TITLE: Integral Extension of a Jacobson Ring
QUESTION [10 upvotes]: 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.

REPLY [10 votes]: Congratulations, you have done a lot yourself: I'll help you conclude. 
The theorem "going-up" is stronger than what you use : it also says that, besides $n\cap A=m$, you can arrange that  $q\subset n$. Now (with your definition  $r=\cap n$)  you have   $ q \subset r$ and $ q\cap A=r\cap A=p $ . 
A useful result ("incomparability", See Atiyah, Corollary 5.9) then allows you to conclude that $q=n$. You are home!
Reminder: incomparability  Let $A\subset B$ be an integral extension of rings. Suppose that $Q \subset J \subset B$ are ideals such that $Q$ is prime and that $A \cap Q=A \cap J $. Then $Q=J $
 [ It is often assumed that  $J$ is also prime but this assumption is unnecessary]