Injective hull of a simple module

Question

Let $M$ be an indecomposable injective right module over a right Artinian ring $R$, so $M$ has exactly one associated prime ideal $P$ (Lectures on Modules and Rings, T.Y. Lam). Now, $R/P$ is a simple Artinian ring having a simple $R/P$-module $V$ (which is also a generator) viewed as a simple right $R$-module. I want to show that $M$ is isomorphic to the injective hull of $V$. Any suggestion would be appreciated!

Answer 1

This answer uses the equivalent characterizations of injective indecomposibles from Theorem 3.52 in T.Y. Lam, Lectures on Modules and Rings. In particular, it follows from this theorem that $M$ is uniform and that $M$ is the injective envelope of each of its nonzero submodules.

Since $P$ is an associated prime of $M$, there exists a nonzero submodule $N \subset M$ such that $\operatorname{ann}(N)=P$. Then $N$ is an $R/P$-module. Since $R/P$ is simple Artinian, it has a unique simple module $V$, and $N \cong_{R/P} V^{(I)}$ for some index set $I$. Viewing $V$ as an $R$-module, also $N \cong_R V^{(I)}$. However, since $M$ is uniform, it cannot contain a submodule which is a proper direct sum. Thus, $N \cong V$. Since $M$ is the injective envelope of each of its nonzero submodules, $M=E(N)\cong E(V)$.