For an integral extension $A \subseteq B$, $\sqrt{IB}\cap A=\sqrt{I}$

Question

Let $A \subseteq B$ be an integral extension and $I \subseteq A$ an ideal. Prove that $\sqrt{IB}\cap A=\sqrt{I}$.

I saw a similar property holds for Jacobson radical where the following two facts were used.
1. Jacobson radical is the intersection of the maximal ideals.
2. $x$ is in the Jacobson radical if and only if $1-xy$ is a uint for all $y$.

I know that $\sqrt{I}$ is the intersection of prime ideals of $A$ containing $I$. I'm wondering what other properties are useful to prove the theorem.

Answer 1

Here's a useful lemma.

Let $\varphi : A \rightarrow B$ be a ring homomorphism (commutative rings with unity). Say $p\in \operatorname{Spec}A$. Then $\exists P \in \operatorname{Spec}B$ such that $\varphi^{-1}(P)=p$ iff $p^{ec}=p$, where for an ideal $I\subset A$ we denote $IB$ by $I^e$ and for an ideal $J\subset B, \ J^c =\varphi^{-1}(J)$.

Proof: If $p=P^c$ then $$p^{ec}=P^{cec}=P^c=p$$ and we are done with one direction. Now assume $p^{ec}=p$ let $B_p$ denote the localization of $B$ at the multiplicative set $\varphi (A\setminus p)$. Then $\varphi$ factors as $$\varphi_p :A_p \rightarrow B_p.$$ Consider a maximal ideal of $B_p$ containing $pA_p^e=p^eB_p$ (note this is not the whole ring), say $m_p$, where $m$ is a prime ideal of $B$. $$m_p^c\supset pA_p$$ and hence by uniqueness of maximal ideal we have $$m_p^c= pA_p.$$ Thus we get $$m^c=p.$$

Now in case of integral extensions we have the Lying-Over property and hence $p^{ec}=p \ \forall p \in \operatorname{Spec}A$. One side inclusion is obvious. For the other side observe if $ p\supset I$ then $$p^e\supset IB$$ and hence $$p^{ec} \supset IB\cap A $$ $$\implies p\supset \sqrt {IB\cap A}= \sqrt{IB} \cap A.$$ Thus any prime containing $I$ contains $\sqrt{IB} \cap A$ and equality follows.