# Injective hull of a simple module

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!

• Look at Theorem 3.52 in Lectures on Modules and Rings. The characterizations in (2) and (3) should get you there. – moonlight Aug 31 '15 at 8:41
• @moonlight Since $R/P$ is simple Artinian we have $R/P≅V^n$ for some $n$ as $R/P$-modules so also as $R$-modules. By taking injective hulls, we get $E(R/P)≅E(V)^n$. But, by the proof of Theorem 3.60, $E(R/P)=M_1⊕...⊕M_k$ where $M_i$ are (isomorphic) indecomposable injective $R$-modules with $Ass(M_i)=P$, so by Krull-Schmidt theorem we have $E(V)≅M_i$. Now, I am not sure why $M_i$ is isomorphic with the module $M$ in my question. – karparvar Aug 31 '15 at 14:10
• @karparvar Can you show such an $M$ has an essential socle? If so, the socle is simple by indecomposability, and then $M$ is the hull of that simple submodule. – rschwieb Sep 1 '15 at 17:27
• I posted my earlier comment as an answer and deleted it. – moonlight Sep 1 '15 at 20:44

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, $M=E(N)\cong E(V)$.