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52 2. Plane Algebraic Curves

Suppose that P 1 ,... , P 6 are six intersection points which lie on a conic C.

Choose P on C and choose P off of C and the line L through P 7 and P 8 .Thenthe

cubic aD +a D +a D going through P , P , and the eight points P 1 ,... , P 8 has

C and thus L as components by (3.1). This contradicts the choice of P off of L and
C. Hence no six intersection points lie on a conic.
Now choose P and P both on the line L through P 1 and P 2 but not on the conic

C through P 3 , P 4 , P 5 , P 6 ,and P 7 . Then the cubic aD +a D +a D , going through

P , P , and the eight points P 1 ,... , P 8 ,has L and thus C as components by (3.1).
This contradicts the fact that P 8 is not on L or C by the analysis in the previous two

paragraphs. Thus we must have D = aD + a D for some point a : a ∈ P 1 (k),and

hence D will go through the ninth intersection point. This proves the theorem.
Exercises
2
1. If two curves of degree m intersect at exactly m points, and if nm of these points lie on
a curve of order n which is irreducible, then show that the remaining (m − n)m points lie
on a curve of degree m − n.
2. (Pascal’s theorem.) The pairs of opposite sides of a hexagon inscribed in an irreducible
conic meet in three points. Show these three points lie on a line.
3. (Pappus’ theorem.) Let P 1 , P 2 ,and P 3 be three points on a line L, and let Q 1 , Q 2 , and Q 3
be three points on a line M with none of these points on L ∩ M. Let L ij be a line through
P i and Q j , and let R k be the intersection point of L ij ∩L ji for {i, j, k}={1, 2, 3}. Show
that R 1 , R 2 ,and R 3 lie on a line.
§4. Multiple or Singular Points
In Theorem (3.1) we showed that two curves of degrees m and n, respectively, with-
out a common component have at most mn intersection points by observing that the
resultant between their equations was a homogeneous form of degree mn. Since the
linear factors of this resultant form of degree mn correspond to intersection points,
we must give a geometric interpretation of the repeated factors. Then we will be in
a position to state the basic intersection theorem of Bezout, namely that the num-
ber of intersections, counted with the appropriate multiplicities, is exactly mn.A
preliminary step is to consider the order of a hypersurface at a point.
(4.1) Definition. Let H f be a hypersurface of degree d, and let P be a point on
H f (k). We can choose affine coordinates such that P is the origin and the equation
of the hypersurface in affine coordinates becomes

0 = f (x 1 ,... , x n ) = f r (x 1 ,... , x n ) +· · · + f d (x 1 ,... , x n ),
where f i (x 1 ,... , x n ) is a homogeneous polynomial of degree i and the forms where
r ≤ d and f r and f d are nonzero. Then P is called a point of order r on H f ,and
f r (x 1 ,... , x n ) is called a leading form at P.