Graphics Reference
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p
Figure 10.12.
Defining the multiplicity of
intersections.
C
1
L
C
2
q
L
q
geometry. Roughly speaking, the theorem states that if two curves in the complex
plane have order m and n, respectively, then they have precisely mn points in the
intersection
if
the intersection points are counted with the appropriate multiplicity.
If m = 1, note the similarity between Bézout's theorem and the fundamental theorem
of algebra that says that a polynomial of degree n has n roots. The latter is also
only true if we have the correct notion of multiplicity of roots. Therefore part of our
problem is going to be to decide how to count. We defined a notion of multiplicity of
points in the intersection of a curve and a line but do not yet have a corresponding
definition when we are intersecting two nonlinear curves. Different approaches to
defining this notion of multiplicity exist. We shall follow the approach used in
[BriK81].
Consider two plane curves
C
1
and
C
2
in
P
2
(
C
) of order m and n, respectively. The
first assumption we make is that these curves have no common component (other-
wise the intersection would contain an infinite number of points). Let
p
be a point
which does not lie on either
C
1
or
C
2
and let
L
be any line that does not contain
p
.
For each point
q
on
L
let
L
q
be the line that contains
p
and
q
. See Figure 10.12. The
central projection of
P
2
(
C
) -
p
onto
L
maps all points of
L
q
-
p
to
q
. Our object will
now be to show two things. First, we show that only a finite number of lines through
p
contain intersection points of
C
1
and
C
2
. Second, we shall define intersection mul-
tiplicities using these lines.
We begin by choosing homogeneous coordinates (X,Y,Z) for
P
2
(
C
) so that
p
has
coordinates (0,0,1) and
L
is the line defined by the equation Z = 0. If we identify points
(X,Y,0) on
L
with the pair of coordinates (X,Y), then the central projection of
P
2
(
C
) -
p
onto
L
is given by
(
)
Æ
(
)
XYZ
,,
XY
, .
Next, let
C
1
and
C
2
be defined by homogeneous polynomials F = F(X,Y,Z) and G =
G(X,Y,Z) of degree m and n, respectively. Let us consider these polynomials in
C
[X,Y].
Then
m
m
-
1
FaZ a Z
=
+
+
.
..
+
a
(10.47a)
m
m
-
1
0
