# Decomposition of a not f.g. module over PID

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?

• @Bernard: That presumably means the torsion submodule, the set of $m\in M$ that are annihilated by some nonzero element of the ring. – Eric Wofsey Jul 26 '18 at 22:25
• @EricWofsey; Why, yes of course! – Bernard Jul 26 '18 at 22:28

## 1 Answer

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 $M/T$.