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.


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]

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  • $\begingroup$ Thank you. I used going-up too weakly. From $p \subseteq m$ we can find $n$ such that $q \subseteq n$ and $n \cap A=m$ and since $m$ is maximal $n$ is also maximal. So $r=\cap_{n^c \supseteq p}n \supseteq q$. I tried the rest, but not sure about the details. Localizing at $p$ induces that $A_p \subseteq B_p$, and $r_p$ contracts to $p_p$. If we find a maximal ideal $s_p$ containing $r_p$, it also contracts to $p_p$ since it is the only maximal ideal in $A_p$. So $s$,$r$ lies over $p$ and by incomparability of prime ideals, $s=r$ so that $q=r$, as desired. $\endgroup$ – Gobi May 12 '11 at 2:20
  • $\begingroup$ Anyway, can I get a reference for the incomparability of any ideals? I can't find it out. $\endgroup$ – Gobi May 12 '11 at 2:32
  • $\begingroup$ I checked details and it is correct. you don't need show me a reference. I leave a proof. Let $A\subseteq B$ be integral and $q\subseteq q'$ in $B$ such that $q$ is prime and $q\cap A=q'\cap A=p$. Then $q=q'$. (Proof) $A_p \subseteq B_p$ is integral. Then $q_p \cap A_p = (q \cap A)_p = p_p$. Let $q''_p$ be maximal and containing $q'_p$. From $q' \subseteq q'' \subseteq p$ we obtain $q'' \cap A=p$ and $q''_p \cap A_p=p_p$. Since $p_p$ is maximal, $q_p, q''_p$ are maximal so $q=q'=q''$. $\endgroup$ – Gobi May 12 '11 at 4:51

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