Let be a field and fix a monomial order on .
- Let be an ideal of . Prove that is a minimal Gröbner basis for if and only if is a minimal generating set for .
- Prove that the leading terms of a minimal Gröbner basis for a given ideal are unique. Conclude that any two minimal Gröbner bases of an ideal have the same cardinality.
Remember that the word “minimal” is being used in two senses here. To wit, a generating set is called minimal if no strict subset of also generates . On the other hand, a Gröbner basis is called minimal if each element has monic leading term and no leading term is divisible by another.
- Let be an ideal, and let .
Suppose is a minimal Gröbner basis for . In particular, each is monic, and no divides for any . Now because is a Gröbner basis, we have that . Suppose that this generating set is not inclusion minimal; then without loss of generality, we have . By this previous exercise, is divisible by some , a contradiction. Thus in fact is a minimal generating set for .
Suppose now that is a minimal generating set for . By Proposition 24 on page 322 of D&F, is a Gröbner basis of . Since is a field, we may assume without loss of generality that each is monic. Moreover, no divides for , as otherwise is not an inclusion-minimal generating set of . Thus is a minimal Gröbner basis.
- Suppose is another minimal Gröbner basis for . By the previous part, both and are inclusion-minimal generating sets for the monomial ideal . By this previous exercise, we have and without loss of generality, for each . Because and and and have the same cardinality, as desired.