3
$\begingroup$

Wikipedia gives the following definitions of the socle of an $R$-module $M$: $$\text{Soc}(M)=\sum \left\{S:S\subseteq M\text{ is simple}\right\}:=S_1$$ and $$\text{Soc}(M)=\bigcap\left\{ E:E\subseteq M\text{ is essential}\right\}:=S_2$$

I'm trying to show that these are equivalent.


I can show that $S_1\subseteq S_2$:

Suppose that $S\subseteq M$ is simple. If $x\in S$ is non-zero then $Rx=S$. For any essential $E\subseteq M$ we have $Rx\cap E\neq0$, and so $Rx\cap E=Rx$ by simplicity. Then $Rx\subseteq E$, so $x\in E$.


However I'm struggling to show the converse. Here is what I have tried so far:

Suppose that $e\in E$ for every essential $E\subseteq M$. I need to show that $e$ can be written as a sum of elements in simple submodules, so I thought I'd try to show that $Re$ is simple.

If not, then we have some $0\subsetneq N\subsetneq Re$, so there exists some $r\in R$ such that $re\notin N$. If $e\in E$ for every essential $E\subseteq M$, then $re$ does also.

Then it would be enough to show that $N\subseteq M$ is essential for a contradiction. Since $$N\subsetneq Re\subseteq E\subseteq M$$ it would then be enough to show that $N\subsetneq Re$ and $Re\subseteq E$ are essential extensions. Unfortunately I can't seem to prove either, and so I'm beginning to doubt that this is the right approach.

Any help would be much appreciated.

$\endgroup$
1
  • 2
    $\begingroup$ I think $Re$ needs not be simple. $\endgroup$
    – Berci
    Nov 8, 2020 at 21:36

1 Answer 1

2
$\begingroup$

This proof follows Proposition 8.8 in these notes.

Let $N$ be any submodule of $S_2$. By Zorn's Lemma, we can find a module $N'\subseteq M$ which is maximal with respect to the property that $N\cap N'=0$.

Then $N\oplus N'\subseteq M$ is essential, since if $L\cap(N\oplus N')=0$ then $N'\oplus L$ would contradict the maximality of $N'$.

This proves that $N\subseteq S_2\subseteq N\oplus N'$, since $S_2$ is the intersection of all essential submodules of $M$.

Then $$S_2=S_2\cap(N\oplus N')=N\oplus(S_2\cap N')$$ so any submodule of $S_2$ is a direct summand.

By the proof linked to here, this shows that $S_2$ is semisimple, and so is the direct sum of its simple submodules.

Then $S_2\subseteq S_1$ and we are done.

$\endgroup$

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.