# $R$ is artinian $\implies$ $(eR)_R$ is local $R$-module for an idempotent $e \in R$

Let $R$ be a ring with unity $1$ and artinian. Consider $R$ as right $R$-module $R_R$. Since it is artinian we have a finite decomposition of $R$ into right ideals:

$$R_R = A_1 \oplus ... \oplus A_n$$ where each $A_i$ is an indecomposable right ideal generated by an idempotent $e \in R$, that is $A_i = e_iR$ for every $i=1,...,n$.

I want to prove that each $e_iR$ is a local right $R$-module. By a local module I mean cyclic, non-zero and has a unique maximal proper submodule.

Do you have any idea?

• The hypotheses are not enough to guarantee it: $e=1$ is idempotent, but $eR$ is not necessarily local. Maybe $A_1,\dots, A_n$ are indecomposable? – egreg Jun 29 '18 at 8:35
• You're right, i've edited. – bozcan Jun 29 '18 at 8:46

Let $J$ be the Jacobson radical of $R$. Then $R/J$ is a semisimple artin ring which can be written as a direct sum of $n$ simple modules. Actually each $A_i/A_iJ$ is simple $R$-module and $rad(A_i)=A_iJ$.