[proofplan] We prove the identities by evaluating polynomials at an arbitrary point of [affine space](/page/Affine%20Space). The union identity follows from the fact that if every product $fg$ vanishes at a point and some $f \in I$ does not vanish there, then every element of $J$ must vanish there. The intersection identity follows because elements of the ideal sum are finite sums of elements from the participating ideals. The closure under finite unions is then obtained by induction from the binary union formula. [/proofplan]

[step:Translate membership in a zero locus into evaluation at a point] Set $R := k[x_1, \ldots, x_n]$. Let $a \in \mathbb{A}^n_k$ be arbitrary. Define the evaluation homomorphism $\operatorname{ev}_a: R \to k$ by $\operatorname{ev}_a(f)=f(a)$ for every $f \in R$. By the definition of $V(I)$, for every ideal $I \trianglelefteq R$, \begin{align*} a \in V(I) \iff \operatorname{ev}_a(f)=0 \text{ for every } f \in I. \end{align*} We will use this equivalence for each ideal under consideration. [/step]

[step:Prove the binary union formula by testing products at an arbitrary point]Let $I,J \trianglelefteq R$. We prove \begin{align*} V(I) \cup V(J) = V(IJ). \end{align*} First suppose $a \in V(I) \cup V(J)$. If $a \in V(I)$, then for every $f \in I$ and every $g \in J$, \begin{align*} \operatorname{ev}_a(fg)=\operatorname{ev}_a(f)\operatorname{ev}_a(g)=0 \cdot \operatorname{ev}_a(g)=0. \end{align*} If $a \in V(J)$, the same computation gives \begin{align*} \operatorname{ev}_a(fg)=\operatorname{ev}_a(f)\operatorname{ev}_a(g)=\operatorname{ev}_a(f)\cdot 0=0. \end{align*} Every element of $IJ$ is a finite sum of elements of the form $fg$ with $f \in I$ and $g \in J$, and $\operatorname{ev}_a$ is additive. Hence every element of $IJ$ vanishes at $a$, so $a \in V(IJ)$. Conversely suppose $a \in V(IJ)$. If $a \in V(I)$, then $a \in V(I) \cup V(J)$ and there is nothing further to prove. Assume instead that $a \notin V(I)$. Then there exists $f_0 \in I$ such that \begin{align*} \operatorname{ev}_a(f_0) \neq 0. \end{align*} For every $g \in J$, the product $f_0g$ belongs to $IJ$, so $a \in V(IJ)$ gives \begin{align*} 0=\operatorname{ev}_a(f_0g)=\operatorname{ev}_a(f_0)\operatorname{ev}_a(g). \end{align*} Since $k$ is a field and $\operatorname{ev}_a(f_0) \neq 0$, cancellation in $k$ gives $\operatorname{ev}_a(g)=0$. Thus every $g \in J$ vanishes at $a$, so $a \in V(J)$. Therefore $a \in V(I) \cup V(J)$. Since $a \in \mathbb{A}^n_k$ was arbitrary, the equality follows.[/step]

[guided]We want to show that the zero locus of the product ideal is exactly the union of the two zero loci. The key point is that evaluation at a point turns multiplication of polynomials into multiplication in the field $k$. Let $a \in \mathbb{A}^n_k$ be arbitrary. Define $\operatorname{ev}_a: R \to k$ by $\operatorname{ev}_a(f)=f(a)$. This is a ring homomorphism, so for all $f,g \in R$, \begin{align*} \operatorname{ev}_a(fg)=\operatorname{ev}_a(f)\operatorname{ev}_a(g). \end{align*} First assume $a \in V(I) \cup V(J)$. If $a \in V(I)$, then every $f \in I$ satisfies $\operatorname{ev}_a(f)=0$. Hence for every $f \in I$ and $g \in J$, \begin{align*} \operatorname{ev}_a(fg)=\operatorname{ev}_a(f)\operatorname{ev}_a(g)=0. \end{align*} If instead $a \in V(J)$, then every $g \in J$ satisfies $\operatorname{ev}_a(g)=0$, and the same product calculation again gives $\operatorname{ev}_a(fg)=0$ for all $f \in I$ and $g \in J$. Now use the definition of the product ideal. An arbitrary element $h \in IJ$ has the form \begin{align*} h=\sum_{r=1}^{m} f_r g_r \end{align*} for some $m \in \mathbb{N}$, some $f_1,\ldots,f_m \in I$, and some $g_1,\ldots,g_m \in J$. Since each product $f_rg_r$ vanishes at $a$ and evaluation is additive, \begin{align*} \operatorname{ev}_a(h)=\sum_{r=1}^{m}\operatorname{ev}_a(f_rg_r)=0. \end{align*} Thus every element of $IJ$ vanishes at $a$, so $a \in V(IJ)$. For the reverse inclusion, assume $a \in V(IJ)$. We must prove that $a$ belongs to at least one of $V(I)$ and $V(J)$. If $a \in V(I)$, this is already done. Suppose therefore that $a \notin V(I)$. By the definition of $V(I)$, there is some polynomial $f_0 \in I$ such that \begin{align*} \operatorname{ev}_a(f_0) \neq 0. \end{align*} Now take an arbitrary $g \in J$. Because $f_0 \in I$ and $g \in J$, the product $f_0g$ belongs to $IJ$. Since $a \in V(IJ)$, this product vanishes at $a$: \begin{align*} 0=\operatorname{ev}_a(f_0g)=\operatorname{ev}_a(f_0)\operatorname{ev}_a(g). \end{align*} The field hypothesis is used exactly here: in a field, a nonzero element can be cancelled. Since $\operatorname{ev}_a(f_0) \neq 0$, we obtain $\operatorname{ev}_a(g)=0$. The element $g \in J$ was arbitrary, so every element of $J$ vanishes at $a$. Therefore $a \in V(J)$. We have shown both inclusions for an arbitrary point $a \in \mathbb{A}^n_k$, so \begin{align*} V(I) \cup V(J) = V(IJ). \end{align*}[/guided]

[step:Prove the arbitrary intersection formula by expanding the ideal sum] Let $(I_\lambda)_{\lambda \in \Lambda}$ be a family of ideals in $R$. We prove \begin{align*} \bigcap_{\lambda \in \Lambda} V(I_\lambda) = V\left(\sum_{\lambda \in \Lambda} I_\lambda\right). \end{align*} If $\Lambda=\varnothing$, then by convention $\sum_{\lambda \in \Lambda} I_\lambda=(0)$ and $\bigcap_{\lambda \in \Lambda} V(I_\lambda)=\mathbb{A}^n_k$. Since every point $a \in \mathbb{A}^n_k$ satisfies $0(a)=0$, we have $V((0))=\mathbb{A}^n_k$. Thus the displayed identity holds when $\Lambda=\varnothing$. For the rest of this step, assume $\Lambda \neq \varnothing$. First suppose \begin{align*} a \in \bigcap_{\lambda \in \Lambda} V(I_\lambda). \end{align*} Let $h \in \sum_{\lambda \in \Lambda} I_\lambda$. By the definition of the ideal sum, there exist $m \in \mathbb{N}$, indices $\lambda_1,\ldots,\lambda_m \in \Lambda$, and elements $h_r \in I_{\lambda_r}$ for $1 \leq r \leq m$ such that \begin{align*} h=\sum_{r=1}^{m} h_r. \end{align*} Since $a \in V(I_{\lambda_r})$ for each $r$, we have $\operatorname{ev}_a(h_r)=0$ for each $r$. Hence \begin{align*} \operatorname{ev}_a(h)=\sum_{r=1}^{m}\operatorname{ev}_a(h_r)=0. \end{align*} Thus $a \in V(\sum_{\lambda \in \Lambda} I_\lambda)$. Conversely suppose \begin{align*} a \in V\left(\sum_{\lambda \in \Lambda} I_\lambda\right). \end{align*} For each $\lambda \in \Lambda$, the inclusion $I_\lambda \subset \sum_{\mu \in \Lambda} I_\mu$ holds by definition of the ideal sum. Therefore every element of $I_\lambda$ vanishes at $a$, so $a \in V(I_\lambda)$ for every $\lambda \in \Lambda$. Hence \begin{align*} a \in \bigcap_{\lambda \in \Lambda} V(I_\lambda). \end{align*} This proves the intersection formula. [/step]

[step:Derive closure under arbitrary intersections and finite unions] The intersection formula shows that an arbitrary intersection of sets of the form $V(I)$ is again of the form $V(K)$, namely with \begin{align*} K=\sum_{\lambda \in \Lambda} I_\lambda. \end{align*} Therefore affine algebraic subsets of $\mathbb{A}^n_k$ are closed under arbitrary intersections. The binary union formula shows that the union of two sets of the form $V(I)$ is again of the form $V(K)$, namely with $K=IJ$. Applying this formula repeatedly gives closure under finite unions. More explicitly, for ideals $I_1,\ldots,I_m \trianglelefteq R$ with $m \in \mathbb{N}$, induction on $m$ gives \begin{align*} V(I_1) \cup \cdots \cup V(I_m)=V(I_1 I_2 \cdots I_m). \end{align*} Thus the affine algebraic subsets, equivalently the Zariski closed subsets, are closed under arbitrary intersections and finite unions. [/step]