Property of remainders of Grobner basis

Question

Let $G$ be a Gröbner basis for an ideal $I \subseteq \mathbb{F}[x_{1}, \ldots x_{n}]$. Let $f,g$ be in $\mathbb{F}[x_{1}, \ldots x_{n}]$ and recall that $\overline{f}^{G}$ is the remainder of $f$ divided by $G$. Why is it the case that $$\overline{fg}^{G} = \overline{\overline{f}^{G} \cdot \overline{g}^{G}}^{G}$$

Answer 1

Just to be clear, let's use the following property.

Let $k$ be a field and $f,g_1,g_2,...,g_s\in k[X_1,..,X_n]$ are nonzero polynomials. There then exists $h_1,...,h_s\in k[X_1,..,X_n]$ such that

$$f=\sum_{i=1}^s h_ig_i + r$$

(1) $LT(h_ig_i)\leq LT(f)$ for all $i$ with $h_i\neq 0$, $LT$ means the leading term.
(2) $r=0$ or no terms of $r$ is divisible by $LT(g_i)$ for all $i$ and $LT(r)\leq LT(f)$.
(3) In particular, if $\{g_1,...,g_s\}$ is a Gröbner basis for an ideal $I$, such $r$ is uniquely determined.

In this problem, by this property it suffices to show that the RHS is a remainder of $fg$ divided by $G$. First, we do division with remainder

$$f=\sum_{i=1}^s h_ig_i + r$$ $$g=\sum_{i=1}^s h_i'g_i + r'.$$

Then $fg=(\sum_{i=1}^s h_ig_i)(\sum_{i=1}^s h_i'g_i) + r'\sum_{i=1}^s h_ig_i + r\sum_{i=1}^s h_i'g_i + rr'$. Do division with remainder to get $rr'=\sum_{i=1}^s r_ig_i + R$ so

$$fg=(\sum_{i=1}^s h_ig_i)(\sum_{i=1}^s h_i'g_i) + r'\sum_{i=1}^s h_ig_i + r\sum_{i=1}^s h_i'g_i + \sum_{i=1}^s r_ig_i + R.$$

$R=RHS$ by definition. Check that the above formula expands and gives division with remainder that satisfies the above (1), (2), (3). Then by the uniqueness of remainders, $RHS=R=LHS$.