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This is from P134 of Rotman's Homological Algebra book.

If R is a PID, then every torsion-free R-module is flat.

Proof. If R is a PID, then every finitely generated R-module M is a direct sum of cyclic modules. If M is torsion-free, then it is a direct sum of copies of R.

So if M is not torsion-free, we cannot conclude that each cyclic module is isomorphic to R? Why? Any help would be appreciated! Thank you

scsnm
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    If $R = \mathbb{Z}$ and $M = \mathbb{Z}\oplus \left( \mathbb{Z} / 2\mathbb{Z}\right)$, then $M$ is a direct sum of cyclic modules over $R$ but is not a sum of copies of $R$ – Didier Jun 25 '20 at 10:57
  • Hi Dldier! Thank you for the example. I understand this. But how being torsion-free gives us each cyclic module isomorphic to R? Would you please shed some lights for me? – scsnm Jun 25 '20 at 11:03
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    If $C$ is a cyclic $R$-module, generated by $a$, you have a surjective morphism $R\to C$, $,\lambda\mapsto \lambda a$, and if $C$ is torsion free, this morphism is also injective. – Bernard Jun 25 '20 at 11:14
  • @Bernard do you mean if R is torsion free? – scsnm Jun 25 '20 at 11:21
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    Why, yes, of course, a ring is torsion free as a module over itself. Actually, any free module is torsion free. – Bernard Jun 25 '20 at 11:23
  • @Bernard Please let me know if I understand you correctly, thank you!!! : C is a cyclic direct summand of M and M is torsion free. Then there is that surjective morphism you mentioned.... – scsnm Jun 25 '20 at 11:27
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    Any submodule of a torsion-free module is torsion-free, whether a direct summand or not. What's the problem with the surjective morphism? – Bernard Jun 25 '20 at 11:29
  • @ Bernard no problem at all. I just didn't copy down what you wrote about the morphism. thx. clear now! – scsnm Jun 25 '20 at 11:31

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