Algebraic Geometry - Computer Graphics and Geometric Modeling - page 774

Proof. See [Walk50].

Next, we describe another way of looking at rational functions.

Definition. Let K be an extension field of the complex numbers. Let f ΠC [X,Y] and
assume that f(X,Y) = 0 is the minimal equation for a curve C Õ K 2 . A point (a,b) Œ K 2
is called a generic point for C if g ΠC [X,Y] and g(a,b) = 0 implies that g(x,y) = 0 for
all points (x,y) ΠK 2 on the curve C .

The parameterizations discussed in the last section provide the context we have
in mind for generic points. It follows that if (a,b) is a generic point for the curve C ,
then g(a,b) = 0 implies that f divides g. Also, the next theorem shows that generic
points are only meaningful in the case of irreducible curves.

10.13.23. Theorem. Let C be a curve as in the definition of a generic point.

(1) If C is reducible, then it does not have any generic points.
(2) If C is irreducible, then every point (a,b) ΠC with not both a and b complex
numbers is a generic point. In particular, C has an infinite number of generic
points.

Proof. See [Seid68].

Definition. Let K be an extension field of k. Two points (x 1 ,y 1 ) and (x 2 ,y 2 ) in K 2 are
said to be isomorphic is there is an isomorphism s :k[x 1 ,y 1 ] Æ k[x 2 ,y 2 ] over k that sends
x 1 to x 2 and y 1 to y 2 .

10.13.24. Theorem. Any two generic points (x 1 ,y 1 ) and (x 2 ,y 2 ) for a curve C as above

are isomorphic.

Proof. See [Seid68].

One can now show ([Seid68]) that if (x,y) is a generic point for an irreducible
curve, then an alternate definition for the field of rational functions on the curve
is to say that it is the field C (x,y). That field is well defined, up to isomorphism
over the complex numbers C , by Theorem 10.13.24. The field C (x,y) will be a
finite extension of C with transcendency degree 1. This fits in with Theorem
10.13.22.
Here are some more useful observations about birational maps on affine varieties.

10.13.25. Proposition. If j : C Æ D is a birational map between irreducible affine
plane curves C and D , then j -1 ( q ) is finite for every q ΠD .

Proof. This is an easy consequence of Theorem 10.5.6. Let y : D Æ C be the inverse
of j. Let U Ã C be the set of points at which j is defined. Let V Ã D be the set of
points at which y is defined. Both C - U and D - V are finite sets. It follows that the
complement of j -1 ( V ) « U in C and the complement of y -1 ( U ) « V in D is finite and
j is a bijection between j -1 ( V ) « U and y -1 ( U ) « V .