Minimal Gröbner Bases' Commonalities (9.6.15)

Dummit and Foote Abstract Algebra, section 9.6, exercise 15:

Fix a monomial ordering on $R=F[x_1,...,x_n]$.
(a) Prove $\{g_1,...,g_n\} \subseteq I \subseteq R$ for $I$ an ideal and monic $g_i$ is a minimal Gröbner basis for $I$ if and only if $\{LT(g_1),...,LT(g_n)\}$ is a minimal generating set for $LT(I)$.
(b) Prove the leading terms and order of a minimal Gröbner basis of $I$ is uniquely determined.

Proof: (a) $⇒$ By the definition of a minimal Gröbner basis, we have $(LT(g_1),...,LT(g_n))=LT(I)$ and $LT(g_i) \not \mid LT(g_j)$ for any $i≠j$. We can see that $(\{LT(g_1),...,LT(g_n)\} \setminus \{LT(g_i)\}) \subset I$ since for there to be equality would be, by exercise 10, to presume an element in $\{LT(g_1),...,LT(g_n)\} \setminus \{LT(g_i)\}$ divides $LT(g_i)$. $\Leftarrow$ By proposition 24 we have $\{g_1,...,g_n\}$ is a Gröbner basis of $I$ and since their leading terms is a minimal generating set for $I$ we must have $LT(g_i) \not \mid LT(g_j)$ for $i≠j$.

(b) Since $LT(I)$ is a monomial ideal, by the previous exercise its minimal generating set of monomials is uniquely determined. By part (a) the leading terms of a minimal Gröbner basis's elements must be uniquely determined, as well as the number of leading terms (which is the order of the basis).$~\square$