Invertible module

Remarks

An $R$-module is invertible exactly when it is finitely generated projective of constant rank $1$. In that case one can take $N = \Hom_R(M, R)$, and the evaluation map $M \otimes_R \Hom_R(M, R) \to R$ is an isomorphism. The isomorphism classes of invertible modules form an abelian group under $\otimes_R$, the Picard group $\mathrm{Pic}(R)$.

Example: let $R$ be a Dedekind domain with field of fractions $K$ and $I$ a nonzero ideal. The inverse fractional ideal $I^{-1} = \set{x \in K : xI \subseteq R}$ satisfies $I I^{-1} = R$, and multiplication gives $I \otimes_R I^{-1} \cong I I^{-1} = R$. So $I$ is invertible.

Non-example: over $R = \Z$, the module $\Z/2\Z$ is not invertible. For every $\Z$-module $N$, the group $\Z/2\Z \otimes_\Z N \cong N/2N$ is killed by $2$, so it is not isomorphic to $\Z$. The free module $R^2$ over a nonzero ring is not invertible either, since it has rank $2$.