Table Of ContentON A QUESTION OF RICKARD ON TENSOR PRODUCT OF STABLY
EQUIVALENT ALGEBRAS
SERGE BOUCAND ALEXANDERZIMMERMANN
Abstract. Let Fp be the algebraic closure of the prime field of characteristic p. After
observing that the principal block B of FpPSU(3,pr) is stably equivalent of Morita type
5 to its Brauer correspondent b, we show however that the centre of B is not isomorphic as
1 an algebra to the centre of b in the cases pr ∈ {3,4,5,8}. As a consequence, the algebra
20 B⊗FpFp[X]/Xp isnot stablyequivalentofMorita typetob⊗FpFp[X]/Xp inthesecases.
This yields a negativeanswer to a question of Rickard.
b
e
F
5 Introduction
] LetK beafield,andletA,B,C andD befinitedimensionalK-algebras. Rickardshowed
R
in [12] that if A and B are derived equivalent, and if C and D are derived equivalent, then
G
also A⊗ C and B ⊗ D are derived equivalent. Rickard asks in [13, Question 3.8] if this
K K
h. still holds when replacing derived equivalence by stable equivalence of Morita type. It is
t clear that we have to suppose that all algebras involved have no semisimple direct factor. A
a
m 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
3
between selfinjective algebras A and B induces a stable equivalence of Morita type between
v
1 A and B. If A and B are not selfinjective, then this implication is not valid. Hence, the
6 natural playground for Rickard’s question are selfinjective algebras.
4
The purpose of this paper is to give a counterexample to Rickard’s question. For an
1
0 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[X]/Xp and B⊗ K[X]/Xp
1 K K
0 are not stably equivalent of Morita type.
5 Note that this answers the general case. Indeed, if A⊗ C is stably equivalent of Morita
K
1
type to B ⊗ C and B ⊗ C is stably equivalent of Morita type to B ⊗ D then A⊗ C
: K K K K
v is stably equivalent of Morita type to B ⊗ D. Hence, we may suppose C = D.
K
i
X In recent years many attempts were proposed to lift a stable equivalence of Morita type
r between selfinjective algebras to a derived equivalence. It is known that this is not possible
a
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]forexample. Inthispaperwegiveanewincidenceofthisfact. Moreover, weprovidetwo
symmetric algebras, which are stably equivalent of Morita type, and have non isomorphic
centres.
Ourexampleistheprincipalp-blockofthegroupPSU(3,pr)anditsBrauercorrespondent
for pr ∈ {3,4,5,8}.
We recall in the first section some basic facts and results which we need for our construc-
tion. 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
Date: February 6, 2015.
2010 Mathematics Subject Classification. Primary 16E35; 18E30; 20C05.
BothauthorsweresupportedbyagrantSTICAsie’Escap’fromtheMinist`eredesAffairesE´trang`eresde
la France.
1
2 SERGEBOUCANDALEXANDERZIMMERMANN
the centre of KN (S) for G= PSU(3,pr) and S one of its Sylow p-subgroupsfor all primes
G
p and integers r.
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, wethankYumingLiuforsuggestingaquestiontousleadingtothepresentpaper,
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⊗ Bop-
K
module M and a B ⊗ Aop-module N such that
K
• M is projective as A-module, and as Bop-module
• N is projective as Aop-module and as B-module
• there is a projective A ⊗ Aop-module P and a projective B ⊗ Bop-module Q
K K
such that M ⊗ N ≃ B ⊕ Q as B ⊗ Bop-modules and N ⊗ M ≃ A ⊕ P as
B K A
A⊗ Aop-modules.
K
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.
Brou´e defined Zst(A) := End (A) and
A⊗KAop
Zpr(A) := ker(EndA⊗KAop(A) → EndA⊗KAop(A))
where we denote by End the endomorphisms taken in the stable module category.
Thecentreofanalgebraisaninvariantofaderivedequivalence, aswasshownbyRickard.
Thestable centreZst(A) isan importantinvariant understable equivalences of Morita type,
as was shown by Brou´e.
Proposition 2. (Brou´e [2]; see also [14, Chapter 5]) If A and B are stably equivalent of
Morita type, then Zst(A) ≃ Zst(B) as algebras.
Now, Liu, Zhou and the second author give a criterion to determine the dimension of
Zst(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 (Zpr(A)) =
K
rank (C ) where C is the Cartan matrix of A and where rank (C ) denotes its rank as
p A A p A
matrix over K.
Moreover, we recall a conjecture of Auslander-Reiten. In [1] Auslander and Reiten con-
jecture 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 non-
projective simple A-modules is equal to the number of non-projective simple B-modules if
and only if dim (HH (A)) = dim (HH (B)), where HH denotes the degree 0 Hochschild
K 0 K 0 0
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.
TENSOR PRODUCT OF STABLY EQUIVALENT ALGEBRAS 3
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 includethe shortproof. 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⊗ B)= J(A)⊗ B +A⊗ J(B).
K K K
Proof. It is clear that J(A)⊗ B+A⊗ J(B) is a nilpotent ideal of A⊗ B, and therefore
K K K
we get
J(A)⊗ B+A⊗ J(B) ⊆ J(A⊗ B).
K K K
Now, (A⊗ B)/(J(A)⊗ B+A⊗ J(B)) = A/J(A)⊗ B/J(B) and both K-algebras
K K K K
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)⊗ B/J(B) is semisimple. Therefore
K
J(A)⊗ B+A⊗ J(B) ⊇ J(A⊗ B).
K K K
This shows the statement. (cid:3)
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 (U) be the field of rational fractions
p
over the finite field F . Let A= K[X]/(Xp−U). Then A is a purely inseparable extension
p
of K, of dimension p. In particular it is a reduced (commutative) algebra, i.e. J(A) = 0.
But A⊗ A∼= K[X,Y]/(Xp−U,Yp−U) contains the non zero element X −Y, such that
K
(X −Y)p = U −U = 0. Hence J(A⊗ A) 6= 0.
K
Lemma 7. Let K be an algebraically closed field, let Λ and ∆ be finite dimensional K-
algebras, and suppose that ∆ is local. Then the projective indecomposable Λ⊗ ∆-modules
K
are precisely the modules P ⊗ ∆ for projective indecomposable Λ-modules P, and if C is
K Λ
the Cartan matrix of Λ, then the Cartan matrix of Λ⊗∆ is C = dim (∆)·C .
Λ⊗K∆ K Λ
Proof. Let P and Q be a indecomposable projective Λ-modules. Then P ⊗ ∆ is a pro-
K
jective indecomposable Λ⊗ ∆-module. Indeed, End (P ⊗ ∆) ≃ End (P)⊗ ∆op.
K Λ⊗K∆ K Λ K
Moreover, since Γ := End (P)op and ∆ are local K-algebras their radical quotient are
Λ
finite-dimensionalskew-fields,andthereforeΓ/J(Γ) ≃ K ≃ ∆/J(∆)sinceK isalgebraically
closed. Moreover, by Lemma 5 we get J(Γ⊗ ∆) = J(Γ)⊗∆+Γ⊗ J(∆). On the other
K K
hand,
(Γ⊗ ∆)/(J(Γ)⊗ ∆+Γ⊗ J(∆)) = K ⊗ K = K
K K K K
and hence we get Γ⊗ ∆ is local, and therefore P ⊗ ∆ is indecomposable. Now,
K K
Hom (P ⊗ ∆,Q⊗ ∆)= Hom (P,Q)⊗ ∆op.
Λ⊗K∆ K K Λ K
Taking K-dimensions proves the lemma. (cid:3)
Remark 8. As a special case of Lemma 7 we get C = p·C for algebraically
A⊗KK[X]/Xp A
closedfieldsK ofcharacteristic p. HencewegetbyTheorem3thatZpr(A⊗ K[X]/Xp)= 0
K
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 Jn+1(A) = 0 6= Jn(A) and Jm+1(B)= 0 6= Jm(B),
then
Jn+m+1(A⊗ B)= 0 6= Jn+m(A⊗ B) = Jn(A)⊗ Jm(B).
K K K
Proof. By Lemma 5, we have J(A⊗ B)= J(A)⊗ B+A⊗ J(B). Therefore
K K K
n+m+1
Jn+m+1(A⊗ B) = Jk(A)⊗ Jn+m+1−k(B) = 0 .
K K
Xk=0
4 SERGEBOUCANDALEXANDERZIMMERMANN
Similarly
n+m
Jn+m(A⊗ B)= Jk(A)⊗ Jn+m−k(B) = Jn(A)⊗ Jm(B)6= 0 ,
K K K
Xk=0
which completes the proof. (cid:3)
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[X]/Xp. It is clear that A⊗ K[X]/Xp ∼= A[X]/Xp.
K K
Now, let a := a +a X +...a Xp−1 ∈ A[X]. Then for b := b ∈ A·1 we get
0 1 p−1 0
ab−ba= (a b−ba )+···+(a b−ba )Xp−1
0 0 p−1 p−1
and so a ∈ Z(A) implies that a commutes with any b ∈ A, and hence a ,...,a are all in
0 p−1
Z(A). Conversely, it is clear that Z(A)[X]/Xp ⊆ Z(A[X]/Xp) since aXn commutes with
all elements of A[X]/Xp 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
Jn(Z(A)) 6= 0 = Jn+1(Z(A)), then
0 6= Jn+p−1 Z(A⊗ K[X]/Xp) = Jn Z(A) ⊗ Xp−1K[X]/Xp
K K
(cid:0) (cid:1) (cid:0) (cid:1)
and
Jn+p Z(A⊗ K[X]/Xp) = 0.
K
Proof. This is an immediate conseq(cid:0)uence of Lemma 9. (cid:1) (cid:3)
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 Jn(Z(A)) 6= 0 =
Jn+1(Z(A)) and Jm(Z(B)) 6= 0 = Jm+1(Z(B)). If dim (Jn(Z(A))) 6= dim (Jm(Z(B)),
K K
then A⊗ K[X]/Xp and B ⊗ K[X]/Xp are not stably equivalent of Morita type.
K K
Proof. If n 6= m, then Z(A⊗ K[X]/Xp) 6≃ Z(B ⊗ K[X]/Xp) by Lemma 11 since the
K K
Loewy lengths of the centres are different. If n = m, then Lemma 11 shows that the centres
of A⊗ K[X]/Xp and of B ⊗ K[X]/Xp are not isomorphic since the dimension of the
K K
lowest Loewy layers of the centres are not of the same dimension. Remark 8 shows that
Z(A⊗ K[X]/Xp) = Zst(A⊗ K[X]/Xp) and Z(B⊗ K[X]/Xp)= Zst(B⊗ K[X]/Xp).
K K K K
Since the stable centre is invariant under stable equivalence of Morita type, we get the
statement. (cid:3)
Remark 13. For a field K and a K-algebra A let n bethe number of isomorphism classes
A
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 = n . [9,
A B
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 = n
A B
is equivalent to dim (HH (A)) = dim (HH (B)). If A is symmetric, then there is a
K 0 K 0
vector space isomorphism HH (A) ≃ HH0(A) = Z(A), we see that the Auslander-Reiten
0
conjecture implies that dim Z(A) = dim Z(B) . More precisely by [9, Corollary 1.2],
K K
for two indecomposable symm(cid:0)etric (cid:1)algebras A(cid:0) and B(cid:1) over an algebraically closed field K
we have n = n ⇔ dim Zpr(A) = dim Zpr(B) , where by definition Zst(A) =
A B K K
Z(A)/Zpr(A). The link to ou(cid:0)r proof(cid:1)is now give(cid:0)n by th(cid:1)e fact that for every algebra the
Higman ideal H(A) of A equal Zpr(A), and for symmetric algebras A over an algebraically
closed field K we have dim (H(A)) equals the p-rank of C .
K A
TENSOR PRODUCT OF STABLY EQUIVALENT ALGEBRAS 5
2. The Example
Let F be the algebraic closure of the prime field F of characteristic p. Let q = pn for
p p
some integer n.
We recall some results on the geometry of PSU(n,q) (cf e.g. [5, II Satz 10.12, page 242]).
The group G := PSU(3,q) acts doubly transitively on the unitary quadric Q of cardinal
q3+1. Note that we use the GAP notation, not the notation used in [5, II Satz 10.12, page
242], namely, PSU(3,q) is defined over a field with q2 elements, and is a natural quotient
of a subgroup of SL (q2) (and not of SL (q) !). The stabiliser of a point X of Q is the
2 2
normaliser in G of a Sylow p-subgroup P of G. Therefore two different conjugate Sylow
p-subgroups P and gP of G fix two different points X and gX of Q. Hence gP ∩P = 1 if
g 6∈ N (P), or in other words, G has a trivial intersection Sylow p-subgroupstructure. This
G
implies that Green correspondence gives a stable equivalence of Morita type between the
principal block B of F G andits Brauer correspondentb (cf e.g. [14, Chapter 2, Proposition
p
2.1.23 and Proposition 2.4.3]).
TheGAP [4] programin Section 3computes theLoewy series of the ringZ(F PSU(3,4))
2
and of Z(F N (S)) for some Sylow 2-subgroup of PSU(3,4). Observe moreover that
2 PSU(3,4)
F PSU(3,4) has two blocks, the principal one and another block of defect 0 (corresponding
2
to the Steinberg character). Moreover, the dimensions of the Loewy series obtained over F
2
also hold by extending the scalars to F , using Lemma 5.
2
We obtain that
dim (Z(B))= 21 = dim (Z(b))
F2 F2
dim (J(Z(B))) = 20 = dim (J(Z(b)))
F2 F2
dim (J2(Z(B))) = 56= 4 = dim (J2(Z(b)))
F2 F2
dim (J3(Z(B))) = 0 = dim (J3(Z(b))).
F2 F2
Similarly we get for the centre of the principal block B of PSU(3,8) and the centre of its
Brauer correspondent b
dim (Z(B))= 27 = dim (Z(b))
F2 F2
dim (J(Z(B))) = 26 = dim (J(Z(b)))
F2 F2
dim (J2(Z(B))) = 36= 2 = dim (J2(Z(b)))
F2 F2
dim (J3(Z(B))) = 0 = dim (J3(Z(b))).
F2 F2
An immediate variant of the program shows that this is a quite general phenomenon in
oddcharacteristic. ThegroupPSU(3,3) givesanexampleincharacteristic 3since,denoting
by B the principal block of F PSU(3,3) and by b its Brauer correspondent,
3
dim (Z(B))= 13 = dim (Z(b))
F3 F3
dim (J(Z(B))) = 12 = dim (J(Z(b)))
F3 F3
dim (J2(Z(B))) = 46= 3 = dim (J2(Z(b)))
F3 F3
dim (J3(Z(B))) = 0 = dim (J3(Z(b))).
F3 F3
Thegroup PSU(3,5) gives an example in characteristic 5 since, denotingby B the principal
block of F PSU(3,5) and by b its Brauer correspondent,
5
dim (Z(B))= 13 = dim (Z(b))
F5 F5
dim (J(Z(B))) = 12 = dim (J(Z(b)))
F5 F5
dim (J2(Z(B))) = 26= 1 = dim (J2(Z(b)))
F5 F5
dim (J3(Z(B))) = 0 = dim (J3(Z(b))).
F5 F5
6 SERGEBOUCANDALEXANDERZIMMERMANN
Theorem 14. Let K be the algebraic closure of F and let B be the principal block of
p
PSU(3,pr). Let b be the Brauer correspondent of B in the group ring of the normaliser of
a 2-Sylow subgroup of PSU(3,pr). Then B and b are stably equivalent of Morita type. If
moreover pr ∈ {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[X]/Xp is not stably equivalent of Morita type
K
to b⊗ K[X]/Xp.
K
Proof. As seen at the beginningof 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[X]/Xp is not stably equivalent of Morita type to
K
b⊗ K[X]/Xp. (cid:3)
K
Remark 15. Theaboveexamples suggestthatingeneral, withthenotation of Theorem14,
the dimension of J2 Z(B) could always be equal to 1+dim J2(Z(b)). By Theorem 41,
K
this is equal to pr +(cid:0)1, wh(cid:1)ere γ is the greatest common divisor of pr +1 and 3.
γ
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;
TENSOR PRODUCT OF STABLY EQUIVALENT ALGEBRAS 7
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 PSU(3,pr)
Recall that we denote by S a Sylow p-subgroup of the projective special unitary group
G = PSU(3,q) over the field with q2 elements, where q = pr, 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 Fq2, 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
0 0 1
0 1 0 .
1 0 0
Notation 17. For x ∈ Fq2, we set x = xq. Then the map x 7→ x is the automorphism of
order 2 of the extension Fq2/Fq. We also set
Ψ = {x ∈ F× | xx = 1}
q2
Let ω be a non zero element of Fq2 such that ω +ω = 0, and τ be an element of Fq2 such
that τ +τ = −1.
8 SERGEBOUCANDALEXANDERZIMMERMANN
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
a b c
M(a,b,c) = 0 a/a −b/a ,
0 0 1/a
where (a,b,c) belongs to the set
Q= {(a,b,c) ∈ F×q2 ×(Fq2)2 |bb+ac+ca = 0} .
Lemma 18. For (a,b,c) ∈ Q, let Mˆ(a,b,c) denote the image of M(a,b,c) in N. Then if
(a′,b′,c′) ∈ Q, we have that Mˆ(a,b,c) = Mˆ(a′,b′,c′) if and only if there is λ ∈ Fq2 with
λq−2 = 1 and (a′,b′,c′)= λ·(a,b,c).
Proof. Mˆ(a,b,c) = Mˆ(a′,b′,c′) if and only if there exists a scalar λ ∈Fq2 such that
a′ b′ c′ a b c
0 a′/a′ −b′/a′ = λ 0 a/a −b/a .
0 0 1/a′ 0 0 1/a
Equivalently (a′,b′,c′) = λ(a,b,c) and λ/λ = λ, i.e. λq−2 = 1. (cid:3)
For two non zero integers s,t denote by (s,t) their greatest common divisor. Observe
that (q−2,q2−1) = (3,q+1), to motivate the following:
Notation 19. We set γ = (3,q+1), and put
Γ = {λ ∈ Fq2 | λγ =1} ≤ Ψ as well as L = {aγ | a ∈ F×q2} .
With this notation, the group N has order q3(q2 − 1)/γ. It is equal to the semidirect
1 b c
productof the group S, consisting of the elements Mˆ(1,b,c) = 0 1 −b , whereb and
0 0 1
c are elements of Fq2 such that bb+c+c = 0, by the cyclic group C of order (q2 −1)/γ
a 0 0
consisting of the elements Mˆ(a,0,0) = 0 a/a 0 , for a ∈ F×/Γ.
q2
0 0 1/a
Lemma 20. (1) Let (a,b,c) and (x,y,z) be elements of Q. Then
bx by c
M(a,b,c)M(x,y,z) = M ax,ay+ ,az− + .
x x x
(cid:16) (cid:17)
1 b
(2) Let (a,b,c) ∈ Q. Then M(a,b,c)−1 = M ,− ,c¯ .
a a
(cid:16) (cid:17)
(3) Let (a,b,c) and (x,y,z) be elements of Q. Then
ab x a2
M(a,b,c)M(x,y,z)M(a,b,c)−1 = M x, −x + y,t ,
(cid:18) a x a (cid:19)
(cid:16) (cid:17)
ac aby bbx
where t = acx+ +ayb− + +aaz.
x x x
Proof. All the assertions follow from straightforward computations. (cid:3)
Proposition 21. (1) Let (x,y,z) and (x′,y′,z′) be elements of Q. If Mˆ(x,y,z) and
Mˆ(x′,y′,z′) are conjugate in N, then x−1x′ ∈ Γ.
(2) The elements Mˆ(x,0,0), for x ∈ Fq2/Γ, lie in distinct conjugacy classes of N.
(3) Let (x,y,z) ∈ Q. Then if x ∈/ Γ, the element Mˆ(x,y,z) of N is conjugate to an
element of the form Mˆ(x,0,xuω), for some u ∈F .
q
TENSOR PRODUCT OF STABLY EQUIVALENT ALGEBRAS 9
(4) Let x ∈ F× and u∈ F . Then if xx 6= 1, the element Mˆ(x,0,xuω) of N is conjugate
q2 q
to Mˆ(x,0,0). If xx = 1, and if u 6= 0, then the element Mˆ(x,0,xuω) is conjugate
to Mˆ(x,0,xω), and not conjugate to Mˆ(x,0,0).
(5) If (1,y,z) ∈ Q, then either y 6= 0 and there exists u∈ F such that z = yy(τ +uω),
q
or y = 0 and there exists u ∈ F such that z = uω. Moreover, if (1,y′,z′) ∈ Q and
q
if Mˆ(1,y′,z′) and Mˆ(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 Mˆ(1,y′,z′)
and Mˆ(1,y,z) are conjugate in N if and only if y′(q2−1)/γ = y(q2−1)/γ in F×, i.e. if
q2
y′/y ∈ L. In particular M(1,y,z) is conjugate to M(1,y,yyτ).
Proof. Assertion (1) follows from Assertion (3) of Lemma 20: if
ab x a2
Mˆ(x′,y′,z′) = Mˆ x, −x + y,t ,
(cid:18) a x a (cid:19)
(cid:16) (cid:17)
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
x y
6= x, and we can set a = 1, b = − , and c = bbτ. Then (a,b,c) ∈ Q and
x
x
−x
x
M(a,b,c)M(x,y,z)M(a,b,c)−1 is of the form M(x,0,t), for some t ∈ Fq2. In particular
(x,0,t) ∈ Q, hence xt+tx = 0. In other words t = vx with v +v = 0. Then v = uω and
u = u, that is u∈ F .
q
For Assertion (4), we have to decidewhentwo elements of theform n= Mˆ(x,0,xuω) and
n′ = Mˆ(x′,0,x′u′ω) are conjugate in N, where x,x′ ∈/ Γ, and u,u′ ∈ F . By Assertion (1),
q
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 ab x −x = 0, hence b = 0. Now xu′ω is equal to
a x
(cid:16) (cid:17)
the element t of Lemma 20, in the case y = b = 0 and z = xuω, that is
ac
xu′ω = acx+ +aaxuω .
x
Moreover ac+ca = 0, since (a,0,c) ∈ Q. So there exists v ∈ F such that c = avω. This
q
gives
aaxvω
xu′ω = −aaxvω+ +aaxuω ,
x
or equivalently
1
u′ = aa u−v(1− ) .
xx
(cid:16) (cid:17)
Thus n and n′ are conjugate in N if and only if there exist a ∈ F× and v ∈ F such that
q2 q
u′ = aa u−v(1− 1 ) . If xx 6= 1, then we can take a = 1 and v = u−u′, so n and n′ are
xx 1
(cid:16) (cid:17) 1−
xx
conjugate. And if xx = 1, then n and n′ are conjugate if and only if there exists a ∈ F×
q2
such that u′ = aau, or equivalently, if there exists λ ∈ F× such that u′ = λu. So either
q
u = u′ = 0, or u and u′ are both non zero. This completes the proof of Assertion (4).
z
For Assertion (5), assumethat(1,y,z) ∈ Q. Thenyy+z+z = 0. Ify 6= 0, setv = −τ.
yy
Then v+v = 0, so there exists u∈ F such that v = uω, thus u= yy(τ+uω). And if y = 0,
q
then z+z = 0, so z = uω for some u∈ F .
q
10 SERGEBOUCANDALEXANDERZIMMERMANN
Now by Assertion (3) of Lemma 20, for (1,y,z) and (1,y′,z′) in Q, the elements n =
Mˆ(1,y,z) and n′ = Mˆ(1,y′,z′) are conjugate in N if and only if there exists (a,b,c) ∈ Q
such that
a2
y′ = y and z′ = ac+ac+ayb−aby+bb+aaz ,
a
that is
a2
y′ = y and z′ = ayb−aby+aaz ,
a
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× such that y′ = a2y = a2−qy. It follows that y′/y belongs to the subgroup
q2 a
×
of F consisting of (q − 2)-th powers, i.e. the subgroup of γ-th powers, i.e. the unique
q2
subgroup of order (q2 −1)/γ of F×. Equivalently (y′/y)(q2−1)/γ = 1. Conversely, suppose
q2
that there exists a ∈ F× such that y′ = a2y. There are elements u and u′ of F such that
q2 a q
z = yy(τ +uω) and z′ = y′y′(τ +u′ω). If we can find b and c such that (a,b,c) ∈ Q and
z′ = aby−aby+aaz, then n and n′ are conjugate in N. This can also be written
aayy(τ +u′ω)= aby−aby+aayy(τ +uω) ,
or equivalently
1 b b
(∗) − = u′−u .
ω ay ay
(cid:16) (cid:17)
1 b b
Now the map b 7→ ω ay − ay is a non zero Fq-linear map from Fq2 to Fq. Hence it is
(cid:16) (cid:17)
bb
surjective, and there exists b ∈ Fq2 such that (*) holds. Now we set c = 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. (cid:3)
Corollary 22. The set
E = Mˆ(x,0,0) |x ∈ F×/Γ Mˆ(x,0,xω) | x ∈ Ψ/Γ Mˆ(1,y,yyτ) |y ∈ F×/L
q2 q2
(cid:8) (cid:9)G(cid:8) (cid:9)G(cid:8) (cid:9)
q2+q
is a set of representatives of conjugacy classes of N. In particular, there are +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
q2−1 q+1 q2+q
|E| = + +γ = +γ .
γ γ γ
(cid:3)
Notation 23.
• For x ∈ F×, we set d = Mˆ(x,0,0) and D = d+ ∈ ZkN.
q2 x x x
• For x ∈ Ψ, we set t = Mˆ(x,0,xω) and T = t+.
x x x
• For y ∈ F×, we set u = Mˆ(1,y,yyτ) and U = u+.
q2 y y y
Proposition 24.
(1) For x ∈ F× −Ψ,
q2
dNx = Mˆ(x,y,z) |y,z ∈ Fq2, yy+xz+zx = 0 .
In particular |dN|(cid:8)= q3. (cid:9)
x
