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I am reading some algebra notes (http://www.math.umn.edu/~garrett/m/algebra/notes/11.pdf).

Corollary 2.0.6 says that if $R$ is an integral domain, any finitely generated module module $M$ is isomorphic to a direct sum of a torsion module and a free module, $$ M\approx M^{\rm tors}\oplus F. $$ The torsion module is unique, but the free part is only unique up to isomorphism. I've seen this in other sources too.

Why is the free part of this decomposition not unique? Is there an example?

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Take $R = \mathbb Z$ and $M = \mathbb Z/2 \oplus \mathbb Z$. So the free part is clearly the submodule generated by $(0, 1)$: $$F = \langle(0, 1)\rangle$$ and the torsion part is $$T(M) = \langle(1, 0)\rangle.$$ But there is also the following submodule: $$F' = \langle(1, 1)\rangle$$ and you can check it is free, that $F' + T(M) = M$, and $F' \cap T(M) = 0$. Thus $M = F' \oplus T(M)$ and $M = F \oplus T(M)$ but $F \neq F'$.

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