The Ideal of a Set of Points
 Definition: Let $K$ be a field and let $X \subseteq \mathbb{A}^n(K)$. The Ideal of $X$ is defined as $I(X) = \{ F \in K[x_1, x_2, ..., x_n] : F(\mathbf{p}) = 0, \: \forall \mathbf{p} \in X \}$.
We verify that $I(X)$ defined above is indeed an ideal.
 Theorem 1: Let $K$ be a field and let $X \subseteq \mathbb{A}^n(K)$. Let $I(X) = \{ F \in K[x_1, x_2, ..., x_n] : F(\mathbf{p}) = 0, \: \forall \mathbf{p} \in X \}$. Then $I(X)$ is an ideal in $K[x_1, x_2, ..., x_n]$.
• Proof: Let $F, G \in I(X)$. Then $F(\mathbf{p}) = 0$ and $G(\mathbf{p}) = 0$ for all $\mathbf{p} \in X$. So $(F + G)(\mathbf{p}) = F(\mathbf{p}) + G(\mathbf{p}) = 0 + 0 = 0$ for all $\mathbf{p} \in X$. Hence $(F + G) \in I(X)$.
• Now let $F \in I(X)$ and let $G \in K[x_1, x_2, ..., x_n]$. Then $F(\mathbf{p}) = 0$ for all $\mathbf{p} \in X$. So $(FG)(\mathbf{p}) = F(\mathbf{p})G(\mathbf{p}) = 0G(\mathbf{p}) = 0$ for all $\mathbf{p} \in X$. Hence $FG \in I(X)$.
• So $I(X)$ is an ideal in $K[x_1, x_2, ..., x_n]$. $\blacksquare$