# $\mathcal{C}$-filtering of modules inherited by submodules

I'll state the question about modules, but I'm open to examples in other contexts. I am not an algebraist, so please forgive any non-conventional terminology.

DEFINITION: Let $$\mathcal{C}$$ be a collection of (countable) modules (for some fixed ring). Given a module $$M$$ of size $$\omega_1$$, let's say that $$M$$ is "strongly $$\mathcal{C}$$-filtered" iff there is a $$\subseteq$$-increasing and continuous sequence $$\langle M_i \ : \ i < \omega_1 \rangle$$ of countable submodules, with union M, such that $$M_i \in \mathcal{C}$$ and $$M_j / M_i \in \mathcal{C}$$ whenever $$i < j < \omega_1$$.

QUESTION: What are some nontrivial examples of classes $$\mathcal{C}$$ that have the following inheritance" property: whenever $$|M|=\omega_1$$ and $$M$$ is strongly $$\mathcal{C}$$-filtered, then every submodule of $$M$$ of size $$\omega_1$$ is also strongly $$\mathcal{C}$$-filtered?

The only example I'm aware of is where $$\mathcal{C}$$ is the collection of (countable) free abelian groups (i.e. countable free $$\mathbb{Z}$$-modules). In this situation, for an abelian group $$M$$ of size $$\omega_1$$, it's easy to see that $$M$$ is strongly $$\mathcal{C}$$-filtered if and only if $$M$$ is free. And, since subgroups of free abelian groups are always free, $$\mathcal{C}$$ has the inheritance property requested in the question.

(More generally, $$\mathcal{C}$$ = "the set of all countable free $$R$$-modules", where $$R$$ is a PID, works for the same reason, since submodules of free modules are free if the ring is a PID. Of course in these examples, one could just as well let $$\mathcal{C} = \{ F_\omega \}$$ where $$F_\omega$$ is the free object on a countably infinite set of generators. But in the question I don't necessarily require that $$\mathcal{C}$$ be a singleton, or even a countable, collection).

• Whitehead's Problem comes to mind... – Asaf Karagila Aug 8 at 16:43
• I would be interested in knowing if Whitehead groups of size $\omega_1$ can be characterized by strong $\mathcal{C}$-filtering, for some $\mathcal{C}$ that satisfies the requirements of the question (assuming we're in a universe where there is a non-free Whitehead group, of course). – Sean Cox Aug 8 at 19:09
• Well, in the proof that $\lozenge$ implies that every Whitehead group is free there is a characterization of what and where things fail. To my understanding that was the origin of club guessing, too. – Asaf Karagila Aug 8 at 20:30