= Radical ideals and closed subsets of the maximal spectrum
{title2=$I\leftrightarrow V(I)\cap\operatorname{MaxSpec}A$}
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.
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