A subset $s \subseteq R$ has finitely generated span if and only if: $\exists$ finite $s' \subseteq s$ such that $\mathsf{span}(s) = \mathsf{span}(s')$
Let $G'' \subseteq R[x_1, \dots, x_n]$ be a set of polynomials such that $$ \forall p \in G'',\ \operatorname{leadingCoeff}(p) \in R^\times. $$ Then, $$ \left\langle \operatorname{lt}(G'') \right\rangle = \left\langle x^{\deg(p)} \mid p \in G'' \right\rangle, $$
$$ \langle \mathrm{lt}(G) \rangle = \left\langle \left\{ x^t : t \in \{ \mathrm{multideg}(p) : p \in G \setminus \{0\} \} \right\} \right\rangle $$
Let $G'' \subseteq R[x_1, \dots, x_n]$, let $I \subseteq R[x_1, \dots, x_n]$ be an ideal, and let $p, r \in R[x_1, \dots, x_n]$. Suppose that:
- $G'' \subseteq I$,
- $r \in I$,
- $r$ is the remainder of $p$ upon division by $G''$.
Then, $$ p \in I. $$
Let $R$ be a commutative ring, and let $G'' \subseteq R[x_1, \dots, x_n]$, $I \subseteq R[x_1, \dots, x_n]$ be an ideal, and $p, r \in R[x_1, \dots, x_n]$. Assume that:
- $G'' \subseteq I$,
- $r$ is the remainder of $p$ upon division by $G''$.
Then, $$ r \in I \quad \Longleftrightarrow \quad p \in I. $$
Let $I \subseteq k[x_i : i \in \sigma]$ be an ideal, and let $G \subseteq I$ be a finite subset. Suppose that $r_1$ and $r_2$ are generalized remainders of a polynomial $p$ upon division by $G$. Then, $$ r_1 - r_2 \in I. $$