Basic Properties of the Ideal of a Set of Points

# Basic Properties of the Ideal of a Set of Points

Recall from The Ideal of a Set of Points page that if $K$ is a field and $X \subseteq \mathbb{A}^n(K)$ then the ideal of $X$ is defined as:

(1)
\begin{align} \quad I(X) = \{ F \in K[x_1, x_2, ..., x_n] : F(\mathbf{p}) = 0, \: \forall \mathbf{p} \in X \} \end{align}

We verified that indeed, $I(X)$ is an ideal of $K[x_1, x_2, ..., x_n]$.

We will now look at some basic properties of such ideals.

 Theorem 1: Let $K$ be a field and let $\mathbb{A}^n(K)$ be the affine $n$-space over $K$. a) If $X, Y \subseteq \mathbb{A}^n(K)$ and $X \subseteq Y$ then $I(X) \supseteq I(Y)$. b) $I(\emptyset) = K[x_1, x_2, ..., x_n]$. c) If $K$ is an infinite field then $I(\mathbb{A}^n(K)) = (0)$.
• Proof of a) Let $F \in I(Y)$. Then $F(\mathbf{p}) = 0$ for all $\mathbf{p} \in Y$. But $X \subseteq Y$. So $F(\mathbf{p}) = 0$ for all $\mathbf{p} \in X$. Therefore $F \in I(X)$ and hence:
(2)
• Proof of b) We have that:
(3)
\begin{align} \quad I(\emptyset) = \{ F \in K[x_1, x_2, ..., x_n] : F(\mathbf{p}) = 0, \: \forall \mathbf{p} \in \emptyset \} = K[x_1, x_2, ..., x_n] \end{align}
• Proof of c) If $K$ is an infinite field then:
(4)
\begin{align} \quad I(\mathbb{A}^n(K)) = \{ F \in K[x_1, x_2, ..., x_n] : F(\mathbf{p}) = 0, \: \forall \mathbf{p} \in \mathbb{A}^n(K) \} = (0) \end{align}
 Theorem 2: Let $K$ be a field and let $\mathbb{A}^n(K)$ be the affine $n$-space over $K$. a) If $S \subseteq K[x_1, x_2, ..., x_n]$ then $S \subseteq I(V(S))$. b) If $X \subseteq \mathbb{A}^n(K)$ then $X \subseteq V(I(X))$.
• Proof of a) By definition we have that $V(S)$ is the set of all points in $\mathbb{A}^n(K)$ such that every polynomial if $S$ simultaneously vanish at these points. So $I(V(S))$ is the set of all functions that vanish for every point in $V(S)$. Hence:
(5)
\begin{align} \quad S \subseteq I(V(S)) \end{align}
• Proof of b) By definition we have that $I(X)$ is the set of all functions in $K[x_1, x_2, ..., x_n]$ such that every point in $X$ if a root of these functions. So $V(I(X))$ is the set of all points such that every polynomial in $I(X)$ simultaneously vanish at these points. Hence:
(6)
\begin{align} \quad X \subseteq V(I(X)) \end{align}