Algebraic Geometry - Computer Graphics and Geometric Modeling - page 784

which have the form

( )

q ¢= Y q , q ,..., q

j j 12 n

in the affine case. We get a unique curve for each basis of K.
All of this may sound good, but we still need to show that C P and C are not simply
sets but varieties, specifically, curves. Various facts have to be established first before
one can prove the following:

10.14.1. Theorem. If C is a variety, then C is a space curve if and only if C is an

irreducible curve.

Proof. The theorem is a consequence of the next two theorems.

We relate the space curve C in k n to an ideal. Define

=Π[

{ ]

( ) =

}

If kXX , ,..., Xf xx , ,..., x 0 .

12 n 12 n

10.14.2. Theorem. The ideal I in k[X 1 ,X 2 ,...,X n ] is a prime ideal and K is
isomorphic to the quotient field of k[X 1 ,X 2 ,...,X n ]/I.

Proof. It is trivial to check that I is an ideal. The rest follows easily by analyzing the

map

[ ] Æ [ ] kX X , ,..., X k xx , ,..., x .

12 n 12 n

X Æ x

i i

This map is onto and has kernel I.

Theorem 10.14.2 shows that I determines the curve C completely. One also has

10.14.3. Theorem.

(1) f ΠI if and only if f( p ) = 0 for all p ΠC .
(2) Let p Πk n . If f( p ) = 0 for all f ΠI, then p ΠC .

Proof. See [Walk50].

In the above we started with a curve and got an ideal that defined it. We can go
the other way and get a curve starting with an ideal.

10.14.4. Theorem. Let k be an algebraically closed field of characteristic 0. An ideal
I in k[X 1 ,X 2 ,...,X n ] is the ideal associated to an irreducible space curve if and only
if I is prime and its transcendence degree over k, that is, tr k (k(I)), is equal to 1.

Proof. See [Walk50].

It follows from Theorem 10.14.3 that C = V(I( C )). This fact and Theorem 10.14.4
show that we could have defined a space curve as being the set of zeros of certain
ideals. This algebraic approach to the definition of a space curve would certainly be
much cleaner than our messy construction for the points of such a curve.