= Radical equality for finitely generated integer algebras
{title2=$\operatorname{Jac}(R)=\sqrt{(0)}$}
For a <finite-type integer algebra>, a nonnilpotent element $r$ gives a nonzero <localization of a ring> $R[1/r]$. A <maximal ideal> there has finite <residue field> by the <finite-field theorem for finitely generated integer algebras>. The image of $R$ in that <field> is a finite <integral domain>, hence a <field>. Its kernel is therefore maximal in $R$ and avoids $r$. Nilpotents lie in every <maximal ideal>, proving the radical equality. Applying the same argument to every prime quotient shows that these algebras are <Jacobson rings>. Merely contracting a localized <maximal ideal> without the finite-image argument would not prove maximality.
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