Graphics Reference
In-Depth Information
()
=+ +
01 2
2
Xt
a
at a t
+
...,
...,
...,
()
=+ +
2
Yt
b
bt bt
+
0
1
2
()
=+ +
01 2
2
Zt
c
ct c t
+
and at least one of ord(X(t)), ord(Y(t)), or ord(Z(t)) is zero. It follows that
(
() () ()
)
=
(
)
π
(
)
cX YZ
=
0
,
0
,
0
abc
,
,
000
, ,
.
000
The point [c] in
P
2
(k) is called the
center
of the parameterization.
Definition.
10.12.1. Proposition.
The center of a parameterization for a plane curve is inde-
pendent of the coordinate system and lies on the curve.
Proof.
The proof of the first part is straightforward and left as an exercise. One must
show that related parameterizations define the same center. Although the proof of the
second part is not hard, the fact may seem more obvious than it actually is. The
problem is that we are dealing with infinite power series and evaluation of such has
to be handled carefully. Specifically, what we know is that
(
()
() ()
)
=
FXt Yt Zt
,
,
0
(10.77)
but all we know immediately is that F(X(t),Y(t),Z(t)) is a power series G(t). There is
no à priori relationship between G(0) and F(c). One way to prove that [c] lies on the
curve is to develop a theory of congruence first (something that is needed in Theorem
B.7.5 to prove that substitution into power series is alright). Equation (10.77) then
implies
(
()
() ()
)
∫
FXt Yt Zt
,
,
0
(mod ).
t
This equation and the fact that
()
∫
()(
)
Xt
X
0
0
0
mod
t
,
()
∫
()(
)
Yt
Y
mod
t
,
()
∫
()(
)
Zt
Z
mod
t
,
imply that F(c) = 0 (mod t), from which the result follows.
Next, given an arbitrary parameterization (X(t),Y(t),Z(t)) of
C
, let h(t) Πk[[t]],
where h π 0 and ord(h) > 0. Let
()
=
(
()
)
( )
=
(
()
)
( )
=
(
()
)
X
t
X h t
,
Y
t
Y h t
,
and
Z
t
Z h t
.
h
h
h
It is easy to show that
10.12.2. Proposition.
(X
h
(t),Y
h
(t),Z
h
(t)) is a parameterization with the same center
as that of (X(t),Y(t),Z(t)).
