2 The Hilbert Symbol 311a 12 (u 1 ,u 1 )+···+a^2 p(up,up)=(u′ 1 ,u′ 1 )= 0 ,and each summand on the left is nonzero. If the sum of the first two terms is zero, then
p>2 and either the sum of the first and third terms is nonzero or the sum of the second
and third terms is nonzero. Hence we may suppose without loss of generality that
a 12 (u 1 ,u 1 )+a 22 (u 2 ,u 2 )= 0.If we put
v 1 =a 1 u 1 +a 2 u 2 ,v 2 =u 1 +bu 2 ,vj=uj for 3≤j≤n,whereb=−a 1 (u 1 ,u 1 )/a 2 (u 2 ,u 2 ),thenB 1 ={v 1 ,...,vn}is an orthogonal basis
andu′ 1 =v 1 +a 3 v 3 +···+apvp. Thusp(B 1 )<p(B). By replacingBbyB 1 and
repeating the procedure, we must arrive afters<nsteps at an orthogonal basisBsfor
whichp(Bs)=1. The induction hypothesis can now be applied toBsin the same way
as forB.
Proposition 33Let F be a Hilbert field. If the non-singular diagonal forms
a 1 ξ 12 +···+anξn^2 and b 1 ξ 12 +···+bnξn^2 are equivalent over F , then
∏
1 ≤j<k≤n(aj,ak)F=∏
1 ≤j<k≤n(bj,bk)F.Proof Suppose first thatn=2. Sincea 1 ξ 12 +a 2 ξ 22 representsb 1 ,ξ 12 +a 1 −^1 a 2 ξ 22 rep-
resentsa− 11 b 1 and hence(−a 1 −^1 a 2 ,a 1 −^1 b 1 )F =1. Thus(a 1 b 1 ,−a 1 a 2 b^21 )F=1and
hence(a 1 b 1 ,a 2 b 1 )F=1. But (Proposition 28 (ii)) the Hilbert symbol is multiplicative,
sinceFis a Hilbert field. It follows that(a 1 ,a 2 )F(b 1 ,a 1 a 2 b 1 )F=1. Since the deter-
minantsa 1 a 2 andb 1 b 2 are in the same square class, this implies(a 1 ,a 2 )F=(b 1 ,b 2 )F,
as we wished to prove.
∏Suppose now thatn >2. Since the Hilbert symbol is symmetric, the product
1 ≤j<k≤n(aj,ak)F is independent of the ordering ofa^1 ,...,an. It follows from
Lemma 32 that we may restrict attention to the case wherea 1 ξ 12 +a 2 ξ 22 is equiva-
lent tob 1 ξ 12 +b 2 ξ 22 andaj=bjfor allj>2. Then(a 1 ,a 2 )F=(b 1 ,b 2 )F,bywhat
we have already proved, and it is enough to show that
(a 1 ,c)F(a 2 ,c)F=(b 1 ,c)F(b 2 ,c)F for anyc∈F×.But this follows from the multiplicativity of the Hilbert symbol and the fact thata 1 a 2
andb 1 b 2 are in the same square class.
Proposition 33 shows that the Hasseinvariant is well-defined.Proposition 34Two non-singular quadratic forms in n variables, with coefficients
from a Hilbert field F of type (B), are equivalent over F if and only if they have the
same Hasse invariant and their determinants are in the same square class.
Proof Only the sufficiency of the conditions needs to be proved. Since this is trivial
forn=1, we suppose first thatn=2. It is enough to show that if
f=a(ξ 12 +dξ 22 ), g=b(η^21 +dη^22 ),