Nonprincipal ideals of an integral domain have rank 1 but are not free

Let R be an integral domain, and let M \subseteq R be a nonprincipal ideal of R, considered as a (left, unital) R-module. Prove that M has rank 1 over R but is not free as an R-module.

Let \{x,y\} \subseteq M be a subset containing two nonzero elements. Since x \cdot y + (-y) \cdot x = 0, \{x,y\} is not linearly independent. Thus the rank of M as an R-module is at most 1. Since R is an integral domain, every (nonzero) singleton set is linearly independent; hence M has rank 1.

Suppose M is free. Then M must have free rank 1, since any free generating set is linearly independent. That is, M = R\alpha for some \alpha. But then M is principal, a contradiction.

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