Pass to the integral extension
$$
A/(I\cap A)\subseteq B/I.
$$
The ring $B/I$ is an <integral domain>, and the ideal $J/I$ has zero contraction to the base. If $J/I$ contained a nonzero element $b$, choose an integral equation for $b$ of least degree:
$$
b^r+a_{r-1}b^{r-1}+\cdots+a_1b+a_0=0.
$$
Its constant term is nonzero, since otherwise the domain property would let us cancel $b$ and obtain an equation of lower degree. But
$$
a_0=-b(b^{r-1}+a_{r-1}b^{r-2}+\cdots+a_1)
$$
is a nonzero element of $(J/I)\cap(A/(I\cap A))$, a contradiction. Therefore $J/I=0$ and $I=J$. This proves <ideal contraction rigidity under an integral extension>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 4 / iii / Solution

Let $A\subseteq B$ be integral and let $I\subseteq J$ be ideals of $B$. If $I$ is prime and $I\cap A=J\cap A$, then $I=J$. After quotienting by $I$, a nonzero element of $J/I$ has an integral equation of least degree whose nonzero constant term lies in $(J/I)\cap(A/(I\cap A))$, a contradiction.


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 Incoming links: Ideal contraction rigidity under an integral extension