Picard group of a ring

Remarks

The operation is well defined, associative and commutative because the tensor product is so up to natural isomorphism. For invertible $P$, the evaluation map $P \otimes_R \Hom_R(P, R) \to R$ is an isomorphism, since it is one after localization at every maximal ideal. An $R$-module $P$ is invertible exactly when there is a module $Q$ with $P \otimes_R Q \cong R$. If $R$ is an integral domain, $\operatorname{Pic}(R)$ is isomorphic to the group of invertible fractional ideals modulo the principal fractional ideals.

Example: for $R = \Z[\sqrt{-5}]$, the ideal $P = (2, 1 + \sqrt{-5})$ is an invertible module that is not free, while $P \otimes_R P \cong P^2 = (2) \cong R$; in fact $\operatorname{Pic}(R)$ has order $2$. If $R$ is a principal ideal domain or a local ring, then $\operatorname{Pic}(R)$ is trivial.

Non-example: the module $R^2$ is finitely generated and projective, but $(R^2)_{\mathfrak{m}} \cong R_{\mathfrak{m}}^2$ is not isomorphic to $R_{\mathfrak{m}}$ for $R \ne 0$; so $R^2$ is not invertible and has no class in $\operatorname{Pic}(R)$.