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.