The principal-divisor exact sequence for a Dedekind domain

The map $\mathrm{cl}$ is well defined and surjective: by [L1], every divisor is a finite sum and therefore determines a unique fractional ideal $\prod {\mathfrak{p}}^{n_{\mathfrak{p}}}$, and every ideal class has such a representative.

If $x\in K^{\times }$, then [L1] applied to the principal fractional ideal $(x)$ gives $(x)={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{v_{\mathfrak{p}}((x))}\mathrm{.}$ Therefore $\mathrm{cl}(\mathrm{div}(x))=[(x)]=0$ in $\mathrm{Cl}(R)$. So $\mathrm{im}(\mathrm{div})\subseteq \ker (\mathrm{cl})$.

Conversely, if $D=\sum n_{\mathfrak{p}}[\mathfrak{p}]$ satisfies $\mathrm{cl}(D)=0$, then the ideal $\prod {\mathfrak{p}}^{n_{\mathfrak{p}}}$ is principal, say equal to $(x)$ with $x\in K^{\times }$. Uniqueness in [L1] then forces $D=\mathrm{div}(x)$. Thus $\ker (\mathrm{cl})=\mathrm{im}(\mathrm{div})$.

If $x\in R^{\times }$, then $(x)=R$, so all valuations of $(x)$ are zero and $\mathrm{div}(x)=0$. Conversely, if $\mathrm{div}(x)=0$, then [L1] gives $(x)=R$, which is exactly the statement that $x$ is a unit of $R$. Hence $\ker (\mathrm{div})=R^{\times }$.

Steps 1.1, 1.2, 1.3, and 2.1 prove exactness of the displayed sequence.