84 IX. TH E TRAC E FOR M I N DIMENSION 7

L E M M A 32.18. Let K be a field with a discrete valuation v, with residue field

k of characteristic ^ 2 . Let c G K* be such that (2 ) -(c) = 0 in H2(K) and that c

is a sum of 4 squares in K(y/2). Then the residue of (c) -(-1) -(-1 ) G H3(K) is 0.

(Recall tha t thi s residu e belong s t o H2(k), se e 7.13.)

P R O O F . W e may assume tha t K i s complete. Pu t m = v(c). Th e residue of

(c) -(—1) '(—1) is equal to 0 if m is even, an d to (—1) '( — 1) i f m is odd. I n the second

case, th e residue o f (2) -(c) i s (2). Sinc e (2 ) -(c) = 0 , this mean s tha t 2 is a squar e

in /c , hence als o i n K, an d we have K(\^2) = K, henc e c is a sum of 4 squares in

K, an d we have (c ) -(-1) -(-1) = 0 . •

T H E O R E M 32.19. If k has property OQ, the same is true for every purely tran-

scendental extension K of k.

P R O O F . I t is enough t o prove thi s fo r K — k(T), wher e T i s an indeterminate.

Let c G K* hav e propertie s 32.10.1 an d 32.10.3, and let e G H3(K) b e the elemen t

(c) •( —1) •(—1). B y the lemma above , al l the residues o f e are 0. Thi s mean s tha t e

is unramifie d i n the sense o f 9.4, henc e constant . T o prove tha t i t is 0, it is enough

to sho w tha t th e value o f e at a suitabl e v o f degree 1 is 0. T o do so, choose v

with k{v) = k an d v(c) = 0 . (Suc h a v exist s sinc e w e may assume b y 32.15 that

the characteristi c o f k is 0, hence tha t k is infinite.) I f c' denotes th e image o f c in

k(v)* = h*, the value of e at v is (cf) -(—1) '( — 1), wher e th e cup product i s computed

in H(k). Sinc e (2 ) -(c) = 0 , we have (2)-(c' ) = 0 , and since (c)-(-1)-(-1) = 0 in

H3(K(V2)), w e hav e (c f) -(-1) •(-! ) = 0 in H3(k(V2)). Sinc e k ha s property Oe ,

these propertie s impl y tha t c 1 is a sum of 4 squares i n /c, hence th e value o f e at v

is 0 . •

COROLLARY 32.20 . A purely transcendental extension of a number field has

property OQ.

This follow s fro m 32.17 an d 32.19.

EXERCISE 32.21. Prove :

(1) I f 2, — 1, o r —2 is a square in k, then k has property OQ. (Not e tha t thi s show s

that ever y field of characteristic ^ 0 has property OQ.)

(2) Le t X2 • T/c— Z2 be the 2-adic cyclotomic character. I f k does not have propert y

OQ, the n % 2 is surjective. [Hint : Us e (1).]

EXERCISE 32.22 . Le t K b e a field which i s complete wit h respec t t o a discrete valu -

ation, wit h residu e field k of characteristic7 ^ 2. Sho w tha t i f k has property OQ, the n so

does K.

[Apply Lemm a 32.18 as in the proof o f Th. 32.19.]

EXERCISE 32.23 . Le t ^ 2 be a 2-Pfister for m an d let q be a quadratic for m o f rank 6

with w\(q) — 0 and W2(q) = ^2(^2)- Sho w that q is a trace for m i f and only if there exist s

c G k* such that :

(1) q*i(l,l)B(c)-q2;

(2) (2)-(c)=0 ;

(3) c is representable b y #2 ove r

[Use the same metho d a s for Th. 32.10, whic h is the special cas e 9 2 = (1,1,1,1).]

32.24. A N EXAMPL E O F A FIEL D WHIC H DOE S N O T HAV E P R O P E R T Y OQ. Le t

k b e the function field ove r Q of the affine quadri c

X\ + X\ + • • • + Xl + 7 = 0.