The Green Rings of Taft algebras
aa r X i v : . [ m a t h . R T ] O c t THE GREEN RINGS OF TAFT ALGEBRAS
HUIXIANG CHEN, FRED VAN OYSTAEYEN, AND YINHUO ZHANG
Abstract.
We compute the Green ring of the Taft algebra H n ( q ), where n is a positive integer greater than 1, and q is an n -th root of unity. It turnsout that the Green ring r ( H n ( q )) of the Taft algebra H n ( q ) is a commutativering generated by two elements subject to certain relations defined recursively.Concrete examples for n = 2 , , ..., Introduction
The tensor product of representations of a Hopf algebra is an important ingredientin the representation theory of Hopf algebras and quantum groups. In particularthe decomposition of the tensor product of indecomposable modules into a directsum of indecomposables has received enormous attention. For modules over a groupalgebra this information is encoded in the structure of the Green ring (or the repre-sentation ring) for finite groups, [1, 2, 3, 4, 9, 11]). For modules over a Hopf algebraor a quantum group there are results by Cibils on a quiver quantum group [7], byWitherspoon on the quantum double of a finite group [18], by Gunnlaugsd´ottir onthe half quantum groups (or Taft algebras) [10], and by Chin on the coordinateHopf algebra of quantum SL(2) at a root of unity [8]. However, the Green rings ofthose Hopf algebras are either equal to the Grothendick rings (in the semisimplecases) or not yet computed because of the complexity.In this paper, we compute the Green rings of Taft algebras. It turns out that theGreen ring of a Taft algebra is much more complicated than its Grothendick ring. InSection 1, we recall some basic definitions and results, and make some preparationsfor the rest of the paper. In Section 2, we recall the indecomposable modules overthe Taft algebra H n ( q ) from [7, 10], using the terminology of matrix representations,where n is a positive integer >
2, and q is a primitive n -th root of unity in the groundfield k . There are n non-isomorphic finite dimensional indecomposable modulesover H n ( q ), and all of them are uniserial. Moreover, for each 1 l n , there areexactly n finite dimensional indecomposable H n ( q )-modules M ( l, r ), r ∈ Z n , upto isomorphism. Every indecomposable projective H n ( q )-module is n -dimensional.In Section 3, we describe the Green ring of Taft algebra H n ( q ). We first recallthe decomposition formula of the tensor product of two indecomposable modulesover H n ( q ) from [7, 10]. From the decomposition formula, we know that the tensorproduct of any two H n ( q )-modules is commutative. Moreover, the tensor productof two indecomposable non-projective modules has a simple summand if and onlyif the two indecomposable modules have the same dimension. Finally we describe Mathematics Subject Classification.
Key words and phrases.
Green ring, indecomposable module, Taft algebra. the structure of the Green ring r ( H n ( q )) of H n ( q ). We show that the Green ring r ( H n ( q )) is generated by two elements subject to certain relations which can bedefined recursively. 1. Preliminaries
Throughout, we work over a fixed field k . Unless otherwise stated, all algebras,Hopf algebras and modules are defined over k ; all modules are left modules andfinite dimensional; all maps are k -linear; dim, ⊗ and Hom stand for dim k , ⊗ k andHom k , respectively. For the theory of Hopf algebras and quantum groups, we referto [12, 14, 15, 16].Let 0 = q ∈ k . For any integer n >
0, set ( n ) q = 1 + q + · · · + q n − . Observe that( n ) q = n when q = 1, and ( n ) q = q n − q − q = 1. Define the q -factorial of n by (0)! q = 1 and ( n )! q = ( n ) q ( n − q · · · (1) q for n >
0. Note that ( n )! q = n ! when q = 1, and( n )! q = ( q n − q n − − · · · ( q − q − n when n > q = 1. The q -binomial coefficients ni ! q are defined inductivelyas follows for 0 i n : n ! q = 1 = nn ! q for n > , ni ! q = q i n − i ! q + n − i − ! q for 0 < i < n. It is well-known that ni ! q is a polynomial in q with integer coefficients and withvalue at q = 1 is equal to the usual binomial coefficient ni ! , and that ni ! q = ( n )! q ( i )! q ( n − i )! q when ( n − q = 0 and 0 < i < n (see [12, Page 74]).Throughout this paper, we fix an integer n > k containsan n -th primitive root q of unity. Then we have ni ! q = 0 , ( i )! q = 0 , for 0 < i < n, and that n is not divisible by the characteristic of k , i.e. 1 n ∈ k . HE GREEN RINGS OF TAFT ALGEBRAS 3
The Taft algebra H n ( q ) is generated by two elements g and h subject to the relations(see [17]) g n = 1 , h n = 0 , hg = qgh.H n ( q ) is a Hopf algebra with coalgebra structure ∆ and antipode S given by∆( g ) = g ⊗ g, ∆( h ) = 1 ⊗ h + h ⊗ g, ε ( g ) = 1 ,ε ( h ) = 0 , S ( g ) = g − = g n − , S ( h ) = − q − g n − h. Note that dim H n ( q ) = n and { g i h j | i, j n − } forms a k -basis for H n ( q ).When n = 2, H ( q ) is exactly Sweedler’s 4-dimensional Hopf algebra.Let H be a Hopf algebra. The representation ring r ( H ) and R ( H ) can be definedas follows. r ( H ) is the abelian group generated by the isomorphism classes [ V ]of finite dimensional H -modules V modulo the relations [ M ⊕ V ] = [ M ] + [ V ].The multiplication of r ( H ) is given by the tensor product of H -modules, that is,[ M ][ V ] = [ M ⊗ V ]. Then r ( H ) is an associative ring. R ( H ) is an associative k -algebra defined by k ⊗ Z r ( H ). Note that r ( H ) is a free abelian group with a Z -basis { [ V ] | V ∈ ind( H ) } , where ind( H ) denotes the category of finite dimensionalindecomposable H -modules.2. Representations of H n ( q )For a module M over a finite dimensional algebra A , let rl( M ) denote the Loewylength (=radical length=socle length) of M , and let l( M ) denote the length of M .Let P ( M ) denote the projective cover of M , and let I ( M ) denote the injective hullof M .Cibils constructed an nd -dimensional Hopf algebra kZ n ( q ) /I d in [7], where q isan n -th root of unity in k with order d . He classified the indecomposable modulesover kZ n ( q ) /I d , and gave the decomposition of the tensor products of two arbitraryindecomposable modules there. When q is a primitive n -th root of unity, kZ n ( q ) /I n is isomorphic to H n ( q ) (see [7]). Therefore, from [7], one can get the classificationof indecomposable modules and the decomposition of the tensor product of twoindecomposable modules over H n ( q ). For the completeness, we will describe theindecomposable modules over H n ( q ) in this section, using the terminology of matrixrepresentation.Let G ( H n ( q )) denote the group of group-like elements in H n ( q ). Then G ( H n ( q )) = { , g, · · · , g n − } is a cyclic group of order n generated by g . The group algebra kG ( H n ( q )) is a Hopf subalgebra of H n ( q ). There is a Hopf algebra epimorphism π : H n ( q ) → kG ( H n ( q )) defined by π ( g ) = g and π ( h ) = 0. Since k containsan n -th primitive root of unity, the group algebra kG ( H n ( q )) is semisimple. Itfollows that Ker π = h h i ⊇ J ( H n ( q )), the Jacobson radical of H n ( q ). On the otherhand, since H n ( q ) h = hH n ( q ) and h n = 0, J ( H n ( q )) ⊇ ( h ) = H n ( q ) h , the idealof H n ( q ) generated by h . Hence Ker π = ( h ) = J ( H n ( q )). Thus, an H n ( q )-module M is semisimple if and only if h · M = 0, and moreover M is simple if and onlyif h · M = 0 and M is simple as a module over the Hopf subalgebra kG ( H n ( q )).Therefore, we have the following lemma. HUIXIANG CHEN, FRED VAN OYSTAEYEN, AND YINHUO ZHANG
Lemma 2.1.
There are n non-isomorphic simple H n ( q ) -modules S i , and each S i is 1-dimensional and determined by g · v = q i v, h · v = 0 , v ∈ S i , where i ∈ Z n := Z / ( n ) . (cid:3) Note that J ( H n ( q )) m = H n ( q ) h m for all m >
1. Hence J ( H n ( q )) n − = 0, but J ( H n ( q )) n = 0. This means that the Loewy length of H n ( q ) is n . Since everysimple H n ( q )-module is 1-dimensional, l( M ) = dim( M ) for any H n ( q )-module M .Now let M be any H n ( q )-module. Since J ( H n ( q )) s = H n ( q ) h s = h s H n ( q ), we haverad s ( M ) = h s · M for all s > Lemma 2.2.
Let l n and i ∈ Z . Then there is an algebra map ρ l,i : H n ( q ) → M l ( k ) given by ρ l,i ( g ) = q i q i − q i − . . . q i − l +1 , ρ l,i ( h ) =
01 01 . . .. . .
01 0 . Let M ( l, i ) denote the corresponding left H n ( q ) -module.Proof. It follows from a straightforward verification. (cid:3)
There is a k -basis { v , v , · · · , v l } of M ( l, i ) such that g · v j = q i − j +1 v j for all1 j l and h · v j = ( v j +1 , j l − , , j = l. Hence we have v j = h j − · v for all 2 j l . Such a basis is called standard basis of M ( l, i ). For any integer i , we will often regard i as its image under the canonicalprojection Z → Z n := Z / ( n ). We have the following lemma. Lemma 2.3.
For any l n and i ∈ Z , let M ( l, i ) be the H n ( q ) -module definedas in Lemma 2.2. Then (1) soc( M ( l, i )) = kv l ∼ = S i − l +1 and M ( l, i ) / rad( M ( l, i )) ∼ = S i . (2) M ( l, i ) is indecomposable and uniserial. (3) If l ′ n and i ′ ∈ Z , then M ( l, i ) ∼ = M ( l ′ , i ′ ) if and only if l ′ = l and i ′ = i in Z n .Proof. (1) Since J ( H n ( q )) = ( h ) = hH n ( q ) = H n ( q ) h , soc( M ( l, i )) = { v ∈ M ( l, i ) | h · v = 0 } = kv l and rad( M ( l, i )) = h · M ( l, i ) = span { v , · · · , v l } . Itfollows that soc( M ( l, i )) ∼ = S i − l +1 and M ( l, i ) / rad( M ( l, i )) ∼ = S i . HE GREEN RINGS OF TAFT ALGEBRAS 5 (2) By (1), soc( M ( l, i )) is simple, and hence M ( l, i ) is indecomposable. Since h l − · M ( l, i ) = 0 and h l · M ( l, i ) = 0, rl( M ( l, i )) = l . Hence l ( M ( l, i )) = rl( M ( l, i )),and so M ( l, i ) is uniserial.(3) Obvious. (cid:3) As a consequence, we obtain the following:
Corollary 2.4.
Let l n and i ∈ Z n . Then (1) M ( l, i ) is simple if and only if l = 1 . In this case, M (1 , i ) ∼ = S i . (2) M ( l, i ) is projective (injective) if and only if l = n . (3) M ( n, i ) ∼ = P ( S i ) ∼ = I ( S i +1 ) .Proof. (1): Follows from Lemma 2.3(1).(2) and (3): Note that any finite dimensional Hopf algebra is a Frobenius algebra,and hence is a self-injective algebra. If l = n , then it follows from [5, Lemma 3.5]that M ( n, i ) is projective and injective.For any 0 i n −
1, let e i = n n − P j =0 q − ij g j . Then { e , e , · · · , e n − } is a setof orthogonal idempotents such that n − P i =0 e i = 1. We also have ge i = q i e i and h n − e i = 0. Therefore, H n ( q ) e i = span { e i , he i , · · · , h n − e i } ∼ = M ( n, i ). Thus, wehave a decomposition of the regular module H n ( q ) as follows: H n ( q ) = n − M i =0 H n ( q ) e i ∼ = n − M i =0 M ( n, i ) . Hence M ( n, i ) ∼ = P ( S i ), and M ( n, M ( n, · · · , M ( n, n −
1) are all non-isomorphic indecomposable projective H n ( q )-modules. So (2) and (3) follow fromLemma 2.3. (cid:3) Since the indecomposable projective H n ( q )-modules are uniserial, any indecompos-able H n ( q )-module is uniserial and is isomorphic to a quotient of a indecomposableprojective module. Thus, we have the following theorem (see [7, Page 467]). Theorem 2.5.
Up to isomorphism, there are n indecomposable finite dimensional H n ( q ) -modules as follows { M ( l, i ) | l n, i n − } . (cid:3) HUIXIANG CHEN, FRED VAN OYSTAEYEN, AND YINHUO ZHANG The Green Ring of Taft Algebra H n ( q )We have already known that there are n non-isomorphic indecomposable modulesover H n ( q ). They are { M ( l, r ) | l n, r ∈ Z n } . The following lemma follows from a straightforward verification.
Lemma 3.1.
Let l n and r, r ′ ∈ Z n . Then M ( l, r ) ⊗ S r ′ ∼ = S r ′ ⊗ M ( l, r ) ∼ = M ( l, r + r ′ ) as H n ( q ) -modules. In particular, S r ⊗ S r ′ ∼ = S r + r ′ and M ( l, r ) ∼ = S r ⊗ M ( l, ∼ = M ( l, ⊗ S r . (cid:3) Cibils and Gunnlaugsd´ottir derived the decomposition formulas of the tensor prod-uct of two indecomposable modules over kZ n ( q ) /I n and the half-quantum group u + q in [7] and [10], respectively. From [7, Theorem 4.1] or [10, Theorem 3.1], onegets the following Propositions 3.2, 3.3 and 3.4. Proposition 3.2.
Let l n and r, r ′ ∈ Z n . Then we have the H n ( q ) -moduleisomorphisms M ( l, r ) ⊗ M ( n, r ′ ) ∼ = M ( n, r ′ ) ⊗ M ( l, r ) ∼ = l M i =1 M ( n, r + r ′ + i − l ) . (cid:3) Proposition 3.3.
Let l, l ′ < n and r, r ′ ∈ Z n . If l + l ′ n , then M ( l, r ) ⊗ M ( l ′ , r ′ ) ∼ = l M i =1 M ( | l − l ′ | − i, r + r ′ + i − l ) , where l = min { l, l ′ } . (cid:3) Proposition 3.4.
Let l, l ′ < n and r, r ′ ∈ Z n . If l + l ′ > n , then M ( l, r ) ⊗ M ( l ′ , r ′ ) ∼ = ( n − l M i =1 M ( | l − l ′ |− i, r + r ′ + i − l )) M ( l + l ′ − n M i =1 M ( n, r + r ′ +1 − i )) , where l = min { l, l ′ } and l = max { l, l ′ } . (cid:3) Following Propositions 3.3 and 3.4, we obtain the following:
Corollary 3.5.
Let l, l ′ n − and r, r ′ ∈ Z n . Then there is a simplesummand in M ( l, r ) ⊗ M ( l ′ , r ′ ) if and only if l = l ′ . (cid:3) The following property of M ( l, r ) can be derived from Lemma 3.1, Propositions3.2, 3.3 and 3.4. Corollary 3.6.
Let l, l ′ n and r, r ′ ∈ Z n . Then M ( l, r ) ⊗ M ( l ′ , r ′ ) ∼ = M ( l ′ , r ′ ) ⊗ M ( l, r ) . (cid:3) From Theorem 2.5 and Corollary 3.6, one can deduce the following known result(see [7, Page 467]).
HE GREEN RINGS OF TAFT ALGEBRAS 7
Corollary 3.7.
For any H n ( q ) -modules M and N , there is an H n ( q ) -module iso-morphism M ⊗ N ∼ = N ⊗ M. (cid:3) In the sequel, we let a = [ S − ] and x = [ M (2 , r ( H n ( q )) of H n ( q ). From Corollary 3.7, we know that r ( H n ( q )) is a commutative ring. Lemma 3.8. (1) a n = 1 and [ M ( l, r )] = a n − r [ M ( l, for all l n and r ∈ Z n . (2) If n > , then [ M ( l +1 , x [ M ( l, − a [ M ( l − , for all l n − . (3) x [ M ( n, a + 1)[ M ( n, . (4) r ( H n ( q )) is generated by a and x as a ring.Proof. (1): Follows from Lemma 3.1 since [ S ] is the identity of the ring r ( H n ( q )).(2): If n > l n −
1, then by Propositions 3.3 and 3.4 and Lemma 3.1,we have M (2 , ⊗ M ( l, ∼ = M ( l − , − ⊕ M ( l + 1 , ∼ = S − ⊗ M ( l − , ⊕ M ( l + 1 , . It follows that [ M ( l + 1 , x [ M ( l, − a [ M ( l − , M (2 , ⊗ M ( n, ∼ = M ( n, − ⊕ M ( n, ∼ = S − ⊗ M ( n, ⊕ M ( n, ∼ = ( S − ⊕ S ) ⊗ M ( n, . It follows that x [ M ( n, a + 1)[ M ( n, (cid:3) Corollary 3.9.
Let u , u , · · · be a series of elements of the ring r ( H n ( q )) definedrecursively by u = 1 , u = x and u l = xu l − − au l − , l > . Then [ M ( l, u l for all l n and ( x − a − u n = 0 .Proof. Follows from Lemma 3.8. (cid:3)
Let Z [ y, z ] be the polynomial algebra over Z in two variables y and z . We define a generalized Fibonacci polynomial f n ( y, z ) ∈ Z [ y, z ], n >
1, recursively as follows: f ( y, z ) = 1 , f ( y, z ) = z, and f n ( y, z ) = zf n − ( y, z ) − yf n − ( y, z ) , n > . Let I be the ideal of Z [ y, z ] generated by polynomials y n − z − y − f n ( y, z ).With the above notations, we have the following main result. Theorem 3.10.
The Green ring r ( H n ( q )) of H n ( q ) is isomorphic to the quotientring Z [ y, z ] /I . HUIXIANG CHEN, FRED VAN OYSTAEYEN, AND YINHUO ZHANG
Proof.
By Lemma 3.8(4), r ( H n ( q )) is generated, as a ring, by a and x . Hence thereis a unique ring epimorphism φ from Z [ y, z ] to r ( H n ( q )) such that φ ( y ) = a and φ ( z ) = x . Since a n = 1 by Lemma 3.8(1), φ ( y n −
1) = 0. Let { u i } i > be the series ofelements of r ( H n ( q )) given in Corollary 3.9. It is easy to see that φ ( f ( y, z )) = u and φ ( f ( y, z )) = u . Now let i > φ ( f i − ( y, z )) = u i − and φ ( f i − ( y, z )) = u i − . Then φ ( f i ( y, z )) = φ ( zf i − ( y, z ) − yf i − ( y, z ))= φ ( z ) φ ( f i − ( y, z )) − φ ( y ) φ ( f i − ( y, z ))= xu i − − au i − = u i . Thus φ ( f i ( y, z )) = u i for all i >
1. In particular, we have φ ( f n ( y, z )) = u n , andhence φ (( z − y − f n ( y, z )) = ( x − a − u n = 0 by Corollary 3.9. It follows that φ ( I ) = 0, and that φ induces a ring epimorphism φ : Z [ y, z ] /I → r ( H n ( q )) suchthat φ ( v ) = φ ( v ) for all v ∈ Z [ y, z ], where v denotes the image of v under thenatural epimorphism Z [ y, z ] → Z [ y, z ] /I .Let A be the subring of r ( H n ( q )) generated by a . A = Z h a i is the group ring of thecyclic group h a i over Z . By Corollary 3.9 we have u = 1 ∈ A and u = x ∈ Ax ⊂ A + Ax . By induction on i one can show that u i ∈ A + Ax + · · · Ax i − for all i > u i ∈ A + Ax + · · · Ax n − for all 1 i n . Thus for all 1 i n and r ∈ Z n ,by Lemma 3.8(1) we have [ M ( i, r )] = a n − r [ M ( i, a n − r u i ∈ A + Ax + · · · Ax n − .It follows that r ( H n ( q )) = A + Ax + · · · Ax n − . Since A is a free Z -module with a Z -basis { a i | i n − } , r ( H n ( q )) is generated by elements a i x j , i, j n − Z -module. Since r ( H n ( q )) is a free Z -module of rank n , { a i x j | i, j n − } forms a Z -basis for r ( H n ( q )). Hence one can define a Z -module homomorphism: ψ : r ( H n ( q )) → Z [ y, z ] /I, a i x j y i z j = y i z j , i, j n − . Obviously, Z [ y, z ] /I is generated by elements, y i z j , i, j n −
1, as a Z -module.Now we have ψφ ( y i z j ) = ψφ ( y i z j ) = ψ ( a i x j ) = y i z j for all 0 i, j n −
1. Hence ψφ = id, and so φ is injective. Thus, φ is a ringisomorphism. (cid:3) The coefficients of the generalized Fibonacci polynomial f n ( y, z ) can be computed.They are quite similar to those of the standard generalized Fibonacci polynomial defined by F ( y, z ) = 1 , F ( y, z ) = z, and F n ( y, z ) = zF n − ( y, z ) + yF n − ( y, z ) , n > . For completeness, we compute f n ( y, z ) in the following lemma, which might befound elsewhere. Lemma 3.11.
Let Z [ y, z ] be the polynomial algebra over Z in two variables y and z . Then for any n > , we have (1) f n ( y, z ) = [( n − / X i =0 ( − i " n − − ii y i z n − − i . HE GREEN RINGS OF TAFT ALGEBRAS 9
Proof.
We prove it by induction on n . It is easy to check that Equation (1) holdsfor 1 n
4. Now let n > n = 2 m + 1 is odd, then we have f n ( y, z ) = zf m ( y, z ) − yf m − = m − P i =0 ( − i " m − − ii y i z m − i − m − P i =0 ( − i " m − − ii y i +1 z m − − i = m − P i =0 ( − i " m − − ii y i z m − i + m P i =1 ( − i " m − − ii − y i z m − i = z m + m − P i =1 ( − i ( " m − − ii + " m − − ii − ) y i z m − i +( − m y m = m P i =0 ( − i " m − ii y i z m − i = [( n − / P i =0 ( − i " n − − ii y i z n − − i . If n = 2( m + 1) is even, then we have f n = zf m +1 ( y, z ) − yf m = m P i =0 ( − i " m − ii y i z m +1 − i − m − P i =0 ( − i " m − − ii y i +1 z m − − i = m P i =0 ( − i " m − ii y i z m +1 − i + m P i =1 ( − i " m − ii − y i z m +1 − i = z m +1 + m P i =1 ( − i ( " m − ii + " m − ii − ) y i z m +1 − i = z m +1 + m P i =1 ( − i " m + 1 − ii y i z m +1 − i = m P i =0 ( − i " m + 1 − ii y i z m +1 − i = [( n − / P i =0 ( − i " n − − ii y i z n − − i . Thus the proof is completed. (cid:3)
Now we can easily derive the Green rings r ( H n ( q )) for n = 2 , , ..., Corollary 3.12.
When n = 2 , r ( H ( q )) ∼ = Z [ y, z ] / ( y − , ( z − y − z ) .When n = 3 , r ( H ( q )) ∼ = Z [ y, z ] / ( y − , ( z − y − z − y )) .When n = 4 , r ( H ( q )) ∼ = Z [ y, z ] / ( y − , ( z − y − z − yz )) .When n = 5 , r ( H ( q )) ∼ = Z [ y, z ] / ( y − , ( z − y − z − yz + y )) .When n = 6 , r ( H ( q )) ∼ = Z [ y, z ] / ( y − , ( z − y − z − yz + 3 y z )) .When n = 7 , r ( H ( q )) ∼ = Z [ y, z ] / ( y − , ( z − y − z − yz + 6 y z − y )) .When n = 8 , r ( H ( q )) ∼ = Z [ y, z ] / ( y − , ( z − y − z − yz + 10 y z − y z )) . (cid:3) Remark 3.13. (1)
One can easily see that the Grothendick ring of H n ( q ) isthe group ring k Z n generated by the simple module M (1 , . From the aboveexamples, we see that the Green ring is much more complicated than theGrothendick ring. (2) The Green rings of generalized Taft algebras and the Green rings of mono-mial Hopf algebras [6] can be computed in a similar way. However, thecomputation of the Green ring of the small quantum group or the Greenring of the quantum double of a Taft algebra seem to be much more com-plicated as they are not finitely generated [5, 13] . (3) Since the module category of a quasitriangular Hopf algebra H is braidedmonoidal, the Green ring of H is commutative. The Taft algebra H n ( q ) isnot quasitriangular in case n > (even not almost cocommutative, see [7] ),but its Green ring is commutative. This leads to the following question: canwe characterize the class of Hopf algebras whose Green ring is commutative? ACKNOWLEDGMENTS
The authors are grateful to the referee for suggesting a formula for the generalizedFibonacci polynomials f n ( y, z ). The first named author would like to thank theDepartment of Mathematics, University of Antwerp for its hospitality during hisvisit in 2011. He is grateful to the Belgium FWO for financial support. He wasalso supported by NSF of China (No. 11171291). References [1] L. Archer, On certain quotients of the Green rings of dihedral 2-groups, J. Pure & Appl.Algebra 212 (2008), 1888-1897.[2] D. J. Benson and J. F. Carlson, Nilpotent elements in the Green ring, J. Algebra 104 (1986),329-350.[3] D. J. Benson and R. A. Parker, The Green ring of a finite group, J. Algebra 87 (1984),290-331.[4] R. M. Bryant and M. Johnson, Periodicity of Adams operations on the Green ring of a finitegroup, arXiv:0912.2933v1[math.RT].
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E-mail address : [email protected] Department of Mathematics and Computer Science, University of Antwerp, Middel-heimlaan 1, B-2020 Antwerp, Belgium
E-mail address : [email protected] Department WNI, University of Hasselt, Universitaire Campus, 3590 Diepeenbeek,Belgium
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