80 IX. TH E TRAC E FOR M I N DIMENSIO N 7
Assume thi s lemma . I f w e writ e g a s (2,1} © Q , w e ma y appl y th e lemm a
with a = 2d, henc e Q = (x,y) 0 g wit h (x)-(2dy) 0 an d rank(g ) = 2 . W e hav e
q ^ q' 0 q" wit h q f = (2 , y) an d q" = (1, x) 0 g. Th e for m g ' i s th e trac e for m o f
an etal e algebr a E'. Th e for m q" ca n b e writte n a s (1) 0 / wher e / = (x) 0 g.
Note tha t (d(f)) = (2dy). Henc e th e relatio n (x)-(2dy) = 0 ca n b e rewritte n a s
(x) -(d(f)) = 0 . Thi s show s tha t / ha s propert y (3 ) o f Prop . 31.12, henc e i s special .
By Theore m 31.18 , ther e exist s a n etal e algebr a E" o f ran k 4 whos e trac e for m i s
q". Henc e q i s isomorphi c t o th e trac e for m o f E' x E".
P R O O F O F L E M MA 31.27. I f a i s a square, Q represent s 1, hence contain s (l,y)
for som e y, an d w e hav e (l)-(ay) = 0 . Assum e a i s no t a square ; pu t k\ = k(u)
with u 2 a. Writ e V fo r th e A:-vecto r spac e underlyin g Q. Th e assumptio n tha t
Q represent s 1 over k\ mean s tha t ther e i s a n elemen t z o f k\ Sk V wit h Q(z) = 1.
If w e writ e z a s z = z' 0 wz", wit h z' ', z " i n V, th e fac t tha t Q(z) = 1 means tha t
z' an d z " ar e orthogona l an d tha t
(31.28) l = Q(z') + a-Q(z").
Put x = Q(^' ) a n ( i ? / = Q{ z")- I f the y ar e bot h no t 0 , Q contain s (x, y) an d (31.28)
shows tha t (ay) = ( 1 x), henc e (x) '(ay) 0, a s required . I f y = 0 , (31.28) show s
that Q represent s 1, henc e contain s (1,6) for som e b G fc*, and w e wi n a s above . I f
x 0, the n Q represent s a , an d the n contain s (6 , a) fo r som e 6 G /c*, an d w e als o
win.
31.29. T H E CASE n = 7. Le t ^ b e a quadrati c for m o f ran k 7 , an d le t d b e it s
discriminant.
T H E O R E M 31.30. For q to be a trace form, it is necessary and sufficient that:
(a) q contains (2,1,1);
(b) q contains (2,1,1,1 ) over the field k(\/2d).
PROOF. Thi s follow s fro m Cor . 31.24 combine d wit h Theore m 31.26. A direc t
proof ca n b e give n b y th e metho d w e followe d i n th e cas e n = 5 , namely : I f q
is a trac e for m g# , w e mak e a n odd-degre e extensio n k' jk ove r whic h E split s a s
E = k x E' wit h rankE 7 = 6 . B y applyin g Theore m 31.26 t o E' ', w e se e tha t q ha s
properties (a ) an d (b ) ove r k 1 ', henc e als o ove r k.
Conversely, i f q ha s propertie s (a ) an d (b) , w e writ e i t a s q = (1) 0 q', an d us e
Theorem 31.26 t o sho w tha t ther e exist s a n etal e algebr a E' suc h tha t qE = q'
For E = k x E' w e hav e q
E
= q.
31.31 . R E M A R K S AN D QUESTIONS .
(1) Le t q b e a quadrati c for m o f ran k n. Suppos e q become s a trac e for m
after a n odd-degre e extensio n o f k. I s i t tru e tha t q i s a trac e for m (ove r
A;)?
The criteri a w e hav e give n abov e sho w tha t thi s i s tru e fo r n 8 .
(2) Le t E b e a n etal e algebr a o f ran k n. I s i t tru e tha t ther e exist s a n etal e
algebra ove r k wit h th e followin g tw o properties :
(2a) q
E
o ^ q
E
-
(2b) i s spli t b y a Galoi s extensio n o f k o f orde r a powe r o f 2 (i n othe r
words, th e imag e o f th e homomorphis m T ^ S
n
associate d wit h
is a 2-group) ?
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