Modules in which its torsion submodule is a direct summand

Question

Let $A$ be a commutative domain.

For every $A$-mod $M$ we can define its torsion submodule as $T(M)=\{m\in M \mid \exists a\in A\setminus \{0\} \text{ such that } am=0 \}$. Hence we have an exact sequence

$$0\rightarrow T(M) \rightarrow M \rightarrow M/T(M)\rightarrow 0$$

Where the last term is torsion free. This sequence split iff $T(M)$ is a direct summand of $M$ and in that case we can write $M$ as the direct sum of a torsion module with a torsion-free module.

This happens for example when $A$ is a PID and $M$ is finitely generated by the structure theorem that we have in this case. Also this happens trivially when $A$ is a field.

My questions are

Is there an example of a sequence that doesn't split in the case that $A$ is PID and $M$ is not finitely generated?

and

Is there known any other "big family" of module for which this exact sequence split?

How about the $\mathbb{Z}$-module $\mathbb{R} / \mathbb{Z}$ which has torsion module $\mathbb{Q} / \mathbb{Z}$? (Not sure if it actually works as an example...) — Daniel Schepler, Oct 03 '17 at 23:39
$\mathbb Q/\mathbb Z$ is injective so the sequence must split, though I can't come up with the splitting map at the moment. — Jim, Oct 03 '17 at 23:56
Doesn't this question answer the first question? Everything there is a $\mathbb{Z}$-module. — Andrew Kelley, Oct 04 '17 at 00:48
@AndrewKelley ooh it's a nice example, I think that's a good answer for my first question. — nowhere dense, Oct 04 '17 at 01:43
Regarding your second question: It seems that this article (which refers to Kaplansky, Irving 1960) suggests that the sequence splits for all fin. gen. $M$ if and only if $A$ is a Prüfer domain. — Bubaya, Sep 12 '19 at 12:38
@Bubaya that's such an interesting article, it addresses exactly this question! Thank you for sharing it. — nowhere dense, Sep 12 '19 at 15:15

Modules in which its torsion submodule is a direct summand

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