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.


  • $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.