Number Theory: An Introduction to Mathematics

2 The Hilbert Symbol 309

witha,b,c ∈ F×. Evidentlyaandcare not squares, and ifdis represented by
ξ 32 −cξ 42 ,thenbdis not represented byξ 12 −aξ 22. Thus(c,d)F=1 implies(a,bd)F=
−1. In particular,(a,b)F =−1 and hence(c,d)F = 1 implies(a,d)F = 1.
By interchanging the roles ofξ 12 −aξ 22 andξ 32 −cξ 42 , we see that(a,d)F=1also
implies(c,d)F=1. Hence(ac,d)F=1foralld∈F×. Thusacis a square andgis
equivalent to

fa,b=ξ 12 −aξ 22 −b(ξ 32 −aξ 42 ).

We now show that fa,band fa′,b′are equivalent if(a,b)F =(a′,b′)F =−1.
Suppose first that(a,b′)F=−1. Then(a,bb′)F=1 and there existx 3 ,x 4 ∈Fsuch
thatb′=b(x 32 −ax 42 ).Since

(x 32 −ax 42 )(ξ 32 −aξ 42 )=η^23 −aη^24 ,

whereη 3 =x 3 ξ 3 +ax 4 ξ 4 ,η 4 =x 4 ξ 3 +x 3 ξ 4 , it follows thatfa,b′is equivalent tofa,b.
For the same reasonfa,b′is equivalent tofa′,b′and thusfa,bis equivalent tofa′,b′.By
symmetry, the same conclusion holds if(a′,b)F=−1. Thus we now suppose

(a,b′)F=(a′,b)F= 1.

But then(a,bb′)F=(a′,bb′)F=−1 and so, by what we have already proved,

fa,b∼fa,bb′∼fa′,bb′∼fa′,b′.

Together, the last two paragraphs show that if F is a Hilbert field, then all
anisotropic quaternary quadratic forms which represent 1 are equivalent. Hence the
Hilbert fieldFis of type (B) if every anisotropic quaternary quadratic form repre-
sents 1.
Suppose now that some anisotropic quaternary quadratic form does not represent 1.
Then some scalar multiple of this form represents 1, but is not universal. Thusfa,bis
not universal for somea,b∈F×with(a,b)F=−1. By Lemma 29, the setGa,bof
allc∈F×which are represented byfa,bis a subgroup ofF×. In factGa,b=Ga,
sinceGa⊆Ga,b,Ga,b=F×andGahas index 2 inF×.Sincefa,b∼fb,a,wehave
alsoGa,b=Gb. Thus(a,c)F=(b,c)Ffor allc∈F×, and hence(ab,c)F=1for
allc∈F×. Thus ab is a square and(a,a)F=(a,b)F=−1. Since(a,−a)F=1, it
follows that(a,− 1 )F=−1. Hencefa,b∼fa,a∼fa,− 1. Replacinga,bby− 1 ,awe
now obtain(− 1 ,− 1 )F=−1andfa,− 1 ∼f− 1 ,− 1.
Thus the form

f=ξ 12 +ξ 22 +ξ 32 +ξ 42

is not universal and the subgroupPof all elements ofF×represented byfcoincides
with the set of all elements ofF×represented byξ^2 +η^2. HenceP+P⊆PandP
is the set of allc∈F×such that(− 1 ,c)F=1. Consequently− 1 ∈/PandFis the
disjoint union of the sets{O},Pand−P. ThusFis an ordered field withPas the set
of positive elements.
For anyc∈ F×,c^2 ∈ P. It follows that ifa,b∈ Pthen(−a,−b)F =−1,
sinceaξ^2 +bη^2 does not represent−1. Hence it follows that, ifa,b ∈ P,