Graphics Reference
In-Depth Information
where m
i
is the multiplicity of the ith singularity and we sum over all singularities of
C
. It turns out that the difference between the maximum number of double points of
a curve and actual number of double points defines an important invariant.
Definition.
Let
C
be a plane curve in
C
2
of order n defined by f(X,Y) = 0 that has
only ordinary singularities. The
genus
g of
C
is defined by
Â
(
)
(
)
-
(
)
gn n
=-
12
-
mm
-
,
i
i
i
where m
i
is the multiplicity of the ith singularity and we sum over all singularities.
For a general definition of the genus of a curve, even if it has nonregular singu-
larities, see, for example, [Walk50]. Note that inequality (10.90) implies that the genus
is a nonnegative integer.
10.15.4. Theorem.
The genus of an irreducible curve is a birational invariant.
Proof.
See [Walk50].
Theorem 10.15.4 has lots of consequences. For example, any nonsingular cubic
curve in
P
2
is not rational because it has genus 1 and the genus of
P
1
is 0. One can
also conclude the following (see [Walk50] for details):
(1) There are an infinite number of birationally nonequivalent curves. One way
to show this is to consider the curves
C
m
defined by
2
()
=
(10.91)
YFX
-
0
,
where F(X) is any polynomial of odd degree n = 2m + 1 with no multiple roots.
The curve
C
m
is nonsingular. The nonsingular projective version of
C
m
is called
a
hyperelliptic curve
. One can show that the hyperelliptic curve has genus m. See
[Shaf94].
(2) No nonsingular irreducible plane curves are equivalent if they have different
orders, the only exception being the case where one is a line and the other a conic.
In general, curves with the same genus need not be birationally equivalent. There
is one case where they are, however.
10.15.5. Theorem.
(Noether's Theorem) A plane curve admits a rational parame-
terization (and is birationally equivalent to a line) if and only if it has genus 0.
Proof.
See [Walk50], [Hoff89], [Abhy90], [Harr92], or [Shaf94].
It follows that any nonsingular cubic curve in
P
2
is not rational because it has
genus 1.