Number Theory: An Introduction to Mathematics

296 VII The Arithmetic of Quadratic Forms

Proof SinceVis non-singular, so also areUandW, and sinceVcontains an isotropic

vectorv′,thereexistu′∈U,w′∈W, not both zero, such that

(u′,u′)=−(w′,w′).

If this common value is nonzero, weare finished. Otherwise eitherU orW is

isotropic. Without loss of generality, supposeUis isotropic. SinceWis non-singular,

it contains a vectorwsuch that(w,w)=0, andUcontains a vectorusuch that

(u,u)=−(w,w), by Proposition 5. 

We now show that the totally isotropic subspaces of a quadratic space are impor-

tant for an understanding of its structure, even though they are themselves trivial as

quadratic spaces.

Proposition 8All maximal totally isotropic subspaces of a quadratic space have the

same dimension.

Proof LetU 1 be a maximal totally isotropic subspace of the quadratic spaceV.Then
U 1 ⊆U 1 ⊥andU 1 ⊥\U 1 contains no isotropic vector. SinceV⊥⊆U 1 ⊥, it follows that
V⊥⊆U 1 .IfV′is a subspace ofVsupplementary toV⊥,thenV′is non-singular
andU 1 =V⊥+U 1 ′,whereU 1 ′⊆V′.SinceU 1 ′is a maximal totally isotropic subspace
ofV′, this shows that it is sufficient to establish the result whenVitself is non-singular.
LetU 2 be another maximal totally isotropic subspace ofV.PutW=U 1 ∩U 2 and
letW 1 ,W 2 be subspaces supplementary toWinU 1 ,U 2 respectively. We are going to
show thatW 2 ∩W 1 ⊥={ 0 }.
Letv∈W 2 ∩W 1 ⊥.SinceW 2 ⊆U 2 ,vis isotropic andv∈U 2 ⊥⊆W⊥. Hence
v∈U 1 ⊥and actuallyv∈U 1 ,sincevis isotropic. SinceW 2 ⊆U 2 this impliesv∈W,
and sinceW∩W 2 ={ 0 }this impliesv=0.
It follows that dimW 2 +dimW 1 ⊥≤dimV.But,sinceVis now assumed non-
singular, dimW 1 =dimV−dimW 1 ⊥, by Proposition 4. Hence dimW 2 ≤dimW 1
and, for the same reason, dimW 1 ≤dimW 2. Thus dimW 2 =dimW 1 , and hence
dimU 2 =dimU 1.

We define theindex,indV, of a quadratic spaceVto be the dimension of any

maximal totally isotropic subspace. ThusVis anisotropic if and only if indV=0.

AfieldFis said to beorderedif it contains a subsetPofpositiveelements, which

is closed under addition and multiplication, such thatFis the disjoint union of the sets

{ 0 },Pand−P={−x:x∈P}. The rational fieldQand the real fieldRare ordered

fields, with the usual interpretation of ‘positive’. For quadratic spaces over an ordered

field there are other useful notions of index.

A subspaceU of a quadratic spaceV over an ordered fieldF is said to be

positive definiteif(u,u)>0 for all nonzerou∈Uandnegative definiteif(u,u)< 0

for all nonzerou∈U. Evidently positive definite and negative definite subspaces are

anisotropic.

Proposition 9All maximal positive definite subspaces of a quadratic space V over an

ordered field F have the same dimension.