[proofplan] We use the unique factorization of nonzero fractional ideals in a Dedekind domain. The divisor map records the prime ideal exponents in the principal fractional ideal $xA$. Its kernel consists exactly of those $x\in F^\times$ for which $xA=A$, which is precisely $A^\times$. Finally, the divisor group maps onto the ideal class group by sending a formal prime divisor to the corresponding fractional ideal class, and its kernel is exactly the subgroup of principal divisors. [/proofplan]

[step:Define the divisor and class maps using fractional ideals] Let $\mathcal I(A)$ denote the abelian group of nonzero fractional ideals of $A$ under ideal multiplication. By the standard fractional ideal factorization theorem for Dedekind domains, the hypothesis that $A$ is Dedekind implies that every $I\in\mathcal I(A)$ has a unique factorization \begin{align*} I=\prod_{\mathfrak p\ne 0}\mathfrak p^{n_{\mathfrak p}(I)}, \end{align*} where $\mathfrak p$ ranges over the nonzero prime ideals of $A$, each $n_{\mathfrak p}(I)\in\mathbb Z$, and all but finitely many $n_{\mathfrak p}(I)$ are zero. For $x\in F^\times$, define $v_{\mathfrak p}(x):=n_{\mathfrak p}(xA)$. This gives the map \begin{align*} \operatorname{div}:F^\times\to\operatorname{Div}(A),\qquad x\mapsto\sum_{\mathfrak p\ne0}v_{\mathfrak p}(x)[\mathfrak p]. \end{align*} The sum is finite because $xA$ is a nonzero fractional ideal. The map is a [group homomorphism](/page/Group%20Homomorphism): for $x,y\in F^\times$, the equality of principal fractional ideals $(xy)A=(xA)(yA)$ and uniqueness of fractional ideal factorization give \begin{align*} v_{\mathfrak p}(xy)=v_{\mathfrak p}(x)+v_{\mathfrak p}(y) \end{align*} for every nonzero prime ideal $\mathfrak p$, hence $\operatorname{div}(xy)=\operatorname{div}(x)+\operatorname{div}(y)$. Define \begin{align*} \theta:\operatorname{Div}(A)\to\operatorname{Cl}(A) \end{align*} by \begin{align*} \theta\left(\sum_{\mathfrak p\ne0}n_{\mathfrak p}[\mathfrak p]\right)=\left[\prod_{\mathfrak p\ne0}\mathfrak p^{n_{\mathfrak p}}\right], \end{align*} where the bracket on the right denotes the class of a nonzero fractional ideal modulo principal fractional ideals. This is a group homomorphism because multiplication of fractional ideals adds the exponents in their prime ideal factorizations. [/step]

[step:Identify the kernel of the divisor map with the unit group]The inclusion $A^\times\to F^\times$ sends a unit $u\in A^\times$ to the same element regarded in $F^\times$. If $u\in A^\times$, then $uA=A$, so $v_{\mathfrak p}(u)=0$ for every nonzero prime ideal $\mathfrak p$. Hence $\operatorname{div}(u)=0$, and the image of $A^\times$ is contained in $\ker(\operatorname{div})$. Conversely, let $x\in F^\times$ satisfy $\operatorname{div}(x)=0$. Then $v_{\mathfrak p}(x)=0$ for every nonzero prime ideal $\mathfrak p$. By the fractional ideal factorization theorem for Dedekind domains, \begin{align*} xA=\prod_{\mathfrak p\ne0}\mathfrak p^{v_{\mathfrak p}(x)}=A. \end{align*} Thus $x\in A$ and $x^{-1}\in A$, so $x\in A^\times$. Therefore \begin{align*} \ker(\operatorname{div})=\operatorname{im}(A^\times\to F^\times). \end{align*}[/step]

[guided]The divisor map measures the failure of a nonzero element of the fraction field to generate the unit fractional ideal. Let $u\in A^\times$. Since $u$ has an inverse in $A$, multiplication by $u$ does not change the ideal $A$: \begin{align*} uA=A. \end{align*} Therefore the prime ideal factorization of $uA$ has exponent $0$ at every nonzero prime ideal $\mathfrak p$, so $v_{\mathfrak p}(u)=0$ for every $\mathfrak p$. Hence \begin{align*} \operatorname{div}(u)=0. \end{align*} This proves that every unit lies in the kernel of $\operatorname{div}$. Now suppose $x\in F^\times$ and $\operatorname{div}(x)=0$. By definition of $\operatorname{div}$, this means that each exponent $v_{\mathfrak p}(x)$ in the factorization of the principal fractional ideal $xA$ is zero. The unique factorization of nonzero fractional ideals in a Dedekind domain then gives \begin{align*} xA=\prod_{\mathfrak p\ne0}\mathfrak p^{v_{\mathfrak p}(x)}=\prod_{\mathfrak p\ne0}\mathfrak p^0=A. \end{align*} Let $1_A$ denote the multiplicative identity of $A$. The equality $xA=A$ has two consequences. Since $x=x\cdot 1_A\in xA=A$, we have $x\in A$. Also, because $1_A\in xA$, there exists $a\in A$ with $xa=1_A$; hence $a=x^{-1}$ in $F$, so $x^{-1}\in A$. Thus $x$ is a unit of $A$. We have proved both inclusions, and therefore \begin{align*} \ker(\operatorname{div})=\operatorname{im}(A^\times\to F^\times). \end{align*}[/guided]

[step:Show that principal divisors are exactly the kernel of the class map] For $x\in F^\times$, the definition of $\theta$ gives \begin{align*} \theta(\operatorname{div}(x))=[xA]. \end{align*} The fractional ideal $xA$ is principal, so its class in $\operatorname{Cl}(A)$ is the identity. Hence \begin{align*} \operatorname{im}(\operatorname{div})\subseteq\ker(\theta). \end{align*} Conversely, let \begin{align*} D=\sum_{\mathfrak p\ne0}n_{\mathfrak p}[\mathfrak p]\in\operatorname{Div}(A) \end{align*} satisfy $\theta(D)=1$ in $\operatorname{Cl}(A)$. Define the associated fractional ideal \begin{align*} I_D:=\prod_{\mathfrak p\ne0}\mathfrak p^{n_{\mathfrak p}}. \end{align*} The condition $\theta(D)=1$ means that $I_D$ is principal, so there exists $x\in F^\times$ such that \begin{align*} I_D=xA. \end{align*} Comparing the unique prime ideal factorizations of $I_D$ and $xA$ supplied by the fractional ideal factorization theorem gives $n_{\mathfrak p}=v_{\mathfrak p}(x)$ for every nonzero prime ideal $\mathfrak p$. Therefore \begin{align*} D=\operatorname{div}(x). \end{align*} Thus $\ker(\theta)\subseteq\operatorname{im}(\operatorname{div})$, and consequently \begin{align*} \ker(\theta)=\operatorname{im}(\operatorname{div}). \end{align*} [/step]

[step:Use fractional ideal factorization to prove surjectivity onto the class group] Let $[I]\in\operatorname{Cl}(A)$ be the class of a nonzero fractional ideal $I\in\mathcal I(A)$. By the fractional ideal factorization theorem for Dedekind domains, there are integers $n_{\mathfrak p}(I)$, all but finitely many zero, such that \begin{align*} I=\prod_{\mathfrak p\ne0}\mathfrak p^{n_{\mathfrak p}(I)}. \end{align*} Define \begin{align*} D_I:=\sum_{\mathfrak p\ne0}n_{\mathfrak p}(I)[\mathfrak p]\in\operatorname{Div}(A). \end{align*} Then \begin{align*} \theta(D_I)=[I]. \end{align*} Hence $\theta$ is surjective. Combining this with the two kernel computations proves exactness of \begin{align*} A^\times \longrightarrow F^\times \xrightarrow{\operatorname{div}} \operatorname{Div}(A) \xrightarrow{\theta} \operatorname{Cl}(A) \longrightarrow 0. \end{align*} [/step]