# Equivalent Definitions of the Socle of a Module

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.

• I think $Re$ needs not be simple. Nov 8, 2020 at 21:36

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.