ON A QUESTION OF RICKARD ON TENSOR PRODUCT OF STABLY EQUIVALENT ALGEBRAS

SERGE BOUC AND ALEXANDER ZIMMERMANN

Abstract. Let r be a positive integer, let p be a prime and \overline{𝔽}_{p} denote an algebraic closure of the prime field 𝔽_{p}. After observing that the principal block B of \overline{𝔽}_{p}P SU(3, p^{r}) is stably equivalent of Morita type to its Brauer correspondent b, we compute the radical series of the center Z(b), and, using GAP, the radical series of Z(B) in the cases p^{r} ∈ {3, 4, 5, 8}. In these cases, the dimensions of the last non zero power of the radical of Z(b) and Z(B) are different, and it follows that the algebra B ⊗_{\overline{𝔽}_{p}} \overline{𝔽}_{p}[X]/X^{p} is not stably equivalent of Morita type to b ⊗_{\overline{𝔽}_{p}} \overline{𝔽}_{p}[X]/X^{p}. This yields a negative answer to a question of Rickard.

Introduction

Let K be a field, and let A, B, C and D be finite dimensional K-algebras. Rickard showed in [12] that if A and B are derived equivalent, and if C and D are derived equivalent, then also A ⊗_{K} C and B ⊗_{K} D are derived equivalent. Rickard asks in [13, Question 3.8] if this still holds when replacing derived equivalence by stable equivalence of Morita type. It is clear that we have to suppose that all algebras involved have no semisimple direct factor. A result due to Liu [8] shows that then we may suppose that all algebras are indecomposable. In [10] Liu, Zhou and the second author showed that the question has a negative solution in case A, B, C and D are not necessarily selfinjective. However, a derived equivalence between selfinjective algebras A and B induces a stable equivalence of Morita type between A and B. If A and B are not selfinjective, then this implication is not valid. Hence, the natural playground for Rickard’s question are selfinjective algebras.

The purpose of this paper is to give a counterexample to Rickard’s question. For an algebraically closed base field K of characteristic p we construct symmetric K-algebras A and B which are stably equivalent of Morita type, but A ⊗_{K} K[X]/X^{p} and B ⊗_{K} K[X]/X^{p} are not stably equivalent of Morita type.

Note that this answers the general case. Indeed, if A ⊗_{K} C is stably equivalent of Morita type to B ⊗_{K} C and B ⊗_{K} C is stably equivalent of Morita type to B ⊗_{K} D then A ⊗_{K} C is stably equivalent of Morita type to B ⊗_{K} D. Hence, we may suppose C = D.

In recent years many attempts were proposed to lift a stable equivalence of Morita type between selfinjective algebras to a derived equivalence. It is known that this is not possible in general, as is seen by the mod 2 group ring of a dihedral group of order 8 and the stable equivalence induced by a uniserial endotrivial module of Loewy length 3. This was used in [10] for example. In this paper we give a new incidence of this fact. Moreover, we provide two symmetric algebras, which are stably equivalent of Morita type, and have non isomorphic centres.

Our example is the principal p-block of the group P SU(3, p^{r}) and its Brauer correspondent for p^{r} ∈ {3, 4, 5, 8}.

We recall in the first section some basic facts and results which we need for our construction. In Section 2 we give our main result and its proof, and in Section 3 we display the GAP program needed for the proof. In Section 4 we determine the algebraic structure of the centre of KN_{G}(S) for G = P SU(3, p^{r}) and S one of its Sylow p-subgroups for all primes p and integers r.

Date: April 1, 2015.

2010 Mathematics Subject Classification. Primary 16E35; 18E30; 20C05.

Both authors were supported by a grant STIC Asie ’Escap’ from the Ministère des Affaires Étrangères de la France.

Acknowledgement: The idea of this paper was born during a visit of both authors in Beijing Normal University. We are very grateful to Yuming Liu for his great hospitality. Moreover, we thank Yuming Liu for suggesting a question to us leading to the present paper, and also for pointing out that it is not sufficient to show an abstract non isomorphism of the centres of the blocks. This led us to complete our proof by adding Lemma 9.

1. Background

Recall the following

Definition 1. [2], (cf also [14, Chapter 5]) Let A and B be two finite dimensional algebras over a field K. Then A and B are stably equivalent of Morita type if there is an A ⊗_{K} B^{op}- module M and a B ⊗_{K} A^{op}-module N such that

• M is projective as A-module, and as B^{op}-module

• N is projective as A^{op}-module and as B-module

• there is a projective A ⊗_{K} A^{op}-module P and a projective B ⊗_{K} B^{op}-module Q such that M ⊗_{B} N ≃ B ⊕ Q as B ⊗_{K} B^{op}-modules and N ⊗_{A} M ≃ A ⊕ P as A ⊗_{K} A^{op}-modules.

Independently Rickard [11] as well as Keller and Vossieck [6], show that if A and B are derived equivalent selfinjective algebras, then A and B are stably equivalent of Morita type.

Broué defined Z^{st}(A) := \underline{End}_{A⊗_{K}A^{op}}(A) and

Z^{pr}(A) := ker(End_{A⊗_{K}A^{op}}(A) → \underline{End}_{A⊗_{K}A^{op}}(A))

where we denote by \underline{End} the endomorphisms taken in the stable module category.

The centre of an algebra is an invariant of a derived equivalence, as was shown by Rickard. The stable centre Z^{st}(A) is an important invariant under stable equivalences of Morita type, as was shown by Broué.

Proposition 2. (Broué [2]; see also [14, Chapter 5]) If A and B are stably equivalent of Morita type, then Z^{st}(A) ≃ Z^{st}(B) as algebras.

Now, Liu, Zhou and the second author give a criterion to determine the dimension of Z^{st}(A).

Theorem 3. [9, Proposition 2.3 and Corollary 2.7] Let A be a finite dimensional symmetric algebra over an algebraically closed field K of characteristic p > 0. Then dim_{K}(Z^{pr}(A)) = rank_{p}(C_{A}) where C_{A} is the Cartan matrix of A and where rank_{p}(C_{A}) denotes its rank as matrix over K.

Moreover, we recall a conjecture of Auslander-Reiten. In [1] Auslander and Reiten conjecture that if A and B are stably equivalent finite dimensional algebras, then the number of simple non-projective A-modules and the number of non-projective simple B-modules coincides. Again in [9] we show

Theorem 4. [9, Theorem 1.1] Let K be an algebraically closed field and let A and B be two finite dimensional K-algebras, which are stably equivalent of Morita type and which do not have any semisimple direct factor. Then the number of isomorphism classes of nonprojective simple A-modules is equal to the number of non-projective simple B-modules if and only if dim_{K}(HH_{0}(A)) = dim_{K}(HH_{0}(B)), where HH_{0} denotes the degree 0 Hochschild homology.

In particular, if A and B are symmetric, then Hochschild homology and cohomology coincides, and the number of non-projective simple A-modules is equal to the number of non-projective simple B-modules if and only if the centres of A and of B have the same dimension.

The following lemma is well-known to the experts, but for the convenience of the reader, and since it is crucial to our arguments, we include the short proof. For an algebra A denote by J(A) its Jacobson radical.

Lemma 5. Let K be a perfect field and let A and B be finite dimensional K-algebras. Then J(A ⊗_{K} B) = J(A) ⊗_{K} B + A ⊗_{K} J(B).

Proof. It is clear that J(A) ⊗_{K} B + A ⊗_{K} J(B) is a nilpotent ideal of A ⊗_{K} B, and therefore we get

J(A) ⊗_{K} B + A ⊗_{K} J(B) ⊆ J(A ⊗_{K} B).

Now, (A ⊗_{K} B)/(J(A) ⊗_{K} B + A ⊗_{K} J(B)) = A/J(A) ⊗_{K} B/J(B) and both K-algebras A/J(A) and B/J(B) are semisimple. Since K is perfect, every finite extension L of K is a separable field extension. By [3, Corollary 7.6] a finite dimensional semisimple K-algebra C is separable if and only if the centres of each of the Wedderburn components is a separable field extension of K. Hence A/J(A) and B/J(B) are both separable K-algebras. By [3, Corollary 7.8] the algebra A/J(A) ⊗_{K} B/J(B) is semisimple. Therefore

J(A) ⊗_{K} B + A ⊗_{K} J(B) ⊇ J(A ⊗_{K} B).

This shows the statement. □

Remark 6. (cf e.g. [14, Example 1.7.17]) Lemma 5 is wrong if we drop the assumption that K is perfect: e.g. let p be a prime, and K = 𝔽_{p}(U) be the field of rational fractions over the finite field 𝔽_{p}. Let A = K[X]/(X^{p} − U). Then A is a purely inseparable extension of K, of dimension p. In particular it is a reduced (commutative) algebra, i.e. J(A) = 0. But A ⊗_{K} A ≅ K[X, Y]/(X^{p} − U, Y^{p} − U) contains the non zero element X − Y, such that (X − Y)^{p} = U − U = 0. Hence J(A ⊗_{K} A) ≠ 0.

Lemma 7. Let K be an algebraically closed field, let Λ and Δ be finite dimensional Kalgebras, and suppose that Δ is local. Then the projective indecomposable Λ ⊗_{K} Δ-modules are precisely the modules P ⊗_{K} Δ for projective indecomposable Λ-modules P, and if C_{Λ} is the Cartan matrix of Λ, then the Cartan matrix of Λ ⊗ Δ is C_{Λ⊗_{K}Δ} = dim_{K}(Δ) · C_{Λ}.

Proof. Let P and Q be a indecomposable projective Λ-modules. Then P ⊗_{K} Δ is a projective indecomposable Λ ⊗_{K} Δ-module. Indeed, End_{Λ⊗_{K}Δ}(P ⊗_{K} Δ) ≃ End_{Λ}(P) ⊗_{K} Δ^{op}. Moreover, since Γ := End_{Λ}(P)^{op} and Δ are local K-algebras their radical quotient are finite-dimensional skew-fields, and therefore Γ/J(Γ) ≃ K ≃ Δ/J(Δ) since K is algebraically closed. Moreover, by Lemma 5 we get J(Γ ⊗_{K} Δ) = J(Γ) ⊗ Δ + Γ ⊗_{K} J(Δ). On the other hand,

(Γ ⊗_{K} Δ)/ (J(Γ) ⊗_{K} Δ + Γ ⊗_{K} J(Δ)) = K ⊗_{K} K = K

and hence we get Γ ⊗_{K} Δ is local, and therefore P ⊗_{K} Δ is indecomposable. Now,

Hom_{Λ⊗_{K}Δ}(P ⊗_{K} Δ, Q ⊗_{K} Δ) = Hom_{Λ}(P, Q) ⊗_{K} Δ^{op}.

Taking K-dimensions proves the lemma. □

Remark 8. As a special case of Lemma 7 we get C_{A⊗_{K}K[X]/X^{p}} = p · C_{A} for algebraically closed fields K of characteristic p. Hence we get by Theorem 3 that Z^{pr}(A⊗_{K} K[X]/X^{p}) = 0 for algebraically closed fields K of characteristic p and symmetric K-algebras A.

Lemma 9. Let K be a perfect field and let n, m be positive integers. Let A and B be finite dimensional commutative K-algebras. If J^{n+1}(A) = 0 ≠ J^{n}(A) and J^{m+1}(B) = 0 ≠ J^{m}(B), then

J^{n+m+1}(A ⊗_{K} B) = 0 ≠ J^{n+m}(A ⊗_{K} B) = J^{n}(A) ⊗_{K} J^{m}(B).

Proof. By Lemma 5, we have J(A ⊗_{K} B) = J(A) ⊗_{K} B + A ⊗_{K} J(B). Therefore

J^{n+m+1}(A ⊗_{K} B) = ∑_{k=0}^{n+m+1} J^{k}(A) ⊗_{K} J^{n+m+1−k}(B) = 0 .

Similarly

J^{n+m}(A ⊗_{K} B) = ∑_{k=0}^{n+m} J^{k}(A) ⊗_{K} J^{n+m−k}(B) = J^{n}(A) ⊗_{K} J^{m}(B) ≠ 0 ,

which completes the proof. □

Remark 10. Let K be any field, and A be a K-algebra. We give an elementary argument to determine the centre of A ⊗_{K} K[X]/X^{p}. It is clear that A ⊗_{K} K[X]/X^{p} ≅ A[X]/X^{p}. Now, let a := a_{0} + a_{1}X + … a_{p−1}X^{p−1} ∈ A[X]. Then for b := b_{0} ∈ A · 1 we get

ab − ba = (a_{0}b − ba_{0}) + ⋯ + (a_{p−1}b − ba_{p−1})X^{p−1}

and so a ∈ Z(A) implies that a commutes with any b ∈ A, and hence a_{0}, …, a_{p−1} are all in Z(A). Conversely, it is clear that Z(A)[X]/X^{p} ⊆ Z(A[X]/X^{p}) since aX^{n} commutes with all elements of A[X]/X^{p} whenever a ∈ A and since sums of elements in the centre are still central.

Lemma 11. If K is a perfect field and A is a finite dimensional K-algebra, and if moreover J^{n}(Z(A)) ≠ 0 = J^{n+1}(Z(A)), then

0 ≠ J^{n+p−1}(Z(A ⊗_{K} K[X]/X^{p})) = J^{n}(Z(A)) ⊗_{K} X^{p−1}K[X]/X^{p}

and

J^{n+p}(Z(A ⊗_{K} K[X]/X^{p})) = 0.

Proof. This is an immediate consequence of Lemma 9. □

Corollary 12. Let K be an algebraically closed field of characteristic p > 0 and let A and B be two finite dimensional K-algebras and let n, m ∈ ℕ such that J^{n}(Z(A)) ≠ 0 = J^{n+1}(Z(A)) and J^{m}(Z(B)) ≠ 0 = J^{m+1}(Z(B)). If dim_{K}(J^{n}(Z(A))) ≠ dim_{K}(J^{m}(Z(B)), then A ⊗_{K} K[X]/X^{p} and B ⊗_{K} K[X]/X^{p} are not stably equivalent of Morita type.

Proof. If n ≠ m, then Z(A ⊗_{K} K[X]/X^{p}) ≄ Z(B ⊗_{K} K[X]/X^{p}) by Lemma 11 since the Loewy lengths of the centres are different. If n = m, then Lemma 11 shows that the centres of A ⊗_{K} K[X]/X^{p} and of B ⊗_{K} K[X]/X^{p} are not isomorphic since the dimension of the lowest Loewy layers of the centres are not of the same dimension. Remark 8 shows that Z(A⊗_{K} K[X]/X^{p}) = Z^{st}(A⊗_{K} K[X]/X^{p}) and Z(B ⊗_{K} K[X]/X^{p}) = Z^{st}(B ⊗_{K} K[X]/X^{p}). Since the stable centre is invariant under stable equivalence of Morita type, we get the statement. □

Remark 13. For a field K and a K-algebra A let n_{A} be the number of isomorphism classes of simple nonprojective A-modules. Auslander-Reiten conjecture [1, page 409, Conjecture (5)] that if A and B are stably equivalent finite dimensional K-algebras, then n_{A} = n_{B}. [9, Theorem 1.1] shows that if K is algebraically closed and if A and B are indecomposable finite dimensional K-algebras which are stably equivalent of Morita type, then n_{A} = n_{B} is equivalent to dim_{K}(HH_{0}(A)) = dim_{K}(HH_{0}(B)). If A is symmetric, then there is a vector space isomorphism HH_{0}(A) ≃ HH^{0}(A) = Z(A), we see that the Auslander-Reiten conjecture implies that dim_{K} (Z(A)) = dim_{K} (Z(B)). More precisely by [9, Corollary 1.2], for two indecomposable symmetric algebras A and B over an algebraically closed field K we have n_{A} = n_{B} ⇔ dim_{K} (Z^{pr}(A)) = dim_{K} (Z^{pr}(B)), where by definition Z^{st}(A) = Z(A)/Z^{pr}(A). The link to our proof is now given by the fact that for every algebra the Higman ideal H(A) of A equal Z^{pr}(A), and for symmetric algebras A over an algebraically closed field K we have dim_{K}(H(A)) equals the p-rank of C_{A}.

2. The Example

Let \overline{𝔽}_{p} be the algebraic closure of the prime field 𝔽_{p} of characteristic p. Let q = p^{n} for some integer n.

We recall some results on the geometry of P SU(n, q) (cf e.g. [5, II Satz 10.12, page 242]). The group G := P SU(3, q) acts doubly transitively on the unitary quadric Q of cardinal q^{3} + 1. Note that we use the GAP notation, not the notation used in [5, II Satz 10.12, page 242], namely, P SU(3, q) is defined over a field with q^{2} elements, and is a natural quotient of a subgroup of SL_{2}(q^{2}) (and not of SL_{2}(q) !). The stabiliser of a point X of Q is the normaliser in G of a Sylow p-subgroup P of G. Therefore two different conjugate Sylow p-subgroups P and ^{g}P of G fix two different points X and gX of Q. Hence ^{g}P ∩ P = 1 if g ∉ N_{G}(P), or in other words, G has a trivial intersection Sylow p-subgroup structure. This implies that Green correspondence gives a stable equivalence of Morita type between the principal block B of \overline{𝔽}_{p}G and its Brauer correspondent b (cf e.g. [14, Chapter 2, Proposition 2.1.23 and Proposition 2.4.3]).

The GAP [4] program in Section 3 computes the Loewy series of the ring Z(𝔽_{2}P SU(3, 4)) and of Z(𝔽_{2}N_{P SU(3,4)}(S)) for some Sylow 2-subgroup of P SU(3, 4). Observe moreover that \overline{𝔽}_{2}P SU(3, 4) has two blocks, the principal one and another block of defect 0 (corresponding to the Steinberg character). Moreover, the dimensions of the Loewy series obtained over 𝔽_{2} also hold by extending the scalars to \overline{𝔽}_{2}, using Lemma 5.

We obtain that

dim_{\overline{𝔽}_{2}}(Z(B)) = 21 = dim_{\overline{𝔽}_{2}}(Z(b))
dim_{\overline{𝔽}_{2}}(J(Z(B))) = 20 = dim_{\overline{𝔽}_{2}}(J(Z(b)))
dim_{\overline{𝔽}_{2}}(J^{2}(Z(B))) = 5 ≠ 4 = dim_{\overline{𝔽}_{2}}(J^{2}(Z(b)))
dim_{\overline{𝔽}_{2}}(J^{3}(Z(B))) = 0 = dim_{\overline{𝔽}_{2}}(J^{3}(Z(b))).

Similarly we get for the centre of the principal block B of P SU(3, 8) and the centre of its Brauer correspondent b

dim_{\overline{𝔽}_{2}}(Z(B)) = 27 = dim_{\overline{𝔽}_{2}}(Z(b))
dim_{\overline{𝔽}_{2}}(J(Z(B))) = 26 = dim_{\overline{𝔽}_{2}}(J(Z(b)))
dim_{\overline{𝔽}_{2}}(J^{2}(Z(B))) = 3 ≠ 2 = dim_{\overline{𝔽}_{2}}(J^{2}(Z(b)))
dim_{\overline{𝔽}_{2}}(J^{3}(Z(B))) = 0 = dim_{\overline{𝔽}_{2}}(J^{3}(Z(b))).

An immediate variant of the program shows that this is a quite general phenomenon in odd characteristic. The group P SU(3, 3) gives an example in characteristic 3 since, denoting by B the principal block of \overline{𝔽}_{3}P SU(3, 3) and by b its Brauer correspondent,

dim_{\overline{𝔽}_{3}}(Z(B)) = 13 = dim_{\overline{𝔽}_{3}}(Z(b))
dim_{\overline{𝔽}_{3}}(J(Z(B))) = 12 = dim_{\overline{𝔽}_{3}}(J(Z(b)))
dim_{\overline{𝔽}_{3}}(J^{2}(Z(B))) = 4 ≠ 3 = dim_{\overline{𝔽}_{3}}(J^{2}(Z(b)))
dim_{\overline{𝔽}_{3}}(J^{3}(Z(B))) = 0 = dim_{\overline{𝔽}_{3}}(J^{3}(Z(b))).

The group P SU(3, 5) gives an example in characteristic 5 since, denoting by B the principal block of 𝔽_{5}P SU(3, 5) and by b its Brauer correspondent,

dim_{\overline{𝔽}_{5}}(Z(B)) = 13 = dim_{\overline{𝔽}_{5}}(Z(b))
dim_{\overline{𝔽}_{5}}(J(Z(B))) = 12 = dim_{\overline{𝔽}_{5}}(J(Z(b)))
dim_{\overline{𝔽}_{5}}(J^{2}(Z(B))) = 2 ≠ 1 = dim_{\overline{𝔽}_{5}}(J^{2}(Z(b)))
dim_{\overline{𝔽}_{5}}(J^{3}(Z(B))) = 0 = dim_{\overline{𝔽}_{5}}(J^{3}(Z(b))).

Theorem 14. Let K be the algebraic closure of 𝔽_{p} and let B be the principal block of P SU(3, p^{r}). Let b be the Brauer correspondent of B in the group ring of the normaliser of a 2-Sylow subgroup of P SU(3, p^{r}). Then B and b are stably equivalent of Morita type. If moreover p^{r} ∈ {3, 4, 5, 8}, then the square of the Jacobson radical of Z(B) is of different dimension than the square of the Jacobson radical of Z(b), whereas Z(B) and Z(b) both have Loewy length 3. In particular B ⊗_{K} K[X]/X^{p} is not stably equivalent of Morita type to b ⊗_{K} K[X]/X^{p}.

Proof. As seen at the beginning of this section, B and b are stably equivalent of Morita type by Green correspondence.

The GAP [4] program in Section 3 shows that the Loewy series of the centres of B and of b are of the same length but the dimensions of the Loewy layers are not equal. In particular the lowest Loewy layers of the algebras Z(B) and Z(b) have different dimension.

Corollary 12 implies that B ⊗_{K} K[X]/X^{p} is not stably equivalent of Morita type to b ⊗_{K} K[X]/X^{p}. □

Remark 15. The above examples suggest that in general, with the notation of Theorem 14, the dimension of J^{2}(Z(B)) could always be equal to 1 + dim_{K} J^{2}(Z(b)). By Theorem 41,

this is equal to \frac{p^{r} + 1}{γ}, where γ is the greatest common divisor of p^{r} + 1 and 3.

3. The GAP program

We display here the GAP program we used.

# the characteristic p
prem:=2;
#
# The group G
g:=PSU(3,prem^2);
#
# the ground field k
corps:=GF(prem);
#
s:=SylowSubgroup(g,prem);
# the normalizer NS of a Sylow p-subgroup
ns:=Normalizer(g,s);
#
# getting a permutation representation of G of smaller degree
f:=FactorCosetAction(g,ns);
g:=Image(f);
ns:=Image(f,ns);
#
# uncomment next line to replace G by NS
#g:=ns;
#
# computing the structure constants of ZkG
c:=ConjugacyClasses(g);
rc:=List(c,Representative);
lc:=Length(c);
ci:=List([1..lc],x->First([1..lc],y->rc[x]^(-1) in c[y]));
l:=List([1..lc],x->NullMat(lc,lc,corps));
for iu in [1..lc] do
    u:=c[iu];
    if rc[iu]=One(g) then
        for iv in [iu..lc] do
            Print("\r",iu,":",iv,"/",lc,"   ");
            v:=List([1..lc],x->Zero(corps));
            v[iv]:=One(corps);
            l[iu][iv]:=v;
            l[iv][iu]:=v;

        od;
    else
        for iv in [iu..lc] do
            Print("\r",iu,":",iv,"/",lc,"   ");
            w:=c[ci[iv]];
            v:=List(List(rc),x->One(corps)*Size(Intersection(u,List(w,y->x*y))));
            l[iu][iv]:=v;
            l[iv][iu]:=v;
        od;
    fi;
od;
Print("\n");
za:=Algebra(corps,l);
Print("Dimension of ZkG \t= ",Dimension(za),"\n");
radza:=RadicalOfAlgebra(za);
Print("Dimension of JZkG \t= ",Dimension(radza),"\n");
bradza:=Basis(radza);
vbradza:=BasisVectors(bradza);
vbr:=vbradza;
#
# Computing the powers of the radical of the center
i:=1;
repeat
    i:=i+1;
    l:=Set(List(Cartesian(vbradza,vbr),x->x[1]*x[2]));
    r:=Ideal(za,l);
    br:=Basis(r);
    vbr:=BasisVectors(br);
    d:=Dimension(r);
    Print("Dimension of (JZkG)^",i,"\t= ",d,"\n");
until  d=0;

4. The centre of the mod p group ring of the normaliser of the Sylow subgroup of P SU(3, p^{r})

Recall that we denote by S a Sylow p-subgroup of the projective special unitary group G = P SU(3, q) over the field with q^{2} elements, where q = p^{r}, and by N the normaliser of S in G. In this section, we determine the ring structure of the center ZkN of the group algebra kN, where k is any commutative ring.

Notation 16. If x ∈ N, we denote by x^{+} ∈ ZkN the sum of the conjugates of x in N.

Then the elements x^{+}, for x in a set of representatives of conjugacy classes of N, form a k-basis of ZkN.

Let V be a three dimensional vector space over the field 𝔽_{q^{2}}, with basis B. We endow V with a non degenerate hermitian product, and without loss of generality, we assume that the matrix of this product in B is equal to

( \begin{matrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix} ) .

Notation 17. For x ∈ 𝔽_{q^{2}}, we set \overline{x} = x^{q}. Then the map x ↦ \overline{x} is the automorphism of order 2 of the extension 𝔽_{q^{2}}/𝔽_{q}. We also set

Ψ = \{x ∈ 𝔽_{q^{2}}^{×} | x\overline{x} = 1\}

Let ω be a non zero element of 𝔽_{q^{2}} such that ω + \overline{ω} = 0, and τ be an element of 𝔽_{q^{2}} such that τ + \overline{τ} = −1.

It follows from [5, II Satz 10.12, page 242] that we can suppose that the group N is equal to the image in G of the group of matrices of the form

M(a, b, c) = ( \begin{matrix}a & b & c \\ 0 & \overline{a}/a & −\overline{b}/a \\ 0 & 0 & 1/\overline{a} \end{matrix} ) ,

where (a, b, c) belongs to the set

Q = \{(a, b, c) ∈ 𝔽_{q^{2}}^{×} × (𝔽_{q^{2}})^{2} | b\overline{b} + a\overline{c} + c\overline{a} = 0\} .

Lemma 18. For (a, b, c) ∈ Q, let M̂(a, b, c) denote the image of M(a, b, c) in N. Then if (a′, b′, c′) ∈ Q, we have that M̂(a, b, c) = M̂(a′, b′, c′) if and only if there is λ ∈ 𝔽_{q^{2}} with λ^{q−2} = 1 and (a′, b′, c′) = λ · (a, b, c).

Proof. M̂(a, b, c) = M̂(a′, b′, c′) if and only if there exists a scalar λ ∈ 𝔽_{q^{2}} such that

( \begin{matrix}a′ & b′ & c′ \\ 0 & \overline{a′}/a′ & −\overline{b′}/a′ \\ 0 & 0 & 1/\overline{a′} \end{matrix} ) = λ ( \begin{matrix}a & b & c \\ 0 & \overline{a}/a & −\overline{b}/a \\ 0 & 0 & 1/\overline{a} \end{matrix} ) .

Equivalently (a′, b′, c′) = λ(a, b, c) and \overline{λ}/λ = λ, i.e. λ^{q−2} = 1. □

For two non zero integers s, t denote by (s, t) their greatest common divisor. Observe that (q − 2, q^{2} − 1) = (3, q + 1), to motivate the following:

Notation 19. We set γ = (3, q + 1), and put

Γ = {λ ∈ 𝔽_{q^{2}} | λ^{γ} = 1} ≤ Ψ as well as L = {a^{γ} | a ∈ 𝔽_{q^{2}}^{×}} .

With this notation, the group N has order q^{3}(q^{2} − 1)/γ. It is equal to the semidirect

product of the group S, consisting of the elements M̂(1, b, c) = ( \begin{matrix}1 & b & c \\ 0 & 1 & −\overline{b} \\ 0 & 0 & 1 \end{matrix} ), where b and

c are elements of 𝔽_{q^{2}} such that b\overline{b} + c + \overline{c} = 0, by the cyclic group C of order (q^{2} − 1)/γ

consisting of the elements M̂(a, 0, 0) = ( \begin{matrix}a & 0 & 0 \\ 0 & \overline{a}/a & 0 \\ 0 & 0 & 1/\overline{a} \end{matrix} ), for a ∈ 𝔽_{q^{2}}^{×}/Γ.

Lemma 20. (1) Let (a, b, c) and (x, y, z) be elements of Q. Then

M(a, b, c)M(x, y, z) = M(ax, ay + \frac{b\overline{x}}{x}, az − \frac{b\overline{y}}{x} + \frac{c}{\overline{x}}) .

(2) Let (a, b, c) ∈ Q. Then M(a, b, c)^{−1} = M(\frac{1}{a}, −\frac{b}{\overline{a}}, c̄).

(3) Let (a, b, c) and (x, y, z) be elements of Q. Then

M(a, b, c)M(x, y, z)M(a, b, c)^{−1} = M (x, \frac{ab}{\overline{a}}(\frac{\overline{x}}{x} − x) + \frac{a^{2}}{\overline{a}} y, t) ,

where t = a\overline{c}x + \frac{\overline{a}c}{\overline{x}} + ay\overline{b} − \frac{\overline{a}b\overline{y}}{x} + \frac{b\overline{b}\overline{x}}{x} + a\overline{a}z.

Proof. All the assertions follow from straightforward computations. □

Proposition 21. (1) Let (x, y, z) and (x′, y′, z′) be elements of Q. If M̂(x, y, z) and

M̂(x′, y′, z′) are conjugate in N, then x^{−1}x′ ∈ Γ.

(2) The elements M̂(x, 0, 0), for x ∈ 𝔽_{q^{2}}/Γ, lie in distinct conjugacy classes of N.

(3) Let (x, y, z) ∈ Q. Then if x ∉ Γ, the element M̂(x, y, z) of N is conjugate to an element of the form M̂(x, 0, xuω), for some u ∈ 𝔽_{q}.

(4) Let x ∈ 𝔽_{q^{2}}^{×} and u ∈ 𝔽_{q}. Then if x\overline{x} ≠ 1, the element M̂(x, 0, xuω) of N is conjugate to M̂(x, 0, 0). If x\overline{x} = 1, and if u ≠ 0, then the element M̂(x, 0, xuω) is conjugate to M̂(x, 0, xω), and not conjugate to M̂(x, 0, 0).

(5) If (1, y, z) ∈ Q, then either y ≠ 0 and there exists u ∈ 𝔽_{q} such that z = y\overline{y}(τ + uω), or y = 0 and there exists u ∈ 𝔽_{q} such that z = uω. Moreover, if (1, y′, z′) ∈ Q and if M̂(1, y′, z′) and M̂(1, y, z) are conjugate in N, then y and y′ are both non zero, or both equal to 0.

(6) If (1, y, z) and (1, y′, z′) are in Q, and if y and y′ are both non zero, then M̂(1, y′, z′) and M̂(1, y, z) are conjugate in N if and only if y^{′(q^{2}−1)/γ} = y^{(q^{2}−1)/γ} in 𝔽_{q^{2}}^{×}, i.e. if y′/y ∈ L. In particular M(1, y, z) is conjugate to M(1, y, y\overline{y}τ).

Proof. Assertion (1) follows from Assertion (3) of Lemma 20: if

M̂(x′, y′, z′) = M̂ (x, \frac{ab}{\overline{a}}(\frac{\overline{x}}{x} − x) + \frac{a^{2}}{\overline{a}} y, t) ,

then there exists λ ∈ Γ such that x′ = λx by Lemma 18.

Assertion (2) is a straightforward consequence of Assertion (1).

For Assertion (3), we use Assertion (3) of Lemma 20 again: since x ∉ Γ, we have \frac{\overline{x}}{x} ≠ x, and we can set a = 1, b = − \frac{y}{\frac{\overline{x}}{x} −x}, and c = b\overline{b}τ. Then (a, b, c) ∈ Q and

M(a, b, c)M(x, y, z)M(a, b, c)^{−1} is of the form M(x, 0, t), for some t ∈ 𝔽_{q^{2}}. In particular (x, 0, t) ∈ Q, hence x\overline{t} + t\overline{x} = 0. In other words t = vx with v + \overline{v} = 0. Then v = uω and u = \overline{u}, that is u ∈ 𝔽_{q}.

For Assertion (4), we have to decide when two elements of the form n = M̂(x, 0, xuω) and n′ = M̂(x′, 0, x′u′ω) are conjugate in N, where x, x′ ∉ Γ, and u, u′ ∈ 𝔽_{q}. By Assertion (1), we can assume that x = x′, and then n and n′ are conjugate if and only if there exists (a, b, c) ∈ Q such that

M(a, b, c)M(x, 0, xuω)M(a, b, c)^{−1} = M(x, 0, xu′ω) .

By Assertion (3) of Lemma 20, we have \frac{ab}{\overline{a}}(\frac{\overline{x}}{x} − x) = 0, hence b = 0. Now xu′ω is equal to the element t of Lemma 20, in the case y = b = 0 and z = xuω, that is

xu′ω = a\overline{c}x + \frac{\overline{a}c}{\overline{x}} + a\overline{a}xuω .

Moreover a\overline{c} + c\overline{a} = 0, since (a, 0, c) ∈ Q. So there exists v ∈ 𝔽_{q} such that c = avω. This gives

xu′ω = −a\overline{a}xvω + \frac{a\overline{a}xvω}{\overline{x}} + a\overline{a}xuω ,

or equivalently

u′ = a\overline{a}(u − v(1 − \frac{1}{x\overline{x}})) .

Thus n and n′ are conjugate in N if and only if there exist a ∈ 𝔽_{q^{2}}^{×} and v ∈ 𝔽_{q} such that

u′ = a\overline{a}(u − v(1 − \frac{1}{x\overline{x}})). If x\overline{x} ≠ 1, then we can take a = 1 and v = \frac{u − u′}{1−\frac{1}{x\overline{x}}}, so n and n′ are

conjugate. And if x\overline{x} = 1, then n and n′ are conjugate if and only if there exists a ∈ 𝔽_{q^{2}}^{×} such that u′ = a\overline{a}u, or equivalently, if there exists λ ∈ 𝔽_{q}^{×} such that u′ = λu. So either u = u′ = 0, or u and u′ are both non zero. This completes the proof of Assertion (4).

For Assertion (5), assume that (1, y, z) ∈ Q. Then y\overline{y}+z +\overline{z} = 0. If y ≠ 0, set v = \frac{z}{y\overline{y}} −τ.

Then v + \overline{v} = 0, so there exists u ∈ 𝔽_{q} such that v = uω, thus u = y\overline{y}(τ + uω). And if y = 0, then z + \overline{z} = 0, so z = uω for some u ∈ 𝔽_{q}.

Now by Assertion (3) of Lemma 20, for (1, y, z) and (1, y′, z′) in Q, the elements n = M̂(1, y, z) and n′ = M̂(1, y′, z′) are conjugate in N if and only if there exists (a, b, c) ∈ Q such that

y′ = \frac{a^{2}}{\overline{a}} y and z′ = a\overline{c} + \overline{a}c + ay\overline{b} − \overline{a}b\overline{y} + b\overline{b} + a\overline{a}z ,

that is

y′ = \frac{a^{2}}{\overline{a}} y and z′ = ay\overline{b} − \overline{a}b\overline{y} + a\overline{a}z ,

In particular y is non zero if and only if y′ is non zero. Assertion (5) follows.

Assume now that both y and y′ are non zero. If n and n′ are conjugate, then there

exists a ∈ 𝔽_{q^{2}}^{×} such that y′ = \frac{a^{2}}{\overline{a}} y = a^{2−q}y. It follows that y′/y belongs to the subgroup of 𝔽_{q^{2}}^{×} consisting of (q − 2)-th powers, i.e. the subgroup of γ-th powers, i.e. the unique

subgroup of order (q^{2} − 1)/γ of 𝔽_{q^{2}}^{×}. Equivalently (y′/y)^{(q^{2}−1)/γ} = 1. Conversely, suppose

that there exists a ∈ 𝔽_{q^{2}}^{×} such that y′ = \frac{a^{2}}{\overline{a}} y. There are elements u and u′ of 𝔽_{q} such that z = y\overline{y}(τ + uω) and z′ = y′\overline{y′}(τ + u′ω). If we can find b and c such that (a, b, c) ∈ Q and z′ = a\overline{b}y − \overline{a}b\overline{y} + a\overline{a}z, then n and n′ are conjugate in N. This can also be written

a\overline{a}y\overline{y}(τ + u′ω) = a\overline{b}y − \overline{a}b\overline{y} + a\overline{a}y\overline{y}(τ + uω) ,

or equivalently

(∗)  \frac{1}{ω} (\frac{\overline{b}}{\overline{a}\overline{y}} − \frac{b}{ay} ) = u′ − u .

Now the map b ↦ \frac{1}{ω} (\frac{\overline{b}}{\overline{a}\overline{y}} − \frac{b}{ay} ) is a non zero 𝔽_{q}-linear map from 𝔽_{q^{2}} to F_{q}. Hence it is

surjective, and there exists b ∈ 𝔽_{q^{2}} such that (*) holds. Now we set c = \frac{b\overline{b}}{\overline{a}}τ, and then (a, b, c) ∈ Q, and the elements n and n′ are conjugate in N. This proves Assertion (6), and completes the proof of Proposition 21. □

Corollary 22. The set

E = \{M̂(x, 0, 0) | x ∈ 𝔽_{q^{2}}^{×}/Γ\} ⨆ \{M̂(x, 0, xω) | x ∈ Ψ/Γ\} ⨆ \{M̂(1, y, y\overline{y}τ) | y ∈ 𝔽_{q^{2}}^{×}/L\}

is a set of representatives of conjugacy classes of N. In particular, there are \frac{q^{2} + q}{γ} + γ conjugacy classes in N.

Proof. Indeed, by Proposition 21, the set E is a set of representatives of conjugacy classes of N. Its cardinality is

|E| = \frac{q^{2} − 1}{γ} + \frac{q + 1}{γ} + γ = \frac{q^{2} + q}{γ} + γ .

□

Notation 23.

• For x ∈ 𝔽_{q^{2}}^{×}, we set d_{x} = M̂(x, 0, 0) and D_{x} = d_{x}^{+} ∈ ZkN.

• For x ∈ Ψ, we set t_{x} = M̂(x, 0, xω) and T_{x} = t_{x}^{+}.
• For y ∈ 𝔽_{q^{2}}^{×}, we set u_{y} = M̂(1, y, y\overline{y}τ) and U_{y} = u_{y}^{+}.

Proposition 24.

(1) For x ∈ 𝔽_{q^{2}}^{×} − Ψ,

d_{x}^{N} = \{M̂(x, y, z) | y, z ∈ 𝔽_{q^{2}}, y\overline{y} + x\overline{z} + z\overline{x} = 0\} .

In particular |d_{x}^{N}| = q^{3}.

(2) For x ∈ Ψ − Γ,

d_{x}^{N} = \{M̂(x, y(\overline{x}^{2} − x), y\overline{y}(\overline{x}^{2} − x)) | y ∈ 𝔽_{q^{2}}\} .

In particular |d_{x}^{N}| = q^{2}.

(3) For x ∈ Γ, the element d_{x} is the identity element of N, and |d_{x}^{N}| = 1.

(4) For x ∈ Ψ, the conjugacy class of t_{x} in N has cardinality q^{2}(q − 1) if x ∉ Γ, and q − 1 otherwise. The conjugacy class of T_{1} consists of the elements M̂(1, 0, λω), for λ ∈ 𝔽_{q}^{×}.

(5) For x ∈ 𝔽_{q^{2}}^{×},

u_{x}^{N} = \{M̂(1, v, v\overline{v}τ + λω) | v ∈ xL, λ ∈ 𝔽_{q}\} .

In particular |u_{x}^{N}| = \frac{q(q^{2} − 1)}{γ}.

Proof. It follows from Proposition 21 that if (x, y, z) ∈ Q and x\overline{x} ≠ 1, then M̂(x, y, z) is conjugate to d_{x}, and that conversely, any conjugate of d_{x} in N is of the form M̂(x, y, z), for some elements y, z ∈ 𝔽_{q^{2}} such that (x, y, z) ∈ Q. This proves Assertion (1).

Now let (a, b, c) and (x, y, z) be elements of Q. By Assertion (1) of Lemma 20, comparing the diagonal elements in the product in the two possible orders, the elements M̂(a, b, c) and M̂(x, y, z) commute if and only if

(∗∗)  ay + \frac{b\overline{x}}{x} = xb + \frac{y\overline{a}}{a} and az − \frac{b\overline{y}}{x} + \frac{c}{\overline{x}} = xc − \frac{y\overline{b}}{a} + \frac{z}{\overline{a}}

• If y = z = 0, this gives \frac{b\overline{x}}{x} = xb and \frac{c}{\overline{x}} = xc. If moreover x\overline{x} = 1 but x^{2} ≠ \overline{x}, then b = 0, but a and c are arbitrary, only subject to a\overline{x} + c\overline{a} = 0. In this case the

centraliser of d_{x} in N has cardinality \frac{q(q^{2} − 1)}{γ}, and the conjugacy class of d_{x} in N has cardinality q^{2}. Now Assertion (2) follows from the fact that the elements

M̂(1, y, y\overline{y}τ)M̂(x, 0, 0)M̂(1, y, y\overline{y}τ)^{−1} = M̂(x, y(\overline{x}^{2} − x), y\overline{y}(\overline{x}^{2} − x)) ,

for y ∈ 𝔽_{q^{2}}, are all distinct.

Finally if x^{2} = \overline{x}, then x ∈ Γ, so d_{x} is the identity element of N, and Assertion (3) follows.

• If x ∈ Ψ, y = 0 and z = xω, then the relations (**) give

b\overline{x}^{2} = xb and axω + cx = xc + \frac{xω}{\overline{a}} ,

that is b(x − \overline{x}^{2}) = 0 and a\overline{a} = 1. If x ≠ \overline{x}^{2}, i.e. if x ∉ Γ, this is equivalent to b = 0 and a\overline{a} = 1. Then c is arbitrary, only subject to a\overline{x} + c\overline{a} = 0. In this case

the centraliser of t_{x} in N has cardinality \frac{q(q + 1)}{γ}, and the conjugacy class of t_{x} in N has cardinality q^{2}(q − 1). Now if x^{2} = \overline{x}, the only condition left is a\overline{a} = 1, so the centraliser of t_{x} in N has cardinality q^{3} (it is equal to S), and the conjugacy class of t_{x} in N has cardinality q − 1. Moreover, by Lemma 20, the conjugates of t_{1} = M̂(1, 0, ω) are the elements M̂(1, 0, a\overline{a}ω), for a ∈ 𝔽_{q^{2}}^{×}. This completes the proof of Assertion (4).

• If x = 1, y ∈ 𝔽_{q^{2}}^{×}, and z = y\overline{y}ω, then the relations (**) give

ay = \frac{y\overline{a}}{a} and ay\overline{y}ω − b\overline{y} = −\frac{y\overline{b}}{a} + \frac{y\overline{y}ω}{\overline{a}} .

Since y ≠ 0, the first relation gives a^{2} = \overline{a}, i.e. a ∈ Γ, so we can assume a = 1 by Lemma 18. Now the second relation reads b\overline{y} = y\overline{b}, i.e. b = uy, for u ∈ 𝔽_{q}. Since c is subject to c + \overline{c} + b\overline{b} = 0, it follows that the centraliser of u_{y} in N has cardinality

q^{2}, and the conjugacy class of u_{y} in N has cardinality \frac{q(q^{2} − 1)}{γ}.

Now Assertion (5) follows from the fact that by Proposition 21, the element M̂(1, v, v\overline{v}τ + λω), for v ∈ xL and λ ∈ 𝔽_{q}, is conjugate to u_{x}, and that there are

\frac{q(q^{2} − 1)}{γ} such elements in N.

□

We recall the following well known fact (cf e.g. [3, (9.28)]):

Lemma 25. Let G be a finite group and k be a commutative ring. For x ∈ G, let x^{+} ∈ ZkG denote the sum of the elements of the conjugacy class x^{G} of x in G. Then for x, y ∈ G

x^{+} · y^{+} = ∑_{z∈[G]} m_{x,y}^{z}z^{+} ,

where [G] denotes a set of representatives of conjugacy classes of G, and

m_{x,y}^{z} = |{(x′, y′) ∈ x^{G} × y^{G} | x′y′ = z}| .

Clearly m_{x,y}^{z} = m_{y,x}^{z} and m_{x^{−1},y^{−1}}^{z^{−1}} = m_{x,y}^{z} for any x, y, z ∈ G, but since

m_{x,y}^{z}|z^{G}| = |{(x′, y′, z′) ∈ x^{G} × y^{G} × z^{G} | x′y′ = z′}| ,

we have also m_{x,y}^{z}|z^{G}| = m_{z,y^{−1}}^{x}|x^{G}| = m_{z,x^{−1}}^{y}|y^{G}|.

Observe that Z(kN) = k ⊗_{ℤ} Z(ℤN) and hence we may and will suppose for the rest of this section that k = ℤ, unless otherwise stated.

Proposition 26. (1) Let x ∈ F_{q^{2}}^{×} − Ψ and y ∈ 𝔽_{q^{2}}^{×} such that xy ∉ Ψ. Then

D_{x}D_{y} = \{ \begin{matrix}q^{3}D_{xy} & if y ∉ Ψ \\ q^{2}D_{xy} & if y ∈ Ψ . \end{matrix}

(2) Let x ∈ F_{q^{2}}^{×} − Ψ and y ∈ Ψ. Then

D_{x}T_{y} = \{ \begin{matrix}q^{2}(q − 1)D_{xy} & if y ∉ Γ \\ (q − 1)D_{xy} & if y ∈ Γ . \end{matrix}

(3) Let x ∈ F_{q^{2}}^{×} − Ψ and y ∈ 𝔽_{q^{2}}^{×}. Then D_{x}U_{y} = \frac{q(q^{2} − 1)}{γ}D_{x}.

Proof. The three assertions follow from the fact that the product of an element in the conjugacy class of r = M̂(x_{1}, y_{1}, z_{1}) of N and an element in the conjugacy class of s = M̂(x_{2}, y_{2}, z_{2}) of N is an element of the form M̂(x_{1}x_{2}, α, β), for some α and β in 𝔽_{q^{2}}. In each assertion, the assumption implies that all these elements are in the conjugacy class of t = d_{x_{1}x_{2}}, since x_{1}x_{2} ∈ 𝔽_{q^{2}} − Ψ. It follows that there exists an integer m such that r^{+}s^{+} = mD_{x_{1}x_{2}}.

Now the augmentation map ε : kN → k restricts to a ring homomorphism ZkN → k, sending x^{+} to |x^{G}|. Hence |r^{N}||s^{N}| = m|t^{N}|. For the three assertions, we can assume that r = d_{x} and x ∉ Ψ, thus |r^{N}| = q^{3}. Similarly t = d_{xy} for Assertions (1) and (2), and xy ∉ Ψ, so |t^{N}| = q^{3}. For Assertion (3), we have t = d_{x}, so |t^{N}| = q^{3} again. It follows that the integer m is equal to |s^{N}|, and s = d_{y} in Assertion (1), s = t_{y} in Assertion (2), and s = u_{y} in Assertion (3). Now Proposition 26 follows from the values of the cardinalities |s^{N}| given by Proposition 24. □

Proposition 27. Let x, y ∈ 𝔽_{q^{2}} − Ψ, such that xy ∈ Ψ − Γ. Then D_{x}D_{y} = q^{3}D_{xy} + q^{3}T_{xy}.

Proof. Any element in the product d_{x}^{N} · d_{y}^{N} is of the form M̂(xy, α, β), for some α, β ∈ 𝔽_{q^{2}}. It follows that there are integers a and b such that D_{x}D_{y} = aD_{xy} +bT_{xy}. Setting z = xy, the integer a is equal to m_{d_{x},d_{y}}^{d_{z}}. Thus a|d_{z}^{N}| = m_{d_{z},d_{y}^{−1}}^{d_{x}}|d_{x}^{N}|, by Lemma 25. But by Proposition 26, we have D_{z}D_{y^{−1}} = q^{2}D_{x}, so m_{d_{z},d_{y}^{−1}}^{d_{x}} = q^{2}. It follows that a|z^{N}| = aq^{2} = q^{2}q^{3}, thus a = q^{3}. Taking augmentation gives

ε(D_{x}D_{y}) = q^{6} = aε(D_{z}) + bε(T_{z}) = aq^{2} + bq^{2}(q − 1) .

It follows that b = \frac{q^{6} − q^{5}}{q^{2}(q − 1)} = q^{3}. □

Proposition 28. Let x ∈ Ψ. Then D_{x}T_{1} = T_{x}.

Proof. If x ∈ Γ, there is nothing to prove, because D_{x} is equal to the identity, in this case. If x ∉ Γ, then D_{x}T_{1} is a sum of elements of the form M̂(x, α, β), so there are natural integers a and b such that D_{x}T_{1} = aD_{x} + bT_{x}. Taking augmentation of this equality gives q^{2}(q − 1) = aq^{2} + bq^{2}(q − 1), that is q − 1 = a + b(q − 1). Since the product d_{x}t_{1} is equal to t_{x}, it follows that b > 0. Hence b = 1 and a = 0. □

Proposition 29. Let x ∈ Ψ − Γ, and y ∈ 𝔽_{q^{2}}^{×}. Then D_{x}U_{y} = \frac{q^{2} − 1}{γ}(D_{x} + T_{x}).

Proof. Again D_{x}U_{y} is a sum of elements of N of the form M̂(x, α, β). Hence there are natural integers a and b such that D_{x}U_{y} = aD_{x} + bT_{x}. The integer a is equal to m_{d_{x},u_{y}}^{d_{x}}, i.e.

a = |{(d′, u′) ∈ d_{x}^{N} × u_{y}^{N} | d′u′ = d_{x}}| .

By Proposition 24, the element d′ ∈ d_{x}^{N} is equal to M̂(x, w(\overline{x}^{2}−x), w\overline{w}(\overline{x}^{2}−x)), for w ∈ 𝔽_{q^{2}}, and the element u′ is equal to M̂(1, v, v\overline{v}τ + λω), for v ∈ xL and λ ∈ 𝔽_{q}. Now

d′u′ = M̂(x, xv + w(\overline{x}^{2} − x), x(v\overline{v}τ + λω) − w\overline{v}(\overline{x}^{2} − x) + w\overline{w}(\overline{x}^{2} − x)) .

This is equal to d_{x} if and only if

xv + w(\overline{x}^{2} − x) = 0 and x(v\overline{v}τ + λω) − w\overline{v}(\overline{x}^{2} − x) + w\overline{w}(\overline{x}^{2} − x) = 0 .

Since x ∉ Γ, the first relation gives w = \frac{v}{1 − \overline{x}^{3}} . Multiplying by \overline{x}, the second one reads

v\overline{v}τ + λω − w\overline{v}(\overline{x}^{3} − 1) + w\overline{w}(\overline{x}^{3} − 1) = 0 .

This gives

v\overline{v}τ + λω + v\overline{v} − \frac{v\overline{v}}{1 − x^{3}} = 0 ,

that is

λ = \frac{1}{ω} (\overline{τ} + \frac{1}{1 − x^{3}} ) .

This defines an element λ of 𝔽_{q}, since τ + \overline{τ} = −1 and

\frac{1}{1 − x^{3}} + \frac{1}{1 − \overline{x}^{3}} = \frac{2 − x^{3} − \overline{x}^{3}}{(1 − x^{3})(1 − \overline{x}^{3})} = 1 .

In other words w and λ are determined by v ∈ xL, which may be chosen arbitrarily. It

follows that a = \frac{q^{2} − 1}{γ}.

Now applying the augmentation to the relation D_{x}U_{y} = aD_{x} + bT_{x} gives

q^{2} \frac{q(q^{2} − 1)}{γ} = aq^{2} + bq^{2}(q − 1) .

It follows that

\frac{q(q^{2} − 1)}{γ} = \frac{q^{2} − 1}{γ} + b(q − 1) ,

hence b = \frac{q^{2} − 1}{γ}. □

Proposition 30.

(1) Let x ∈ Ψ − Γ. Then

D_{x}D_{x^{−1}} = q^{2}Id + q ∑_{y∈𝔽_{q^{2}}^{×}/L} U_{y} .

(2) Let x ∈ 𝔽_{q^{2}}^{×} − Ψ. Then

D_{x}D_{x^{−1}} = q^{3}Id + q^{3}T_{1} + q^{3} ∑_{y∈𝔽_{q^{2}}^{×}/L} U_{y} .

Proof. For x ∈ 𝔽_{q^{2}}^{×}, the product D_{x}D_{x^{−1}} is a sum of elements of the form M̂(1, α, β) of N. So there are integers a, b, c_{y} ∈ ℕ, for y ∈ 𝔽_{q^{2}}^{×}/L such that

(∗ ∗ ∗)  D_{x}D_{x^{−1}} = aId + bT_{1} + ∑_{y∈𝔽_{q^{2}}^{×}/L} c_{y}U_{y} .

Then a = m_{d_{x},d_{x^{−1}}}^{Id} = |{(d′, d′′) ∈ d_{x}^{N} × d_{x^{−1}}^{N} | d′d′′ = Id}| = |d_{x}^{N}|. Thus a = q^{2} if x ∈ Ψ − Γ, and a = q^{3} if x ∈ 𝔽_{q^{2}}^{×} − Ψ.

On the other hand, by Lemma 25, for y ∈ 𝔽_{q^{2}}^{×},

c_{y}|u_{y}^{N}| = m_{d_{x},d_{x^{−}1}}^{u_{y}}|u_{y}^{N}| = m_{u_{y},d_{x}}^{d_{x}}|d_{x}^{N}|

• If x ∈ Ψ − Γ, then m_{u_{y},d_{x}}^{d_{x}} = \frac{q^{2} − 1}{γ}, by Proposition 29. It follows that

c_{y} \frac{q(q^{2} − 1)}{γ} = \frac{q^{2} − 1}{γ}q^{2} ,

hence c_{y} = q.

Applying augmentation to equation (***), we get q^{2}q^{2} = a+b(q−1)+q·γq \frac{q^{2} − 1}{γ}.

This gives b(q − 1) = q^{4} − q^{2} − q^{2}(q^{2} − 1) = 0, which proves Assertion (1).

• If x ∈ 𝔽_{q^{2}}^{×} − Ψ, then m_{u_{y},d_{x}}^{d_{x}} = \frac{q(q^{2} − 1)}{γ} by Proposition 26. Thus c_{y} = q^{3} in this case. Applying augmentation to equation (***) gives

q^{3} · q^{3} = q^{3} + b(q − 1) + q^{3}γ · \frac{q(q^{2} − 1)}{γ} ,

that is b(q − 1) = q^{6} − q^{3} − q^{4}(q^{2} − 1) = q^{3}(q − 1), hence b = q^{3}, which proves Assertion (2).

□

Proposition 31. Let x, y ∈ Ψ − Γ such that xy ∉ Γ. Then D_{x}D_{y} = D_{xy} + (q + 1)T_{xy}.

Proof. The product D_{x}D_{y} is a sum of elements of N of the form M̂(xy, α, β), so there are integers a and b such that D_{x}D_{y} = aD_{xy} +bT_{xy}. The integer a is the number of pairs (d′, d′′) in d_{x}^{N} × d_{y}^{N} such that d′d′′ = d_{xy}.

By Proposition 24, the class d_{x}^{N} consists of the elements M̂(x, α(\overline{x}^{2} − x), α\overline{α}(\overline{x}^{2} − x)), for α ∈ 𝔽_{q^{2}}. Equivalently, in a form that will be more convenient for computation, it consists of the elements d′ = M̂(x, u, v), for u ∈ 𝔽_{q^{2}} and v = \frac{u\overline{u}}{x^{2} − \overline{x}}. Similarly, the class d_{y}^{N} consist

of the elements d′′ = M̂(y, r, s), for r ∈ 𝔽_{q^{2}} and s = \frac{r\overline{r}}{y^{2} − \overline{y}}. Since x\overline{x} = 1 = y\overline{y}, we have

d′d′′ = ( \begin{matrix}x & u & v \\ 0 & \overline{x}^{2} & −\overline{u} \overline{x} \\ 0 & 0 & x \end{matrix} ) ( \begin{matrix}y & r & s \\ 0 & \overline{y}^{2} & −\overline{r} \overline{y} \\ 0 & 0 & y \end{matrix} ) .

The product d′d′′ is equal to d_{xy} if and only if

xr + u\overline{y}^{2} = 0 and xs − u\overline{r} \overline{y} + vy = 0 .

The first equation gives r = −u\overline{x} \overline{y}^{2}, thus r\overline{r} = u\overline{u}. Now the second equation becomes

\frac{xu\overline{u}}{y^{2} − \overline{y}} + u\overline{u}xy + \frac{yu\overline{u}}{x^{2} − \overline{x}} = 0 .

Then either u = 0, hence r = s = v = 0, or

\frac{x}{y^{2} − \overline{y}} + xy + \frac{y}{x^{2} − \overline{x}} = 0 .

Equivalently (x^{3} − 1) + (x^{3} − 1)(y^{3} − 1) + (y^{3} − 1) = 0, thus x^{3}y^{3} = 1, which doesn’t hold since xy ∉ Γ, using the remark after Lemma 18.

So the only pair (d′, d′′) ∈ d_{x}^{N} × d_{y}^{N} such that d′d′′ = d_{xy} is the pair (d_{x}, d_{y}). It follows that a = 1.

Applying augmentation to the equality D_{x}D_{y} = aD_{xy}+bT_{xy} now gives q^{4} = q^{2}+bq^{2}(q−1), hence b = q + 1 □

Proposition 32. Let x ∈ Ψ − Γ and y ∈ 𝔽_{q^{2}}^{×} with xy ∉ Γ. Then

D_{x}T_{y} = (q^{2} − 1)D_{xy} + (q^{2} − q − 1)T_{xy} .

Proof. The product D_{x}T_{y} is a sum of elements of the form M̂(xy, α, β), so there are integers a and b such that D_{x}T_{y} = aD_{xy} + bT_{xy}. By Lemma 25, Proposition 24, and Proposition 31, we have

aq^{2} = m_{d_{x},t_{y}}^{d_{xy}}|d_{xy}^{N}| = m_{d_{xy},d_{x}^{−1}}^{t_{y}}q^{2}(q − 1) = q^{2}(q^{2} − 1) ,

hence a = q^{2} − 1. Taking augmentation gives

ε(D_{x}T_{y}) = q^{2}q^{2}(q − 1) = aε(D_{xy}) + bε(T_{xy}) = (q^{2} − 1)q^{2} + bq^{2}(q − 1) ,

hence b = q^{2} − q − 1. □

Proposition 33. (1) T_{1}^{2} = (q − 1)Id + (q − 2)T_{1}.

(2) If x ∈ Ψ − Γ, then T_{x}T_{1} = (q − 1)D_{x} + (q − 2)T_{x}.

Proof. By Proposition 24, the product of any two conjugates of t_{1} is either the identity, or again a conjugate of t_{1}. It follows that there are integers a and b such that T_{1}^{2} = aId + bT_{1}. Moreover a is equal to the cardinality of the conjugacy class of t_{1}, that is a = q − 1. Now taking augmentation gives (q − 1)^{2} = a + (q − 1)b, hence b = q − 2. Now for x ∈ Ψ − Γ,

T_{x}T_{1} = D_{x}T_{1}^{2} = (q − 1)D_{x} + (q − 2)T_{x} ,

since D_{x}T_{1} = T_{x} by Proposition 28. □

Proposition 34. Let x ∈ Ψ − Γ. Then D_{x}T_{x^{−1}} = q^{2}T_{1} + q(q − 1) ∑_{y∈𝔽_{q^{2}}^{×}/L} U_{y}.

Proof. Again D_{x}T_{x^{−1}} is a sum of elements of the form M̂(1, α, β), so there are integers a, b,

and c_{y}, for y ∈ 𝔽_{q^{2}}^{×}/L, such that D_{x}T_{x^{−1}} = aId + bT_{1} + ∑_{y∈𝔽_{q^{2}}^{×}/L} c_{y}U_{y}. Since t_{x^{−1}} = t_{x}^{−1},

and since no conjugate of d_{x} is a conjugate of t_{x}, we have a = 0. Then b = m_{d_{x},t_{x^{−}1}}^{t_{1}}, hence b(q − 1) = m_{t_{1},d_{x}}^{t_{x}}q^{2}(q − 1) = q^{2}(q − 1), by Proposition 28. Hence b = q^{2}. Similarly

c_{y} = m_{d_{x},t_{x^{−1}}}^{u_{y}}, so c_{y} \frac{q(q^{2} − 1)}{γ} = m_{u_{y},d_{x^{−1}}}^{t_{x^{−1}}} q^{2}(q − 1), hence c_{y} \frac{q(q^{2} − 1)}{γ} = \frac{q^{2} − 1}{γ}q^{2}(q − 1),

thus c_{y} = q(q − 1). □

Proposition 35. Let x ∈ Ψ − Γ and y ∈ 𝔽_{q^{2}}. Then T_{x}U_{y} = \frac{(q^{2} − 1)(q − 1)}{γ}(D_{x} + T_{x}).

16 SERGE BOUC AND ALEXANDER ZIMMERMANN

Proof. By Proposition 28 and Proposition 29, we have that

T_{x}U_{y} = D_{x}T_{1}U_{y} = \frac{(q^{2} − 1)}{γ}(D_{x} + T_{x})T_{1}
= \frac{(q^{2} − 1)}{γ}(T_{x} + (q − 1)D_{x} + (q − 2)T_{x})
= \frac{(q^{2} − 1)(q − 1)}{γ}(D_{x} + T_{x})
□

Proposition 36. Let x ∈ Ψ − Γ. Then

T_{x}T_{x^{−1}} = q^{2}(q − 1)Id + q^{2}(q − 2)T_{1} + q(q − 1)^{2} ∑_{y∈𝔽_{q^{2}}^{×}/L} U_{y} .

Proof. Indeed by Proposition 30, Proposition 28, Proposition 33 and Proposition 34

T_{x}T_{x^{−1}} = D_{x}T_{1}D_{x^{−1}}T_{1}
= D_{x}D_{x^{−1}}T_{1}^{2}
= D_{x}D_{x^{−1}}((q − 1)Id + (q − 2)T_{1})
= D_{x}((q − 1)D_{x^{−1}} + (q − 2)T_{x^{−1}})
= (q − 1)(q^{2}Id + q ∑_{y∈𝔽_{q^{2}}^{×}/L} U_{y}) + (q − 2)(q^{2}T_{1} + q(q − 1) ∑_{y∈𝔽_{q^{2}}^{×}/L} U_{y})
= q^{2}(q − 1)Id + q^{2}(q − 2)T_{1} + q(q − 1)^{2} ∑_{y∈𝔽_{q^{2}}^{×}/L} U_{y} .
□

Proposition 37. Let x, y ∈ Ψ − Γ such that xy ∉ Γ. Then

T_{x}T_{y} = (q − 1)(q^{2} − q − 1)D_{xy} + (q(q − 1)^{2} + 1)T_{xy} .

Proof. Indeed, by Proposition 31, Proposition 28 and Proposition 33

T_{x}T_{y} = D_{x}T_{1}D_{y}T_{1}
= (D_{xy} + (q + 1)T_{xy})((q − 1)Id + (q − 2)T_{1})
= (q − 1)D_{xy} + (q − 2)T_{xy} + (q^{2} − 1)T_{xy} + (q − 2)(q + 1)((q − 1)D_{xy} + (q − 2)T_{xy})
= (q − 1)(q^{2} − q − 1)D_{xy} + (q(q − 1)^{2} + 1)T_{xy}
□

Proposition 38. Let x ∈ 𝔽_{q^{2}}^{×}. Then T_{1}U_{x} = (q − 1)U_{x}. Proof. The product T_{1}U_{x} is a linear combination of elements of N of the form M̂(1, α, β). Hence there are integers a, b and c_{y}, for y ∈ 𝔽_{q^{2}}^{×}/L, such that

(#)  T_{1}U_{x} = aId + bT_{1} + ∑_{y∈𝔽_{q^{2}}^{×}/L} c_{y}U_{y} .

Observe now that t_{1} and u_{x}^{−1} are not conjugate in N, e.g. because the conjugacy class of t_{1} has cardinality q − 1, and the conjugacy class of u_{x} has cardinality \frac{q(q^{2} − 1)}{γ} ≠ q − 1. It follows that a = 0.

Now by Proposition 24, the conjugacy class of T_{1} consists of the elements M̂(1, 0, λω), for λ ∈ 𝔽_{q}^{×}, and the conjugacy class of u_{x} consists of the elements M̂(1, v, v\overline{v}τ + μω), for v ∈ xL and μ ∈ 𝔽_{q}. The product π = M̂(1, 0, λω)M̂(1, v, v\overline{v}τ + μω) is equal to u_{y} = M̂(1, y, y\overline{y}τ) if and only if v = y and v\overline{v}τ + μω + λω = y\overline{y}τ. It follows that c_{y} = 0 unless y ∈ xL, i.e. unless yL = xL. If yL = xL, then u_{y} is conjugate to u_{x} in N, and we can assume that y = x. In this case π = u_{x} if and only if v = x and μ = −λ. It follows that c_{x} = q − 1.

Applying augmentation to Equation (#) now gives ’

(q − 1) · \frac{q(q^{2} − 1)}{γ} = b(q − 1) + (q − 1) · \frac{q(q^{2} − 1)}{γ} ,

hence b = 0. □

Proposition 39. (1) If 3 ∤ q + 1, then L = 𝔽_{q^{2}}^{×}, and

U_{1}^{2} = q(q^{2} − 1)Id + q(q^{2} − 1)T_{1} + q(q^{2} − 2)U_{1} .

(2) If 3 | q + 1, then 𝔽_{q^{2}}/L = {L, tL, t^{2}L}, where t is any non cube element of 𝔽_{q^{2}}^{×}. Let l = |{v ∈ L | 1 − v ∈ L}|, m = |{v ∈ L | t − v ∈ L}|, and n = |{v ∈ L | t − v/t ∈ L}|. Then for x ∈ 𝔽_{q^{2}}^{×}/L,

U_{x}^{2} = \frac{q(q^{2} − 1)}{γ}(Id + T_{1}) + qlU_{x} + qm(U_{tx} + U_{t^{2}x})
U_{x}U_{tx} = qnU_{t^{2}x} + qm(U_{x} + U_{tx}) .

Proof. By Proposition 24, for x ∈ 𝔽_{q^{2}}^{×}, the conjugacy class of u_{x} in N consists of the elements M̂(1, v, v\overline{v}τ + λω), for v ∈ xL and λ ∈ 𝔽_{q}. Since the inverse of u_{x} = M̂(1, x, x\overline{x}τ) is M̂(1, −x, x\overline{x}τ), and since −x ∈ xL as −1 = (−1)^{γ} ∈ L, we have that u_{x}^{−1} is conjugate to u_{x}.

For x, y ∈ 𝔽_{q^{2}}^{×}, the product U_{x}U_{y} is a sum of elements of the form M̂(1, α, β), hence there are integers a, b and c_{x,y}^{z}, for z ∈ 𝔽_{q^{2}}^{×}/L, such that

(##)  U_{x}U_{y} = aId + bT_{1} + ∑_{z∈𝔽_{q^{2}}^{×}/L} c_{x,y}^{z}U_{z} .

Note that for x, y, z ∈ 𝔽_{q^{2}}^{×}, we have

c_{x,y}^{z}|u_{z}^{N}| = m_{u_{x},u_{y}}^{u_{z}} \frac{q(q^{2} − 1)}{γ} = m_{u_{z},u_{x}^{−1}}^{u_{y}} \frac{q(q^{2} − 1)}{γ} = c_{z,x}^{y} \frac{q(q^{2} − 1)}{γ} = c_{z,x}^{y}|u_{z}^{N}| ,

as u_{x}^{−1} is conjugate to u_{x}. So c_{x,y,z} is a symmetric function of x, y, z.

If xL ≠ yL, then no conjugate of u_{x}^{−1} is conjugate to u_{y}, so a = 0. In this case, we also have

(###)  b|t_{1}^{N}| = m_{u_{x},u_{y}}^{t_{1}}(q − 1) = m_{t_{1},u_{x}^{−1}}^{u_{y}}|u_{y}^{N}| ,

and m_{t_{1},u_{x}^{−1}}^{u_{y}} = 0 by Proposition 38. It follows that b = 0 in this case.

If xL = yL, i.e. U_{x} = U_{y}, then clearly a = |u_{x}^{N}| = \frac{q(q^{2} − 1)}{γ}. Moreover Equation (###)

gives b(q − 1) = (q − 1)|u_{y}^{N}|, hence b = \frac{q(q^{2} − 1)}{γ}.

In the case 3 ∤ q + 1, we have γ = 1 and L = 𝔽_{q^{2}}^{×}. Then

U_{1}^{2} = q(q^{2} − 1)(Id + T_{1}) + c_{1,1}^{1}U_{1} .

Taking augmentation gives

(q(q^{2} − 1))^{2} = q(q^{2} − 1)(1 + q − 1) + c_{1,1}^{1}q(q^{2} − 1) ,

hence c_{1,1}^{1} = q(q^{2} − 2), which completes the proof of Assertion (1).

In the case 3 | q + 1, then γ = 3, and L has index 3 in 𝔽_{q^{2}}^{×}, so 𝔽_{q^{2}}^{×}/L = {1, tL, t^{2}L} for any non cube element t of 𝔽_{q^{2}}^{×}.

For x, y, z ∈ 𝔽_{q^{2}}^{×}, the product of the element u′ = M̂(1, v, v\overline{v}τ +λω) in the conjugacy class of u_{x} (where v ∈ xL and λ ∈ 𝔽_{q}) by the element u′′ = M̂(1, r, r\overline{r}τ + μω) in the conjugacy class of u_{y} (where r ∈ yL and μ ∈ 𝔽_{q}) is equal to u_{z} if and only if

v + r = z and r\overline{r}τ + μω − v\overline{r} + v\overline{v}τ + λω = z\overline{z}τ .

The second equation determines μ once v, r and λ are known, and λ can be chosen arbitrarily in 𝔽_{q}, once v and r satisfy v + r = z. Hence in Equation (##), we have

c_{x,y}^{z} = q |{v ∈ xL | z − v ∈ yL}| .

In particular for any x ∈ 𝔽_{q^{2}}^{×}

c_{x,x}^{x} = q |{v ∈ xL | x − v ∈ xL}| = q |{w ∈ L | x − xw ∈ xL}| = l .

Similarly

c_{x,x}^{xt} = q |{v ∈ xL | xt − v ∈ xL}| = q |{w ∈ L | xt − xw ∈ xL}| = m .

Finally

c_{x,xt}^{xt^{2}} = q |{v ∈ xL | xt^{2} − v ∈ xtL}| = q |{w ∈ L | t^{2} − w ∈ tL}| = n .

This completes the proof, since c_{x,y}^{z} is symmetric in x, y, z. □

Remark 40. Applying augmentation to the equations of Proposition 39 gives that n = l+1

and n + 2m = \frac{q^{2} − 1}{3}. So it suffices to know l, and then m and n can be computed.

By definition l = |{v ∈ L | 1 − v ∈ L}|. Since 3 | q + 1 | q^{2} − 1, the field 𝔽_{q^{2}} contains all cubic roots of unity. Now clearly

l = |{(x, y) ∈ 𝔽_{q^{2}}^{×} × 𝔽_{q^{2}}^{×} | x^{3} + y^{3} = 1}|/9 ,

since multiplying x or y by any cubic root of unity doesn’t change x^{3} nor y^{3}. It follows that 9l is almost equal to the number of points of the elliptic curve x^{3} + y^{3} = z^{3} over 𝔽_{q^{2}}: the difference consists of three points (θ, 0, 1) of the projective plane over 𝔽_{q^{2}}, where θ is any cubic root of 1, three points (0, θ, 1), and three points (θ, −1, 0). It follows that 9l = N_{2} − 9, where N_{2} is the number of points over 𝔽_{q^{2}} of the Fermat cubic E with equation x^{3} +y^{3} = z^{3}.

Now this is an elliptic curve, and by [7, (2.6)], the zeta function of E can be defined as

Z_{E}(u) = exp( ∑_{m≥1} N_{m} \frac{u^{m}}{m} ) ,

where N_{m} is the number of points of E over 𝔽_{q^{m}}. By [7, Theorem 2.8], it has the following form

Z_{E}(u) = \frac{1 − au + qu^{2}}{(1 − u)(1 − qu)} ,

where a = 1 + q − N_{1}. Comparing the terms of degree 2 in u in the expansion of those two expressions of Z_{E}(u) as series in u gives N_{2} = N_{1}(2(q + 1) − N_{1}).

Now since 3 | q + 1, it follows that 3 ∤ q − 1, and x ↦ x^{3} is a bijection of 𝔽_{q}. Hence E has as many points over 𝔽_{q} as the projective line with equation x + y = z, that is N_{1} = q + 1. Hence N_{2} = (q + 1)^{2}, which gives the following values for l, n and m:

l = (\frac{q + 1}{3})^{2} − 1, m = \frac{q^{2} − q − 2}{9}, n = (\frac{q + 1}{3})^{2} .

Theorem 41. Let k be a field of characteristic p. Then:

(1) The radical J(ZkN) of the center of the group algebra kN has a k-basis consisting of the elements D_{x}, for x ∈ 𝔽_{q^{2}}^{×}/Γ − {Γ}, T_{x}, for x ∈ Ψ/Γ − {Γ}, T_{1} + Id, and U_{x},

for x ∈ 𝔽_{q^{2}}^{×}/L. In particular, the dimension of J(ZkN) is equal to \frac{q^{2} + q}{γ} + γ − 1.

(2) The square J^{2}(ZkN) of J(ZkN) has a k basis consisting of the elements D_{x} + T_{x}, where x ∈ Ψ/Γ−\{Γ\}. In particular, the dimension of J^{2}(ZkN) is equal to \frac{q + 1}{γ} −1.

(3) The cube J^{3}(ZkN) of J(ZkN) is equal to 0.

Proof. As the group algebra kN is indecomposable when k is a field of characteristic p, the radical J(ZkN) is equal to the kernel of the augmentation ε : ZkN → k. If X is the sum of the elements of a conjugacy class C of N, then ε(X) = |C|, and by Proposition 24, this is a multiple of p, unless C is the class of the identity element of N, or C is the class of t_{1}, and |C| = q−1 in this case. It follows that the elements listed in Assertion (1) generate J(ZkN). Moreover, they are obviously linearly independent, so they form a basis B of J(ZkN).

Now by Proposition 28, for x ∈ Ψ − Γ, we have that D_{x}(Id + T_{1}) = D_{x} + T_{x} in ZkN, so the elements D_{x} + T_{x}, where x ∈ Ψ/Γ − {Γ}, are indeed in J^{2}(ZkN), and they are clearly linearly independent. Moreover, reducing mod p the formulas for products stated in Propositions 26 to 39, one checks easily that any product of two elements of the basis B is equal to a (possibly zero) scalar multiple of an element D_{x} + T_{x}, for some x ∈ Ψ − Γ, and that the product of any three elements of B vanishes. This completes the proof of Theorem 41. □

If k is a field of characteristic p it is not difficult to give the explicit structure of Z(kN) as a quotient of a polynomial ring in several variables.

Proposition 42. Let γ be the greatest common divisor of 3 and q + 1, and let

Γ := {x ∈ 𝔽_{q^{2}} | x^{γ} = 1}, Ψ := {x ∈ 𝔽_{q^{2}} | x^{q+1} = 1}, L := {a^{γ} | a ∈ 𝔽_{q^{2}}^{×}} .

Let 𝔘 := 𝔽_{q^{2}}^{×}/Γ, let 𝔙 := Ψ/Γ and let 𝔚 := 𝔽_{q^{2}}^{×}/L. Let k be a field of characteristic p > 0 and let N be the normaliser of a Sylow p-subgroup of P SU(3, q), where p divides q. Then,

Z(kN) ≃ k[T, X_{n}, Y_{m} | n ∈ 𝔚, m ∈ 𝔘]/I

where I is the ideal generated by

T^{2}, T X_{n_{1}}, T Y_{m_{1}}, X_{n_{1}}X_{n_{2}}, X_{n_{1}}Y_{m_{1}}, Y_{m_{1}}Y_{m_{2}},

X_{n_{1}}Y_{m_{2}} + \frac{1}{γ} X_{n_{1}}T, Y_{m_{2}}Y_{m_{3}} − (1 − δ_{m_{2},m_{3}^{−1}})X_{m_{2}m_{3}}T

where

n_{1}, n_{2} ∈ 𝔚, m_{1} ∈ 𝔘 − 𝔙, m_{2}, m_{3} ∈ 𝔙

and δ_{a,b} is the Kronecker symbol.

Proof. We have a basis of Z(kN) given in Theorem 41 by the elements D_{x}, for xΓ ∈ 𝔽_{q^{2}}^{×}/Γ, T_{x}, for xΓ ∈ Ψ/Γ − {Γ}, T_{1} + Id, and U_{x}, for xL ∈ 𝔽_{q^{2}}^{×}/L. Observe that D_{1} = 1. Moreover, by Proposition 28 we do not need to include T_{x} as variable of the polynomial ring. This element is already the product of T_{1} and U_{x}.

We obtain the following multiplication table.

T_{1} + id  U_{y}  D_{y} (y ∉ Ψ)  D_{y} (y ∈ Ψ − Γ)
T_{1} + id  0  0  0  T_{y} + D_{y}
Prop. 33  Prop. 38  Prop. 26(2)  Props. 28
U_{x}  0  0  0  −\frac{1}{γ} (D_{x} + T_{x})
Prop. 38  Prop. 39  Prop. 26(3)  Prop. 29
D_{x} (x ∉ Ψ)  0  0  0  0
Prop. 26(2)  Prop. 26(3)  Props. 27,26(1),30  Prop. 26(1)
D_{x} (x ∈ Ψ − Γ) T_{x} + D_{x} −\frac{1}{γ} (D_{x} + T_{x})  0  (1 − δ_{xyΓ,Γ})(T_{xy} + D_{xy})
Prop. 28  Prop. 29  Prop. 26(1)  Props. 31,30

Now, mapping T to T_{1} + id, X_{n} to U_{n} and Y_{m} to D_{m} gives an algebra homomorphism of the corresponding polynomial ring with kernel precisely the ideal I. □

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[8] Yuming Liu, Summands of stable equivalences of Morita type. Communications of Algebra 36 (2008), no. 10, 3778-3782.

[9] Yuming Liu, Guodong Zhou and Alexander Zimmermann, Higman ideal, stable Hochschild homology and Auslander-Reiten conjecture, Mathematische Zeitschrift 270 (2012) 759-781.

[10] Yuming Liu, Guodong Zhou and Alexander Zimmermann, Two questions on stable equivalences of Morita type, preprint 2014.

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[14] Alexander Zimmermann, Representation Theory; A homological algebra point of view, Springer Verlag London 2014.

S.B.: Université de Picardie, LAMFA (UMR 7352 du CNRS), 33 rue St Leu, F-80039 Amiens Cedex 1, France

   E-mail address: serge.bouc@u-picardie.fr

A.Z.: Université de Picardie, Département de Mathématiques et LAMFA (UMR 7352 du CNRS), 33 rue St Leu, F-80039 Amiens Cedex 1, France

   E-mail address: alexander.zimmermann@u-picardie.fr