Non-finitely-generated module is union of countable chain?

Question

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.

Answer 1

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 $M$.