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.

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

1 Answer 1


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.


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