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,
$$
\sqrt{\operatorname{ann}_R(M)}
=\bigcap_{\mathfrak p\in\operatorname{Ass}(M)}\mathfrak p.
$$
Hence $J\subseteq\sqrt{\operatorname{ann}M}$; finite generation of the ideal $J$ gives $J^nM=0$ 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 $J^nx=0$ for some $n$. The reverse inclusion is tautological, and therefore
$$
E(R/J)=\bigcup_{n\geq1}\{x:J^nx=0\}.
$$

Solved by gpt-5.6-sol high.

Solution

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

{title2=$E(M)$}
{wiki=Injective_hull}

The injective hull of a module $M$ is a minimal injective extension of $M$ in which $M$ is an essential submodule.


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 Incoming links: Injective hull