Let $R$ be a ring with $1$ and $M_R$ any right $R$-module. Assume that $M$ has the descending chain condition on essential submodules. Let $\mathrm{soc}(M)= \bigcap_{i\in I}A_i$ be the intersection of all essential submodules of $M$. How can I prove that $\mathrm{soc}(M)$ is a finite intersection of essential submodules?!.
This is my attempt: if $I$ is countable, say $I=\lbrace i_1,i_2,\ldots \rbrace$, then we have the chain \begin{align} A_{i_1} \supseteq A_{i_1} \cap A_{i_2} \supseteq A_{i_1} \cap A_{i_2} \cap A_{i_3} \supseteq \ldots \end{align} and so, by the hypothesis, there exists $m\in \mathbb{N}$ such that $A_{i_1} \cap \ldots \cap A_{i_m} = \bigcap_{k=1}^{\infty}A_{i_k}$.
But if $I$ is not countable, what can I do ?!.
The question in another form may be as follows: How to prove that $\mathrm{soc}(M)$ is essential in $M$ ?!. In fact, this is the required. I try to prove the intersection of all essential submodules is finite in order to prove this statement.
Thanks in advance.
The statement of the descending chain condition is rooted in countable chains (well-founded ones, in fact) to begin with, so $I=\mathbb N$. There is no need to concern yourself with uncountable chains or chains with other order structures.
Your argument already suffices.