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I know this is the case if $R$ is a PID, but PID's are special instances of Integral Domains, so I am wondering if there is a counter-example to the case where R is an integral domain.

This post shows the result when $R$ is a Principal Ideal Domain: Proof about finitely generated torsion-free R-module M is free, where R is a PID

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  • $\begingroup$ Geometrically, you are asking if all line bundles over an (irreducible) affine variety are trivial. The simplest counterexample is to take the line bundle $L(P - Q)$ on an elliptic curve, punctured at the origin, where $P$ and $Q$ are distinct points not equal to the origin. $\endgroup$ – hunter May 17 '19 at 2:07
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Let $R$ be a Dedekind domain, which is not a principal ideal domain, for example the ring of integers in a number field such as $R=\Bbb Z[\sqrt{-6}]$. Then $R$ will have a non-principal ideal, $I$ say. Then, as an $R$-module, $I$ is finitely generated, torsion free, but not principal. In the above example, we could take $I=2R+\sqrt{-6}R$.

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