In a finite-type algebra over a field, radical ideals are intersections of maximal ideals

If $J=A$, then the family of maximal ideals containing $J$ is empty, and the stated convention makes the displayed intersection equal to $A=J$. So only the proper case needs proof.

Assume now that $J$ is proper, and put $B=A\mathrm{/}J$. Then $B$ is a reduced finite-type $k$-algebra. It is enough to prove that the intersection of the maximal ideals of $B$ is $0$, because pulling those ideals back along $A\to B$ then gives the displayed formula for $J$.

Let $0\mathrm{\ne }b\in B$. Because $B$ is reduced, the localization $B_b$ is nonzero. By [L1], the zero ideal of $B_b$ lies in some maximal ideal $\mathfrak{n}$. Let $\mathfrak{p}=\mathfrak{n}\cap B$. By [L2], $\mathfrak{p}$ is a prime ideal of $B$ that does not contain $b$.

The composed map $k\to B\to B_b\to B_b\mathrm{/}\mathfrak{n}$ is a finite-type $k$-algebra map to a field. By [L4], the field $B_b\mathrm{/}\mathfrak{n}$ is finite over $k$. The image of $B\mathrm{/}\mathfrak{p}$ inside that field is a finite-type $k$-domain contained in a finite-dimensional $k$-vector space, so it is itself a field. Therefore $\mathfrak{p}$ is maximal in $B$.

Step 4.1 gives a maximal ideal of $B$ that avoids the chosen nonzero element $b$. Hence the intersection of all maximal ideals of $B$ is $0$. Returning to step 2.1 proves that every radical ideal of $A$ is the intersection of the maximal ideals containing it.