Let $I \subseteq k[x_1, \ldots, x_n]$ be an ideal. Then there exists a finite subset $G = \{g_1, \ldots, g_t\}$ of $I$ such that $G$ is a Gröbner basis for $I$.
Let $G = \{g_1, \dots, g_t\}$ be a Gröbner basis for an ideal $I \subseteq k[x_1, \dots, x_n]$ and let $f \in k[x_1, \dots, x_n]$. Then $f \in I$ if and only if the remainder on division of $f$ by $G$ is zero.
Let $G = \{g_1, \ldots, g_t\}$ be a finite subset of $k[x_1, \ldots, x_n]$. Then $G$ is a Gröbner basis for the ideal $I = \langle G \rangle$ if and only if for every $f \in I$, the remainder of $f$ on division by $G$ is zero.
Let $G = \{g_1, \ldots, g_t\}$ be a Gröbner basis for an ideal $I \subseteq k[x_1, \ldots, x_n]$. Then $G$ is a basis for the vector space $I$ over $k$.
Let $f, h_1, \dots, h_m \in k[\mathbf{x}] \setminus \{0\}$, and suppose $$f = c_1 h_1 + \cdots + c_m h_m, \quad \text{with } c_i \in k.$$ If $$\mathrm{lm}(h_1) = \mathrm{lm}(h_2) = \cdots = \mathrm{lm}(h_i) > \mathrm{lm}(f),$$ then $$f = \sum_{1 \leq i < j \leq m} c_{i,j} S(h_i, h_j), \quad c_{i,j} \in k.$$ Furthermore, if $S(h_i, h_j) \ne 0$, then $\mathrm{lm}(h_i) > \mathrm{lm}(S(h_i, h_j))$.
$h_1, h_2 \in k[\mathbf{x}], lm(h_1) = lm(h_2), S(h_1, h_2) \ne 0$, then $lm(S(h_1, h_2)) < lm(h_1)$.
A basis $G = \{ g_1, \ldots, g_t \}$ for an ideal $I$ is a Gröbner basis if and only if $S(g_i, g_j) \to_G 0$ for all $i \neq j$.