298 VII The Arithmetic of Quadratic Forms
Proof By choosing an orthogonal basis forVwe are reduced to showing that ifα,β,
γ∈F×q, then there existξ,η∈Fqsuch thatαξ^2 +βη^2 =γ.Asξruns throughFq,
αξ^2 takes(q+ 1 )/ 2 = 1 +(q− 1 )/2 distinct values. Similarly, asηruns throughFq,
γ−βη^2 takes(q+ 1 )/2 distinct values. Since(q+ 1 )/ 2 +(q+ 1 )/ 2 >q,thereexist
ξ,η∈Fqfor whichαξ^2 andγ−βη^2 take the same value.
Proposition 12Any non-singular quadratic form f in n variables over a finite fieldFq
is equivalent overFqto the quadratic form
ξ 12 +···+ξn^2 − 1 +δξn^2 ,whereδ=detf is the determinant of f.
There are exactly two equivalence classes of non-singular quadratic forms in n
variables overFq, one consisting of those forms f whose determinantdetf is a square
inF×q, and the other those for whichdetf is not a square inF×q.
Proof Since the first statement of the proposition is trivial forn=1, we assume that
n>1 and it holds for all smaller values ofn. It follows from Lemma 11 thatfrepre-
sents 1 and hence, by the remark after the proof of Proposition 2,fis equivalent over
Fqto a quadratic formξ 12 +g(ξ 2 ,...,ξn).Sincefandghave the same determinant,
the first statement of the proposition now follows from the induction hypothesis.
SinceF×qcontains(q− 1 )/2 distinct squares, every element ofF×qis either a square
or a square times a fixed non-square. The second statement of the proposition now fol-
lows from the first.
We now return to quadratic spaces over an arbitrary field. A 2-dimensional quadratic
space is said to be ahyperbolic planeif it is non-singular and isotropic.
Proposition 13Fo r a 2 -dimensional quadratic space V , the following statements are
equivalent:
(i)V is a hyperbolic plane;
(ii)V has a basis u 1 ,u 2 such that(u 1 ,u 1 )=(u 2 ,u 2 )= 0 ,(u 1 ,u 2 )= 1 ;
(iii)V has a basisv 1 ,v 2 such that(v 1 ,v 1 )= 1 ,(v 2 ,v 2 )=− 1 ,(v 1 ,v 2 )= 0 ;
(iv)−detV is a square in F×.
Proof Suppose first that V is a hyperbolic plane and letu 1 beanyisotropic
vector inV.Ifvis any linearly independent vector, then(u 1 ,v)=0, sinceV is
non-singular. By replacingvby a scalar multiple we may assume that(u 1 ,v)=1. If
we putu 2 =v+αu 1 ,whereα=−(v,v)/2, then
(u 2 ,u 2 )=(v,v)+ 2 α= 0 ,(u 1 ,u 2 )=(u 1 ,v)= 1 ,andu 1 ,u 2 is a basis forV.
Ifu 1 ,u 2 are isotropic vectors inVsuch that(u 1 ,u 2 )=1, then the vectorsv 1 =
u 1 +u 2 /2andv 2 =u 1 −u 2 /2 satisfy (iii), and ifv 1 ,v 2 satisfy (iii) then detV=−1.
Finally, if (iv) holds thenVis certainly non-singular. Letw 1 ,w 2 be an orthogonal
basis forVand putδj=(wj,wj)(j= 1 , 2 ). By hypothesis,δ 1 δ 2 =−γ^2 ,where
γ∈F×.Sinceγw 1 +δ 1 w 2 is an isotropic vector, this proves that (iv) implies (i).