Let $R$ be a ring, $\Sigma$ be a multiplicatively closed subset of $R$. $M$ is an $R$-module. Denote the injective hull of $M$ by $E(M)$.


$M$ is $\Sigma$-torsion if for any $m$ in $M$, there is $\sigma \in \Sigma$ such that $\sigma m = 0$. we say that $\Sigma$ operates regularly on $M$ if $\sigma m =0$ implies $m =0$.


i want to show that


(i) $\Sigma$ operates regularly on every uniform injective $\Sigma$-torsionfree module 


implies 


(ii) for all $x \in \Sigma$, $R/Rx$ is $\Sigma$-torsion.


Proof:


Assume $R/Rx$ is not $\Sigma$-torsion. write $T(R/Rx) = N/Rx$, where $T$ is the torsion radical. Then $N$ $\neq$ $R$. Note that $R/N = (R/Rx) / T(R/Rx)$ which is $\Sigma$-torsionfree. 


Consider $R/Rx \rightarrow R/N \rightarrow E(R/N) = \oplus E\_i$ where the $E\_i$ are indecomposable and injective. **(Why????)** 


then $x+Rx = x(1+Rx)$ is mapped to $(xe\_i)\_i$, where $e\_i$ is non zero for some $i$


but $x+Rx = 0$ so that $(xe\_i)\_i = (0)\_i$. 


Since $x$ operates regularly on $E\_i$, $e\_i =0$ for all $i$, contradiction.


End of Proof


Remark:
$E\_i$ is $\Sigma$-torsionfree, because the class of $\Sigma$-torsionfree modules is closed under essential extensions and submodules.



decomposition of the injective hull of a torsion free module
Here is an answer as I understood the question (it was changed in the meantime). The $R$-module $A$ has an injective hull (or envelope) $E(A)$ containing $A$. Then we ask about conditions ensuring that the
injective module $E(A)$ has a decomposition (as a direct sum of indecomposable submodules).
One possibility to ensure this is the following result of Bass, and Papp:


Theorem (Bass, Papp): Every injective (left)-module has a decomposition as a direct sum of indecomposable, injective submodules if and only if $R$ is left Noetherian (i.e., every left ideal is finitely generated).


