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


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


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.

  • $\begingroup$ If $R$ is Noetherian then for every $R$-module $A$ we have $E(A)=\oplus_i E_i(A)$ with a family $\lbrace E_i(A)\rbrace $ of injective submodules. $\endgroup$ – Dietrich Burde May 18 '13 at 20:26
  • $\begingroup$ You can always take $I$ to be a singleton. $\endgroup$ – Fernando Muro May 19 '13 at 21:57
  • $\begingroup$ Thank you Fernando. i have explained myself better in my reply below. would you be able to tell me why i can decompose the injective hull in my case? Thank you very much. $\endgroup$ – Paslig Keir May 20 '13 at 19:40

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).

  • $\begingroup$ Thank you for your reply. i have explained the situation better in my own reply. your help is very much appreciated.:) $\endgroup$ – Paslig Keir May 20 '13 at 19:37

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.