306 VII The Arithmetic of Quadratic Forms
wherea 0 ,b 0 ∈{ 3 , 5 , 7 }anda′,b′∈Z 2. Hence
1 =ax^2 +by^2 = 22 α(a 0 +b 0 y 02 + 8 z′),wherez′∈Z 2 .Sincea 0 ,b 0 are odd andy^20 ≡ 0 ,1or4mod8,wemusthaveα=0,
y^20 ≡4 mod 8 anda 0 =5. Thus, by Proposition VI.16 again,a−4 is a square inQ 2.
This proves that the condition in (i) is necessary.
Ifais a square inQ 2 , then certainly(a,b) 2 =1. Ifa−4 is a square, then
a= 5 + 8 a′,wherea′∈Z 2 ,anda+ 4 b= 1 + 8 c′,wherec′∈Z 2. Hencea+ 4 b
is a square inQ 2 and the quadratic formaξ^2 +bη^2 represents 1. This proves that the
condition in (i) is sufficient.
Suppose next that there existx,y∈Q 2 such thatax^2 + 2 by^2 =1. By the same
argument as for oddpin Proposition 25, we must have|x| 2 =1,|y| 2 ≤1. Thus
x=x 0 + 4 x′,y=y 0 + 4 y′,wherex 0 ∈{ 1 , 3 },y 0 ∈{ 0 , 1 , 2 , 3 }andx′,y′∈Z 2.
Writinga=a 0 + 8 a′,b=b 0 + 8 b′,wherea 0 ,b 0 ∈{ 1 , 3 , 5 , 7 }anda′,b′∈Z 2 ,we
obtaina 0 x 02 + 2 b 0 y^20 ≡1 mod 8. Since 2y^20 ≡0 or 2 mod 8, this implies eithera 0 ≡ 1
ora 0 + 2 b 0 ≡1 mod 8. Hence eitheraora+ 2 bis a square inQ 2. It is obvious that,
conversely,(a, 2 b) 2 =1 if eitheraora+ 2 bis a square inQ 2.
The special case whereaandbare integers again follows from Corollary VI.17.
ForF = R, the factor groupF×/F×^2 is of order 2, with 1 and−1asrep-
resentatives of the two square classes. ForF =Qp, withpodd, it follows from
Corollary VI.17 that the factor groupF×/F×^2 is of order 4. Moreover, ifr is
an integer such that(r/p) =−1, then 1,r,p,rpare representatives of the four
square classes. Similarly forF=Q 2 , the factor groupF×/F×^2 is of order 8 and
1 , 3 , 5 , 7 , 2 , 6 , 10 ,14 are representatives of the eight square classes. The Hilbert sym-
bol(a,b)Ffor these representatives, and hence for alla,b∈F×, may be determined
directly from Propositions 24, 25 and 27. The values obtained are listed in Table 1,
whereε=(− 1 /p)and thusε=±1 according asp≡±1mod4.
It will be observed that each of the three symmetric matrices in Table 1 is a
Hadamard matrix! In particular, in each row after the first row of+’s there are equally
many+and−signs. This property turns out to be of basic importance and prompts
the following definition:
AfieldFis aHilbert fieldif somea∈F×is not a square and if, for every sucha,
the subgroupGahas index 2 inF×.
Thus the real fieldR=Q∞and thep-adic fieldsQpare all Hilbert fields. We now
show that in Hilbert fields further properties of the Hilbert symbol may be derived.
Proposition 28A field F is a Hilbert field if and only if some a∈F×is not a square
and the Hilbert symbol has the following additional properties:
(i)if(a,b)F= 1 for every b∈F×, then a is a square in F×;
(ii)(a,bc)F=(a,b)F(a,c)Ffor all a,b,c∈F×.
Proof LetFbe a Hilbert field. Then (i) holds, sinceGa=F×ifais not a square.
If(a,b)F =1or(a,c)F =1, then (ii) follows from Proposition 23(v). Suppose
now that(a,b)F=−1and(a,c)F=−1. Thenais not a square and fadoes not
representborc.SinceFis a Hilbert field andb,c∈/Ga, it follows thatbc∈Ga. Thus
(a,bc)F=1. The converse is equally simple.