Let $A \subseteq B$ be an integral extension. Show that if $A$ is a Jacobson ring, then $B$ is also a Jacobson ring.
My trial: Let $q$ be a prime ideal in $B$, and let $p:=q^c=q \cap A$. Since $A$ is Jacobson, $p=\cap_{m\supseteq p}m$. By going-up, we can find a maximal ideal $n$ in $B$ such that $m=n^c=n \cap A$. Let $r:=\cap_{n^c \supseteq p}n$, then $r \cap A = \cap_{m \supseteq p}m = p$.
But now how can I get $q=r$ so that $B$ is Jacobson? I found a link explaning this, but I couldn't understand it.
Also I found another link, where hint for another approach is suggested in problem 1.