The ideal class group is the Picard group of rank-one projectives

Send the class of an invertible fractional ideal $I$ to the isomorphism class of $I$ as an $R$-module. If $I=(a)J$ with $a\in K^{\times }$, then multiplication by $a$ is an $R$-module isomorphism $J\to I$. Therefore the map depends only on the class of $I$ in [F1].

For invertible fractional ideals $I$ and $J$, multiplication gives an $R$-bilinear map $I\times J\to IJ$, hence an $R$-linear map $\mu :I{\otimes }_{R}J\to IJ$. After localising at a maximal ideal $\mathfrak{m}$, both $I_{\mathfrak{m}}$ and $J_{\mathfrak{m}}$ are principal by [L1], so ${\mu }_{\mathfrak{m}}$ identifies with the tensor-unit isomorphism $R_{\mathfrak{m}}{\otimes }_{R_{\mathfrak{m}}}{R}_{\mathfrak{m}}\cong R_{\mathfrak{m}}$ from [L4]. Thus every ${\mu }_{\mathfrak{m}}$ is an isomorphism, and [L3] makes $\mu$ an isomorphism. Hence the map of step 1.1 is a group homomorphism.

The map is surjective by [L1], because every rank-one projective module is represented by an invertible fractional ideal. If the class of $I$ maps to the free module class, choose an isomorphism $R\to I$ and let $x$ be the image of $1$. Then every element of $I$ is $rx$, so $I=xR$ is principal. Therefore the kernel is trivial.

Steps 1.1, 2.1, and 2.2 prove the asserted canonical isomorphism $\mathrm{Cl}(R)\cong \mathrm{Pic}(R)$.