Suppose $G$ and $G'$ are grobner bases for the ideal $I$. Show that $\overline{f}^{G} = \overline{f}^{G'}$ for $f \in k[x_1, \cdots, x_n]$
Suppose $G$ and $G'$ are grobner bases for the ideal $I$. Show that $\overline{f}^{G} = \overline{f}^{G'}$ for $f \in k[x_1, \cdots, x_n]$
By the division algorithm and Proposition $1$, we may write $f \in k[x_1, \cdots, x_n]$ as: $$
f = \underbrace{q_1 g_1 + q_2 g_2 + \cdots + q_n g_n}_{qg} + r
$$ such that the remainder $r \in k[\overline{x}]$ is unique. We may also do the exact division process with $G'$:
$$
f = \underbrace{q_1' g_1' + \cdots + q_n' g_n'}_{q'g'} + r'
$$
and as before $r' \in k[\overline{x}]$ is unique by proposition $1$. We wish to show the two remainder must be the same. Therefore, consider their difference:
$$
f - f = qg - q' g' = r - r' = 0 \in k[\overline{x}]
$$ Note that $qg$ and $q'g'$ are both contained in the ideal. Thus, $r - r'$ must be contained in the ideal and therefore $\overline{(r-r')}^{G,G'}$ is zero.
But I know that the leading terms $LT(g)$ do NOT divide $r$. Likewise, the leading terms $LT(g')$ do not divide $r'$. What can I say about the leading term of $r - r'$ ?
My attempt:
There are three cases: $LT(r-r') = r$, $LT(r -r') = r'$, and $LT(r -r') = \text{ something other than } r,r'$.
Case 1. Suppose $LT(r-r') = r$. Since $r-r' \in I$, then $LT(r-r') \in I$. So $r \in I$. But by Proposition $1$, $LT(g)\nmid r$. This is a contradiction ? But I am unsure of its implications.
Solution 1:
Write $f=g+r$ and $f=g'+r'$ where $r=\overline{f}^G$ and $r'=\overline{f}^{G'}$ and $g,g'\in I$. We wish to show $r=r'$. Suppose otherwise, so $r-r'\not=0$ and since $r=f-g$ and $r'=f-g'$ we have $$ r-r'=f-g-(f-g')=g'-g\in I $$ and thus $LT(r-r')\in \langle LT(I)\rangle=\langle LT(G)\rangle=\langle LT(G')\rangle$. By properties of monomial ideals, we get that $LT(r-r')$ is divisible by an element of $LT(G)$ and an element of $LT(G')$. But any monomial of $r-r'$ is either a monomial of $r$ or a monomial of $r'$. So in particular, the leading monomial of $r-r'$ is either a monomial of $r$ or a monomial of $r'$. However, since by your Proposition 1, no monomial of $r$ is divisible by any element of $LT(G)$ and no monomial of $r'$ is divisible by any element of $LT(G')$, it is impossible for $LT(r-r')$ to simultaneously be divisible by an element of $LT(G)$ and $LT(G')$, a contradiction. Hence $r-r'=0$ and thus $r=r'$ as desired.