We switch now to the case of cubics:
Definition 7.11.1 By a cubic we mean an algebraic curve of degree 3.
The Bezout Theorem7.7.4tells us in particular that two curves of degree 3 with- out a common component have at most nine intersection points. Furthermore, this bound nine can easily be reached.
Example 7.11.2 Two cubics admitting exactly nine intersection points.
Proof Simply consider inR2the two curves with equations Y =X(X−1)(X+1), X=Y (Y−1)(Y+1)
pictured in Fig.7.16. They already have nine intersection points, thus these become intersection points of their projective extensions
Y Z2=X(X−Z)(X+Z), XZ2=Y (Y−Z)(Y +Z)
further viewed as curves inP2(C).
On the other hand Theorem7.8.2tells us that through nine points always passes a cubic. This bound of nine cannot possibly be improved.
312 7 Algebraic Curves
Fig. 7.16
Counterexample 7.11.3 Ten points of P2(C)through which it is impossible to draw a cubic.
Proof We are looking for a cubicCpassing through the pointsA1, . . . , A10as pic- tured in Fig.7.17. The four pointsA1, A2, A3, A4are on the same lined, thus by the Bezout Theorem7.7.4, that linedmust be a component ofC. For the same reason, the linedthroughA1, A5, A6, A7must be a component ofC. But then the simple equation ofCis necessarily the product of three linear factors andCis the union of danddand a third lined. SinceA8andA9are not ondord, they are ond; thus dis the line throughA8andA9. But the last pointA10is not onC=d∪d∪d.
Putting together Example7.11.2and Counterexample7.11.3we get:
The Cramer paradox
Nine points do not suffice to characterize a cubic, but ten points are too many.
If we consider the case of a curve of degree 4, which is called a quartic, we have this time in Theorems7.7.4and7.8.2
n2=16, n(n+3)
2 =14.
Moreover adapting in a straightforward way Example 7.11.2 and Counterexam- ple7.11.3, we get:
The Cramer paradox
Sixteen points do not suffice to characterize a quartic, but fifteen points are too many.
Fig. 7.17
This paradox was first pointed out by Colin MacLaurin (1698–1746). To dissipate this paradox, Gabriel Cramer (1704–1752) created the theory of determinants.
So: how do we resolve this paradox? To understand what is happening, it suffices to go back to the easier case of conics. We have this time in Theorems7.7.4and7.8.2
n2=4, n(n+3)
2 =5
Four points do not suffice to characterize a conic, and through five points one can always draw a conic.
In this case, the paradox does not appear. Furthermore, Proposition7.10.6provides additional precision, which we could rephrase intuitively as:
Five points “in general position” determine a unique conic.
Indeed, the precise nature of the family of points is certainly needed to characterize a unique conic: even a million points are not sufficient to characterize a conic, if you put all these points on the same line!
Thus clearly, if we intend to improve Theorem7.8.2in order to obtain the exis- tence of a unique curve of degreenpassing through the given points, it is necessary to give a precise meaning to the expression “a family of points in general position”.
Notice that Warning7.8.3tells us at once the answer to this question: the system of equations considered in the proof of Theorem7.8.2must be of rank n(n2+3).
With these observations and this terminology in mind, we can rephrase Cramer’s paradox as:
Nine points do not suffice to characterize a cubic, but nine points “in general position” do characterize a unique cubic.
This time we are no longer amazed. The explanation is most probably that in the first part of this “paradox”, the nine points are not “in general position”. Notice that indeed, in Example7.11.2used to illustrate the first line of the paradox, the nine
314 7 Algebraic Curves points are by no means an “arbitrary family of points”, but a family of points ob- tained as the intersection of two cubics. As the following theorem indicates, passing through all these nine points is a redundant condition.
Theorem 7.11.4 If two cubics intersect in exactly nine points, every cubic passing through eight of these points necessarily passes through the ninth point.
Proof Let us writeP (X, Y, Z)=0 andQ(X, Y, Z)=0 for the simple equations of two cubicsCandCintersecting in the nine pointsA1, . . . , A9. Consider further a cubicCwith simple equationH (X, Y, Z)=0 passing throughA1, . . . , A8. We shall prove by reductio ad absurdum thatCalso passes throughA9.
Suppose thatC does not contain A9. Consider all the cubics C admitting an equation of the form
α P (X, Y, Z)+β Q(X, Y, Z)+γ H (X, Y, Z)=0.
Among these we haveC andC whose equations are not proportional, since they have only nine intersection points. ButCis also a cubic of this form and its equation is not a linear combination of the equations ofCandC, sinceA9is onCandC, but not onC. Thus identifying a homogeneous equation of degree 3 inX,Y,Z with its list of coefficients, we conclude that all the equations above constitute a vector space of dimension at least 3. Imposing that a cubic of this formCpasses through two specified additional points is equivalent to imposing two linear equations on the coefficients ofC, and since we are in a vector space of dimension 3, there is always a non-zero solution, thus there always exists a cubicC passing throughP1, . . . , P8 and two arbitrarily specified pointsBandC.
We observe next that four of the pointsA1, . . . , A9 can never be on the same lined, otherwise by the Bezout Theorem7.7.4, the lined would be a component of bothCandC, which would then have infinitely many intersection points. In an analogous way, seven of the pointsA1, . . . , A9 can never be on a conic, because again this conic would be a component of bothCandC.
Let us now split the problem into three cases, which we shall treat separately:
1. three of the pointsA1, . . . , A9are on the same line;
2. six of the pointsA1, . . . , A9are on the same conic;
3. the first two possibilities never occur.
Thus in each case we aim to reach a contradiction.
First, assume thatA1, A2, A3are on the same lined. By Proposition7.10.4, we consider a conicQpassing throughA4, A5, A6, A7, A8. Such a conic is necessarily unique. Indeed ifQis another such conic, once more by the Bezout Theorem7.7.4, QandQhave five common points, thus a common component. Therefore
Q=d0∪d1, Q=d0∪d2
for three linesd0, d1, d2. Notice that outsided0, the only possibility for one of the five pointsA4, . . . , A8to be on bothQandQis to be at the intersection ofd1and
d2. This means that all the other points—thus at least four—are ond0. But as we have seen earlier in this proof, it is impossible to have four of the pointsA1, . . . , A9 on the same line. Thus indeed, the conicQas above is unique.
Let us now choose a pointBon the lined, distinct fromA1, A2, A3, and a point Cwhich is not on the linedor on the conicQ. Choose a cubicCas above which con- tains these two additional pointsB andC. SinceCcontainsBand alsoA1, A2, A3
it intersects the lined at four distinct points, thus by the Bezout Theorem 7.7.4, the lined must be a component of the cubicC. ThusC is the union of the lined and a conicQ. SinceA4, . . . , A8 andC are not on d, they must be onQ. But sinceQcontainsA4, . . . , A8, by uniqueness ofQ, we must haveQ=Q. This is a contradiction, sinceC∈Q.
Second, let us now assume thatA1, . . . , A6 are on a conicQ and let us write d for the line throughA7andA8. Choose now forB a point onQdistinct from A1, . . . , A6and forC, a point which is not onQor ond. We consider again a cubic Cas above containing these two additional pointsBandC. The conicQthen has at least seven intersection points with the cubicC, thus by the Bezout Theorem7.7.4, Qis a component ofC. The second component ofCmust then be a line containing the remaining pointsA7,A8,C, which is impossible sinceC is not on the lined throughA7andA8.
It remains to consider the last case: three of the pointsA1, . . . , A9 are never on the same line and six of these points are never on the same conic. This time choosed to be the line throughA1andA2. Choose further the pointsB,C ond, distinct fromA1,A2. The conicCas above, containing the pointsBandC, has four intersection points withd; thusdis a component ofCby the Bezout Theorem7.7.4.
By assumption of this third case, no other pointA3, . . . , A8can be ond. The other component ofCmust therefore be a conic containing the six pointsA3, . . . , A8. But again, this contradicts the assumption of this third case.
As corollary we get at once:
Corollary 7.11.5 If two cubics intersect in exactly nine points and if exactly six of these points are on a conicQ, the remaining three points are on a lined.
Proof Let A1, . . . , A9 be the nine intersection points, with A1, . . . , A6 on the conic Q. Writed for the line throughP7 andP8. ThenQ∪d is a cubic which, by Theorem7.11.4, passes through the pointP9. By assumption onQ,P9cannot be
onQ, thus it is ond.
Corollary 7.11.6 (Pascal’s Theorem) InP2(C), Pascal’s Theorem6.19.2is a spe- cial case of Corollary7.11.5and holds even without the assumption of the conic being irreducible.
Proof In Fig.6.15, choose the six pointsA, B, C, D, E, F on an arbitrary conicQ.
Define
• C=dAB∪dCD∪dEF;
316 7 Algebraic Curves
• C=dBC∪dDE∪dF A,
whose nine intersection points are the six given points and the pointsX,Y,Z. By Corollary7.11.5, these last three points are on a line.
Corollary 7.11.7 (Pappus’ Theorem) InP2(C), Pappus’ Theorem6.9.1is a special case of Corollary7.11.5.
Proof In Fig.6.6, the six pointsA,B,C,A,B,Care on the conicd∪d. Define
• C=dAB∪dBC∪dAC;
• C=dBA∪dBC∪dAC,
whose nine intersection points are the six given points and the pointsL,M,N. By Corollary7.11.5, these last three points are on a line since the first six points are on
the conicd∪d.