Suppose that $R$ is a commutative ring, an $R$-module $M$ is said to be finitely embedded if $M$ has a finitely generated essential socle. Now Let $M$ be finitely embedded and not artinian, let $S$ be the set of all submodules $N$ of $M$ such that $M/N$ is not finitely embedded. Is it true that $S$ has a minimal element by Zorn's lemma?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Call $\mathcal F$ the family of all the submodules $N\subseteq M$ such that $M/N$ is not f.e.
How could you prove that given a chain $N_1 \subseteq N_2\subseteq\dots\subseteq N_n\subseteq \dots$ of elements of $\mathcal F$ then $N=\bigcup_n N_n$ is in $\mathcal F$? It seems not to be true by I cannot actually see counter examples now...
A characterization of f.e. modules can be obtained using Zorn's lemma, that is:
Theorem. E is f.e. iff any inverse system of non-zero submodules is bounded (from below) by a non-zero submodule.
(the above theorem is due to prof. Vamos and the proof is essentially a dualization of an analogous property for f.g. modules)
So finally the answer is, could you prove that $\mathcal F$ is an inverse system?
-
$\begingroup$ I don't understand a lot of the aforementioned algebraic concepts but in the past I played with some weird abelian groups and Zorn's lemma... and I'd say that it's not necessary to prove that $N=\bigcup_n N_n$ is in $\mathcal F$, it suffices to prove that there is another $N'\in\mathcal F$ such that $N\subseteq N'$ (Zorn's lemma only requires the existence of an upper bound for any chain, it doesn't ask for that upper bound to be exactly the union...) $\endgroup$ Mar 23, 2011 at 21:31