Radical ideals and closed subsets of the maximal spectrum ($I\leftrightarrow V(I)\cap \mathrm{MaxSpec}A$)

For a finitely generated algebra over an algebraically closed field $k$, radical ideals correspond, with inclusion reversed, to closed subsets of the maximal spectrum. If $f\notin \sqrt{I}$, the nonzero algebra $(A/I)[1/f]$ is still finitely generated over $k$. Its maximal residue field is $k$ by the Zariski lemma, so contracting a maximal ideal gives a maximal ideal of $A$ containing $I$ and avoiding $f$. Hence the intersection of all maximal ideals containing $I$ is $\sqrt{I}$. This proves the inverse correspondence, including the empty closed set and unit ideal.