Algebraic Geometry - Computer Graphics and Geometric Modeling - page 746

The next theorem answers the question about when extensions exist in one case.

10.11.3. Theorem. (The Extension Theorem) Assume that k is an algebraically
closed field. Let I =<f 1 ,f 2 ,...,f m >Õk[X 1 ,X 2 ,...,X n ] and let I 1 be the first elimination
ideal of I. For each f i , 1 £ i £ m, write f i in the form

N i

( )

fgX = ,..., XX + terms in which X has degree less than N ,

i i 2 n 1 i

1

where N i ≥ 0 and g i Œ k [X 2 ,...,X n ] is nonzero. Suppose that we have a partial solu-
tion (a 2 ,..., a n ) Œ V(I 1 ). If (a 2 ,..., a n ) œ V(g 1 ,g 2 ,...,g m ), then there exists an a 1 Œ k,
such that (a 1 ,a 2 ,..., a n ) ΠV(I).

Proof. See [CoLO97].

The reader should look at Examples 10.10.17 and 10.10.18 again to see how the
last two theorems justify the Gröbner basis approach to solving the problems in those
examples. A geometric interpretation of Theorem 10.11.3 can be found in [CoLO97].
Next, we address a question raised in the solution for Example 10.10.18 that had
to do with the nature of the implicit representation that we obtain for a set X from a
given parametric one using the Gröbner basis approach. Let V be the variety that one
obtains. The set X is presumably contained in V by construction, but since the set X
might not even be a variety, one should not expect V to be the same as X in general.
On the other hand, the variety V could easily be much larger than necessary. There-
fore, the implicitization problem should be stated more precisely as asking for the
smallest variety that contains X . The following questions need to be answered:

(1) Is the variety V obtained via the Gröbner basis approach this smallest variety?
(2) If V π X , then how does one determine those points of V not in X ?

The first of these questions is answered by the following theorem.

10.11.4. Theorem. (The Polynomial Implicitization Theorem) Let k be an infinite
field. Consider the parameterization defined in equation (10.55) and define F : k n Æ k m
by

( ) = ( ( ) ( ) ( ) ) F t , t ,..., t f t , t ,..., t , f t , t ,..., t ,..., f t , t ,..., t .

12 n 112 n 212 n m 12 n

Let I be the ideal

->Õ [

]

<- xfx f , - ,..., x f kt ,..., tx , ,... x

112 2 mm 1 n 1 m

and let

=« [

]

I I kxx , ,..., x

n 12 m

be the nth elimination ideal. Then V(I n ) is the smallest variety in k m containing F(k n ).

Proof. See [CoLO97].