why is torsion free needed

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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