We continue to adopt the
Convention In this section,Kis a field of characteristic distinct from 2 and equa- tion of a quadric always means equation of degree 2.
Let us conclude this chapter by proving that with respect to a given basis, a non-degenerate quadric can only have “one” equation (of course, up to a non-zero multiplicative constant). This section relies heavily on the theory of the resultant of two polynomials (see AppendixD).
Lemma 2.26.1 LetQbe a quadric in an affine space of dimensionn. The following conditions are equivalent:
1. two equations of this quadric with respect to a given basis are proportional;
2. these two equations, transformed via a change of basis, are proportional.
2.26 The Equation of a Non-degenerate Quadric 101 Proof Trivially if in one basis two equations take the form
p(X1, . . . , Xn)=0, kp(X1, . . . , Xn)=0, 0=k∈K
withpa polynomial, every change of coordinates respects this proportionality, since
it occurs “inside”p(X1, . . . , Xn).
Lemma 2.26.2 LetQbe a non-degenerate quadric in an affine space of dimen- sionn. Given an equation
p(X1, . . . , Xn)=0
of this quadric in some affine basis, one of the two following possibilities holds:
1. p(X1, . . . , Xn)is irreducible;
2. p(X1, . . . , Xn)is the product of two non-proportional factors of degree 1.
Proof Ifpis not irreducible, it factors uniquely as a product of irreducible factors (see TheoremB.4.9), which are thus necessarily of degree 1. If these factors were proportional, the equation of the quadric would take the form
k(a0+a1X1+ ã ã ã +anXn)2=0, 0=k∈K.
Thus the quadric would be the hyperplane with equation a0+a1X1+ ã ã ã +anXn=0.
This would contradict the non-degeneracy of the quadric.
Lemma 2.26.3 LetQbe a non-degenerate quadric in an affine space of dimen- sionn. If for some affine basis, an equation ofQdoes not contain the variableXn, thenXndoes not appear in any equation ofQwith respect to this same affine basis.
Proof The assumption implies that given a pointP of the quadric and an arbitrary scalark∈K, the corresponding pointPk
P=
⎛
⎜⎜
⎜⎝ k1
... kn−1
kn
⎞
⎟⎟
⎟⎠ Pk=
⎛
⎜⎜
⎜⎝ k1
... kn−1
k
⎞
⎟⎟
⎟⎠
is still on the quadric.
Consider an arbitrary equation n i,j=1
aijXiXj+ n
i=1
biXi+c=0
of the quadric with respect to the given affine basis. GivenP∈Q, the pointsP1and P−1are still on the quadric. This means that each point of the quadric satisfies the equations
n−1
i,j=1
aijXiXj+2
n−1
i=1
ainXi+ann+
n−1
i=1
biXi+bn+c=0
n−1
i,j=1
aijXiXj−2
n−1
i=1
ainXi+ann+
n−1
i=1
biXi−bn+c=0.
Subtracting these two equations, we conclude that every point of the quadric satisfies the equation
2
n−1
i=1
ainXi+bn=0.
If one coefficientain is non-zero, this is the equation of a hyperplane in which the quadric is thus contained. This contradicts the non-degeneracy of the quadric. Thus all coefficientsain are equal to zero. As a consequence, since the quadric is not empty,bn=0 as well. Thus the equation of the quadric has the form
n−1
i,j=1
aijXiXj+annXn2+
n−1
i=1
biXi+c=0.
Let us next express the fact that given a pointP of the quadric, the two pointsP1
andP0are still on the quadric. This means that every point of the quadric satisfies the equations
n−1
i,j=1
aijXiXj+ann+
n−1
i=1
biXi+c=0
n−1
i,j=1
aijXiXj+
n−1
i=1
biXi+c=0.
Subtracting, we conclude that every point of the quadric satisfies the equationann= 0, or in other words, since the quadric is non-empty,ann=0.
Now comes the key step for proving the uniqueness of the equation of a non- degenerate quadric:
Lemma 2.26.4 LetQbe a non-degenerate quadric in an affine space(E, V )of dimensionn. Consider another quadric Q⊆QE. Assume the existence of an affine basis with respect to which the quadricQ admits an equation of the form
2.26 The Equation of a Non-degenerate Quadric 103 p(X1, . . . , Xn)=0, where one of the coordinates appears with exponent 1 but not with exponent 2. ThenQ=Qand two equations of this quadric with respect to the same basis are necessarily proportional.
Proof By Lemma 2.26.1, it suffices to prove the result for the basis where the quadric Q admits the equation p(X1, . . . , Xn)=0 as in the statement. Let q(X1, . . . , Xn)=0 be an equation of the quadricQwith respect to this same basis.
Notice first thatQis a non-degenerate quadric. Indeed it is not the whole space Eby assumption and it is not contained in a hyperplane, since this is already the case forQ. Thus the equations ofQandQare actual equations of degree 2.
By the assumptionQ⊆Q,
p(a1, . . . , an)=0 =⇒ q(a1, . . . , an)=0.
Ifp(X1, . . . , Xn)is irreducible, by PropositionD.2.2,pdividesq. Since both have degree 2, this implies the proportionality of p and q. This forces in particular Q=Q.
Ifp(X1, . . . , Xn)is not irreducible, by Lemma2.26.2, it is the product of two non-proportional factors of degree 1.
p(X1, . . . , Xn)=r(X0, . . . , Xn)s(X0, . . . , Xn).
Ifr(a1, . . . , an)=0, thenp(a1, . . . , an)=0 and thusq(a1, . . . , an)=0. By Corol- laryD.2.3,r(X1, . . . , Xn)is then a factor ofq(X1, . . . , Xn). An analogous argument holds fors(X1, . . . , Xn). Since these two factors are not proportional and the decom- position ofq(X1, . . . , Xn)into irreducible factors is unique (see TheoremB.4.9), it follows that
q(X1, . . . , Xn)=kr(X1, . . . , Xn)s(X1, . . . , Xn)=kp(X1, . . . , Xn).
Again the proportionality ofpandqforcesQ=Q.
Let us recall that for a non-degenerate quadric, the type of a reduced equation is fixed (see Theorems2.25.5and2.24.2).
Proposition 2.26.5 LetQbe a non-degenerate quadric of type 3 in an affine space of dimensionn. Two equations of this quadric with respect to the same basis are necessarily proportional.
Proof By Lemma2.26.4, choosingQ=Q.
Perhaps more surprising is that the quadrics of type 2 also enter into the context of Lemma2.26.4.
Proposition 2.26.6 LetQbe a non-degenerate quadric of type 2 in an affine space of dimensionn. Two equations of this quadric with respect to the same basis are necessarily proportional.
Proof Choose as origin a centerP0of symmetry, which is thus a point of the quadric (see Theorem2.25.5). Since the quadric is non-degenerate, there arenother points P1, . . . , Pnsuch thatP0, P1, . . . , Pnare affinely independent, that is,
(P0;e1=−−→
P0P1, . . . , en=−−→
P0Pn)
is an affine basis. With respect to this basis, an equation takes the form n
i,j=1
aijXiXj=0
because the origin is a center of symmetry belonging to the quadric (see Proposi- tion2.25.3). Introducing the coordinates of the vectorei into this equation, we get aii=0. Thus finally the equation takes the form
i=j
aijXiXj=0
where all appearing variablesXi have the exponent 1. The result again follows by
Lemma2.26.4, choosingQ=Q.
It remains to handle the case of quadrics of type 1. In this case, it is generally impossible to find an affine basis such that the equation contains one of the variables with only the exponent 1 (see Problem2.27.14). But nevertheless, this case can once more be reduced to the situation in Lemma2.26.4.
Lemma 2.26.7 LetQbe a non-degenerate quadric of type 1 in an affine space of dimensionn. Suppose that with respect to some affine basis,Qadmits a reduced equation of the form
n i=1
aiX2i =1 where the quadratic form
φ (X1, . . . , Xn)= n
i=1
aiXi2
does not admit any non-zero isotropic vectors (see DefinitionG.2.1). Then two equa- tions ofQwith respect to the same affine basis are necessarily proportional.
Proof By Lemma2.26.1, it suffices to consider the case of a second equation ofQ with respect to the same affine basis. By Proposition2.25.3, the origin is a center of symmetry thus this second equation takes the form
n i,j=1
bijXiXj=1.
2.26 The Equation of a Non-degenerate Quadric 105 Let us then consider the following two quadricsQ1andQ2inKn+1
n i=1
aiXi2=Xn2+1,
n i,j=1
bijXiXj=X2n+1. Notice that
P+=
⎛
⎜⎜
⎜⎝ k1
... kn
1
⎞
⎟⎟
⎟⎠∈Q1 ⇐⇒ P =
⎛
⎜⎝ k1
... kn
⎞
⎟⎠∈Q ⇐⇒ P+=
⎛
⎜⎜
⎜⎝ k1
... kn
1
⎞
⎟⎟
⎟⎠∈Q2.
Up to multiplication by an arbitrary scalark, these are thus all the points ofQ1and Q2admitting a non-zero last component.
Let us next observe thatQ1does not have any other point, that is, a point with coordinates
⎛
⎜⎜
⎜⎝ k1
... kn
0
⎞
⎟⎟
⎟⎠=
⎛
⎜⎜
⎜⎝ 0
... 0 0
⎞
⎟⎟
⎟⎠.
Indeed having such a point onQ1would mean φ (k1, . . . , kn)=
n i=1
aik2i =0.
This cannot exist because by assumption, the form φ does not admit a non-zero isotropic vector.
The arguments above thus prove already that Q1⊆Q2. They also prove that these quadrics are not the whole spaceKn+1: given a point not onQ, it suffices to add a last coordinate 1 to get a point which is neither onQ1nor onQ2.
Our next concern is to observe thatQ1andQ2are non-degenerate. OnQ, there existn+1 pointsPi which are affinely independent. This implies the non-existence of a first degree equation
c1X1+ ã ã ã +cnXn+cn+1=0
satisfied by the coordinates of all the pointsPi. This can be rephrased as the non- existence of a homogeneous first degree equation
c1X1+ ã ã ã +cnXn+cn+1Xn+1=0
satisfied by the coordinates of all the pointsPi+. In other words the vectorsPi+∈ Kn+1are not contained in any vector hyperplane and are therefore linearly indepen- dent. Together with the origin ofKn+1, this yieldsn+2 points, affinely independent, belonging to bothQ1andQ2. So indeed,Q1andQ2are non-degenerate.
Since Q1 is non-degenerate of type 2, as observed in the proof of Proposi- tion2.26.6, there is a basis with respect to which the equation no longer contains a variable with exponent 2. We can therefore apply Proposition2.26.4toQ1andQ2: they are equal and their equations are proportional. This also means that the two
equations ofQare proportional.
Proposition 2.26.8 LetQbe a non-degenerate quadric of type 1 in an affine space of dimensionn. Two equations ofQwith respect to the same basis are necessarily proportional.
Proof The proof shall be by induction on the dimensionnof the space.
In dimension 1, two equations of the quadric with respect to a basis whose origin is a center of symmetry therefore have the forms (see Proposition2.25.3)
aX2=1, bX2=1
with of coursea andbnon-zero. Fixing a pointP of coordinatek on the quadric, we thus have
ak2=1=bk2. This forcesk=0 and thereforea=b.
Let us now assume that the result is valid in dimensionn−1 and let us prove it in dimensionn. Let us fix a basis with respect to which the quadric admits a reduced equation
n i=1
aiX2i =1.
Let us write
φ (X1, . . . , Xn)= n
i=1
aiXi2
for the quadratic form in this equation. If the quadratic formφ does not have a non-zero isotropic vector, the result follows by Lemma2.26.7.
Ifφ admits a non-zero isotropic vector−→v, let us construct a new affine basis (O;e1, . . . , en).
• We choose the same originOas in the first basis; thusOis a center of symmetry not belonging to the quadric.
• Since the quadric is non-degenerate, we can findnpointsPi on the quadric such that the vectors−−→
OPiare linearly independent. Then the isotropic vector−→v can be written uniquely as a linear combination of these vectors−−→
OPi. Since−→v =0, at least one vector−−→
OPi0appears in this linear combination, that is−→v is not a linear combination of the vectors−−→
OPi,i=i0. Up to possibly renumbering the points,
2.26 The Equation of a Non-degenerate Quadric 107 there is no loss of generality in choosingi0=n. The vectors
e1=−−→
OP1, . . . , en−1=−−−−→
OPn−1, en= −→v
are then linearly independent and we choose them as vectors of the new affine basis.
With respect to this new basis, the reduced equation transforms as n
i,j=1
aijXiXj=1
again by Proposition2.25.3.
The left hand side of this equation is thus the matrix expression ofφwith respect to the new basis. Sinceen is an isotropic vector of φ, the left hand side takes the value 0 when we substitute the variablesXi by the coordinates of
en=
⎛
⎜⎜
⎜⎝ 0
... 0 1
⎞
⎟⎟
⎟⎠.
This reduces precisely toann=0. ThusXndoes not appear with an exponent 2.
IfXn appears with an exponent 1 in the equation, the result again follows by Lemma2.26.4, choosingQ=Q.
IfXndoes not appear in the equation, then by Lemma2.26.3, it does not appear in any equation ofQwith respect to the basis(O;e1, . . . , en). By Proposition2.25.3, a second equation of the quadric with respect to the same affine basis indicated thus has the form
n−1
i,j=1
bijXiXj=1.
Now view
n−1
i,j=1
aijXiXj=1,
n−1
i,j=1
bijXiXj=1
as two equations of the intersection of the quadric Q with the subspace of di- mensionn−1 admitting the affine basis (0;e1, . . . , en−1). This therefore yields a quadricQin this subspace of dimensionn−1. Let us prove that it remains non- degenerate.
By construction of the basis, we haveei=−−→
OPi, 1≤i≤n−1, with eachPi on the quadricQtherefore also on the quadricQ. But since the origin is a center of
symmetry, the pointPisymmetric to eachPiwith respect to the originOis also on the quadricQ, thus onQ. If the quadricQ is degenerate, it is contained in some hyperplane with equation
k0+k1X1+ ã ã ã +kn−1Xn−1=0
thus with at least oneki, 1≤i≤n−1 non-zero. Saying thatPi satisfies this equa- tion means
k0+ki=0
thuski= −k0. Saying thatPisatisfies this equation means k0−ki=0
thuski=k0. Thus eachki is zero, which is a contradiction. So indeed the quadric Q is non-degenerate and by the inductive assumption, we conclude that the two equations ofQ—which are also the two equations ofQ—are proportional.
Theorem 2.26.9 LetQbe a non-degenerate quadric in an affine space of dimen- sionn. Two equations ofQwith respect to the same basis are necessarily propor- tional.
Proof This follows by Theorem 2.24.2 and Propositions 2.26.5, 2.26.6 and
2.26.8.
Problem2.27.15presents an alternative answer to the question treated in Theo- rem2.26.9.