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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.

Asaf Karagila
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wxu
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1 Answers1

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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.

Qiaochu Yuan
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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 05 '12 at 23:42
  • @Asaf: I don't know what it means to take a reduced product by a filter. – Qiaochu Yuan Mar 05 '12 at 23:45
  • http://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 05 '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 05 '12 at 23:58