (K, λ) = K 〈X,Y 〉/(Y X − λXY ). Jordanian plane over K is the K−algebra Λ2 (K) given by generators X and Y and defining relation Y X = XY + Y 2, i.e. Λ2 (K) = K 〈X,Y 〉 / ( Y X −XY − Y 2 ) . Proposition 1.2. The basis of Λ2 (K) is { XiY j |i, j ∈ N0 } . In particu- lar, Y mXn = n ∑ l=0 ( n l ) (m+ n− l − 1)! (m− 1)! X lY m+n−l, m, n ∈ N, and Λ2 (K) is a domain. Proof. 1. We claim that the monomials X•Y • are linear independent. Let us consider the linear space L = 〈U iV j |i, j ∈ N0〉 with basis U•V •. Denote by ρ : Λ2 (K) → L (L) the linear map such that ρ (X) (UmV n) = Um+1V n, ρ (Y ) (V n) = V n+1, ρ (Y ) ( Um+1V n ) = ρ (Y ) ρ (Y ) (UmV n) + ρ (X) ρ (Y ) (UmV n) , m, n ∈ N0. We shall check that the map ρ is well defined. It is enough to prove that ρ (Y ) (UmV n) ⊆ 〈 U iV j ∣ ∣ i ≤ m 〉 for m ∈ N and n ∈ N0. We shall proceed by induction on m. If m = 1, then ρ (Y ) (UV n) = V n+2 + UV n+1 ⊆ 〈 U iV j ∣ ∣ i ≤ 1 〉 . Jo u rn al A lg eb ra D is cr et e M at h .62 Two-generated graded algebras We now assume that ρ (Y ) (UmV n) ⊆ 〈 U iV j ∣ ∣ i ≤ m 〉 for all m = 0, ..., l. If m = l + 1, then by the inductive assumption we have ρ (Y ) ( U l+1V n ) = ρ (Y ) ρ (Y ) ( U lV n ) + ρ (X) ρ (Y ) ( U lV n ) = ρ (Y ) ( ∑ i≤l,j αijU iV j) + ρ (X) ( ∑ i≤l,j αijU iV j) = ∑ i≤l,j αij ∑ i′≤i,j′ δiji′j′U i′V j′ + ∑ i≤l,j αijU i+1V j ⊆ 〈 U iV j ∣ ∣ i ≤ l + 1 〉 . Thus, the linear map ρ is well defined and ρ (Y ) ρ (X) = ρ (X) ρ (Y ) + ρ (Y ) ρ (Y ) . In fact, for basic elements we have ρ (Y ) ρ (X) (UmV n) = ρ (Y ) ( Um+1V n ) = ρ (X) ρ (Y ) (UmV n) + ρ (Y ) ρ (Y ) (UmV n) . Thus, ρ is an algebra homomorphism. Note that ρ ( XmY d ) (1) = UmV d. Now assume that monomials X•Y • are linear dependent, i.e. ∑ i,j αijX iY j = 0 for some coefficients αij ∈ K. Then ∑ i,j αijU iV j = ∑ i,j αijρ ( Xi ) ρ ( Y j ) (1) = 0, which is impossible since the monomials U•V • are linear independent. So, the monomials X•Y • are linear independent too. 2. We claim that the monomials X•Y • span Λ2 (K). It is enough to check that Y mXn = n ∑ l=0 ( n l ) (m+n−l−1)! (m−1)! X lY m+n−l for all m,n ∈ N. We shall proceed by induction on n. If n = 1, then an easy induction on m shows that Y mX = XY m +mY m+1, m ∈ N. Assume that for n = l our statement is holds. If n = l+1, then using the inductive assumption and Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 63 the equality Y mX = XY m +mY m+1, m ∈ N, we obtain Y mX l+1 = (Y mX l)X = l ∑ k=0 ( l k ) (m+ l − k − 1)! (m− 1)! XkY m+l−kX = l ∑ k=0 ( l k ) (m+l−k−1)! (m−1)! Xk ( XY m+l−k + (m+ l − k)Y m+l+1−k ) = l+1 ∑ k=1 ( l k − 1 ) (m+l−k)! (m−1)! X kY m+l+1−k + l ∑ k=0 ( l k ) (m+ l − k)! (m− 1)! XkY m+l+1−k = l ∑ k=1 (( l k − 1 ) + ( l k )) (m+l−k)! (m−1)! X kY m+l+1−k +X l+1Y m + (m+ l)! (m− 1)! Y m+l+1 = l+1 ∑ k=0 ( l + 1 k ) (m+l−k)! (m−1)! X kY m+l+1−k. In the same way one can prove Proposition 1.3. The basis of Λ1 (K, λ) is { XiY j |i, j ∈ N0 } . In parti- cular, Y mXn = λmnXnY m for all m,n ∈ N and Λ1 (K, λ) is a domain. In what follows we assume that elements of Λ1 (K, λ) or Λ2 (K) are presented in the canonical form ∑ αijX iY j . TheX−degree of an element w = n ∑ i=0 Xiϕi (Y ) of Λ1 (K, λ) or of Λ2 (K) is equal to n, provided ϕn 6= 0. We shall write degX w = n. It is clear that the family of linear spans of monomials of degree n, n ∈ N0, in X and Y induces the gradings of Λ1 (K, λ) and of Λ2 (K). So, these algebras are graded. Note that the algebras Λ1 (K, λ) and Λ2 (K) have the structure of iterated skew polynomial rings [4], [7]. Theorem 1.4. Λ1 (K, λ) ≇ Λ2 (K), λ ∈ K∗. Proof. Put A = K 〈X1, Y1〉/(Y1X1 − λX1Y1), B = K 〈X2, Y2〉 / ( Y2X2 −X2Y2 − Y 2 2 ) . Jo u rn al A lg eb ra D is cr et e M at h .64 Two-generated graded algebras Assume that there exists an isomorphism ψ : B → A. Let ψ (X2) = ∑ i,j αijX i 1Y j 1 , ψ (Y2) = ∑ i,j βijX i 1Y j 1 . Since Y2X2 = X2Y2 + Y 2 2 , we have ( ∑ i,j βijX i 1Y j 1 )( ∑ i,j αijX i 1Y j 1 ) = ( ∑ i,j αijX i 1Y j 1 )( ∑ i,j βijX i 1Y j 1 ) + ( ∑ i,j βijX i 1Y j 1 )2. Since the algebra A is graded, we can conclude that α00β00 = β2 00+α00β00 and (β10X1 + β01Y1) (α10X1 + α01Y1) + ( β20X 2 1 + β11X1Y1 + β02Y 2 1 ) α00 = (α10X1 + α01Y1) (β10X1 + β01Y1) + α00 ( β20X 2 1 + β11X1Y1 + β02Y 2 1 ) + (β10X1 + β01Y1) 2 . Therefore, β00 = 0 and β10X 2 1 +ξX1Y1+β01Y 2 1 = 0 for some ξ ∈ K. Since the monomials X2 1 , X1Y1 and Y 2 1 are linear independent, we can conclude that β10 = β01 = 0. Consider the inverse isomorphism ϕ = ψ−1 : A→ B. Let ϕ (X1) = ∑ i,j cijX i 2Y j 2 , ϕ (Y1) = ∑ i,j dijX i 2Y j 2 . Since Y1X1 = λX1Y1, we have ( ∑ i,j dijX i 2Y j 2 )( ∑ i,j cijX i 2Y j 2 ) = λ( ∑ i,j cijX i 2Y j 2 )( ∑ i,j dijX i 2Y j 2 ). Since the algebra B is graded, we have c00d00 = λc00d00. The algebra B is non-commutative, so λ 6= 1. Hence c00d00 = 0. Suppose that c00 = 0 and d00 6= 0. Since ϕ(X1) 6= 0, there exists a positive integer n such that cij = 0 provided that i + j < n and ci′j′ 6= 0 for some i′, j′ ∈ N, i′+j′ = n. Since the algebra B is graded, we have d00( ∑ i,j:i+j=n cijX i 2Y j 2 ) = λd00( ∑ i,j:i+j=n cijX i 2Y j 2 ). The monomialsX• 2Y • 2 are linear independent and λ 6= 1, thus ci′j′ = 0, a contradiction. Similarly, the case c00 6= 0 and d00 = 0 is impossible too. Thus, c00 = d00 = 0. Finally, we obtain ψ(Y2) = ∑ i,j:i+j≥2 βijX i 1Y j 1 , and Y2 = ϕ (ψ (Y2)) = ∑ i,j:i+j≥2 βij   ∑ i′,j′:i′+j′≥1 cijX i′ 2 Y j′ 2   i   ∑ i′,j′:i′+j′≥1 dijX i′ 2 Y j′ 2   j . But the polynomial ϕ (ψ (Y2)) either vanishes or the degree of each mono- mial of ϕ (ψ (Y2)) is at least 2. This contradiction proves Theorem 1.4. Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 65 2. Centre and Spectrum of Λ1 (K, λ) and Λ2 (K) As in Proposition 1.2 we have Proposition 2.1. Let w = ∑ i≥0,j≥0 αijX iY j ∈ Λ2 (K). Then Y w = ∑ k≥0 w (k) X Y k+1, where w (k) X = ∑ i≥k,j≥0 αij i! (i−k)!X i−kY j is the formal partial derivative w by X and wX = Xw + w′ Y Y 2, where w′ Y = ∑ i≥0,j≥1 jαijX iY j−1 is the formal partial derivative w by Y . The following Theorem 2.2 describes the centre Z (Λ2 (K)) of Λ2 (K) depending on the characteristic of K and the centre Z (Λ1 (K, λ)) of Λ1 (K, λ) depending on the parameter λ. Theorem 2.2. (i) If charK = 0, then Z (Λ2 (K)) = K; if charK = p > 0, then Z (Λ2 (K)) is the subalgebra generated by Xp, Y p. (ii) If λ is not a root of unity, then Z (Λ1 (K, λ)) = K; if λ is a root of unity of the degree m, m ∈ N, then Z (Λ1 (K, λ)) is the subalgebra generated by Xm, Y m. Proof. (i) Let f ∈ Z (Λ2 (K)) \ {0}, f = n ∑ i=0 Xiψi(Y ). Then Xf = fX and Y f = fY . By Proposition 2.1 we have fX = Xf + f ′Y Y 2 and Y f(X) = ∑ k≥0 f (k)(X)Y k+1. Therefore, f ′Y Y 2 = 0 and ∑ k≥1 f (k) X (X)Y k+1 = 0. Since the algebra Λ2 (K) is a domain, we conclude that f ′Y = 0. We shall consider two cases. Let first charK = 0. Since f ′Y = 0, we have ψ′ i(Y ) = 0, i.e. ψi(Y ) = ai ∈ K for all i = 0, ..., n. Hence f = n ∑ i=0 aiX i, an 6= 0. If n ≥ 1 then the coefficient in Xn−1Y 2 in ∑ k≥1 f (k) X (X)Y k+1 = 0 is equal to nan 6= 0, which is impossible. Hence n = 0 and f = a0. Thus, Z (Λ2 (K)) = K. Suppose secondly that charK = p > 0. By Proposition 2.1 elements Xp, Y p are central in Λ2 (K). Since f ′Y = 0, we have ψi(Y ) = ψ̃i(Y p) for all i = 0, ..., n and ψ̃i ∈ K [V ]. Set f1 = ∑ 1X iψ̃i (Y p), where the sum Σ1 Jo u rn al A lg eb ra D is cr et e M at h .66 Two-generated graded algebras is taken over all i = 0, ..., p such that p ∤ i. Since f = ∑ 2X iψ̃i (Y p) + f1, where the sum Σ2 is taken over all i = 0, ..., p such that p | i, we see that f1 ∈ Z (Λ2 (K)), i.e. ∑ k≥1 (f1) (k) X (X)Y k+1 = 0. Assume that f1 6= 0. Then degX f1 = n1, where p ∤ n1. Let ψ̃n1 (Y p) = l ∑ j=0 ajY pj , where al 6= 0. Then the coefficient in Xn1−1Y pl+2 in ∑ k≥1 (f1) (k) X (X)Y k+1 = 0 is equal to n1al 6= 0, a contradiction. Therefore, f1 = 0 and f = ∑ 2X iψ̃i (Y p). Now the proof follows. (ii)It is easy to check that if w = ∑ i≥0,j≥0 αijX iY j ∈ Λ1 (K, λ), then wX = ∑ i≥0,j≥0 αijX iY jX = ∑ i≥0,j≥0 αijλ jXi+1Y j , Y w = ∑ i≥0,j≥0 αijY X iY j = ∑ i≥0,j≥0 αijλ iXiY j+1. Let f = n ∑ i=0 Xiψi(Y ), ψn (Y ) 6= 0, be a non-zero central element in Λ1 (K, λ). Assume that λ is not a root of unity. Since Y f = fY , we have n ∑ i=0 λiXiψi(Y )Y = n ∑ i=0 Xiψi(Y )Y . In particular, λnψn (Y ) = ψn (Y ), i.e. λn = 1. Since λ is not a root of unity, we can conclude that n = 0, i.e. f = ψ0 (Y ). Let f = m ∑ j=0 αjY j , where αm 6= 0. Since Xf = fX, we have m ∑ j=0 αjλ jXY j = m ∑ j=0 αjXY j . In particular, αmλ m = αm, i.e. λm = 1. Since λ is not a root of unity, we can conclude that m = 0, f = α0. Thus, Z (Λ1 (K, λ)) = K. The case when λ is a root of unity of degree m is similar. Proposition 2.3. If charK = 0 and I is a proper two-sided ideal of the algebra Λ2 (K), then I ∩ K [Y ] = (Y n) for some n ∈ N. Proof. We first claim that I ∩ K [Y ] 6= (0). Choose an element f = m ∑ i=0 Xiψi(Y ) ∈ I\ {0} of least possible X−degree, m say, i.e. ψm(Y ) 6= 0. Assume that m 6= 0. Consider the element Y f − fY = ∑ k≥1 f (k) X Y k+1 ∈ I, degX(Y f−fY ) ≤ m−1. Because of the choice of f , we have Y f−fY = 0. The coefficient in Xm−1 in Y f − fY = 0 is equal to mψm(Y )Y 2 6= 0, a contradiction. So, m = 0 and f = ψ0(Y ) ∈ I ∩ K [Y ]. Clearly, I ∩ K [Y ] ⊳ K [Y ]. Since K [Y ] is a principal ideal ring, we have that Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 67 I ∩ K [Y ] = (ψ(Y )). Let ψ (Y ) = n ∑ i=0 aiY i and an 6= 0, n ≥ 1. Consider the element ψ(Y )X −Xψ(Y ) = ψ′(Y )Y 2 ∈ I. Then ψ′(Y )Y 2 ∈ (ψ(Y )), i.e. ψ′(Y )Y 2 = ψ(Y ) (αY + β). Let a0 = ... = aj−1 = 0 and aj 6= 0. Considering the coefficients in Y j , we get (j − 1)aj−2 = αaj−1 + βaj , i.e. β = 0. Therefore, ψ′(Y )Y 2 = αψ(Y )Y . Comparing coefficients in Y r in both sides, we get (r − 1) ar−1 = αar−1. Thus (r − 1 − α) ar−1 = 0. In particular α = n and therefore ai = 0 for all i < n. Theorem 2.4. If charK = 0 and I is a proper prime ideal of Λ2 (K), then either I = (Y ), or I = (Y, ψ(X)) for some irreducible polynomial ψ(X) ∈ K [X]. Proof. It is easy to check that the ideals (Y ) and (Y, ψ(X)), ψ(X) ∈ K [X] is irreducible, are prime. We claim that there is no other prime ideal in Λ2 (K). It follows from Proposition 2.3 that if I is a proper prime ideal of Λ2 (K), then I∩K [Y ] = (Y n)⊳K [Y ] for some n ∈ N. If n ≥ 2, then Y /∈ I, Y n−1 /∈ I and Y Λ2(K)Y n−1 ⊆ Λ2(K)Y n ⊆ I, i.e. the ideal I is not prime. Therefore, n = 1, i.e. Y ∈ I. Suppose that I 6= (Y ). Each element of Λ2 (K) can be represented in the form w (X,Y )Y + g (X) and therefore I ∩ K [X] = (ψ (X)) 6= 0 and I = (Y, ψ(X)). Now Λ2 (K)/I is isomorphic to K [X]/I ∩ K [X] and therefore ψ(X) ∈ K [X] is irreducible. Similarly, one can prove Theorem 2.5. If λ ∈ K∗ is not a root of unity and I is a proper prime ideal of Λ1 (K, λ), then I is one of ideals (X), (Y ), (X,ψ (Y )), (Y, ψ(X)) for some irreducible polynomial ψ in one variable. 3. Endomorphisms of Λ2 (K) Now we shall study endomorphisms of algebra Λ2 (K). Theorem 3.1. If charK = 0 and ϕ is an automorphism of the algebra Λ2 (K), then ϕ (X) = γX + g (Y ), ϕ (Y ) = γY for some γ ∈ K∗ and g (Y ) ∈ K [Y ]. Proof. From Theorem 2.4 it follows that (Y ) is a minimal nonzero prime ideal of Λ2 (K). Therefore, ϕ ((Y )) = (Y ), i.e. ϕ (Y ) = γY for some γ ∈ K∗. If ϕ−1 (X) = n ∑ i=0 Xiψi (Y ) thenX = ϕ ( ϕ−1 (X) ) = n ∑ i=0 (ϕ (X))i ψi (γY ). It is clear that degX ϕ (X) ≥ 1 and n ≥ 1. Then degX ϕ(X) = 1 and n = 1, i.e. ϕ(X) = Xf(Y ) + g(Y ) for some f 6= 0. Then X = ϕ ( ϕ−1 (X) ) = (Xf (Y ) + g (Y ))ψ1 (γY ) + ψ0 (γY ). In particular, 1 = Jo u rn al A lg eb ra D is cr et e M at h .68 Two-generated graded algebras f (Y )ψ1 (γY ). Then f = α ∈ K∗, i.e. ϕ (X) = αX + g (Y ). Since Y X = XY + Y 2, we have ϕ (Y )ϕ (X) = ϕ (X)ϕ (Y ) + (ϕ (Y ))2. Then we have γY (αX + g (Y )) = (αX + g (Y )) γY + γ2Y 2. Therefore, αY 2 = γY 2, i.e. α = γ. Proposition 3.2. If charK = 0 and ϕ is an endomorphism of the algebra Λ2 (K) with nonzero kernel, then ϕ (Y ) = 0. Proof. Since the algebra Λ2 (K) is a domain, we can conclude that the ideal kerϕ is prime. From Theorem 2.4 it follows that Y ∈ kerϕ, i.e. ϕ (Y ) = 0. Note that ϕ can take X to any element of Λ2 (K). Notes. So, in the case charK = 0 we have described the group of automorphisms of Λ2 (K) and all endomorphisms ϕ of Λ2 (K) when kerϕ 6= 0. It is clear from Theorem 3.1 that AutΛ2 (K) ∼= K∗ × K [Y ] with respect to the operation ◦ such that (γ2, g2 (Y )) ◦ (γ1, g1 (Y )) = (γ1γ2, γ1g2 (Y ) + g1 (γ2Y )). The semigroup EndΛ2 (K) has not been de- scribed yet. Note that there exist some endomorphisms ϕ : Λ2 (K) → Λ2 (K) such that kerϕ = 0 and Imϕ 6= Λ2 (K). It is easy to check that maps X 7→ n−1XY n−1 + g (Y ), Y 7→ Y n for all g (Y ) ∈ K [Y ], n ∈ N and X 7→ αX2 + βXY , Y 7→ 2αXY + 2 (α+ β)Y 2 for all α, β ∈ K satisfy these properties. One can prove that if ϕ ∈ EndΛ2 (K), then ϕ (Y ) = w (X,Y )Y for some w (X,Y ) ∈ Λ2 (K). Note that endomor- phisms of Λ1 (K, λ) are classified in [3]. 4. Derivations of Λ2 (K) In this section we shall consider derivations of the algebra Λ2 (K). All derivations of Λ1 (K, λ) in the case when λ is not a root of unity were classified in [2]. Notes. Let Λ be an algebra over field K. Recall that a K−linear map ∂ : Λ → Λ is a derivation of Λ if for all a, b ∈ Λ we have ∂ (ab) = ∂ (a) b+ a∂ (b). Given an element w ∈ Λ consider the inner derivation adw such that adw (a) = wa−aw, a ∈ Λ. The space of all derivations of Λ is a Lie algebra with respect to the operation of commutation. Denote this algebra by DerΛ. The subspace DerintΛ of inner derivations is always an ideal in DerΛ. Let L = DerΛ/DerintΛ be an algebra of outer derivations of Λ. Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 69 Proposition 4.1. Let w = XaY b ∈ Λ2 (K). Then adw (X) = bXaY b+1 adw (Y ) = − ∑ k≥1 (Xa)(k) Y b+k+1. Proof. By Proposition 2.1 we have adw(X) = wX −Xw = XaY bX −Xa+1Y b = Xa(XY b + bY b+1) −Xa+1Y b = bXaY b+1, adw(Y ) = wY − Y w = XaY b+1 − Y XaY b = XaY b+1 − ( ∑ k≥0 (Xa)(k) Y k+1)Y b = − ∑ k≥1 (Xa)(k)Y b+k+1. Proposition 4.2 (Derivations of Λ2(K)). (I) If charK = 0, then each derivation ∂ of Λ2 (K) can be represented in the form ∂ (X) = αY + ψ (X) + adw (X) , ∂ (Y ) = ψ′ (X)Y + adw (Y ) for some α ∈ K, ψ ∈ K [X], w ∈ Λ2 (K). (II) If charK = p > 2, then each derivation ∂ of Λ2 (K) can be represented in the form ∂ (X) = ψ (X) + T (Xp, Y p)Y + adw (X) , ∂ (Y ) = ψ′ (X)Y + S (Xp, Y p)Y Xp−1Y + adw (Y ) for some ψ ∈ K [X], T, S ∈ Z (Λ2 (K)), w ∈ Λ2 (K). (III) If charK = 2, the each derivation ∂ of Λ2 (K) can be represented in the form ∂ (X) = ψ (X) + T ( X2, Y 2 ) Y + adw (X) , ∂ (Y ) = ϕ (X) + ( ϕ′ (X) + ψ′ (X) ) Y + S ( X2, Y 2 ) Y XY + adw (Y ) for some ϕ,ψ ∈ K [X], T, S ∈ Z (Λ2 (K)), w ∈ Λ2 (K). Proof. The linear map ∂ : Λ2 (K) → Λ2 (K) is a derivation of Λ2 (K) if and only if ∂ (Y X) = ∂ (XY ) + ∂ ( Y 2 ) , i.e. ∂ (Y )X + Y ∂ (X) = ∂ (X)Y + X∂ (Y ) + ∂ (Y )Y + Y ∂ (Y ). It can easily be checked that if the linear map ∂ : Λ2 (K) → Λ2 (K) satisfies the conditions of Proposition 4.1, then ∂ is the derivation of Λ2 (K). We claim that there is no other derivations of Λ2 (K). Let ∂ ∈ DerΛ2 (K). Put U = ∂ (X), V = ∂ (Y ). Jo u rn al A lg eb ra D is cr et e M at h .70 Two-generated graded algebras Then V X + Y U = UY + XV + V Y + Y V . If U = m ∑ i=0 ψi (X)Y i and V = n ∑ i=0 ϕi (X)Y i, then by Proposition 2.1 we get V X = XV + V ′ Y Y 2, Y V = ∑ k≥0 V (k) X Y k+1, Y U = ∑ k≥0 U (k) X Y k+1. Hence, V ′ Y Y 2 + ∑ k≥1 U (k) X Y k+1 = 2V Y + ∑ k≥1 V (k) X Y k+1. (4.2.1) We shall consider three cases. (I) Let charK = 0. If m ≥ 2, then put w = m ∑ k=2 (k − 1)−1 ψk(X)Y k−1, ∂1 = ∂ − adw. From Proposition 4.1 we get ∂1 (X) = U − adw (X) = ψ1 (X)Y +ψ0 (Y ). Without loss of generality we can assume that ∂ = ∂1, i.e. ∂ (X) = ψ1 (X)Y + ψ0 (Y ). Consider the coefficients in Y , Y 2 and Y 3 in (4.2.1). We have, respectively, 2ϕ0 = 0, ϕ1 + ψ′ 0 = 2ϕ1 + ϕ′ 0, 2ϕ2 + ψ′ 1 + ψ′′ 0 = 2ϕ2 + ϕ′ 1 + ϕ′′ 0. Then ϕ0 = 0, ψ′ 0 = ϕ1, ψ ′ 1 = 0. Lemma. ϕr = ϕ (r−2) 2 for all r ≥ 2. Proof. We shall proceed by induction on r. The case r = 2 is clear. Assume that ϕr = ϕ (r−2) 2 for all r = 2, ..., k and consider the coefficients of Y k+2 in (4.2.1). We have (k+1)ϕk+1+ k ∑ i=0 ψ (k+1−i) i = 2ϕk+1+ k ∑ i=0 ϕ (k+1−i) i , where ψ2 = ... = ψk = 0, ϕ0 = 0 and ψ (k+1) 0 = ϕ (k) 1 . By induction ϕ (k+1−i) i = ϕ (k−1) 2 for all i = 2, ..., k. Therefore, (k + 1)ϕk+1 = 2ϕk+1 + (k − 1)ϕ (k−1) 2 , i.e. ϕk+1 = ϕ (k−1) 2 . So, ∂ (Y ) = ψ′ 0 (X)Y + ∑ k≥0 ϕ (k) 2 (X)Y k+2. From Proposition 4.1 it follows that ∑ k≥0 ϕ (k) 2 (X)Y k+2 = − (adΦ2 (X)) (Y ), where Φ2 ∈ K [X] and Φ′ 2 (X) = ϕ2 (X). Clearly, (ad Φ2 (X)) (X) = 0 and so ∂ (X) = ψ1 (X)Y + ψ0 (Y ) − (ad Φ2 (X)) (X) , ∂ (Y ) = ψ′ 0(X)Y − (adΦ2 (X)) (Y ) . (II) Let charK = p > 2. Then for any ψ ∈ K [X] we have ψ(p) (X) = 0. From (4.2.1) we get V ′ Y Y 2 + ∑ 1≤k≤p−1 U (k) X Y k+1 = 2V Y + ∑ 1≤k≤p−1 V (k) X Y k+1. (4.2.2) Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 71 From Proposition 4.1 it follows that ψk (X)Y k = ( ad ψk(X) k−1 Y k−1 ) (X) when p ∤ (k − 1). Put w = ∑ 1 ψk(X) k−1 Y k−1, where the sum ∑ 1 is taken over all k ≥ 1 such that p ∤ (k − 1), ∂1 = ∂ − adw. Then ∂1 (X) = ψ0 (X)+ l ∑ k=0 ψkp+1 (X)Y kp+1 for some l ∈ N0. Without loss of generality we can assume that ∂ = ∂1. As in the case charK = 0 it follows from (4.2.2) that ϕ0 = 0 and ψ′ 0 = ϕ1. Lemma. ψ′ np+1 = 0 and ϕnp+2+i = ϕ (i) np+2 for all n ≥ 0 and i = 0, . . . , p− 1. Proof. We shall proceed by induction on n. Let n = 0. As in the case charK = 0 by (4.2.2) we get ψ′ 1 = 0. Let us check by induction on i that ϕ2+i = ϕ (i) 2 for all i = 0, ..., p− 1. The case i = 0 is trivial. Assume that ϕ2+i = ϕ (i) 2 when 0 ≤ i ≤ k ≤ p− 2 and consider the coefficients in Y k+4 in (4.2.2). We have (k + 3)ϕk+3 + k+2 ∑ j=0 ψ (k+3−j) j = 2ϕk+3 + k+2 ∑ j=0 ϕ (k+3−j) j , where ϕ0 = 0, ψ′ 1 = 0, ψ (k+3) 0 = ϕ (k+2) 1 . Since k + 2 ≤ p, we have ψ2 = ... = ψk+2 = 0. By inductive assumption ϕ (k+3−j) j = ϕ (k+1) 2 when 2 ≤ j ≤ k+2. Combining these results, we get (k+1)ϕk+1 = (k+1)ϕ (k−1) 2 . Since 1 ≤ k + 1 ≤ p − 1, we have k + 1 6= 0 in the field K. So, ϕk+1 = ϕ (k−1) 2 . Assume that the statement of lemma holds for all n = 0, ...,m and consider the coefficients of Y p(m+1)+3 in (4.2.2). We have (2 + p(m+ 1))ϕp(m+1)+2 + p(m+1)+1 ∑ j=pm+3 ψ (p(m+1)+2−j) j = 2ϕp(m+1)+2 + p(m+1)+1 ∑ j=pm+3 ϕ (p(m+1)+2−j) j , where ψpm+3 = ... = ψp(m+1) = 0. By inductive assumption ϕ (p(m+1)+2−j) j = ϕ (p) pm+2 = 0 when pm+3 ≤ j ≤ p(m+1)+1. So, ψ′ p(m+1)+1 = (p−1)ϕ (p) pm+2 = 0. Now one can check by induction on i as above that ϕ(m+1)p+2+i = ϕ (i) (m+1)p+2 for all i = 0, ..., p− 1. Jo u rn al A lg eb ra D is cr et e M at h .72 Two-generated graded algebras Since ψ′ pj+1 = 0, 0 ≤ j ≤ l, we have ψpj+1 (X) = ψ̃j (Xp). So, U = ψ0(X) + l ∑ j=0 ψ̃j(X p)Y pj+1 = ψ0 (X) + Y T (Xp, Y p), where l ∑ j=0 ψ̃j(X p)Y pj = T (Xp, Y p) ∈ Z (Λ2 (K)). We have already proved that V = ψ′ 0(X)Y + ∑ k≥0 p−1 ∑ i=0 ϕ (i) pk+2(X)Y pk+2+i. If ψ(X) = Xpr+i, Ψ(X) = Xpr+i+1 i+1 , r ∈ N0, 0 ≤ i ≤ p− 2, then Ψ′ (X) = ψ (X) and from Proposi- tion 4.1 we get p−1 ∑ j=0 ψ(j) (X)Y pk+2+j = p ∑ j=1 Ψ(j) (X)Y pk+1+j = − ( adΨ (X)Y pk ) (Y ) , where ( adΨ (X)Y pk ) (X) = 0 for all k ≥ 0. Let ϕpk+2(X) = sk ∑ i=0 αkiX i, w = l ∑ k=0 ∑ 0≤i≤sk,p(i+1) αki Xi+1 i+1 Y pk, ∂ = ∂2 − adw. Then ∂2 (X) = ∂ (X) and ∂2 (Y ) = ψ′ 0 (X)Y + ∑ k≥0 ∑ i≥1,pi−1≤sk αkpi−1Y pk p ∑ j=1 (p−1)! (p−j)!X ip−jY j+1 = ψ′ 0 (X)Y + ( ∑ k≥0 ∑ i≥1,pi−1≤sk αkpi−1X i(p−1)Y pk)( p ∑ j=1 (p−1)! (p−j)!X p−jY j+1) = ψ′ 0 (X)Y +R (X,Y )S (Xp, Y p) , where S (Xp, Y p) = ∑ k≥0 ∑ i≥1,pi−1≤sk αkpi−1X i(p−1)Y pk ∈ Z (Λ2 (K)) , R (X,Y ) = p ∑ j=1 (p− 1)! (p− j)! Xp−jY j+1 = Y Xp−1Y. (III) Let charK = 2. Then for any ψ ∈ K [X] we have ψ′′ (X) = 0. From (4.2.1) we get V ′ Y Y 2 + U ′ XY 2 = V ′ XY 2, i.e. V ′ Y + U ′ X = V ′ X (4.2.3) From Proposition 4.1 it follows that ψk (X)Y k = ( ad ψk(X) k−1 Y k−1 ) (X) when k ∈ N, 2 | k. Put w = ∑ 1 ψk(X) k−1 Y k−1, where the sum ∑ 1 is Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 73 taken over all k ≥ 1 such that 2 | k, ∂1 = ∂ − adw. Then ∂1 (X) = ψ0 (X)+ l ∑ k=0 ψ2k+1 (X)Y 2k+1 for some l ∈ N0. As above we shall assume that ∂ = ∂1. From (4.2.3) it follows that ϕ0 can be equal to any element of K [X]. Considering monomials in X in (4.2.3) we have ϕ1 + ψ′ 0 = ϕ′ 0, i.e. ϕ1 = ψ′ 0 + ϕ′ 0. Then ϕ′ 1 = ψ′′ 0 + ϕ′′ 0 = 0. Consider the coefficients in Y 2n, n ∈ N, in (4.2.3). We have ϕ2n+1 + ψ′ 2n = ϕ′ 2n, where ψ2n = 0. Then ϕ2n+1 = ϕ′ 2n and ϕ′ 2n+1 = ϕ′′ 2n = 0. Considering the coefficients of Y 2n−1, n ∈ N, in (4.2.3) we have ψ′ 2n−1 = ϕ′ 2n−1 = 0. Thus, ∂(X) = ψ0(X) + l ∑ j=0 ψ̃j(X 2)Y 2j+1 = ψ0(X) + T (X2, Y 2)Y, ∂(Y ) = ϕ0(X) + (ϕ′ 0(X) + ψ′ 0(X))Y + ∑ j≥1 (ϕ2j(X)Y 2j + ϕ′ 2j(X)Y 2j+1), where T ( X2, Y 2 ) = l ∑ j=0 ψ̃j ( X2 ) Y 2j ∈ Z (Λ2 (K)). As in the case charK = p > 2 it is easily shown that ∂ (Y ) = ϕ0 (X)+ ( ϕ′ 0 (X) + ψ′ 0 (X) ) Y +R (X,Y )S ( X2, Y 2 ) +adw (Y ) , where S ( X2, Y 2 ) ∈ Z (Λ2 (K)), R (X,Y ) = XY 2 + Y 3 = Y XY , w ∈ Λ2 (K) and adw (X) = 0. The next propositions are technical. We briefly indicate their proofs. Proposition 4.3. For any n ∈ N Qn(X,Y ) = n−1 ∑ k=0 XkY Xn−1−k = n−1 ∑ k=0 (n− 1 − k)! ( n k ) XkY n−k = n−1 ∑ k=0 n! k! (n− k) XkY n−k. The proof is a direct calculation based on Proposition 2.1. Proposition 4.4. If ∂ ∈ DerΛ2 (K) and ϕ ∈ K [X], then ∂ (ϕ (X)) = ϕ′ (X) ∂ (X) + adwϕ (X) for some wϕ ∈ Λ2 (K). Proof. From Proposition 4.2 it follows that ∂(X) = zY +ψ(X) for some z ∈ Z (Λ2 (K)) and ψ ∈ K [X]. It is enough to check our statement for Jo u rn al A lg eb ra D is cr et e M at h .74 Two-generated graded algebras ϕ (X) = Xn, n ∈ N0. The cases n = 0 and n = 1 are clear. If n ≥ 2, then from Proposition 4.3 we obtain ∂ (Xn) = nψ (X)Xn−1 + znXn−1Y + z n−2 ∑ k=0 (n− 1 − k)! ( n k ) XkY n−k. From Proposition 4.1 it follows that z n−2 ∑ k=0 (n− 1 − k)! ( n k ) XkY n−k = adwϕ (X) , where wϕ = z n−2 ∑ k=0 (n− 2 − k)! ( n k ) XkY n−k−1. Thus, ∂ (Xn) = (Xn)′ ∂(X) + adwϕ (X) . Proposition 4.5. Let charK = 0 and ∂1, ∂2 ∈ DerΛ2 (K), where ∂i (X) = αiY + ψi (X) , ∂i (Y ) = ψ′ i (X)Y for some αi ∈ K, ψi ∈ K [X] , i = 1, 2. Then there exists an element w ∈ Λ2 (K) such that [∂1, ∂2] (X) = ψ1 (X)ψ′ 2 (X) − ψ′ 1 (X)ψ2 (X) + adw (X) , [∂1, ∂2] (Y ) = ( ψ1 (X)ψ′′ 2 (X) − ψ′′ 1 (X)ψ2 (X) ) Y + adw (Y ) . Proof. From Proposition 4.4 we get [∂1, ∂2] (X) = ψ1 (X)ψ′ 2 (X) − ψ′ 1 (X)ψ2 (X) + ad w̃ (X) for some w̃ ∈ Λ2 (K). If ∂ = [∂1, ∂2] − ad w̃, then ∂ (X) = ψ1 (X)ψ′ 2 (X) − ψ′ 1 (X)ψ2 (X) . As in the proof of Proposition 4.2 it is easily shown that ∂ (Y ) = ( ψ1 (X)ψ′ 2 (X) − ψ′ 1 (X)ψ2 (X) )′ Y + ad ˜̃w (Y ) for some ˜̃w ∈ Λ2 (K), where ad ˜̃w (X) = 0. Let w = w̃ + ˜̃w. Finally, we obtain [∂1, ∂2] (X) = ψ1 (X)ψ′ 2 (X) − ψ′ 1 (X)ψ2 (X) + adw (X) , [∂1, ∂2] (Y ) = ψ1 (X)ψ′′ 2 (X) − ψ′′ 1 (X)ψ2 (X) + adw (Y ) . Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 75 Combining Propositions 4.2 and 4.5 we obtain Theorem 4.6. If charK = 0, then each derivation ∂ of Λ2 (K) can be represented in the form ∂ (X) = αY + ψ (X) + adw (X), ∂ (Y ) = ψ′ (X)Y +adw (Y ), where α ∈ K, ψ ∈ K [X], w ∈ Λ2 (K) and the Lie al- gebra of outer derivations of Λ2 (K) is isomorphic to the algebra K⊕K [X] with respect to the operation [·, ·] such that [(α1, ψ1 (X)) , (α2, ψ2 (X))] = ( 0, ψ1 (X)ψ′ 2 (X) − ψ′ 1 (X)ψ2 (X) ) . Note that if Λ is an algebra over field K, ∂ ∈ DerΛ and z ∈ Z (Λ), then ∂ (z) ∈ Z (Λ). Indeed, for any w ∈ Λ we have ∂ (w) z+w∂ (z) = ∂ (wz) = ∂ (zw) = ∂ (z)w+z∂ (w) = ∂ (z)w+∂ (w) z, i.e. ∂ (z)w = w∂ (z). From Theorem 2.2 it follows that if z ∈ Z (Λ2 (K)), then z = ∑ i≥0,j≥0 αijX ipY jp. Put z′Xp = ∑ i≥1,j≥0 iαijX (i−1)pY jp, z′Y p = ∑ i≥0,j≥1 jαijX ipY (j−1)p. Proposition 4.7. If charK = p > 0, z ∈ Z (Λ2 (K)), ∂ ∈ DerΛ2 (K), ∂ (X) = ψ + Y T , ∂ (Y ) = ψ′Y + Y Xp−1Y S, where ψ ∈ K [X], T, S ∈ Z (Λ2(K)), then ∂(z) = − ( z′XpT (Xp, Y p)Y p + z′Y pS (Xp, Y p)Y 2p ) . The proof is a direct calculation based on Propositions 1.2 and 4.3. Proposition 4.8. If charK = p > 0, ∂ ∈ DerΛ2 (K), ∂ (X) = ψ + Y T , ∂ (Y ) = ψ′Y + Y Xp−1Y S, where ψ ∈ K [X], T, S ∈ Z (Λ2 (K)), then for some coefficients αij ∈ K ∂ ( Y Xp−1Y ) = ψ′Y Xp−1Y + 2Y Xp−1Y Xp−1Y S + ∑ αijX iY j , where the sum is taken over all i, j such that j ≥ 2, p ∤ i+ 1. The proof is a direct calculation based on Propositions 2.1 and 4.4. Proposition 4.9. If charK = p > 0, ∂i ∈ DerΛ2 (K), i = 1, 2, ∂i (X) = ψi + Y Ti, ∂i (Y ) = ψ′ iY + Y Xp−1Y Si, where ψi ∈ K [X], Ti, Si ∈ Z (Λ2 (K)), then for some w ∈ Λ2 (K) [∂1, ∂2] (X) = adw (X) + ψ1ψ ′ 2 − ψ′ 1ψ2 − Y p+1 (S1T2 − S2T1) + Y ( Y p ( (T1) ′ Xp T2 − (T2) ′ Xp T1 ) +Y 2p ( (T1) ′ Y p S2 − (T2) ′ Y p S1 )) , [∂1, ∂2] (Y ) = adw (Y ) + ( ψ1ψ ′′ 2 − ψ′′ 1ψ2 ) Y + Y Xp−1Y (( (S1) ′ Xp T2 − (S2) ′ Xp T1 ) Y p + ( (S1) ′ Y p S2 − (S2) ′ Y p S1 ) Y 2p ) . Jo u rn al A lg eb ra D is cr et e M at h .76 Two-generated graded algebras The proof is a direct calculation based on Propositions 2.1, 4.1, 4.2, 4.4, 4.7, 4.8. Combining Propositions 4.2 and 4.9 we obtain Theorem 4.10. If charK = p > 2, then each derivation ∂ of Λ2 (K) can be written in the form ∂ (X) = ψ (X) + T (Xp, Y p)Y + adw (X) , ∂ (Y ) = ψ′ (X)Y + S (Xp, Y p)Y Xp−1Y + adw (Y ) , where ψ ∈ K [X], T, S ∈ Z (Λ2 (K)), w ∈ Λ2 (K) and the Lie alge- bra of outer derivations of Λ2 (K) is isomorphic to the algebra K [X] ⊕ K [Xp, Y p] ⊕ K [Xp, Y p] with respect to the operation [·, ·] such that [(ψ1, T1, S1) , (ψ2, T2, S2)] = (ψ, T, S) , where ψ = ψ1ψ ′ 2 − ψ′ 1ψ2, T = −Y p (S1T2 − S2T1) + Y p ( (T1) ′ Xp T2 − (T2) ′ XP T1 ) + Y 2p ( (T1) ′ Y p S2 − (T2) ′ Y p S1 ) , S = Y p ( (S1) ′ XpT2 − (S2) ′ XpT1 ) + Y 2p ( (S1) ′ Y pS2 − (S2) ′ Y pS1 ) . Now we shall consider the Lie algebra of outer derivations of Λ2 (K) in the case charK = 2. As above we shall omit technical details. Proposition 4.11. If charK = 2 and θ ∈ K [X], then θ′ ∈ Z (Λ2 (K)). Proof. It is enough to consider the case θ = Xn, n ∈ N . If n = 2k, then θ′ = nXn−1 = 0. If n = 2k+ 1, then θ′ = nXn−1 = X2k. From Theorem 2.2 we get θ′ ∈ Z (Λ2 (K)). Proposition 4.12. If charK = 2, z ∈ Z (Λ2 (K)), ∂ ∈ DerΛ2 (K), ∂ (X) = ψ+Y T , ∂ (Y ) = ϕ+ (ψ′ + ϕ′)Y +Y XY S, where ψ,ϕ ∈ K [X], T, S ∈ Z (Λ2 (K)), then ∂(z) = z′ X2TY 2 + z′ Y 2 ( SY 2 + ϕ′ ) Y 2. Proof. Consider derivations ∂i ∈ DerΛ2(K), i = 1, 2, ∂1 (X) = ψ + Y T , ∂1 (Y ) = ψ′Y +Y XY S, ∂2 (X) = 0, ∂ (Y ) = ϕ+ϕ′Y . Then ∂ = ∂1 +∂2. It remains to use Propositions 2.1 and 4.7. Proposition 4.13. If charK = 2, ∂ ∈ DerΛ2 (K), ∂ (X) = ψ + Y T , ∂ (Y ) = ϕ+(ψ′ + ϕ′)Y +Y XY S, where ψ,ϕ ∈ K [X], T, S ∈ Z (Λ2 (K)), then for some coefficients αij ∈ K ∂ (Y XY ) = ψ′Y XY + ∑ αijX iY j, where the sum is taken over all i, j such that j ≥ 2, 2 | i. Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 77 Proof. Consider derivations ∂i ∈ DerΛ2(K), i = 1, 2, ∂1 (X) = ψ + Y T , ∂1 (Y ) = ψ′Y + Y XY S, ∂2 (X) = 0, ∂ (Y ) = ϕ + ϕ′Y . Then ∂ = ∂1 +∂2. It follows from Proposition 4.8 that for some coefficients βij ∈ K ∂1 (Y XY ) = ψ′Y XY + ∑ 1 βijX iY j , where the sum ∑ 1 is taken over all i, j such that j ≥ 2, 2 | i. From Propositions 2.1 and 4.11 we get for some coefficients γij ∈ K ∂2 (Y XY ) = ∂2 (Y )XY + Y X∂2 (Y ) = ( ϕ+ ϕ′Y ) XY + Y X ( ϕ+ ϕ′Y ) = ϕXY + Y ϕX = ϕXY + ϕXY + (ϕX)′ Y 2 = (ϕX)′ Y 2 = ∑ 1 γijX iY j Finally, we obtain for some coefficients αij ∈ K ∂(Y XY ) = ∂1(Y XY ) + ∂2(Y XY ) = ψ′Y XY + ∑ 1 αijX iY j . Proposition 4.14. If charK = 2, θ ∈ K [X], ∂ ∈ DerΛ2 (K), ∂ (X) = ψ + Y T , ∂ (Y ) = ϕ + (ψ′ + ϕ′)Y + Y XY S, where ψ,ϕ ∈ K [X], T = P +QY 2, P = P ( X2 ) , P,Q, S ∈ Z (Λ2 (K)), then for some coefficients γij ∈ K ∂ (θ) = ψθ′+Y Pθ′+Y XY (θ′)′X2 T + ∑ γijX iY j, where the sum is taken over all i, j such that j ≥ 2, 2 | i. Proof. Put z = (Xθ)′. From Proposition 4.11 we get z ∈ Z (Λ2 (K)). It is clear that θ = Xθ′ + z. From Proposition 4.12 we obtain for some coefficients αij ∈ K ∂ (z) = z′ X2TY 2 + z′ Y 2 ( SY 2 + ϕ′ ) Y 2 = z′ X2TY 2 = ∑ 1 αijX iY j , where the sum ∑ 1 is taken over all i, j such that j ≥ 2, 2 | i. Since XY 2 = Y XY + Y 3, we can conclude that for some coefficients βij ∈ K ∂ ( Xθ′ ) = ∂ (X) θ′ +X∂ ( θ′ ) = ( ψ + Y ( P +QY 2 )) θ′ +XY 2 ( θ′ )′ X2 T = ψθ′ + Y Pθ′ + Y XY ( θ′ )′ X2 T + ( Qθ′ + ( θ′ )′ X2 T ) Y 3 = ψθ′ + Y Pθ′ + Y XY (θ′)′X2 T + ∑ 1 βijX iY j Finally, we obtain for some coefficients γij ∈ K ∂ (θ) = ψθ′ + Y Pθ′ + Y XY (θ′)′X2 T + ∑ 1 γijX iY j . Jo u rn al A lg eb ra D is cr et e M at h .78 Two-generated graded algebras Proposition 4.15. If charK = 2, ∂i ∈ DerΛ2 (K), i = 1, 2, ∂i (X) = ψi + Y Ti, ∂i (Y ) = ϕi + (ψ′ i + ϕ′ i)Y + Y XY Si, where ψi, ϕi ∈ K [X], Ti = Pi +QiY 2, Pi = Pi ( X2 ) , Pi, Qi, Si ∈ Z (Λ2 (K)), then [∂1, ∂2] (X) = ψ + Y T + adw (X) , [∂1, ∂2] (Y ) = ϕ+ ( ψ′ + ϕ′ ) Y + Y XY S + adw (Y ) , where ψ = (ψ1ψ2) ′ + ϕ1P2 + ϕ2P1, ϕ = (ψ1ϕ2 + ψ2ϕ1 + ϕ1ϕ2) ′ , T = ϕ′ 1T2 + ϕ′ 2T1 + ( (T2) ′ Y 2 ϕ ′ 1 + (T1) ′ Y 2 ϕ ′ 2 ) Y 2 + (S1T2 + S2T1)Y 2 + ( (T2) ′ X2 T1 + (T1) ′ X2 T2 ) Y 2 + ( (T2) ′ Y 2 S1 + (T1) ′ Y 2 S2 ) Y 4, S = ( ϕ′ 2 )′ X2 T1 + ( ϕ′ 1 )′ X2 T2 + ϕ′ 2S1 + ϕ′ 1S2 + ( (S2) ′ Y 2 ϕ ′ 1 + (S1) ′ Y 2 ϕ ′ 2 ) Y 2 + ( (S2) ′ X2 T1 + (S1) ′ X2 T2 ) Y 2 + ( (S2) ′ Y 2 S1 + (S1) ′ Y 2 S2 ) Y 4. Proof. Apply Proposition 4.1, 4.4, 4.12, 4.13, 4.14. Combining Propositions 4.2 and 4.15 we obtain Theorem 4.16. If charK = 2, then each derivation ∂ of Λ2 (K) can be represented in the form ∂ (X) = ψ (X) + T ( X2, Y 2 ) Y + adw (X) , ∂ (Y ) = ϕ (X) + ( ϕ′ (X) + ψ′ (X) ) Y + S ( X2, Y 2 ) Y XY + adw (Y ) , where ϕ,ψ ∈ K [X], T = P + QY 2, P = P ( X2 ) , P,Q, S ∈ Z (Λ2 (K)), w ∈ Λ2 (K), and the Lie algebra of outer derivations of Λ2 (K) is isomor- phic to the algebra K [X] ⊕ K [X] ⊕ K [ X2 ] ⊕ K [ X2, Y 2 ] ⊕ K [ X2, Y 2 ] with respect to the operation [·, ·] such that [(ψ1, ϕ1, P1, Q1, S1) , (ψ1, ϕ1, P1, Q1, S1)] = (ψ,ϕ, P,Q, S) , where ψ = (ψ1ψ2) ′ + ϕ1P2 + ϕ2P1, ϕ = (ψ1ϕ2 + ψ2ϕ1 + ϕ1ϕ2) ′ , P = ϕ′ 1P2 + ϕ′ 2P1, Q = ϕ′ 1Q2 + ϕ′ 2Q1 + (T2) ′ Y 2 ϕ ′ 1 + (T1) ′ Y 2 ϕ ′ 1 + S1T2 + S2T1 + (T2) ′ X2 T1 + (T1) ′ X2 T2 + ( (T2) ′ Y 2 S1 + (T1) ′ Y 2 S2 ) Y 2, S = ( ϕ′ 2 )′ X2 T1 + ( ϕ′ 1 )′ X2 T2 + ϕ′ 2S1 + ϕ′ 1S2 + ( (S2) ′ Y 2 ϕ′ 1 + (S1) ′ Y 2 ϕ′ 2 ) Y 2 + ( (S2) ′ X2 T1 + (S1) ′ X2 T2 ) Y 2 + ( (S2) ′ Y 2 S1 + (S1) ′ Y 2 S2 ) Y 4, where Ti = Pi +QiY 2, i = 1, 2. Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 79 The Lie algebra of derivations of Λ2 (K) in characteristic p > 0 is a p−algebra. We shall consider this structure. Let d dx : k [X] → K [X] be the operator of formal differentiation, i.e. d dx (ψ (X)) = ψ′ (X) and mθ : K [X] → K [X] be the operator of multiplication by θ(X) ∈ K [X], i.e. mθ (ψ (X)) = θ (X)ψ (X). Proposition 4.17. If ∂ ∈ DerΛ2 (K), ∂ (X) = ψ, ∂ (Y ) = ψ′Y , where ψ ∈ K [X], then for all n ∈ N ∂n(X) = (mψ ◦ d dx )n−1(ψ), ∂n(Y ) = ( d dx ◦mψ)n(1)Y. The proof is a direct calculation. Suppose that charK = p > 0. Let ∂ ∂Xp : Z (Λ2 (K)) → Z (Λ2 (K)) be the operator of formal differentiation by Xp, i.e. ∂ ∂Xp (z) = zXp , ∂ ∂Y p : Z (Λ2 (K)) → Z (Λ2 (K)) be the operator of formal differentiation by Y p, i.e. ∂ ∂Y p (z) = zY p and mw : Z (Λ2 (K)) → Z (Λ2 (K)) be the operator of multiplication by w ∈ Z (Λ2 (K)), i.e. mw (z) = wz. Put d = mT ◦ ∂ ∂Xp +mS ◦mY p ◦ ∂ ∂Y p : Z (Λ2 (K)) → Z (Λ2 (K)). Proposition 4.18. If charK = p > 2, ∂ ∈ DerΛ2 (K), ∂ (X) = Y T , ∂ (Y ) = Y Xp−1Y S, where T, S ∈ Z (Λ2 (K)), then for some w ∈ Λ2 (K) ∂p (X) = Y (d ◦mY p)p−1 (T ) + adw (X) , ∂p (Y ) = Y Xp−1Y (mS + d) ◦ (d ◦mY p)p−1 (1) + adw (Y ) . The proof is a direct calculation based on Propositions 4.2, 4.7. As in [5] it is easy to prove Proposition 4.19. If Λ is a K-algebra, chark = p > 0, ∂i ∈ DerΛ, i = 1, 2, [∂1, ∂2] = adw for some w ∈ Λ, then (∂1 + ∂2) p = ∂p1 +∂p2 +adu for some u ∈ Λ. Theorem 4.20. If charK = p > 2, ∂ ∈ DerΛ2 (K), ∂(X) = ψ + Y T , ∂(Y ) = ψ′Y + Y Xp−1Y S, where ψ ∈ K [X], T, S ∈ Z (Λ2 (K)), then ∂p (X) = ψ̃ + Y T̃ + adw (X) , ∂p (Y ) = ψ̃′Y + Y Xp−1Y S̃ + adw (Y ) , where w ∈ Λ2 (K), ψ̃ = ( mψ ◦ d dx )p−1 (ψ), T̃ = (d ◦mY p)p−1 (T ), S̃ = (mS + d) ◦ (d ◦mY p)p−1 (1). Proof. Consider derivations ∂i ∈ DerΛ2 (K), i = 1, 2, where ∂1 (X) = ψ, ∂1 (Y ) = ψ′Y , ∂2 (X) = Y T , ∂ (Y ) = Y Xp−1Y S. Then ∂ = ∂1 + ∂2. It follows from Proposition 4.9 that [∂1, ∂2] = adw for some w ∈ Λ2(K). Now the statement of Proposition 4.20 follows from Propositions 4.17, 4.18 and 4.19. Jo u rn al A lg eb ra D is cr et e M at h .80 Two-generated graded algebras Proposition 4.21. If charK = 2, ∂ ∈ DerΛ2 (K), ∂ (X) = Y T , ∂ (Y ) = Y XY S, where T, S ∈ Z (Λ2 (K)), then for some w ∈ Λ2 (K) ∂2 (X) = Y (d ◦mY 2) (T ) + adw (X) , ∂2 (Y ) = Y XY (mS + d) ◦ (d ◦mY 2) (1) + adw (Y ) . Proof. Since ∂2 ∈ DerΛ2 (K), it follows from Proposition 4.2 that ∂2 (X) = ψ̃ + Y T̃ + adw (X) , ∂2 (Y ) = ϕ̃+ ( ψ̃′ + ϕ̃′ ) Y + Y XY S̃ + adw (Y ) , here w ∈ Λ2 (K), ψ̃, ϕ̃ ∈ K [X], T̃ , S̃ ∈ Z (Λ2 (K)). As in the proof of Proposition 4.18 it is easily shown that ψ̃ = 0. Since ∂2 (Y ) = ∂ ((Y XS)Y ) = (∂ (Y XS) + Y XSY XS)Y , we get ϕ̃ = 0. The following argumentation is the same as in the proof of Proposition 4.18. Proposition 4.22. If charK = 2, ∂ ∈ DerΛ2 (K), ∂ (X) = 0, ∂ (Y ) = ϕ+ ϕ′Y , where ϕ ∈ K [X], then ∂2 (X) = 0, ∂2 (Y ) = ϕϕ′ + (ϕ′)2 Y . Proof. We have ∂2 (X) = 0, ∂2 (Y ) = ∂ (ϕ+ ϕ′Y ) = ϕ′∂ (Y ) = ϕϕ′ + (ϕ′)2 Y . Theorem 4.23. If charK = 2, ∂ ∈ DerΛ2 (K), ∂ (X) = ψ+Y T , ∂ (Y ) = ϕ+(ψ′ + ϕ′)Y +Y XY S, where ϕ,ψ ∈ K [X], T = P+QY 2, P = P ( X2 ) , P,Q, S ∈ Z (Λ2 (K)), then ∂2 (X) = ψ̃ + Y T̃ + adw (X) , ∂2 (Y ) = ϕ̃+ (ψ̃′ + ϕ̃′)Y + Y XY S̃ + adw (Y ) , where T̃ = (d ◦mY 2) (T ) +ϕ′ ( T + T ′ Y 2 ) , S̃ = (mS + d) ◦ (d ◦mY 2) (1) + (ϕ′)′X2 T + ϕ′S + ϕ′S′ Y 2Y 2, ψ̃ = ( mψ ◦ d dx )2 (X) + ϕP , w ∈ Λ2 (K), ϕ̃ = ϕϕ′. Proof. Consider derivations ∂i ∈ DerΛ2(K), i = 1, 2, 3, where ∂1 (X) = ψ, ∂1 (Y ) = ψ′Y , ∂2 (X) = Y T , ∂ (Y ) = Y Xp−1Y S, ∂3 (X) = 0, ∂3 (Y ) = ϕ + ϕ′Y . Then ∂ = ∂1 + ∂2 + ∂3. From Proposition 4.15 we get [∂1, ∂2] = adw12 for some w12 ∈ Λ2 (K), [∂1, ∂3] = ∂13 + adw13, where ∂13 (X) = 0, ∂13 (Y ) = (ψϕ)′ and w13 ∈ Λ2 (K), [∂2, ∂3] = ∂23 + adw23, where ∂23 (X) = ϕP + ϕ′ ( T + T ′ Y 2 ) Y , ∂23 (Y ) = (ϕP )′ Y + Y XY ( (ϕ′)′X2 T + ϕ′S + ϕ′S′ Y 2Y 2 ) , w23 ∈ Λ2 (K). We have ∂2 = (∂1 + ∂2 + ∂3) 2 = ∂2 1 + ∂2 2 + ∂2 3 + ∂1∂2 + ∂2∂1 + ∂1∂3 + ∂3∂1 + ∂2∂3 + ∂3∂2 = ∂2 1 + ∂2 2 + ∂2 3 + ∂13 + ∂23 + ad(w12 + w13 + w23). It remains to use Propositions 4.17, 4.21 and 4.22. Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 81 5. Classification Theorems Let A = ∞ ⊕ n=0 An be an associative graded algebra over field A0 = K generated by elements X,Y ∈ A1. Suppose that dimA2 = 3. Then monomials X2, Y 2, XY and Y X are linear dependent over K, so there exists a unique to proportionality set of coefficients (α, β, γ, δ) ∈ K4\{0} such that αX2 + βY 2 + γXY + δY X = 0. (5.0.4) Note that similar algebras over field of zero characteristic are considered in [8]. Proposition 5.1. If A is an algebra without zero divisors, then αβ−γδ 6= 0. Proof. Assume that αβ − γδ = 0. If δ = 0 then either α = 0 or β = 0. Consider the case α = 0. Since (α, β, γ, δ) ∈ K4\{0}, we see that either β 6= 0 or γ 6= 0. Since dimA2 = 3, we can conclude that X and Y are linear independent over K. Thus, βY + γX 6= 0 and 0 = βY 2 + γXY = (βY +γX)Y , which is impossible since A has no zero divisors. Similarly, if β = 0, then αX + γY 6= 0 and 0 = αX2 + γXY = X (αX + γY ), a contradiction. Therefore δ 6= 0. Then αX + δY 6= 0, δX + βY 6= 0 and (αX + δY )(δX + βY ) = δ(αX2 + βY 2 + γXY + δY X) = 0. This contradiction proves the 5.1. Proposition 5.2. Suppose that K has no quadratic extensions and αβ− γδ 6= 0. Then there exist generators X1 and Y1 such that either Y1X1 = λX1Y1 for some λ ∈ K∗ or Y1X1 = X1Y1 + Y 2 1 . Proof. We shall consider two cases. Let first α 6= 0, β = 0. Put X = Y1, Y = X1. Suppose secondly that α 6= 0, β 6= 0. We shall find X1 and Y1 such that X = X1, Y = ξX1 + Y1, where ξ ∈ K. We shall latter specify the value of parameter ξ. If we replace X by X1 and Y by ξX1 + Y1 in (5.0.4), then we get αX2 1 + β (ξX1 + Y1) 2 + γX1 (ξX1 + Y1) + δ (ξX1 + Y1)X1 = = α̃X2 1 + β̃Y 2 1 + γ̃X1Y1 + δ̃Y1X1 = 0, where α̃ = α + βξ2 + γξ + δξ. Since β 6= 0, it follows that there exists an element ξ ∈ K such that α̃ = 0. We claim that α̃β̃ − γ̃δ̃ 6= 0. Indeed, Jo u rn al A lg eb ra D is cr et e M at h .82 Two-generated graded algebras the coefficients of quadratic form αX2 +βY 2 +γXY + δY X are changed under the substitution X = X1, Y = ξX1 + Y1 according to the rule ( α γ δ β ) 7→ ( α̃ γ̃ δ̃ β̃ ) = ( 1 0 ξ 1 )T ( α γ δ β ) ( 1 0 ξ 1 ) . Then α̃β̃ − γ̃δ̃ = −γ̃δ̃ = det ( α̃ γ̃ δ̃ β̃ ) = (αβ − γδ) 6= 0. Thus, without loss of generality, we can assume that α = 0. Then γδ 6= 0 and Y X = −γδ−1XY − βδ−1Y 2 = γ1Y X + β1Y 2, (5.2.5) where γ1 6= 0. If we replace X by X1 + ζY1 for some ζ ∈ K and Y by Y1 in (5.2.5), then we get Y1 (X1 + ζY1) = γ1 (X1 + ζY1)Y1 + β1Y 2 1 . Therefore, Y1X1 = γ1X1Y1+(β1 + γ1ζ − ζ)Y 2 1 . If either γ1 6= 1 or β1 = 0, then there exists an element ζ ∈ K such that β1 + γ1ζ − ζ = 0. Thus we have Y1X1 = γ1X1Y1. In the converse case γ1 = 1 and β1 6= 0, i.e. Y X = XY+β1Y 2. LetX = β1X1, Y = Y1. Then Y1X1 = X1Y1+Y 2 1 . So, without loss of generality, we can assume that generators X and Y satisfy either the equality Y X = λXY , λ ∈ K∗, or the equality Y X = XY + Y 2. The following theorems shows that if some additional conditions hold true, then there exist only two classes of these algebras, namely quantum polynomials in two variables and Jordanian plane. Theorem 5.3. If K has no quadratic extensions, A is a central algebra and αβ − γδ 6= 0, then either A = Λ1 (K, λ) and λ ∈ K∗ is not a root of unity, or A = Λ2 (K) and charK = 0. In particular, A is a domain and dimAn = n+ 1, n ∈ N. Proof. From Proposition 5.2 it follows that A = B/I, where either B = Λ1 (K, λ) or B = Λ2 (K) and I is a homogeneous prime ideal of the algebra B, I 6= B. We are going to prove that I = 0. Assume the converse and consider two cases. Case 1: let B = Λ1 (K, λ), where λ ∈ K∗. If λm = 1 for some m ∈ N, then from Theorem 2.2 we can conclude that Xm, Y m are central in Λ1 (K, λ). Consider the canonical homomorphism π : Λ1 (K, λ) → Λ1 (K, λ)/I. Since π is surjective we have π (Xp) , π (Y p) ∈ Z ( Λ1 (K, λ)/I ) . But the algebra Λ1 (K, λ)/I is central and so π (Xm) = α ∈ K and π (Y m) = β ∈ Jo u rn al A lg eb ra D is cr et e M at h .E. N. Shirikov 83 K. Therefore, Xm−α, Y m−β ∈ I. Since the ideal I is homogeneous, we get α = β = 0, i.e. Xm, Y m ∈ I. But the ideal I is prime, so X,Y ∈ I and I = (X,Y ). But dimA2 = 3. This contradiction shows that λ is not a root of unity. Then by Theorem 2.5 it follows that I is one of ideals (X), (Y ) or (X,Y ). In each case dimA2 < 3. Thus, if B = Λ1 (K, λ), then I = 0. Case 2: B = Λ2 (K). If charK = p > 0, then from Theorem 2.2 we get Y p is central in Λ2 (K). Consider the canonical homomor- phism π : Λ2 (K) → Λ2 (K)/I. Since π is surjective we have π (Y p) ∈ Z ( Λ1 (K, λ)/I ) . Then we can apply the same arguments as in the pre- ceding case. Theorem 5.4. If K has no quadratic extensions, dimAn = n+1, n ∈ N, and αβ−γδ 6= 0, then either A = Λ1 (K, λ) or A = Λ2 (K). In particular, A is a domain. Proof. Without loss of generality, we can assume that generators X and Y satisfy either the equality Y X = λXY , λ ∈ K∗, or the equality Y X = XY + Y 2. Consider the case Y X = XY + Y 2. Put Λ2 (K) = K 〈 X̃, Ỹ 〉 / ( Ỹ X̃ − X̃Ỹ − Ỹ 2 ) = ∞ ⊕ n=0 Ãn, where Ã0 = K, Ãn, n ∈ N, is a linear span of monomials of degree n in X̃, Ỹ . From Proposition 1.2 we get dim Ãn = n + 1. There exists a graded algebra homomorphism ϕ : Λ2 (K) → A, X̃ 7→ X, Ỹ 7→ Y . Then kerϕ = 0, i.e. A = Λ2 (K). In the case Y X = λXY , λ ∈ K∗, using the same arguments we get A = Λ1 (K, λ). Corollary 5.5. If dimAn = n + 1, n ∈ N, αβ − γδ 6= 0, then A is a domain. Proof. Put Ā = K̄ ⊗K A. Then Ā is generated over K̄ by elements X̄ = 1⊗X and Ȳ = 1⊗Y and Ā = ∞ ⊕ n=0 Ān, where Ān, n ∈ N, is the linear span of all monomials of degree n in X̄ and Ȳ . In particular, dim Ān = n+ 1. It is evident that αX̄2 + βȲ 2 + γX̄Ȳ + δȲ X̄ = 0. Then from Theorem 5.4 it follows that Ā is a domain. References [1] Artamonov V. A., Quantum Serre conjecture, Uspehi mat. nauk 53(1998), N 4, 3-77. Jo u rn al A lg eb ra D is cr et e M at h .84 Two-generated graded algebras [2] Artamonov V. A., Automorphisms and derivations of quantum polynomials. In the book: Ignacio Bajo and Esperanza Sanmartin (eds.) Recent Advances in Lie Theory v. 25. Heldermann Verlag, 2002, 109-120. [3] Artamonov V. A. Action of Hopf algebras on generic quantum Malcev power series and quantum planes, J. Math. Sci. 2005 (to appear) [4] Ken A. Brown, Ken A. Goodearl, Lectures on Algebraic Quantum Groups, Birkhäuser, Basel, Boston, 2002. [5] Jacobson N., Lie Algebras, Inerscience Publ. N.Y.-London, 1962. [6] Manin Yu. I., Topics in Non-Commutative Geometry, Princeton: Princeton Univ. Press, Univ. of Tokyo Press, 1991. [7] J. C. McConnell, J. C. Robson, Noncommutative Noetherian Rings, John Mi- ley& Sons, Chichester - N. Y. - Birsbane - Toronto -Singapore, 1987. [8] Golovashkin A. V., Maximov V. M. Algebras of skew polynomials, generated by quaratic homogeneous defining relations. In the book: Representation theory, dynamic systems, combinatorial and algorithmic methods, IX (Zapiski nauchnih seminarov POMI), v. 301, S-Petersbourg, 2003, 144-171. Contact information E. N. Shirikov E-Mail: evgenii_shirikov@mail.ru Received by the editors: 15.06.2005 and final form in 13.07.2005.

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