1
$\begingroup$

Let $M$ be any $R$-module. Let $A$ be the submodule generated by all artinian submodules of $M$. Show any nonzero submodule of $A$ contains at least one simple submodule.

My attempt: By definition, $A=\sum_{S \leq M} S$, where $S$ is artinian. But this is also $A=\bigcap_{K \subseteq T,K \text{ artinian}}T$; that is, it is the intersection of all submodules of $M$ containing every artinian submodule of $M$. So if $S \leq A$, then it is a submodule of every submodule of $M$ containing all artinian submodules. So in particular, it contains an artinian submodule. This submodule contains a simple submodule, which is then a submodule of $S$.

Is this really that simple?

$\endgroup$
1
$\begingroup$

Your statement about the intersection is quite dubious and the argument is unfortunately wrong.


Prove that, if $L$ is a nonzero submodule of $A$ and $0\ne x\in L$, then $xR$ (or $Rx$ if you're dealing with left modules) is artinian.

Every nonzero artinian module has a simple submodule.

$\endgroup$
  • $\begingroup$ Yes, it felt far to swift to me. Using your idea, would it be just the fact that $x=s_1+\cdots+s_n$, where $s_i \in S_i$ is artinian. Then I have the exact sequence: $0 \to Rx \to S_1+\cdots+S_n \to (\sum S_i)/(Rx)$. Then $Rx$ is artinian being the submodule of an artinian module? $\endgroup$ – NPH Jan 9 '17 at 21:45
  • $\begingroup$ @NPH Yes: $x$ belongs to a finite sum of artinian submodules of $M$ and a sum of a finite number of artinian submodules is artinian. $\endgroup$ – egreg Jan 9 '17 at 21:47
  • $\begingroup$ I didn't try your line of thought because trying to show a submodule of a submodule of a module generated by artinian submodules was artinian started to kill my head and seemed like a train of thought not worth pursuing. But taking your suggestion, took me all of 30 seconds! How wrong I was! Thanks! $\endgroup$ – NPH Jan 9 '17 at 21:55

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.