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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.

Hussein Eid
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1 Answers1

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But if is not countable, what can I do ?!.

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.

rschwieb
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  • Excuse me, this statement 'There is no need to concern yourself with uncountable chains or chains with other order structures' is not clear to me!. I still don't know why my argument suffices. – Hussein Eid Jun 15 '23 at 14:20
  • @HusseinEid One can certainly imagine arbitrary linearly ordered subsets of the submodules of a module indexed by a well-founded totally ordered set reflecting the total order of the chain. That could be done with any ordinal. But the definition of the chain used in the ascending chain condition is, as far as I've ever seen, always taken to be one indexed by $\mathbb N$. – rschwieb Jun 15 '23 at 14:37
  • @HusseinEid If it makes you feel better, a strictly decreasing chain exists iff a countable strictly decreasing chain exists (you just define it inductively, noting that each element you come up with must be followed by another somewhere.). Therefore checking for the smaller class of chains suffices. This is probably why nobody frets about arbitrary chains. – rschwieb Jun 15 '23 at 17:43