Integral Extension of a Jacobson Ring

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.

Solution 1:

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]