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Suppose $M$ is a $A$-module, $A$ is a commutative ring with 1, such that for every countably generated submodule $N$ of $M$, there exists a finitely generated submodule $L$ which contains $N$.

Must $M$ be finitely generated?

(Maybe it should be tagged by set-theory? )

Thanks.

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I don't see why it's [set-theory], to be honest. It's an interesting question, though. –  Asaf Karagila Mar 5 '12 at 23:26
    
By the way, not assuming the axiom of choice I can give some interesting examples for such $M$ which is not finitely generated! :-) –  Asaf Karagila Mar 5 '12 at 23:30

1 Answer 1

up vote 4 down vote accepted

Let

  • $X$ be an uncountable set,
  • $F$ be a field,
  • $A$ be the ring of functions $X \to F$ which are constant except possibly on a countable subset of $X$,
  • $M$ be the left $A$-module of functions $X \to F$ which are zero except possibly on a countable subset of $X$.

Then every countably generated submodule of $M$ is in fact contained in a submodule generated by one element (given a sequence $m_1, m_2, ... \in M$, the submodule they generate is contained in the submodule generated by an $m$ which is nonzero whenever any of the $m_i$ is nonzero), but $M$ itself is uncountably generated. Both properties follow from the fact that a countable union of countable subsets of $X$ is countable.

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$A$ is just the reduced product of $\aleph_1$ many copies of $R$ by the filter of cocountable sets, right? –  Asaf Karagila Mar 5 '12 at 23:42
    
@Asaf: I don't know what it means to take a reduced product by a filter. –  Qiaochu Yuan Mar 5 '12 at 23:45
    
en.wikipedia.org/wiki/Reduced_product - I mean that we identify two sequences in the product if they are the same on a cocountable set. However when I think about it, $A$ is not this sort of subring. –  Asaf Karagila Mar 5 '12 at 23:49
    
@Asaf: on further reflection, you are right that I don't use the ordinal structure on $\omega_1$. I've edited to reflect the structure I'm actually using. –  Qiaochu Yuan Mar 5 '12 at 23:58

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