## The leading terms of a minimal Gröbner basis are unique

Let $F$ be a field and fix a monomial order on $R = F[x_1,\ldots,x_n]$.

1. Let $I$ be an ideal of $R$. Prove that $\{g_1, \ldots, g_m\} \subseteq I$ is a minimal Gröbner basis for $I$ if and only if $\{ LT(g_1), \ldots, LT(g_m) \}$ is a minimal generating set for $LT(I)$.
2. Prove that the leading terms of a minimal Gröbner basis for a given ideal $I$ are unique. Conclude that any two minimal Gröbner bases of an ideal $I$ have the same cardinality.

Remember that the word “minimal” is being used in two senses here. To wit, a generating set $A \subseteq I$ is called minimal if no strict subset of $A$ also generates $I$. 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.

1. Let $I \subseteq R$ be an ideal, and let $G = \{g_1, \ldots, g_m\} \subseteq I$.

$(\Rightarrow)$ Suppose $G$ is a minimal Gröbner basis for $I$. In particular, each $LT(g_i)$ is monic, and no $LT(g_i)$ divides $LT(g_j)$ for any $j \neq i$. Now because $G$ is a Gröbner basis, we have that $LT(I) = (LT(g_1), \ldots, LT(g_m))$. Suppose that this generating set is not inclusion minimal; then without loss of generality, we have $LT(g_1) \in (LT(g_2), \ldots, LT(g_m))$. By this previous exercise, $LT(g_1)$ is divisible by some $LT(g_j)$, a contradiction. Thus in fact $\{LT(g_1), \ldots, LT(g_m)\}$ is a minimal generating set for $LT(I)$.

$(\Leftarrow)$ Suppose now that $\{LT(g_1), \ldots, LT(g_m)\}$ is a minimal generating set for $LT(I)$. By Proposition 24 on page 322 of D&F, $G$ is a Gröbner basis of $I$. Since $F$ is a field, we may assume without loss of generality that each $LT(g_i)$ is monic. Moreover, no $LT(g_i)$ divides $LT(g_j)$ for $j \neq i$, as otherwise $\{LT(g_1), \ldots, LT(g_m)\}$ is not an inclusion-minimal generating set of $LT(I)$. Thus $G$ is a minimal Gröbner basis.

2. Suppose $H = \{h_1, \ldots, h_\ell\}$ is another minimal Gröbner basis for $I$. By the previous part, both $\{LT(g_1), \ldots, LT(g_m)\}$ and $\{LT(h_1), \ldots, LT(h_\ell)\}$ are inclusion-minimal generating sets for the monomial ideal $LT(I)$. By this previous exercise, we have $m = \ell$ and without loss of generality, $LT(h_i) = LT(g_i)$ for each $i$. Because $G$ and $\{LT(g_1), \ldots, LT(g_m)\}$ and $H$ and $\{LT(h_1), \ldots, LT(h_m)\}$ have the same cardinality, $|G| = |H|$ as desired.

• Confused  On February 23, 2012 at 10:19 am

Proposition 24 on page 322 of D&F???? What?? At least give the edition number, or link the prop/proof.

• nbloomf  On March 17, 2012 at 2:51 pm

Since you asked so nicely…

‘D&F’ is short for ‘Dummit and Foote’, the authors of a book whose title is ‘Abstract Algebra’. Because there are so many books with that title, it is customary to refer to them by their author’s last names.

These exercises come from that book, so I frequently will refer back to results in that text under the assumption that readers also have a copy and are following along.

Third edition.