Basic Properties of Affine Algebraic Sets

Recall from the Affine Algebraic Sets page that if $K$ is a field and $S \subseteq K[x_1, x_2, ..., x_n]$ then the zero locus of $S$ is defined to be:

(1)

\begin{align} \quad V(S) = \{ \mathbf{p} \in \mathbb{A}^n(K) : F(\mathbf{p}) = 0, \: \forall F \in S \} \end{align}

We said that a set $X \subseteq \mathbb{A}^n(K)$ is an affine algebraic set if there exists an $S \subseteq K[x_1, x_2, ..., x_n]$ such that:

(2)

\begin{align} \quad X = V(S) \end{align}

We now state some basic properties of affine algebraic sets.

Theorem 1: Let $K$ be a field and let $\mathbb{A}^n(K)$ be the affine $n$-space over $K$. a) If $S, T \subseteq K[x_1, x_2, ..., x_n]$ and $S \subseteq T$ then $V(S) \supseteq V(T)$. b) The union of a finite collection of affine algebraic sets is an affine algebraic set. That is, if $S, T \subseteq K[x_1, x_2, ..., x_n]$ then $V(S) \cup F(T) = V(ST)$ where $ST := \{ FG : F \in S, G \in T \}$. . c) The intersection of an arbitrary collection of affine algebraic sets is an affine algebraic set. That is, if $\{ S_{\alpha} \}_{\alpha \in \Gamma}$ is an arbitrary collection of subsets of $K[x_1, x_2, ..., x_n]$ such that $\displaystyle{S = \bigcup_{\alpha \in \Gamma} S_{\alpha}}$ then $\displaystyle{V(S) = \bigcap_{\alpha \in \Gamma} V(S_{\alpha})}$. d) If $0$ denotes the additive identity in $K$ and $1$ denotes the multiplicative identity in $K$ then $V(0) = \mathbb{A}^n(K)$ and $V(1) = \emptyset$.

• Proof of a) Let $\mathbf{p} \in V(T)$. Then $F(\mathbf{p}) = 0$ for all $F \in T$. But $S \subseteq T$. So $F(\mathbf{p}) = 0$ for all $F \in S$. So $\mathbf{p} \in V(S)$. Hence:

(3)

\begin{align} \quad V(S) \supseteq V(T) \quad \blacksquare \end{align}

• Proof of b) Let $\mathbf{p} \in V(S) \cup V(T)$. Then $\mathbf{p} \in V(S)$ or $\mathbf{p} \in V(T)$. So $F(\mathbf{p}) = 0$ for all $F \in S$ or $G(\mathbf{p}) = 0$ for all $G \in T$. So $(FG)(p) = F(p)G(p) = 0$ for all $FG \in ST$. Therefore $\mathbf{p} \in V(ST)$. So $V(S) \cup V(T) \subseteq V(ST)$. Now let $\mathbf{p} \in V(ST)$. Then $(FG)(\mathbf{p}) = F(\mathbf{p})G(\mathbf{p}) = 0$ for all $FG \in ST$. So either $F(\mathbf{p}) = 0$ or $G(\mathbf{p}) = 0$ for every $F \in S$ and for every $G \in T$. So $\mathbf{p} \in V(S)$ or $\mathbf{p} \in V(T)$, i.e., $\mathbf{p} \in V(S) \cup V(T)$. So $V(S) \cup V(T) \supseteq V(ST)$. Hence:

(4)

\begin{align} \quad V(S) \cup V(T) = V(ST) \quad \blacksquare \end{align}

• Proof of c) Let $\mathbf{p} \in V(S)$. Then $F(\mathbf{p}) = 0$ for all $F \in S$. Since $S = \bigcup_{\alpha \in \Gamma} S_{\alpha}$ we have that $F(\mathbf{p}) = 0$ for all $F \in S_{\alpha}$ and for all $\alpha \in \Gamma$. Therefore $\displaystyle{\mathbf{p} \in \bigcap_{\alpha \in \Gamma} V(S_{\alpha})}$. So $\displaystyle{V(S) \subseteq \bigcap_{\alpha \in \Gamma} V(S_{\alpha})}$. Now let $\displaystyle{\mathbf{p} \in \bigcap_{\alpha \in \Gamma} V(S_{\alpha})}$. Then $F(\mathbf{p}) = 0$ for all $F \in S_{\alpha}$ and for all $\alpha \in \Gamma$. So $F(\mathbf{p}) = 0$ for all $F \in S$. Therefore $\mathbf{p} \in V(S)$. So $\displaystyle{V(S) \supseteq \bigcap_{\alpha \in \Gamma} V(S_{\alpha})}$. We conclude that:

(5)

\begin{align} \quad V(S) = \bigcap_{\alpha \in \Gamma} V(S_{\alpha}) \quad \blacksquare \end{align}

• Proof of d) We have that:

(6)

\begin{align} \quad V(0) = \{ \mathbf{p} \in \mathbb{A}^n(K) : 0(\mathbf{p}) = 0 \} = \mathbb{A}^n(K) \end{align}

(7)

\begin{align} \quad V(1) = \{ \mathbf{p} \in \mathbb{A}^n(K) : 1 = 0 \} = \emptyset \end{align}

Theorem 2: Let $K$ be a field and let $\mathbb{A}^n(K)$ be the affine $n$-space over $K$. Then every finite subset of $\mathbb{A}^n(K)$ is an affine algebraic set.

• Proof: Let $\mathbf{p} = (p_1, p_2, ..., p_n) \in \mathbb{A}^n(K)$. For each $k \in \{ 1, 2, ..., n \}$ consider the polynomial:

(8)

\begin{align} \quad F_k(x_1, x_2, ..., x_n) = x_k - p_k \end{align}

• And consider the zero locus of $F_1, F_2, ..., F_n$:

(9)

\begin{align} \quad V(F_1, F_2, ..., F_n) &= V(x_1 - p_1, x_2 - p_2, ..., x_n - p_n) \\ &= \{ \mathbf{p} \in \mathbb{A}^n(K) : x_1 = p_1, x_2 = p_2, ..., x_n = p_n \} \\ &= \{ \mathbf{p} = (p_1, p_2, ..., p_n) \} \end{align}

• Therefore every singleton set $\{ \mathbf{p} \}$ is an affine algebraic set. By Theorem 1 (b), the union of a finite collection of affine algebraic sets is an affine algebraic set. Therefore, every finite subset of $\mathbb{A}^n(K)$ is an affine algebraic set. $\blacksquare$

Corollary 3: If $K$ is a finite field then every subset of $\mathbb{A}^n(K)$ is an affine algebraic set.

• Proof: If $K$ is a finite field of order $m$ then $\mathbb{A}^n(K)$ contains $mn$ distinct elements. So every subset of $\mathbb{A}^n(K)$ contains at most $mn$ elements, i.e., every subset of $\mathbb{A}^n(K)$ is finite. By Theorem 2, we have that every subset of $\mathbb{A}^n(K)$ is an affine algebraic set. $\blacksquare$