2 The Hilbert Symbol 303The subspaceHhas twom-dimensional totally isotropic subspacesU 1 ,U 1 ′such
that
H=U 1 +U 1 ′, U 1 ∩U 1 ′={ 0 }.EvidentlyV 1 :=V⊥+U 1 is a totally isotropic subspace ofV. In factV 1 is maximal,
since any isotropic vector inU 1 ′⊥V 0 is contained inU 1 ′. Thusm=indV−dimV⊥is
uniquely determined andHis uniquely determined up to an isometry. If also
V=V⊥⊥H′⊥V 0 ′,whereH′is the orthogonal sum ofmhyperbolic planes andV 0 ′is anisotropic then,
by Proposition 18,V 0 is isometric toV 0 ′.
Proposition 21 reduces the problem of equivalence for quadratic forms over an ar-
bitrary field to the case of anisotropic forms. As we will see, this can still be a difficult
problem, even for the field of rational numbers.
Two quadratic spaces V,V′ over the same field F maybesaidtobe
Witt-equivalent, in symbolsV≈V′, if their anisotropic componentsV 0 ,V 0 ′are iso-
metric. This is certainly an equivalence relation. The cancellation law makes it pos-
sible to define various algebraic operations on the setW(F)of all quadratic spaces
over the fieldF, with equality replaced by Witt-equivalence. If we define−Vto be the
quadratic space with the same underlying vector space asVbut with(v 1 ,v 2 )replaced
by−(v 1 ,v 2 ),then
V⊥(−V)≈{O}.If we define thesumof two quadratic spacesVandWto beV⊥W,then
V≈V′,W≈W′⇒V⊥W≈V′⊥W′.Similarly, if we define theproductofVandWto be the tensor productV⊗Wof the
underlying vector spaces with the quadratic space structure defined by
({v 1 ,w 1 },{v 2 ,w 2 })=(v 1 ,v 2 )(w 1 ,w 2 ),then
V≈V′,W≈W′⇒V⊗W≈V′⊗W′.It is readily seen that in this wayW(F)acquires the structure of a commutative ring,
theWitt ringof the fieldF.
2 TheHilbertSymbol
Again letFbe any field of characteristic=2andF×the multiplicative group of all
nonzero elements ofF.WedefinetheHilbert symbol(a,b)F,wherea,b∈F×,by
(a,b)F=1ifthereexistx,y∈Fsuch thatax^2 +by^2 = 1 ,
=−1otherwise.By Proposition 6,(a,b)F= 1 if and only if the ternary quadratic form aξ^2 +bη^2 −ζ^2
is isotropic.