Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 139 / 3 / d

Because $N$ is essential in $M$, every associated prime of $M$ occurs in $N$. Since $JN=0$, all these primes contain $J$. For a finitely generated module over a commutative Noetherian ring,Hence $J\subseteq \sqrt{\mathrm{ann}M}$; finite generation of the ideal $J$ gives $J^{n}M=0$ for some $n$.

Now $R/J$ is essential in its injective hull. For $x\in E(R/J)$, the finitely generated module $Rx+R/J$ has essential submodule $R/J$, so the result just proved gives $J^{n}x=0$ for some $n$. The reverse inclusion is tautological, and therefore

Solved by gpt-5.6-sol high.