Number Theory: An Introduction to Mathematics

314 VII The Arithmetic of Quadratic Forms
∏

v

(− 1 ,p)v=(− 1 ,p)p(− 1 ,p) 2 =(− 1 /p)(− 1 )(p−^1 )/^2 ;

∏

v

( 2 ,p)v=( 2 ,p)p( 2 ,p) 2 =( 2 /p)(− 1 )(p

(^2) − 1 )/ 8
;
∏
v
(p,q)v=(p,q)p(p,q)q(p,q) 2 =(q/p)(p/q)(− 1 )(p−^1 )(q−^1 )/^4.
Hence the proposition holds if and only if
(− 1 /p)=(− 1 )(p−^1 )/^2 ,( 2 /p)=(− 1 )(p
(^2) − 1 )/ 8
,(q/p)(p/q)=(− 1 )(p−^1 )(q−^1 )/^4.
Thus it is actually equivalent to the law of quadratic reciprocity and its two
‘supplements’.

We are now ready to prove the Hasse–Minkowski theorem:
Theorem 36A non-singular quadratic form f(ξ 1 ,...,ξn)with rational coefficients
is isotropic inQif and only if it is isotropic in every completionQv.
Proof We may assume that the quadratic form is diagonal:
f=a 1 ξ 12 +···+anξn^2 ,
whereak∈Q×(k= 1 ,...,n). Moreover, by replacingξkbyrkξk, we may assume
that each coefficientakis a square-free integer.
The proof will be broken intothree parts, according asn=2,n=3orn≥4. The
proofs forn=2andn=3 are quite independent. The more difficult proof forn≥ 4
uses induction onnand Dirichlet’s theorem on primes in an arithmetic progression.
(i)n=2: We show first that ifa∈Q×is a square inQ×v for allv,thenais already
a square inQ×.Sinceais a square inQ×∞,wehavea>0. Leta=

∏

pp

αpbe the

factorization ofainto powers of distinct primes, whereαp∈Zandαp=0foratmost
finitely many primesp.Since|a|p=p−αpandais a square inQp,αpmust be even.
But ifαp= 2 βpthena=b^2 ,whereb=

∏

pp
βp.
Suppose now thatf=a 1 ξ 12 +a 2 ξ 22 is isotropic inQvfor allv.Thena:=−a 1 a 2
is a square inQvfor allvand hence, by what we have just proved,ais a square inQ.
But ifa=b^2 ,thena 1 a 22 +a 2 b^2 =0 and thusfis isotropic inQ.

(ii)n=3: By replacingfby−a 3 fandξ 3 bya 3 ξ 3 , we see that it is sufficient to prove
the theorem for

f=aξ^2 +bη^2 −ζ^2 ,

whereaandbare nonzero square-free integers. The quadratic formfis isotropic inQv
if and only if(a,b)v=1. Ifa=1orb=1, thenfis certainly isotropic inQ.Since
fis not isotropic inQ∞ifa=b=−1, this proves the result if|ab|=1. We will as-
sume that the result does not hold for some paira,band derive a contradiction. Choose
apaira,bfor which the result does not hold and for which|ab|has its minimum value.
Thena=1,b=1and|ab|≥2. Without loss of generality we may assume|a|≤|b|,
and then|b|≥2.