Number Theory: An Introduction to Mathematics

2 The Hilbert Symbol 307

Table 1.Values of the Hilbert symbol(a,b)FforF=Qv

Q∞=RQp:podd

a\b 1 − 1 a\b 1 prpr

1 ++ 1 ++++

− 1 +− p + ε −ε −

rp +−εε−

r +−−+

whereris a primitive root modpand

ε=(− 1 )(p−^1 )/^2

Q 2

a\b 136 2141057

1 ++++ ++ ++

3 +−+− +− +−

6 ++−− ++ −−

2 +−−+ +− −+

14 ++++ −− −−

10 +−+− −+ −+

5 ++−− −− ++

7 +−−+ −+ +−

The definition of a Hilbert field can be reformulated in terms of quadratic forms. If
fis an anisotropic binary quadratic form with determinantd,then−dis not a square
andfis equivalent to a diagonal forma(ξ^2 +dη^2 ). It follows thatFis a Hilbert field
if and only if there exists an anisotropic binary quadratic form and for each such form
there is, apart from equivalent forms, exactly one other whose determinant is in the
same square class. We are going to show that Hilbert fields can also be characterized
by means of quadratic forms in 4 variables.

Lemma 29Let F be an arbitrary field and a,belementsofF×with(a,b)F=− 1.
Then the quadratic form

fa,b=ξ 12 −aξ 22 −b(ξ 32 −aξ 42 )

is anisotropic. Morover, the set Ga,bof all elements of F×which are represented by
fa,bis a subgroup of F×.

Proof Since(a,b)F =−1,ais not a square and hence the binary form fais
anisotropic. Iffa,bwere isotropic, somec ∈F×would be represented by both fa
andbfa.Butthen(a,c)F=1and(a,bc)F=1. Since(a,b)F=−1, this contradicts
Proposition 23.
Clearly ifc∈Ga,b,thenalsoc−^1 ∈Ga,b, and it is easily verified that if

ζ 1 =ξ 1 η 1 +aξ 2 η 2 +bξ 3 η 3 −abξ 4 η 4 ,ζ 2 =ξ 1 η 2 +ξ 2 η 1 −bξ 3 η 4 +bξ 4 η 3 ,

ζ 3 =ξ 1 η 3 +ξ 3 η 1 +aξ 2 η 4 −aξ 4 η 2 ,ζ 4 =ξ 1 η 4 +ξ 4 η 1 +ξ 2 η 3 −ξ 3 η 2 ,