3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Look at Theorem 3.52 in Lectures on Modules and Rings. The characterizations in (2) and (3) should get you there. $\endgroup$ – moonlight Aug 31 '15 at 8:41
  • $\begingroup$ @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. $\endgroup$ – karparvar Aug 31 '15 at 14:10
  • $\begingroup$ @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. $\endgroup$ – rschwieb Sep 1 '15 at 17:27
  • $\begingroup$ I posted my earlier comment as an answer and deleted it. $\endgroup$ – moonlight Sep 1 '15 at 20:44
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.