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For a module $M$ over PID which is not finitely generated, do we still have the isomorphism $M\cong Tor(M)\oplus M/Tor(M)$? If not can you give a counter-example?

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    $\begingroup$ @Bernard: That presumably means the torsion submodule, the set of $m\in M$ that are annihilated by some nonzero element of the ring. $\endgroup$ – Eric Wofsey Jul 26 '18 at 22:25
  • $\begingroup$ @EricWofsey; Why, yes of course! $\endgroup$ – Bernard Jul 26 '18 at 22:28
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No. For instance, over the ring $\mathbb{Z}$, let $M=\prod\mathbb{Z}/(p)$ where the product ranges over all primes $p$. Then the torsion subgroup $T$ of $M$ is just the direct sum $\bigoplus \mathbb{Z}/(p)$, set set of elements of $M$ which have only finitely many nonzero coordinates. The quotient $M/T$ is then divisible, since for any $m\in M$ and any nonzero $a\in\mathbb{Z}$, we can divide $m$ by $a$ after modifying only finitely many coordinates of $m$ (namely, those corresponding to primes that divide $a$).

However, the only element of $M$ which is divisible by every nonzero integer is $0$, since if $m\in M$ is divisible by a prime $p$ then its $p$-coordinate must be $0$. So $M$ has no nontrivial divisible subgroups, and in particular it does not have a subgroup isomorphic to $M/T$.

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