Let $M$ be an $R$-module and $N$ be a simple $R$-module (that is, isomorphic to $R/m$ for some maximal ideal $m$ of $R$). I guess that the following statement is true, but I can't prove or disprove it. Can anyone help me?
$M$ is an essential extension of $N$ if and only if $M$ is a submodule of $E(N)$, where $E(N)$ is the injective hull of $N$.
Suppose $M$ is an essential extension of $N$ and consider the injective hull $E(N)$ of $N$. Denote by $f\colon N\to M$ an injective homomorphism and by $i\colon N\to E(N)$ the embedding in the injective hull.
By definition of injective module, there exists $g\colon M\to E(N)$ such that $gf=i$. Since $$ \ker g\cap f(N)=\{0\} $$ and $f(N)$ is essential in $M$ (by assumption), we conclude $\ker g=\{0\}$, so $g$ is an embedding.
It's an abuse of language saying that $M$ is a submodule of $E(N)$; what you can say is that $M$ is isomorphic to a submodule of $E(N)$.
The converse is obvious: if $N\subseteq M\subseteq E(N)$, $N$ is essential in $E(N)$ and so also in $M$.
Note that the simplicity of $N$ has not been used. Indeed, the result holds for every $R$-module $N$. The key fact is that $E(N)$ is an essential extension of $N$.