Work modulo $I$. Put
$$
C=R/(I+J),
$$
and let $A$ be the image of $I'/I$ in $C$. The separating condition $I'=I'\cap(I+J)$ modulo $I$ says that $I'/I\to C$ is injective, so we may regard $A$ as a submodule of $C$.

Since $r\in\sqrt{(I:I')}$, some power $r^q$ annihilates $A$. Apply the <Artin-Rees lemma> to $A\subseteq C$ and the principal ideal $(r)$. There is $s$ such that for every $m\ge s$,
$$
A\cap r^mC=r^{m-s}(A\cap r^sC).
$$
For $m\ge s+q$, the right side is zero. Pulling the equality $A\cap r^mC=0$ back to $R$ gives
$$
I'\cap(I+J+(r^m))=I,
$$
as required.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 4 / d / Solution

{c}
{wiki=Artin–Rees_lemma}

If $R$ is Noetherian, $I$ is an ideal, and $N\subseteq M$ are finitely generated modules, then some $k$ satisfies
$$
I^nM\cap N=I^{n-k}(I^kM\cap N)
$$
for every $n\ge k$.


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 Incoming links: Artin-Rees lemma