Algebraic Geometry - Computer Graphics and Geometric Modeling - page 739

(2) fi (3): This is a straightforward consequence of Theorem 10.10.5.
(3) fi (4): Obviously, <lt(P)>Õ<lt(I)>. We need to show the reverse inclusion. Let
g Œ<lt(I)>. Then g is a linear combination of leading terms of polynomials in I. But
if f ΠI, then (3) implies that

Â

() = ()() lt f lt a lt p ,

i i

i

where the sum is taken over all i with the property that

() = () ( . lpp f lpp a lpp p

i i

This shows that g is a linear combination of lt(p i ), that is, g Œ<lt(P)>.
(4) fi (1): Let f ΠI. By (4), lt(f) can be expressed in the form

m

Â

() = ()

lt f b lt p ,

i i

i = 1

where b i Πk[X 1 ,X 2 ,...,X n ]. Consideration of the right-hand side of this equation after
expanding the polynomials b i into their monomial parts shows that lpp(f) must be
divisible by lpp(p j ) for some j, which is what we had to prove.

10.10.9. Corollary. If P = {p 1 ,p 2 ,...,p m } is a Gröbner basis for an ideal I, then I =

<p 1 ,p 2 ,...,p m >.

Proof. Since each p i belongs to I, we clearly have <p 1 ,p 2 ,...,p m >ÕI. On the other
hand, if f Œ I, then Theorem 10.10.8(3) implies that f Œ<p 1 ,p 2 ,...,p m >.

We have listed some properties of Gröbner bases, but do they exist?

10.10.10. Theorem. Every nonzero ideal in k[X 1 ,X 2 ,...,X n ] has a Gröbner basis.

Proof. See [CoLO97] or [AdaL94]. One needs the Hilbert basis theorem for this.

The next question is how one can find a Gröbner basis. Consider an ideal I =<p 1 ,p 2 ,
...,p m > and let P = {p 1 ,p 2 ,...,p m }. What might cause P not to be a Gröbner basis for
I? First of all, although every element f in I is a linear combination of the p i , the leading
terms might cancel out leaving f with a leading term that is smaller than all of the
leading terms of the p i .

10.10.11. Example. Using the deglex order of k[X,Y] with X > Y, let

2 2

I =< p = X Y + X p , = XY + 1 > and f = Yp - Xp = XY - X ΠI .

1 2 1 2

The existence of such a polynomial f in I shows that the set P = {p 1 ,p 2 } does not form
a Gröbner basis.

The next observation strikes more at the heart of the problem with picking any
old basis for an ideal. The problem is that one usually has many choices when reduc-