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Let $R$ be a ring. Let $M$ be a left $R$-module.

Then: $M$ is not finitely generated <=> $M$ is the union of a set of proper submodules closed under binary sums. To recall why: (<=) If $M$ would be f.g., then chosen finitely many generators appear in certain members of the given set of proper submodules. Hence, this set being closed under binary sums, they all appear in one such submodule. So this submodule is not properly contained in $M$, contradiction. (=>) Pick the set of finitely generated submodules of $M$.

Strengthening the RHS of this equivalence a bit led me to the question: Can it happen that $M$ is not f.g., but also not the union of an $\mathbb{N}$-indexed chain of proper submodules? I.e. "does (=>) break down if one wants an $\mathbb{N}$-indexed chain"?

If $R$ is a field and $M$ an infinite-dimensional vector space, such a chain exists: pick a basis of $M$, write it as $B\sqcup \{ b_i : i \in \mathbb{N} \}$ and form the chain $\langle B\sqcup\{b_1\}\rangle\subset\langle B\sqcup\{b_1,b_2\}\rangle\subset\dots$.

But for general $R$, I suspect a different behaviour.

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This chain condition is called "dually slender" and was already discussed on MO:… . The counterexample mentioned my a-fortiori also appears there. – Martin Brandenburg Feb 16 '12 at 9:07
Thanks! (I didn't look into that discussion when it was new, and it didn't show up automatically when I posted this question.) It is exactly what I was after. – Matthias Künzer Feb 17 '12 at 17:18
up vote 3 down vote accepted

Counterexample: let $R$ be a valuation ring with value group $\Gamma=\mathbf Z^{\omega_1}$ with the lexicographic ordering and $M$ the maximal ideal of $R$. Any proper submodule of $M$ is contained in an ideal of the form $\{r\in R\mid v(r)>\gamma\}$ for some $\gamma>0$, but every countable collection of positive elements of $\Gamma$ has a positive lower bound, so no countable union of proper submodules can be $M$.

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Thanks! Looks already pretty big, compared with what I had tried so far. – Matthias Künzer Feb 17 '12 at 17:13

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