Prove that two follow conditions are equivalent.

Question

Problem: Let $M$ be a $R-$module and $N$ is a submodule of $M$ satisfies $\textit{Soc} (M) \subset N$. Prove that two follow conditions are equivalent.

For each element $x \in M, x \notin N$, there is a essential submodule $B$ of $M$ includes $N$ but not include $x$.
$N$ is the intersection of family of all essential submodule $B$ of $M$ such that $N \subset B$.

I have no idea to solve this problem, it so hard with me. Could you give me some hint to solve this problem. Thank you!

@Dbchatto67 $\textit{Soc} (M)$ denote for socle of module $M$. I think this is a international notion, because I see this notion in many module book. — Minh Nguyễn Hoàng, Jun 04 '19 at 04:57
Odd problem. It seems to just be a re-interpretation of the definition of the intersection of sets, and has nothing to do with modules or socles in any way. — rschwieb, Jun 04 '19 at 13:11

score 1 · Accepted Answer · answered Jun 04 '19 at 06:07

This is a straightforward argument that doesn't depend on the underlying algebraic structures in any essential way. I'll write $E$ to denote the set of essential submodules of $M$ containing $N$.

$(2) \implies (1):$ Suppose $(2)$, so $N = \cap_{B \in E} B$. If $x \in M \backslash N$, then $x \notin \cap_{B \in E} B$, so this implies that $\exists B \in E$ such that $x \notin B$ (or else $x$ would be in the above intersection). This is the statement of $(1)$.

$(1) \implies (2):$ Suppose $(1)$, so $x \in M \backslash N \implies \exists B \in E$ such that $x \notin B$. The contrapositive of this statement is $(\forall B \in E$ $x \in B)$ $\implies x \in N$, or $\cap_{B \in E} B \subset N$. Since the reverse inclusion is true by definition of $E$, this gives $N = \cap_{B \in E} B$, the statement of $(2)$.

Prove that two follow conditions are equivalent.

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