34. FURTHE R APPLICATION S T O NOETHER' S PROBLE M
91
There i s a differen t identificatio n o f H/Z(H) wit h D4 whic h give s th e characte r
\' suc h tha t x'( r ) = 1 a n d x'{ s) 1- Cal l if; the produc t X'X'i o n e n a s ^ ( r ) = 1
and t/j(s) = —1.
In th e proo f o f Th. 33.26 , we defined a n elemen t e of Inv(iJ, Z/2Z) . Le t u s recal l
its definition : i f fc is a field o f characteristi c no t 2 , an d x belong s t o H 1 (k, H), the n
X define s a n elemen t X4 o f H 1 (k1D±), henc e a ran k 4 etal e algebr a E±(x) m, o n th e
other hand , x define s vi a x '• H {1, —1} a n elemen t o f H 1 ^, {1, —1}), henc e a
quadratic algebr a E
2
(x). Th e produc t
(34.2) E
6
(x) = E
A
(x) xE
2
(x)
is a ran k 6 etal e algebr a ove r fc, wit h trivia l w\ an d u 2 (se e (33.20 ) an d (33.21)).
Hence b y Th . 32. 5 it s trac e for m q
x
qE
6
(x) m a Y b e writte n a s (1, l,c, c , c, c) fo r
some c G h*. T/i e invariant e(x) is then defined as
(34.3) e(x ) = (-1 ) -(-1) -(c ) z n H 3(k),
and w e hav e see n tha t e i s a nonzer o elemen t o f Inv/
Co
(i7, Z/2Z) i f th e groun d field
fco i s chose n t o b e Q .
We no w sho w tha t th e invarian t e ha s propert y (2 ) o f Prop . 15.9:
PROPOSITION 34.4 . Res ^ (e ) = 0 for every proper subgroup H' of H.
PROOF. W e hav e t o sho w that , i f x G iJ 1(fc,iJ) correspond s t o a homomor -
phism (f'.Tk H whic h i s no t surjective , the n e(x) = 0 , i.e. , th e trac e for m q
x
i s
isomorphic t o (1,1,1,1,1,1)An . y prope r subgrou p o f H i s contained i n a subgrou p
of inde x 2 , henc e i n th e kerne l o f x , x'\ o r i ;- Thi s give s u s thre e case s t o consider .
To d o so , w e writ e q
x
explicitl y a s i n §32 : th e algebr a E±(x) ca n b e define d b y a
polynome bicarr e X 4-2AX2+B, henc e ha s trace for m (1, A, A 2 - B, AB{A 2 - B)),
see Lemm a 31.19; th e algebr a E
2
{x) ha s trac e for m (2 , 2JB), S O tha t w e have :
qx = (1, 2, A, 2B, A 2 - B, AB(A 2 - B)).
By composin g ip:Tk -^ H wit h a quadrati c characte r e o f H, on e get s a quadrati c
character o f T^ , henc e a n elemen t x
£
o f /c*//c* 2. On e finds:
xx = B, x^ = B(A 2-B), an d x
x
= A 2 - B.
We ca n no w handl e th e thre e cases :
Case I: im(0 ) is contained in ker(x) - Thi s mean s tha t B i s a square , henc e q
x
may b e writte n a s
(1,2,A,2,A2-BJA(A2-B)),
which contain s (1,2,2) = (1,1,1)thi ; s implie s tha t "c " ca n b e chose n t o b e equa l
to 1, an d henc e e(x) = 0 .
Case II: im(0 ) is contained in ker(xO - Thi s mean s tha t A 2 B i s a square ,
hence q
x
contain s (1,1,2 ) an d "c " ca n b e chose n t o b e equa l t o 2 , an d w e hav e
e(x) 0 sinc e 2 i s a su m o f 4 squares .
Case III: im(/ ) is contained in ker(^) . Thi s mean s th&tB(A 2-B) i s a square ,
i.e., w e hav e A 2 - B = Bt 2 fo r som e t G fc*. Thi s implie s tha t B = A 2/(I + t 2 ),
hence B i s a su m o f tw o square s an d w e hav e (-l)-(B) = 0 . Th e trac e for m q
x
may b e writte n
qx = (1, 2, A, 2B, B, A) = (1, 2, B) 0 (A, A, 2B).
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