Full text of DOI:10.1007/PL00005881

manuscripta math. 106, 1 – 12 (2001)
© Springer-Verlag 2001
Grégory Berhuy · Christoph Frings
On the second trace form of central simple algebras
in characteristic two
Received: 14 September 2000 / Revised version: 18 January 2001
Abstract. Let K be a field and let A be a central simple algebra over K. Let T and T
A
2,A
be respectively the trace form and the second trace form of A. If K has characteristic not
two, it is shown in [U] that T2 does not give much more information than the usual trace
,A
form. If K has characteristic two, the quadratic form T has rank zero. In this article, we
A
show that the second trace form of a central simple algebra A of even degree over a field of
characteristic two is non-degenerate and we compute its classical invariants.
Introduction
Let K be a field and let A be a central simple algebra over K. The quadratic forms
2
TA : x ∈ A
ꢀ
→ TrdA(x ) and T2,A : x ∈ A
ꢀ
→ SrdA(x) are called respectively
the trace form and the second trace form of A. If K has characteristic not two,
the trace form has been studied by many authors (see [B1,B2,L,LM,Se,Ti] for
example). In particular, its classical invariants are well-known (see [L,LM,Se] and
[Ti]). The second trace form has also been studied in [U] when K has characteristic
not two, but it is shown that this form does not give much more information than
the usual trace form. When K has characteristic two, the trace form has rank zero.
In this article, we show that the second trace form of a central simple algebra A of
even degree over a field of characteristic two is non-degenerate and we compute
its classical invariants. In the first part, we compute the second trace form of a split
algebra. In the second one, we consider the case of cyclic algebras. Finally, we
compute the Arf invariant and the Clifford invariant of the second trace form in the
general case. The reader will also find in Appendix the proof of an unpublished
result of Saltman which is the main ingredient of our work.
Preliminaries
Let A be a central simple algebra over a field K of arbitrary characteristic. If
a ∈ A, the reduced characteristic polynomial of a, denoted by PrdA(a), is defined
as follows: let L be a splitting field of A and ϕ : A⊗L → Mn(L) a K-isomorphism.
Then PrdA(a) := det(XIn − ϕ(a ⊗ 1)) is an element of K[X] and is independent
G. Berhuy, C. Frings: UMR 6623 du CNRS, Laboratoire de Mathématiques, Bureau 401B,
1
6 route de Gray, 25030 Besançon Cedex, France. e-mail: berhuy@math.univ-fcomte.fr;
frings@math.univ-fcomte.fr

2
G. Berhuy, C. Frings
of the choice of L and ϕ (cf. [Sc, Chapter 8, §5] for example). Write PrdA(a) =
n
n−1
n−2
X − s1X
+ s2X
+ . . . , then TrdA(a) := s1 and SrdA(a) := s2 are called
respectively the reduced trace and the reduced second trace of a. If x1, · · · , xn are
the roots of PrdA(a) in an algebraic closure, we have TrdA(a) = x1 + · · · + xn and
ꢀ
SrdA(a) =
xixj . This easily implies the following equality for the bilinear form
i<j
b2,A associated to the second trace form when the ground field has characteristic
two:
b2,A(x, y) := SrdA(x + y) + SrdA(x) + SrdA(y) = TrdA(x)TrdA(y) + TrdA(xy)
Finally, if A has degree n over K and if L is a maximal commutative subfield
of A of degree n (if there is any), then it is well-known that A can be endowed
with a structure of right L-vector space and that the map A ⊗ L → EndL(A),
a ⊗λ ꢀ→ (z ꢀ→ azλ) is an isomorphism. In particular, PrdA(a) is the characteristic
polynomial of the left multiplication by a in the right L-vector space A.
2
Assume that char K = 2. We denote by ℘(K) the set {x + x, x ∈ K}. If
∗
α ∈ K and β ∈ K, we denote by (α, β] the class of the corresponding quaternion
algebra in the Brauer group. This algebra has a K-basis 1, e, f, ef satisfying the
2
2
relations e = α, f + f = β and ef + f e = e. Moreover, the map (α, β) ∈
∗
∗2
K /K × K/℘(K)
we denote by Pa,b the quadratic form (x, y) ∈ K
of the Clifford algebra of Pa,b in Br(K) is denoted by ((a, b)). It is easy to see that
ꢀ→ (α, β] ∈ Br(K) is well defined and bilinear. If a, b ∈ K,
2 2 2
ꢀ
→ ax + xy + by . The class
(
(a, b)) = 0 if a = 0 and ((a, b)) = (a, ab] if a ꢁ= 0. A non-degenerate quadratic
form over K has even rank and is isomorphic to an orthogonal sum of some Pa,b.
If q ꢂ Pa ,b ⊥ · · · ⊥ Pa ,b , then the Arf invariant of q is the element of K/℘(K)
1
1
r
r
defined by Arf(q) := a1b1 + · · · + arbr. We also define the Clifford invariant of
q, denoted by c(q), to be the class of the Clifford algebra of q in the Brauer group.
It is easy to see that
c(q) = ((a1, b1)) + · · · + ((ar, br)) ∈ Br(K)
if q ꢂ Pa ,b ⊥ · · · ⊥ Pa ,b . If L/K is any field extension, ResL/K denotes the
1
1
r
r
homomorphism [A] ∈ Br(K) ꢀ→ [A ⊗ L] ∈ Br(L). Then c(qL) = ResL/K(c(q)).
1
. Motivations
Let K be a field of any characteristic.There are two interesting K-algebra structures,
namely étale algebras and central simple algebras. In order to classify these algebras
up to isomorphism, we need invariants. Since it is quite simple to deal with quadratic
forms, one often searches for GrW-invariants.
Definition 1. Let K be a field. A GrW-invariant of étale algebras of rank n over
K (resp. of central simple algebras of degree n over K) is a map E
ꢀ→ qE (resp.
A
ꢀ→ qA), which sends every étale F-algebra of rank n (resp. every central simple
F-algebra of degree n) to an element of GrW(F) (the Grothendieck–Witt group of
F) for every field extension F/K, and which commutes with scalar extensions.

Second trace form of central simple algebras in characteristic two
3
For example, E
invariants as soon as these forms are non-degenerate.
ꢀ
→ TE, E
ꢀ
→ T2,E, A
ꢀ
→ TA and A
ꢀ
→ T2,A induce GrW-
If char K
ꢁ= 2, the trace form invariants have been studied extensively, and the
second trace form of central simple algebras has been studied by T. Unger in [U].
The second trace form of étale algebras has not been studied in characteristic not
two, but it is easy to show as in [U] that
T2,E ꢂ T2,F ⇐⇒ TE ꢂ TF
Moreover, we have the following result, proved by J.-P. Serre (unpublished):
Theorem 1 (Serre, unpublished). Let K be a field of characteristic not two and
let E
ꢀ
→ qE be a GrW-invariant of étale algebras of rank n over K. Then
m
ꢁ
i
qE ꢂ
λi# TE
i=0
ꢂ ꢃ
n
n
where m =
denotes the integral part of 2 and λi is an element of GrW(K).
2
In other words, the trace form is essentially the only GrW-invariant of étale
algebras in characteristic not two.
Ifchar K = 2, TE andTA haverankzero. Inthecaseofétalealgebras, thesecond
trace form is non-degenerate if and only if n is even (see [BM, Proposition 2.1 (ii)]).
Then Bergé and Martinet proved the following result (see [BM, Theorem 5.1]):
Theorem 2 (Bergé-Martinet, [BM]). Let K be a field of characteristic 2, and let
E be an étale algebra of rank 2m over K. Then
T2,E ꢂ P1,Arf(E/K) ⊥ (m − 1) × P0,0
where Arf(E/K) is the Arf invariant of the second trace form of E/K.
For two reasons, this theorem says that the second trace form is a very poor
substitute for the trace form in characteristic two. The first reason is that this result
implies in particular that c(T2,E) = 0 for any étale algebra of even rank. This
is no longer true if char K
ꢁ= 2. For example, an easy computation shows that
c(T2,E) is the class of the quaternion algebra (a, b) if E is the biquadratic extension
√
√
k( a, b).
The second, obvious, reason is that the second trace form is uniquely determined
by itsArf-invariant (up to isomorphism), contrary to what happens in characteristic
= 2.
ꢁ
Remark 1. In [Be] Berlekamp defined an invariant for finite separable extensions
of fields of characteristic 2 – similar to the usual discriminant in characteristic
= 2
in the following way: let L/K be a separable field extension of finite degree n and
ꢁ
–
η0 a primitive element of L over K (such that L = K(η0)). Consider the conjugates
ηi, 0 ≤ i ≤ n − 1 of η0 and define the Berlekamp-invariant of L/K (or additive
+
discriminant as we will call it according to [BM]) d
by
L/K
ꢁ (
ηi + ηj )
ηiηj
+
d
=
∈ K/℘(K).
L/K
2
i<j

4
G. Berhuy, C. Frings
This discriminant can be extended to étale algebras in an obvious way and is related
to the Arf-invariant of the second trace form by the following formula:
+
ꢃ
d
= Arf(E /K) + εn
E/K
ꢃ
where E = E (resp. E × K) if n is even (resp. n is odd), and εn is the element in
n
4
n+1
{
0, 1} representing the congruence class of [ ] modulo 2 (resp. [
]) if n is even
4
(
resp. if n is odd). See [BM,W] or [Wa] for more details.
+
Thus the additive discriminant d
determines uniquely the isomorphism class
E/K
of the second trace form of E/K.
Now consider the GrW-invariants of central simple algebras.
Let A
for any central simple algebra of degree n. Indeed, in this case A has a splitting
field L/K of odd degree, so we have
ꢀ
→ qA be a GrW-invariant. If n is an odd integer, we have qA ꢂ q
Mn(K)
qA ⊗ L ꢂ qA⊗L ꢂ qMn(L) ꢂ qMn(K) ⊗ L
and we conclude by Springer’s theorem (which also holds in characteristic two, see
[Re, lemma p. 231], for example).
Now assume that n is even. In [Ti,LM,U] and [Se], it is shown that
n
c(qA) = c(qMn(K)) + [A]
2
when qA = TA or T2,A and char K ꢁ= 2. In Sect. 4, we show that this formula holds
for the second trace form in characteristic two. This means that the second trace
form is an equivalent substitute to the trace form in characteristic two, in opposition
to the case of étale algebras. It can be explained by the fact that the crucial point in
the proof of the result of Bergé-Martinet is that an étale algebra is commutative.
2
. The split case
Since we are in the split case, the reduced characteristic polynomial of a matrix M
coincides with its usual characteristic polynomial, and will be denoted by χ(M).
Proposition 1. Let K be a field of characteristic two, n = 2m ≥ 2 an even integer
and A = Mn(K). Then
ꢄ
ꢅ
ꢄ
ꢅ
m
m
2
T2,A ꢂ
× P1,1 ⊥ (2m −
) × P0,0.
2
2
Proof. Let (Ei,j ) be the standard basis of A. We will write Ei instead of Ei,i. For
1
≤ k ≤ m, let
F2k−1 := E1 + · · · + E2k−2 + E2k−1 and F2k := E1 + · · · + E2k−2 + E2k.
Using the fact that the bilinear form b2,A associated to the second trace form satisfies
b2,A(x, y) = TrdA(x)TrdA(y)+TrdA(xy), it is easy to see that putting together the

Second trace form of central simple algebras in characteristic two
5
symplectic pairs (Ei,j , Ej,i), i < j, (F2k−1, F2k), 1 ≤ k ≤ m gives a symplectic
basis for T2,A.
n
Moreover, we have χ(Ei,j ) = X , so SrdA(Ei,j ) = 0. We also get
2
k−1 n−2k+1
χ(F2k) = χ(F2k−1) = (X + 1)
X , so we have
ꢆ
ꢇ
ꢆ
ꢇ
2
k − 1
k − 3
2k − 1
T2,A(F2k−1) = T2,A(F2k) =
This finishes the proof. ꢄꢅ
=
= (2k − 1)(k − 1) = k − 1.
2
2
Corollary 1. Let K be a field of characteristic two, n ≥ 2 an even integer and A a
central simple algebra of degree n. Then T2,A is a non-degenerate quadratic form
over K.
Proof. Let L be any splitting field of A. We have
T2,A ⊗ L ꢂ T2,A⊗L ꢂ T2,Mn(L)
By Proposition 1, the latter form is non-degenerate, so is T2,A.
3
. The cyclic case
We recall first the definition of a cyclic algebra. Let E/K be a cyclic extension of
∗
degree n, σ a generator of the Galois group and a ∈ K . The K-vector space
n−1
ꢈ
i
(
a, E/K, σ) := i=0 Ee
n
σ
with the multiplication law e = a and eλ = λ e, λ ∈ E is a central simple
algebra of degree n over K, called a cyclic algebra, which contains E as a maximal
commutative subfield. The cyclic algebra (1, E /K, σ)is split (see [Sc], Chapter 8,
§
12 for example).
Proposition 2. Let K be a field of characteristic two, n = 2m ≥ 2 and A =
(a, E/K, σ) a cyclic algebra of degree n. Then we have
ꢄ
ꢅ
ꢉ
ꢄ
ꢅꢊ
m
m
2
T2,A ꢂ
×Pa−1 ⊥P1,Arf(E/K) ⊥Pa−1
⊥ 2m − 2 −
×P0,0
,a
,a Arf(E/K)
2
2
where Arf(E/K) is the Arf invariant of the second trace form of the field extension
E/K.
n−1
Proof. • If x = λ0 + λ1e + · · · + λn−1e
, then TrdA(x) = TrE/K (λ0). Indeed,
we have seen that TrdA(x) is the trace of left mutiplication by x in A, considered
n−j
j
j
j
σ
0
as a right E-vector space. Since we have xe = λ0e + · · · = e λ
+ · · · , we
get
n−1
ꢁ
n−j
σ
TrdA(x) =
λ
= TrE/K (λ0).
0
j=0

6
G. Berhuy, C. Frings
It follows easily that we have the following orthogonal decomposition of the K-
m
vector space A with respect to T2,A: A = E ⊕ Ee ⊕ M,
k
where M = ꢆλe , λ ∈ E , k
ꢁ
= 0, mꢇ, using the formula
b2,A(x, y) = TrdA(x)TrdA(y) + TrdA(xy).
•
Now we study the restriction of the second trace form to the three spaces
m
E, Ee and M. For this, we first compute the matrix S = (sij )0≤i,j≤n−1 of
k
left multiplication by λe , 0 ≤ k ≤ m, λ ∈ E. If k = 0, then we have S =
n−1
σ
σ
diagꢆλ, λ
, . . . , λ ꢇ. Thus we get SrdA(λ) = SrdE(λ), i.e. T2,A|E ꢂ T2,E.
−k−j
σ
k
j
k+j
Assume now that 1 ≤ k ≤ m. We have λe e = e
λ
, thus

−
σ
k−j


sk+j,j = λ
if 0 ≤ j ≤ n − k − 1,
k−j
−
σ
sk+j−n,j = aλ
if n − k ≤ j ≤ n − 1,


si,j = 0
otherwise.
For any matrix C = (ci,j )0≤i,j≤n−1, we know that
ꢁ
τ∈Sn
det C =
ε(τ)c
· · · cn−1,τ(n−1)
0
,τ(0)
where ε(τ) denotes the signature of τ. Since we want to compute the coefficient
n−2
corresponding to X
in the expansion of det(XIn − S), we have to sum over the
elements of Sn which have exactly n − 2 fixed points, namely the transpositions.
So we get
n−k−1
ꢁ
i>j
ꢁ
k
SrdA(λe ) =
si,j sj,i =
sk+j,j sj,j+k.
j=0
If i < j, we have si,j
ꢁ
= 0 if and only if i = k + j − n. In particular, sj,j+k
ꢁ
= 0 if
k
and only if j = 2k + j − n, i.e. k = m. Thus SrdA(λe ) = 0 for 1 ≤ k ≤ m − 1.
i
j
Since we have b2,A(λe , µe ) = 0 for λ, µ ∈ E and 1 ≤ i, j ≤ m − 1, we finally
k
get that the restriction of the second trace form to the subspace H := ꢆλe , λ ∈
E , k = 1, . . . , m − 1ꢇ is zero. So M is metabolic because H is a subspace of M
1
2
satisfying dimKH = dimK M. In particular, T2,A|M is hyperbolic. Moreover we
m−1
ꢀ
−j
−m−j
σ
m
σ
have SrdA(λe ) = a
λ
λ
.
j=0

Second trace form of central simple algebras in characteristic two
Finally we have obtained
7
T2,A ꢂ T2,E ⊥ aq ⊥ h,
m−1
ꢀ
−j
−m−j
where q is the quadratic form λ ∈ E
ꢀ
→
λσ λσ
and h is hyperbolic.
j=0
ꢂ ꢃ
m
•
If a = 1, the algebra A is split, so we get T2,E ⊥ q ∼
× P1,1 by
2
Proposition 1 and the previous point, where ∼ denotes the Witt-equivalence of
quadratic forms. By Theorem 2, we have T2,E ∼ P
, so
,Arf(E/K)
1
ꢄ
ꢅ
m
q ∼ P1,Arf(E/K)
Using the fact that aPu,v ꢂ P
⊥
× P1,1.
2
∗
u
a
if a ∈ K and u, v, ∈ K, we get the result.
,av
4
. The general case
Theorem 3. Let K be a field of characteristic two, n ≥ 2 an even integer and A a
central simple algebra of degree n over K. Then we have:
ꢂ ꢃ
n
4
(
(
1) Arf(T2,A) =
n
2
2) c(T2,A) = [A]
Before proving the theorem, we want to recall further results. Let K be a field
and A a central simple algebra of degree n over K. Fix a K-basis e1, . . . , en
A and let nA(X1, . . . , Xn en ). This polynomial is
) := NrdA(X1e1 + · · · + X
absolutely irreducible, so RA := K[X1, . . . , X ]/(nA) is a domain.
2
of
2
2
2
n
2
n
Proposition 3. The quotient field K(A) of RA has the following properties:
(
(
(
a) K(A) splits A,
b) K is algebraically closed in K(A),
c) Ker ResK(A)/K = ꢆ[A]ꢇ.
The proof of (a) can be found in [S1], and (b) is proved in [L, p. 369]. Moreover,
it is shown in [S1] that K(A) is a rational extension of the field K(νA) of rational
functions of the Severi-Brauer variety of A, so ResK(A)/K(νA) is an injection (see
[
J, Theorem 3.8.6] for example).
Since Ker ResK(νA)/K = ꢆ[A]ꢇ (see [Am, Theorem 9.3 and Theorem 12.1]) and
, we get assertion (c).
ResK(A)/K = Res
◦ Res
K(A)/K(νA)
K(νA)/K
The following theorem is due to Saltman, and will be proved in Appendix:
Theorem 4 (Saltman, unpublished). Let K be a field, A a central simple algebra
of degree n over K and G a finite group of order n. Then there exists a field extension
LG/K such that:
(
(
1) A ⊗ LG is isomorphic to a G-crossed product,
2) ResL /K is an injection.
G

8
G. Berhuy, C. Frings
Proof of Theorem 3. • Let us prove assertion (1). If L/K is a field extension, the
inclusion K ⊆ L induces a map ιL/K : K/℘(K) → L/℘(L). Then we have
Arf(qL) = ιL/K(Arf(q))
for any quadratic form over K. Moreover ι
is an injection. Indeed, if x ∈
K(A)/K
2
K(A) satisfies x = λ+λ for some λ ∈ K(A), then λ is an element of K(A) which
is algebraic over K, so λ ∈ K by Proposition 3 (b), i.e. x ∈ ℘(K).
Since we have
ιK(A)/K(Arf(T2,A)) = Arf(T2,A ⊗ K(A)) = Arf(T
)
2,A⊗K(A)
=
=
=
Arf(T
) (by Proposition 3 (a))
2
,Mn(K(A))
Arf(T2,Mn(K)⊗K(A)) = Arf(T2,Mn(K) ⊗ K(A))
K(A)/K(Arf(T2,Mn(K))),
ι
we get the result using Proposition 1 and the injectivity of ιK(A)/K
.
•
Now we prove (2) for cyclic algebras. Using Proposition 2, we get
ꢄ ꢅ
n
−
c(T2,A) = (1, Arf(E/K)] +
(a 1, 1] + (a−1, Arf(E/K)]
4
ꢄ ꢅ
n
=
(a,
+ Arf(E/K)].
4
ꢂ ꢃ
n
4
Since n is even, we have
= εn + 2l, for a suitable integer l, so
+
c(T2,A) = (a, d + 2l]
E
+
=
=
(a, d ] + 2(a, l]
E
+
(a, d ],
E
since a quaternion algebra has order at most 2 in Br(K). By [J], Corollary 2.13.20,
n
2
we have [A] = (a, F/K, σ|F ), where F is the unique quadratic subfield of E.
We now recall how to associate a separable field extension of degree at most
2
to an étale algebra over a field K of any characteristic: if E is an étale algebra
over K of rank n, let H be the set of the n K-homomorphisms from E to Ks. Then
˜
Gal(Ks/K) acts on H by left multiplication. Now define E to be the subfield of Ks
fixed by the elements s ∈ Gal(Ks/K) inducing an even permutation on H. Then
˜
E/K is a separable field extension of degree at most 2. If char K = 2, it is shown
+
˜
in [BM, Theorem 2.6.], that this extension is defined by d , i.e. E is generated by
E
2
+
an element x ∈ Ks satisfying x + x + d = 0 (if char K
ꢁ
= 2, one can show that
E
√
˜
E = K( dE), where dE := det TE is the classical discriminant).
˜
We now prove that in our case, we have E = F. Here E/K is a Galois field
extension, so h(E) ⊆ E for every h ∈ H and we have in fact H = Gal(E/K). In
particular, sh = h for s ∈ Gal(Ks/E) and h ∈ H, so every element of Gal(Ks/E)
induced the trivial permutation on H, which is even. This implies that any element
of E is fixed by Gal(Ks/E), so E is a subfield of E. Since E/K is cyclic, the
generator σ of Gal(E/K) permutes cyclically the elements of Gal(E/K), and this
permutation is odd (since it is a n-cycle with n even). In particular, the subgroup of
˜
˜
˜
Gal(Ks/K) which is used to define E is not the full absolute Galois group (since it

Second trace form of central simple algebras in characteristic two
9
˜
˜
does not contain any extension of σ to Ks), so [E : K] = 2. Hence E is a quadratic
subfield of E. Since E/K is cyclic of even degree, E contains a unique quadratic
˜
subfield, so E = F.
n
2
˜
We finally obtain [A] = (a, E/K, σ | ). It is immediate to check that this algebra
˜
E
+
is (a, d ].
E
Now let A be any central simple algebra of degree n over K. Using Theorem 2
with G cyclic, we get a field extension LG/K such that A ⊗ LG is a cyclic algebra
and ResL /K is an injection. Since we have
G
ResL /K(c(T2,A)) = c(T2,A ⊗ LG) = c(T2,A⊗L )
G
G
ꢉ
ꢊ
n
n
= 2
[A ⊗ LG] = ResL /K
[A] ,
G
2
we get the result. ꢄꢅ
Remark 2. As in [U,LM,Ti] and [Se], we obtain
c(T2,A) = c(T2,Mn(K)) + rn[A]
for any central simple algebra of even degree n, where rn is an integer which only
depends on n. This is not very surprising, and can be explained as follows: let
A
ꢀ
→ qA be a GrW-invariant of central simple K-algebras, where K is a field of
any characteristic. Assume that qA is a quadratic form for every central simple
algebra A of degree n. We easily get that c(qA) − c(q ) ∈ Ker ResK(A)/K,
Mn(K)
so c(qA) = c(q
) + r(A)[A]. Replacing A by the generic division algebra
Mn(K)
UD := UD(K, n, r) of degree n over K (see for example [S2, Sect. 14]) and K by
itscenter, wegetc(qUD) = c(q
)+r(UD)[UD]. By[Ro,Theorem1], wehave
Mn(K)
n
2
exp(UD) = n. So r(UD) is a multiple of and we have r(UD)[UD] = rn[UD],
n
2
with rn = 0 or , since [UD] is killed by n. Thus c(qA) = c(q
) + rn[UD].
Mn(k)
Since any central simple algebra can be obtained by specialization of UD (see [S2,
Sect. 14]), we get c(qA) = c(qM (K)) + rn[A] for any central simple algebra of
n
n
2
degree n over K, where rn = 0 or . It is also easy to show that det qA = det q
Mn(k)
(
or Arf(qA) = Arf(q
) if char K = 2). This method has first been applied by
Mn(K)
Saltman to compute the Clifford invariant of the trace form of a central simple
algebra when char K
= 2 (unpublished).
ꢁ
Appendix: Proof of Theorem 4
In this appendix, we want to give a proof of Theorem 4, since Saltman never
published his result, which is nevertheless of independent interest.
We first recall the notion of generic G-crossed product, defined by Saltman in
[S2, Sect. 12, p. 84].
Let K be any field and G a finite group of order n. Consider the following short
exact sequence
f
→ M → ⊕ Z[G]dg → Z[G]
g∈G
0

1
0
G. Berhuy, C. Frings
where f is Z[G]-linear and maps dg to g −1. Then M is a finitely generated Z[G]-
module, which is free as a Z-module. We will write it multiplicatively. Now let
c(g, h) := gdh + dg − dgh ∈ M for g, h ∈ G. We extend the action of G on the
group algebra K[M] by K-linearity. This action extends naturally on the quotient
∗
field K(M) of K[M]. Then M ⊆ K(M) and c is a 2-cocycle of G with values in
∗
ꢃ
G
ꢃ
K(M) . If K := K(M) , the crossed product EG := (K(M)/K , G, c) is called
the generic G-crossed product over K. Before proving Theorem 4, we need the
following proposition:
Proposition 4. The generic G-crossed product has exponent n, i.e. [EG] has order
ꢃ
n in Br(K ).
2
∗
Proof. This is equivalent to show that [c] has order n in H (G, K(M) ).
2
2
∗
•
We first prove that the map H (G, M) → H (G, K(M) ) is injective. Since
r
M ꢂ Z as an abelian group, we have
K[M] = K[m1, m−1 , · · · , mr, m−r 1
1
]
where (mi) is a basis of M. In particular, K[M] is a unique factorization domain.
∗
∗
So the primes of K[M] form a basis of K(M) /K[M] . Moreover, it is easy to
see that G preserves the set of primes up to units. So G permutes the elements of
the basis of K(M) /K[M] . For each orbit ω of K(M) /K[M] under this action,
choose a representative pω and let Hω be the stabilizer of pω. Then we have
∗
∗
∗
∗
ꢈ
∗
∗
K(M) /K[M] ꢂ
Z[G/Hω]
ω
as G-modules.
1
1
By Shapiro’s lemma, we have H (G, Z[G/Hω]) = H (Hω, Z). Here G acts
1
trivially on Z, so H (Z[G/Hω]) = Hom(Hω, Z) = 0 since Hω is finite. Finally
we get
1
∗
∗
H (G, K(M) /K[M] ) = 0.
Using the long exact sequence of group cohomology induced by the short exact
sequence
∗
∗
∗
∗
0
→ K[M] → K(M) → K(M) /K[M] → 0
2
∗
2
∗
we get that the map H (G, K[M] ) → H (G, K(M) ) is injective. Moreover
∗
∗
K[M] is the set of monomials with leading coefficient in K . Then it is easy to
∗
∗
see that K[M] = K ⊕ M as G-modules. So we have
2
∗
2
∗
2
H (G, K[M] ) = H (G, K ) ⊕ H (G, M)
2
2
∗
and the map H (G, M) → H (G, K[M] ) is injective. We get the desired conclu-
sion by composition with the previous injective map. Notice that the injectivity of
2
∗
2
∗
the map H (G, K[M] ) → H (G, K(M) ) can also be obtained as a particular
case of [S2], Theorem 12.4 (a) (untwisted case).

Second trace form of central simple algebras in characteristic two
11
2
•
By the previous point, it suffices to show that [c] has order n in H (G, M).
Let t : Z[G] → Z be the Z-linear map which sends g to 1. Then (g − 1)g∈G is a
basis of Ker t and we have the following exact sequence
f
0
→ M → ⊕ Z[G]dg → Ker t → 0.
g∈G
Since every free Z[G]-module is cohomologically trivial (apply Shapiro’s lemma
to the trivial subgroup), the long exact sequence of cohomology gives that the
connecting map
1
2
∂
: H (G, Ker t) → H (G, M)
is an isomorphism. If α is the 1-cocycle g
ꢀ
→ g−1, then ∂([α]) = [c], so it remains
1
to show that [α] has order n in H (G, Ker t).
•
Taking the long exact sequence of cohomology associated to
0
→ Ker t → Z[G] → Z → 0
we get
∗
t
∂
0
0
1
H (G, Z[G]) → H (G, Z) → H (G, Ker t) → 0.
ꢀ
0
∗
It is easy to see that H (G, Z[G]) = Z
g, so Imt = nZ and we finally get
g∈G
1
H (G, Ker t) ꢂ Z/nZ. But [α] corresponds precisely to the image of 1 in Z/nZ,
which has order n. This finishes the proof. ꢄꢅ
op
ꢃ
ꢃ
Now we are ready to prove Theorem 4. Let LG = K ((A⊗K K )⊗
ꢃ
E ). By
K
G
Proposition 3 (a), we have A ⊗K LG ꢂ EG ⊗
ꢃ
LG. By [J], Theorem 2.13.16 for
K
ꢃ
example, we know that EG ⊗
ꢃ
LG is Brauer-equivalent to a G -crossed product
K
ꢃ
ꢃ
over LG, where G = Gal(LGK(M)/LG). Since K is algebraically closed in
ꢋ
ꢃ
LG, we have LG K(M) = K , since an element of LG which belongs to this
ꢃ
ꢃ
intersection is algebraic over K . Since K(M)/K is a Galois extension, this implies
ꢃ
that LG and K(M) are linearly disjoint over K . In particular, [LGK(M) : LG] = n
and Gal(LGK(M)/LG) ꢂ G.
Finally, A ⊗K LG is Brauer-equivalent to a G-crossed product. Since the degrees
are equal, we get the desired isomorphism.
We know prove that ResL /K is an injection. We have
G
ResK(M)/K = ResK(M)/K
ꢃ
◦ Res
ꢃ
.
K /K
l
Notice that K(M)/K is rational. Indeed, since M ꢂ Z as a Z-module, we have
−
1
−1
l
K[M] ꢂ K[X1, X , · · · , Xl, X ], so K(M) ꢂ K(X1, · · · , Xl). Consequently
1
Res
is an injection, so Res
ꢃ
is an injection.
, we get
K(M)/K
K /K
Since ResL /K = Res
ꢃ
◦ Res
ꢃ
G
LG/K
K /K
ꢃ
op
E ]ꢇ
Ker ResL /K = Br(K) ∩ ꢆ[(A ⊗K K ) ⊗
ꢃ
G
K
G
op
G
ꢃ
by Proposition 3 (c). Let [B] = r[A ⊗K K ) ⊗
ꢃ
E ] be an element of this
K
ꢃ
G
˜
kernel. Let K = K K = K(M) , where K is an algebraic closure of K. Since
˜ ˜
[
B] ∈ Br(K) and K contains K, we have [B ⊗K K] = 0. On the other hand, we

1
2
G. Berhuy, C. Frings
op
G
˜
˜
˜ ˜
K], since K splits A ∈ Br(K). So
op
have [B ⊗K K] = r[E
⊗
ꢃ
K] = −r[EG ⊗
ꢃ
K
K
˜
˜
we have r[EG ⊗
ꢃ
K] = 0. Since EG ⊗
ꢃ
K is the generic G-crossed product over
K
K
ꢃ
˜
K, which has order n in Br(K), we get n|r. Now A ⊗K K and E have degree n
G
ꢃ
over K , so [B] = 0.
Acknowledgements. This work is supported by the TMR research network (ERB FMRX
CT-97-0107) on “K-theory and algebraic groups”.
The authors are very grateful to Pr. Saltman for patiently answering their numerous
questions on his unpublished result and several other topics.
References
[
[
[
[
[
[
Am] Amitsur, S.A.: Generic splitting fields of central simple algebras. Ann. Math. 62,
8
–43 (1955)
B1]
B2]
Be]
Berhuy, G.:Autour des formes trace des algèbres cycliques. to appear in Pub. Math.
de Besançon, Théorie des nombres
Berhuy, G.: Trace forms of central simple algebras over a local field or a global
field. Preprint
Berlekamp, E.: An Analogue to the Discriminant over Fields of Characteristic Two,
J. Algebra 38, 315–317 (1976)
BM] Bergé, A.-M., Martinet, J.: Formes quadratiques et extensions en caractéristique 2.
Ann. Inst. Fourier Grenoble 35, 57–77 (1985)
J]
Jacobson, A.: Finite-Dimensional Algebras over Fields. Berlin–Heidelberg–New
York: Springer-Verlag 1996
[L]
Lewis, D.W.: Trace forms of central simple algebras. Math.Z. 215, 367–375 (1994)
[LM] Lewis, D.W., Morales, J.: The Hasse invariant of the trace form of a central simple
algebra. Pub. Math. de Besançon, Théorie des nombres, 1–6 (1993/94)
[
[
Re]
Ro]
Revoy, Ph.: Remarques sur la forme trace. Linear Mult.Algebra 10, 223–233 (1981)
Rowen, L.H: Universal PI-algebras and algebras of generic matrices, Israel J. Math.
(18), 65–74 (1974)
[
[
S1]
S2]
Saltman, D.: Norm polynomials and algebras. J. of Algebra 62, 333–345 (1980)
Saltman, D.: Lectures on division algebras. Conference board of the mathematical
science: regional conference series in maths. Providence, RI: AMS, 1999
Scharlau,W.:Quadraticandhermitianforms. GrundlehrenMath.Wiss. 270. Berlin–
Heidelberg–New York: Springer-Verlag, 1985
Serre, J.-P.: Cohomologie galoisienne. Cinquième édition, Lecture Notes in Math-
ematics 5. Berlin–Heidelberg–New York: Springer-Verlag, 1994
Tignol, J.-P.: La norme des espaces quadratiques et la forme trace des algèbres
simples centrales. Pub.Math.Besançon, Théorie des nombres (92/93–93/94)
Unger, T.: A note on surrogate forms of central simple algebras. Mathematical
Proceedings of the Royal Irish Academy (to appear)
[
[
[
[
[
[
Sc]
Se]
Ti]
U]
W]
Wadsworth,A.: Discriminants in characteristic 2. Linear. Mult.Algebra 17, 235–263
(1985)
Wa] Waterhouse, W.C.: Discriminants of étale algebras and related structures. J. reine
angew. Math. 379, 209–220 (1987)