Zero loci in affine space are the closed sets of the classical Zariski topology

If $\{S_{\alpha }{\}}_{\alpha \in A}$ is any family of subsets of $k[x_1,\dots ,x_n]$, then a point lies in every $V(S_{\alpha })$ exactly when it annihilates every polynomial in every $S_{\alpha }$, that is, exactly when it lies in $V({\bigcup }_{\alpha \in A}{S}_{\alpha })$. Hence ${\bigcap }_{\alpha \in A}V(S_{\alpha })=V\text{\hspace{0.17em}⁣}\left({\bigcup }_{\alpha \in A}{S}_{\alpha }\right)\mathrm{.}$

By definition, $V(\mathrm{\varnothing })={\mathbf{A}}_k^n$ and $V(1)=\mathrm{\varnothing }$.

Let $I,J\subseteq k[x_1,\dots ,x_n]$ be ideals. If $a\in V(I)\cup V(J)$, then every product $fg$ with $f\in I$ and $g\in J$ vanishes at $a$, so $a\in V(IJ)$. Conversely, if $a\in V(IJ)$ but $a\mathrm{\notin }V(I)$ and $a\mathrm{\notin }V(J)$, choose $f\in I$ and $g\in J$ with $f(a)\mathrm{\ne }0$ and $g(a)\mathrm{\ne }0$. Then $(fg)(a)\mathrm{\ne }0$, contradicting $a\in V(IJ)$. Thus $V(I)\cup V(J)=V(IJ)$.

For arbitrary subsets $S,T$, step 1.3 and [L2] give $V(S)\cup V(T)=V((S))\cup V((T))=V((S)(T)),$ so finite unions of zero loci are zero loci. Repeating this argument proves the same for any finite union.

Steps 1.1, 1.2, and 2.1 are exactly the topology axioms for the closed subsets of ${\mathbf{A}}_k^n$, and step 1.3 gives the displayed formula $V(I)\cup V(J)=V(IJ)$.