<< . .

. 26
( : 75)



. . >>

1 ’ ’ ’ F — ’ ’ ’ “(A, σ, f ) ’ ’ ’ O+ (A, σ, f ) ’ ’ ’ 1
’’ ’’ ’’ ’’
¦ ¦ ¦
¦ ¦µ ¦
2 Sn

1 ’ ’ ’ F —2 ’ ’ ’ F— F — /F —2
’’ ’’ ’’’
’’ ’ ’ ’ 1.
’’
We also de¬ne the group of spinor norms:
Sn(A, σ, f ) = { µ(γ) | γ ∈ “(A, σ, f ) } ‚ F — ,
so Sn O+ (A, σ, f ) = Sn(A, σ, f )/F —2 , and the spin group:
Spin(A, σ, f ) = { γ ∈ “(A, σ, f ) | µ(γ) = 1 } ‚ “(A, σ, f ).
In the split case, if (A, σ, f ) = EndF (V ), σq , fq for some nonsingular quadratic
space (V, q) of even dimension, the standard identi¬cations associated to q yield
Spin(A, σ, f ) = Spin(V, q) = { c ∈ “+ (V, q) | „ (c) · c = 1 },
where „ is the canonical involution of C(V, q) which is the identity on V . From
the description of the spinor norm in Scharlau [?, Chap. 9, §3], it follows that the
group of spinor norms Sn(V, q) = Sn(A, σ, f ) consists of the products of any even
number of represented values of q.
§13. QUADRATIC PAIRS 187


The vector representation χ induces by restriction a homomorphism
Spin(A, σ, f ) ’ O+ (A, σ, f )
which we also denote χ.
(13.31) Proposition. The vector representation χ ¬ts into an exact sequence:
χ Sn
1 ’ {±1} ’’ Spin(A, σ, f ) ’’ O+ (A, σ, f ) ’’ F — /F —2 .
Proof : This follows from the exactness of the top sequence in (??) and the de¬nition
of Sn.
Assume now deg A = 2m ≥ 4. We may then use the extended Cli¬ord group
„¦(A, σ, f ) to de¬ne an analogue of the spinor norm on the group PGO+ (A, σ, f ), as
we proceed to show. The map S de¬ned below may be obtained as a connecting map
in a cohomology sequence, see §??. Its target group is the ¬rst cohomology group of
the absolute Galois group of F with coe¬cients in the center of the algebraic group
Spin(A, σ, f ). The approach we follow in this subsection does not use cohomology,
but since the structure of the center of Spin(A, σ, f ) depends on the parity of m,
we divide the construction into two parts, starting with the case where the degree
of A is divisible by 4. As above, we let Z = Z(A, σ, f ) be the center of the Cli¬ord
algebra C(A, σ, f ).
(13.32) De¬nition. Assume deg A ≡ 0 mod 4. We de¬ne a homomorphism
S : PGO+ (A, σ, f ) ’ Z — /Z —2
as follows: for g · F — ∈ PGO+ (A, σ, f ), pick ω ∈ „¦(A, σ, f ) such that χ (ω) = g · F —
and set
S(g · F — ) = σ(ω)ω · Z —2 = µ(ω) · Z —2 .
Since ω is determined by g · F — up to a factor in Z — and σ is of the ¬rst kind, by
(??), the element µ(ω)ω · Z —2 depends only on g · F — . The map S thus makes the
following diagram commute:
χ
1 ’ ’ ’ Z — ’ ’ ’ „¦(A, σ, f ) ’ ’ ’ PGO+ (A, σ, f ) ’ ’ ’ 1
’’ ’’ ’’ ’’
¦ ¦ ¦
¦ ¦µ ¦
2 S

1 ’ ’ ’ Z —2 ’ ’ ’ Z— Z — /Z —2
’’ ’’ ’’’
’’ ’ ’ ’ 1.
’’
Besides the formal analogy between the de¬nition of S and that of the spinor
norm Sn, there is also an explicit relationship demonstrated in the following propo-
sition:
(13.33) Proposition. Assume deg A ≡ 0 mod 4. Let
π : O+ (A, σ, f ) ’ PGO+ (A, σ, f )
be the canonical map. Then, the following diagram is commutative with exact rows:
µ
π
O+ (A, σ, f ) ’ ’ ’ PGO+ (A, σ, f ) ’ ’ ’ F — /F —2
’’ ’’
¦ ¦
¦ ¦
Sn S

NZ/F
F — /F —2 Z — /Z —2 ’ ’ ’ F — /F —2 .
’’’
’’ ’’
188 III. SIMILITUDES


Proof : Consider g ∈ O+ (A, σ, f ) and γ ∈ “(A, σ, f ) such that χ(γ) = g. We then
have Sn(g) = µ(γ) · F —2 . On the other hand, we also have χ (γ) = g · F — , hence
S(g · F — ) = µ(γ) · Z —2 . This proves that the left square is commutative.
Consider next g · F — ∈ PGO+ (A, σ, f ) and ω ∈ „¦(A, σ, f ) such that χ (ω) =
g · F — , so that S(g · F — ) = µ(ω) · Z —2 . By (??) we have µ(ω) · F — = κ(ω) and, by
(??), NZ/F κ(ω) = µ(g) · F —2 . Therefore, NZ/F S(g · F — ) = µ(g) · F —2 and the
right square is commutative.
Exactness of the lower sequence is a consequence of Hilbert™s Theorem 90 (??):
if z ∈ Z — is such that zι(z) = x2 for some x ∈ F — , then NZ/F (zx’1 ) = 1, hence
by (??) there exists some y ∈ Z — such that zx’1 = ι(y)y ’1 . (Explicitly, we may
take y = t + xz ’1 ι(t), where t ∈ Z is such that t + xz ’1 ι(t) is invertible.) Then
zy 2 = xNZ/F (y), hence z · Z —2 lies in the image of F — /F —2 .
Note that the spin group may also be de¬ned as a subgroup of the extended
Cli¬ord group: for deg A ≡ 0 mod 4,
Spin(A, σ, f ) = { ω ∈ „¦(A, σ, f ) | µ(ω) = 1 }.
Indeed, (??) shows that the right side is contained in “(A, σ, f ).
The restriction of the homomorphism
χ : „¦(A, σ, f ) ’ PGO+ (A, σ, f )
to Spin(A, σ, f ), also denoted χ , ¬ts into an exact sequence:
(13.34) Proposition. Assume deg A ≡ 0 mod 4. The sequence
χ S
1 ’ µ2 (Z) ’ Spin(A, σ, f ) ’ PGO+ (A, σ, f ) ’ Z — /Z —2
’ ’
is exact, where µ2 (Z) = { z ∈ Z | z 2 = 1 }.
Proof : Let g · F — be in the kernel of S. Then there exists ω ∈ „¦(A, σ, f ), with
σ(ω)ω = z 2 for some z ∈ Z — , such that χ (ω) = g · F — . By replacing ω by ωz ’1 ,
we get ω ∈ Spin(A, σ, f ). Exactness at Spin(A, σ, f ) follows from the fact that
Z — © Spin(A, σ, f ) = µ2 (Z) in „¦(A, σ, f ).
In the case where deg A ≡ 2 mod 4, the involution σ is of the second kind,
hence µ(ω) ∈ F — for all ω ∈ „¦(A, σ, f ). The rˆle played by µ in the case where
o
deg A ≡ 0 mod 4 is now played by a map which combines µ and κ.
Consider the following subgroup U of F — — Z — :
U = { (±, z) | ±4 = NZ/F (z) } ‚ F — — Z —
and its subgroup U0 = { NZ/F (z), z 4 | z ∈ Z — }. Let22
H 1 (F, µ4 [Z] ) = U/U0 ,
and let [±, z] be the image of (±, z) ∈ U in H 1 (F, µ4 [Z] ).
For ω ∈ „¦(A, σ, f ), let k ∈ Z — be a representative of κ(ω) ∈ Z — /F — . The
element kι(k)’1 is independent of the choice of the representative k and we de¬ne
µ— (ω) = µ(ω), kι(k)’1 µ(ω)2 ∈ U.

22 It
will be seen in Chapter ?? (see (??)) that this factor group may indeed be regarded as
a Galois cohomology group if char F = 2. This viewpoint is not needed here, however, and this
de¬nition should be viewed purely as a convenient notation.
§13. QUADRATIC PAIRS 189


For z ∈ Z — , we have κ(z) = z 2 · F — and µ(z) = NZ/F (z), hence
µ— (z) = NZ/F (z), z 4 ∈ U0 .
(13.35) De¬nition. Assume deg A ≡ 2 mod 4. De¬ne a homomorphism
S : PGO+ (A, σ, f ) ’ H 1 (F, µ4 [Z] )
as follows: for g · F — ∈ PGO+ (A, σ, f ), pick ω ∈ „¦(A, σ, f ) such that χ (ω) = g · F —
and let S(g · F — ) be the image of µ— (ω) in H 1 (F, µ4 [Z] ). Since ω is determined up
to a factor in Z — and µ— (Z — ) ‚ U0 , the de¬nition of S(g · F — ) does not depend on
the choice of ω. In other words, S is the map which makes the following diagram
commute:
χ
1 ’ ’ ’ Z — ’ ’ ’ „¦(A, σ, f ) ’ ’ ’ PGO+ (A, σ, f ) ’ ’ ’ 1
’’ ’’ ’’ ’’
¦ ¦ ¦
¦ µ— ¦ µ— ¦
S

’ ’ ’ H 1 (F, µ4 [Z] ) ’ ’ ’ 1.
1 ’ ’ ’ U0 ’ ’ ’
’’ ’’ U ’’ ’’
In order to relate the map S to the spinor norm, we de¬ne maps i and j which
¬t into an exact sequence
j
i
F — /F —2 ’ H 1 (F, µ4 [Z] ) ’ F — /F —2 .
’ ’
For ± · F —2 ∈ F — /F —2 , we let i(± · F —2 ) = [±, ±2 ]. For [±, z] ∈ H 1 (F, µ4 [Z] ),
we pick z0 ∈ Z — such that ±’2 z = z0 ι(z0 )’1 , and let j[±, z] = NZ/F (z0 ) · F —2 ∈
F — /F —2 . If NZ/F (z0 ) = β 2 for some β ∈ F — , then we may ¬nd z1 ∈ Z — such that
z0 β ’1 = z1 ι(z1 )’1 . It follows that ±’2 z = z1 ι(z1 )’2 , hence
2


(±, z) = ±NZ/F (z1 )’1 , ±2 NZ/F (z1 )’2 · NZ/F (z1 ), z1 ,
4


and therefore [±, z] = i ±NZ/F (z1 )’1 · F —2 . This proves exactness of the sequence
above.
(13.36) Proposition. Assume deg A ≡ 2 mod 4. Let
π : O+ (A, σ, f ) ’ PGO+ (A, σ, f )
be the canonical map. Then, the following diagram is commutative with exact rows:
µ
π
O+ (A, σ, f ) ’ ’ ’ PGO+ (A, σ, f ) ’ ’ ’ F — /F —2
’’ ’’
¦ ¦
¦ ¦
Sn S

j
i
F — /F —2 ’ ’ ’ H 1 (F, µ4 [Z] ) ’ ’ ’ F — /F —2 .
’’ ’’

Proof : Let g ∈ O+ (A, σ, f ) and let γ ∈ “(A, σ, f ) be such that χ(γ) = g. We then
have Sn(g) = µ(γ) · F —2 . On the other hand, we also have κ(γ) = 1, by (??), hence
µ— (γ) = µ(γ), µ(γ)2 = i µ(γ) · F —2 .
Since χ (γ) = g · F — , this proves commutativity of the left square.
Consider next g · F — ∈ PGO+ (A, σ, f ) and ω ∈ „¦(A, σ, f ) such that χ (ω) =
g · F — . We have j —¦ S(g · F — ) = NZ/F (k) · F — , where k ∈ Z — is a representative of
κ(ω) ∈ Z — /F — . Proposition (??) shows that NZ/F (k) · F —2 = µ(g · F — ), hence the
right square is commutative. Exactness of the bottom row was proved above.
190 III. SIMILITUDES


As in the preceding case, the spin group may also be de¬ned as a subgroup of
„¦(A, σ, f ): we have for deg A ≡ 2 mod 4,
Spin(A, σ, f ) = { ω ∈ „¦(A, σ, f ) | µ— (ω) = (1, 1) },
since (??) shows that the right-hand group lies in “(A, σ, f ). Furthermore we have
a sequence corresponding to the sequence (??):
(13.37) Proposition. Assume deg A ≡ 2 mod 4 and deg A ≥ 4. The sequence
χ S
1 ’ µ4 [Z] (F ) ’ Spin(A, σ, f ) ’ PGO+ (A, σ, f ) ’ H 1 (F, µ4 [Z] ),
’ ’
is exact, where µ4 [Z] (F ) = { z ∈ Z — | z 4 = 1 and ι(z)z = 1 }.
Proof : As in the proof of (??) the kernel of S is the image of Spin(A, σ, f ) under
χ . Furthermore we have by (??)
Z — © Spin(A, σ, f ) = { z ∈ Z — | z 2 ∈ F — and σ(z)z = 1 } = µ4 [Z] (F )
in „¦(A, σ, f ).

13.C. Multipliers of similitudes. This section is devoted to a generalization
of Dieudonn´™s theorem on the multipliers of similitudes (??). As in the preceding
e
sections, let (σ, f ) be a quadratic pair on a central simple algebra A of even degree
over an arbitrary ¬eld F , and let Z = Z(A, σ, f ) be the center of the Cli¬ord algebra
C(A, σ, f ). The nontrivial automorphism of Z/F is denoted by ι.
For ± ∈ F — , let (Z, ±)F be the quaternion algebra Z • Zj where multiplication
is de¬ned by jz = ι(z)j for z ∈ Z and j 2 = ±. In other words,
F [X]/(X 2 ’ δ);
(δ, ±)F if char F = 2 and Z
(Z, ±)F =
F [X]/(X 2 + X + δ).
[δ, ±)F if char F = 2 and Z
(Compare with §??).
(13.38) Theorem. Let g ∈ GO(A, σ, f ) be a similitude of (A, σ, f ).
(1) If g is proper, then Z, µ(g) F splits.
(2) If g is improper, then Z, µ(g) F is Brauer-equivalent to A.

When A splits, the algebra Z, µ(g) F splits in each case, so µ(g) is a norm
from Z/F for every similitude g. We thus recover Dieudonn´™s theorem (??).
e
In the case where g is proper, the theorem follows from (??) (or, equivalently,
from (??) and (??)). For the rest of this section, we ¬x some improper similitude g.
According to (??), the automorphism C(g) of C(A, σ) then restricts to ι on Z,

so C(g) induces a Z-algebra isomorphism C(A, σ, f ) ’ ι C(A, σ, f ) by mapping

c ∈ C(A, σ, f ) to ι C(g)(c) ∈ ι C(A, σ, f ). When we view C(A, σ, f ) as a left
Z-module, we have the canonical isomorphism
C(A, σ, f )op —Z C(A, σ, f ) = EndZ C(A, σ, f )
which identi¬es cop — c2 with the endomorphism de¬ned by
1
op
c c1 —c2
= c1 cc2 for c, c1 , c2 ∈ C(A, σ, f ).
We then have Z-algebra isomorphisms:
EndZ C(A, σ, f ) = C(A, σ, f )op —Z C(A, σ, f ) ι
C(A, σ, f )op —Z C(A, σ, f ).
§13. QUADRATIC PAIRS 191


The embedding ν of ι C(A, σ, f )op —Z C(A, σ, f ) into the endomorphism algebra of
the Cli¬ord bimodule B(A, σ, f ) (see (??)) yields an embedding
νg : EndZ C(A, σ, f ) ’ EndA B(A, σ, f )
de¬ned by
op
xνg (c1 —c2 )
= C(g)(c1 ) — x · c2 for c1 , c2 ∈ C(A, σ, f ), x ∈ B(A, σ, f ).
Let γg ∈ EndF C(A, σ, f ) be the endomorphism C(g ’1 ), i.e.,
cγg = C(g ’1 )(c) for c ∈ C(A, σ, f ).
Since g is improper, γg is not Z-linear, but ι-semilinear. Thus Int(γg ), which maps
’1
f ∈ EndZ C(A, σ, f ) to γg —¦ f —¦ γg , is an automorphism of EndZ C(A, σ, f ).
De¬ne an F -algebra Eg as follows:
Eg = EndZ C(A, σ, f ) • EndZ C(A, σ, f ) · y
where y is subject to the following relations:
’1
yf = (γg —¦ f —¦ γg )y for f ∈ EndZ C(A, σ, f ),
y 2 = µ(g)γg .
2


The algebra Eg is thus a generalized cyclic algebra (see Albert [?, Theorem 11.11],
Jacobson [?, § 1.4]); the same arguments as for the usual cyclic algebras show E g
is central simple over F .
(13.39) Proposition. The homomorphism νg extends to an isomorphism of F -
algebras:

νg : Eg ’ EndA B(A, σ, f )

by mapping y to the endomorphism β(g) of (??).
Proof : Since Eg and EndA B(A, σ, f ) are central simple F -algebras of the same
dimension, it su¬ces to show that the map de¬ned above is a homomorphism, i.e.,
that β(g) satis¬es the same relations as y:
’1
β(g) —¦ νg (f ) = νg (γg —¦ f —¦ γg ) —¦ β(g) for f ∈ EndZ C(A, σ, f ),
(13.40)
β(g)2 = µ(g)νg (γg ).
2

It su¬ces to check these relations over an extension of the base ¬eld. We may
thus assume that F is algebraically closed and (A, σ, f ) = EndF (V ), σq , fq for
some nonsingular quadratic space (V, q). Choose » ∈ F — satisfying »2 = µ(g).
Then q »’1 g(v) = q(v) for all v ∈ V , hence there is an automorphism of C(V, q)
which maps v to »’1 g(v) for all v ∈ V . By the Skolem-Noether theorem we may
thus ¬nd b ∈ C(V, q)— such that
b · v · b’1 = »’1 g(v) for v ∈ V .
Then C(g) is the restriction of Int(b) to C0 (V, q) = C(A, σ, f ), hence γg = Int(b’1 )
and
γg = (b2 )op — b’2 ∈ C(A, σ, f )op —Z C(A, σ, f ).
2

On the other hand, for v, w1 , . . . , w2r’1 ∈ V ,
β(g)
= µ(g)r v — g ’1 (w1 ) · · · g ’1 (w2r’1 )
v — (w1 · · · w2r’1 )
= » v — b’1 · (w1 · · · w2r’1 ) · b .
192 III. SIMILITUDES


The equations (??) then follow by explicit computation.
To complete the proof of (??) we have to show that the algebra Eg is Brauer-
equivalent to the quaternion algebra Z, µ(g) F . As pointed out by A. Wadsworth,
this is a consequence of the following proposition:
(13.41) Proposition. Let S be a central simple F -algebra, let Z be a quadratic
Galois ¬eld extension of F contained in S, with nontrivial automorphism ι. Let
s ∈ S be such that Int(s)|Z = ι. Let E = CS (Z) and ¬x t ∈ F — . Let T be the
F -algebra with presentation
where yey ’1 = ses’1 for all e ∈ E, and y 2 = ts2 .
T = E • Ey,
Then M2 (T ) (Z, t) —F S.
Proof : Let j be the standard generator of (Z, t) with jzj ’1 = ι(z) for all z ∈ Z
and j 2 = t. Let R = (Z, t) —F S, and let
T = (1 — E) + (1 — E)y ‚ R, where y = j — s.
Then T is isomorphic to T , since y satis¬es the same relations as y. (That is, for
’1 2
= 1 — (ses’1 ) and y = 1 — ts2 .) By the double centralizer
any e ∈ E, y (1 — e)y
theorem (see (??)) R T —F Q where Q = CR (T ), and Q is a quaternion algebra
over F by a dimension count. It su¬ces to show that Q is split. For, then
R T —F Q T —F M2 (F ) M2 (T ).
Consider Z —F Z ‚ R; Z —F Z centralizes 1 — E and Int(y ) restricts to ι — ι on
Z—F Z. Now Z—F Z has two primitive idempotents e1 and e2 , since Z—F Z Z•Z.
The automorphisms Id — ι and ι — Id permute them, so ι — ι maps each ei to itself.
Hence e1 and e2 lie in Q since they centralize 1 — E and also y . Because e1 e2 = 0,
Q is not a division algebra, so Q is split, as desired.
(13.42) Remark. We have assumed in the above proposition that Z is a ¬eld.
The argument still works, with slight modi¬cation, if Z F — F .
As a consequence of Theorem (??), we may compare the group G(A, σ, f ) of
multipliers of similitudes with the subgroup G+ (A, σ, f ) of multipliers of proper
similitudes. Since the index of GO+ (A, σ, f ) in GO(A, σ, f ) is 1 or 2, it is clear
that either G(A, σ, f ) = G+ (A, σ, f ) or G+ (A, σ, f ) is a subgroup of index 2 in
G(A, σ, f ). If A is split, then
[GO(A, σ, f ) : GO+ (A, σ, f )] = [O(A, σ, f ) : O+ (A, σ, f )] = 2

and

G(A, σ, f ) = G+ (A, σ, f )
since hyperplane re¬‚ections are improper isometries (see (??)). If A is not split, we
deduce from (??):
(13.43) Corollary. Suppose that (σ, f ) is a quadratic pair on a central simple
F -algebra A of even degree. If A is not split, then O(A, σ, f ) = O+ (A, σ, f ) and
[GO(A, σ, f ) : GO+ (A, σ, f )] = [G(A, σ, f ) : G+ (A, σ, f )].
If G(A, σ, f ) = G+ (A, σ, f ), then A is split by Z = Z(A, σ, f ).
§14. UNITARY INVOLUTIONS 193


Proof : If g ∈ O(A, σ, f ) is an improper isometry, then (??) shows that A is Brauer-
equivalent to Z, µ(g) F , which is split since µ(g) = 1. This contradiction shows
that O(A, σ, f ) = O+ (A, σ, f ).
If [GO(A, σ, f ) : GO+ (A, σ, f )] = [G(A, σ, f ) : G+ (A, σ, f )], then necessarily
GO(A, σ, f ) = GO+ (A, σ, f ) and G(A, σ, f ) = G+ (A, σ, f ).
Therefore, A contains an improper similitude g, and µ(g) = µ(g ) for some proper
similitude g . It follows that g ’1 g is an improper isometry, contrary to the equality
O(A, σ, f ) = O+ (A, σ, f ).
Finally, if µ is the multiplier of an improper similitude, then (??) shows that
A is Brauer-equivalent to (Z, µ)F , hence it is split by Z.
(13.44) Corollary. If disc(σ, f ) is trivial, then G(A, σ, f ) = G+ (A, σ, f ).
Proof : It su¬ces to consider the case where A is not split. Then, if G(A, σ, f ) =
G+ (A, σ, f ), the preceding corollary shows that A is split by Z; this is impossible
if disc(σ, f ) is trivial, for then Z F — F .

§14. Unitary Involutions
In this section, we let (B, „ ) be a central simple algebra with involution of the
second kind over an arbitrary ¬eld F . Let K be the center of B and ι the nontrivial
automorphism of K/F .
We will investigate the group GU(B, „ ) of similitudes of (B, „ ) and the unitary
group U(B, „ ), which is the kernel of the multiplier map µ (see §??). The group
GU(B, „ ) has di¬erent properties depending on the parity of the degree of B. When
deg B is even, we relate this group to the group of similitudes of the discriminant
algebra D(B, „ ).
14.A. Odd degree.
(14.1) Proposition. If deg B is odd, the group G(B, „ ) of multipliers of simili-
tudes of (B, „ ) is the group of norms of K/F :
G(B, „ ) = NK/F (K — ).
Moreover, GU(B, „ ) = K — · U(B, „ ).
Proof : The inclusion NK/F (K — ) ‚ G(B, „ ) is clear, since K — ‚ GU(B, „ ) and
µ(±) = NK/F (±) for ± ∈ K — . In order to prove the reverse inclusion, let deg B =
2m + 1 and let g ∈ GU(B, „ ). By applying the reduced norm to the equation
„ (g)g = µ(g) we obtain
NK/F NrdB (g) = µ(g)2m+1 .
Therefore,
µ(g) = NK/F µ(g)’m NrdB (g) ∈ NK/F (K — ),
hence G(B, „ ) ‚ NK/F (K — ). This proves the ¬rst assertion.
The preceding equation shows moreover that µ(g)m NrdB (g)’1 g ∈ U(B, „ ).
Therefore, letting u = µ(g)m NrdB (g)’1 g and ± = µ(g)’m NrdB (g) ∈ K — , we get
g = ±u. Thus, GU(B, „ ) = K — · U(B, „ ).
Note that in the decomposition g = ±u above, the elements ± ∈ K — and
u ∈ U(B, „ ) are uniquely determined up to a factor in the group K 1 of norm 1
elements, since K — © U(B, „ ) = K 1 .
194 III. SIMILITUDES


14.B. Even degree. Suppose now that deg B = 2m and let g ∈ GU(B, „ ).
By applying the reduced norm to the equation „ (g)g = µ(g), we obtain

NK/F NrdB (g) = µ(g)2m ,

hence µ(g)m NrdB (g)’1 is in the group of elements of norm 1. By Hilbert™s The-
orem 90, there is an element ± ∈ K — , uniquely determined up to a factor in F — ,

<< . .

. 26
( : 75)



. . >>