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I've been trying to show that, given a ring R and a R-module M, the two properties below are equivalent.

$1)$ M is finitely co-generated (that is for a given family $ \{N_i\}_i$ of submodules of M such that $ \bigcap_iN_i = \{0\} $ there exists a finite number of elements $ \{N_{i_1},...,N_{i_k}\} $ of the family such that $ \bigcap_{j=1}^{k} N_{i_j} = \{0\} $);

$2)$ For every countable chain of non-zero submodules $ N_1 > N_2 > ... $ we have $ \bigcap_{i\in\omega}N_i \neq \{0\}$.

The implication $1)\rightarrow2)$ is imediate, but I can't see a way of proving the other one.

Thanks.

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  • $\begingroup$ Is there a typo? How could $ \bigcap_{i\in\omega}N_i $ be anything but $N_1$ at the bottom of the chain? Did you mean for inclusions to run the opposite direction? $\endgroup$ – rschwieb Oct 1 '15 at 3:25
  • $\begingroup$ Where was this problem proposed? $\endgroup$ – rschwieb Oct 17 '15 at 19:07
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It's no surprise you're having trouble proving that (2) implies (1), because it isn't true! For a counterexample, let $G$ be the free abelian group on the first uncountable ordinal $\omega_1$ (or more generally, any ordinal with uncountable cofinality). Make $G$ into a totally ordered group by ordering it reverse-lexicographically (that is, if $\alpha<\beta<\omega_1$, then every integer multiple of $\alpha$ is less than $\beta$ as elements of $G$). Let $R$ be any valuation ring with value group $G$ and let $M=R$. Then $M$ satisfies (2) but not (1). Indeed, submodules of $M$ are in bijection with upward-closed subsets of the positive part of $G$, and every countable subset of $G$ has an upper bound, so (2) holds. But the collection of all nonzero ideals is a counterexample to (1).

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