[proofplan] We compare the two groups through the same objects: rank-one invertible $A$-modules. A nonzero fractional ideal over a Dedekind domain is invertible, so it defines an element of $\operatorname{Pic}(A)$, and ideal multiplication matches [tensor product](/page/Tensor%20Product). Principal fractional ideals map to the identity, isomorphic fractional ideals differ by multiplication by an element of $K^\times$, and every invertible module can be embedded into $K$ as a fractional ideal. [/proofplan] [step:Define the map from fractional ideals to invertible modules] Let $K=\operatorname{Frac}(A)$. Let $\mathcal I(A)$ denote the abelian group of nonzero fractional ideals of $A$ under ideal multiplication, and let $\mathcal P(A)\subset \mathcal I(A)$ denote the subgroup of principal fractional ideals $aA$ with $a\in K^\times$. Thus \begin{align*} \operatorname{Cl}(A)=\mathcal I(A)/\mathcal P(A). \end{align*} For a nonzero fractional ideal $I\subset K$, choose $d\in A\setminus\{0\}$ with $dI\subset A$. Then $dI$ is a nonzero ideal of the Dedekind domain $A$. It is finitely generated because $A$ is Noetherian. To see it is projective of rank $1$, localize at each maximal ideal $\mathfrak m\subset A$. The local ring $A_{\mathfrak m}$ is a discrete valuation ring, and every nonzero ideal in a discrete valuation ring is principal, so $(dI)_{\mathfrak m}$ is a free $A_{\mathfrak m}$-module of rank $1$. Hence $dI$ is finite locally free of rank $1$, and therefore finitely generated projective of rank $1$. Multiplication by $d^{-1}\in K^\times$ identifies $dI$ with $I$ as an $A$-module, so $I$ is a finitely generated projective $A$-module of rank $1$. For commutative rings, finitely generated projective modules of constant rank $1$ are precisely invertible modules; hence $I$ is an invertible $A$-module. Define \begin{align*} \Phi:\mathcal I(A)\to \operatorname{Pic}(A),\qquad I\mapsto [I]. \end{align*} This map is well-defined because every nonzero fractional ideal is an invertible $A$-module. [/step] [step:Check that ideal multiplication agrees with tensor product] Let $I,J\in \mathcal I(A)$. Define the $A$-bilinear multiplication map \begin{align*} \mu_{I,J}:I\times J\to IJ,\qquad (x,y)\mapsto xy. \end{align*} By the [universal property of tensor products](/theorems/3990), $\mu_{I,J}$ induces an $A$-[linear map](/page/Linear%20Map) \begin{align*} \widetilde{\mu}_{I,J}:I\otimes_A J\to IJ,\qquad x\otimes y\mapsto xy. \end{align*} We prove that $\widetilde{\mu}_{I,J}$ is an isomorphism. Since $A$ is a Dedekind domain, every nonzero fractional ideal is locally principal. For a maximal ideal $\mathfrak m \subset A$ and an $A$-module $M$, write $M_{\mathfrak m}:=A_{\mathfrak m}\otimes_A M$ for localization at $\mathfrak m$; in particular, $I_{\mathfrak m}$ and $J_{\mathfrak m}$ are the localizations of $I$ and $J$. Indeed, choose $d,e\in A\setminus\{0\}$ with $dI\subset A$ and $eJ\subset A$. The ordinary nonzero ideals $dI$ and $eJ$ become nonzero ideals of the discrete valuation ring $A_{\mathfrak m}$ after localizing at $\mathfrak m$, and hence become principal. Multiplying the local generators back by $d^{-1}$ and $e^{-1}$ inside $K$ shows that $I_{\mathfrak m}$ and $J_{\mathfrak m}$ are principal fractional ideals over $A_{\mathfrak m}$. Hence for every maximal ideal $\mathfrak m\subset A$, there exist elements $u_{\mathfrak m},v_{\mathfrak m}\in K^\times$ such that \begin{align*} I_{\mathfrak m}=u_{\mathfrak m}A_{\mathfrak m}\quad\text{and}\quad J_{\mathfrak m}=v_{\mathfrak m}A_{\mathfrak m}. \end{align*} Localization commutes with products of fractional ideals inside $K$, so $(IJ)_{\mathfrak m}=I_{\mathfrak m}J_{\mathfrak m}=u_{\mathfrak m}v_{\mathfrak m}A_{\mathfrak m}$. After localizing at $\mathfrak m$, the map $\widetilde{\mu}_{I,J}$ becomes \begin{align*} u_{\mathfrak m}A_{\mathfrak m}\otimes_{A_{\mathfrak m}}v_{\mathfrak m}A_{\mathfrak m}\to u_{\mathfrak m}v_{\mathfrak m}A_{\mathfrak m},\qquad u_{\mathfrak m}a\otimes v_{\mathfrak m}b\mapsto u_{\mathfrak m}v_{\mathfrak m}ab. \end{align*} This localized map is an isomorphism, with inverse \begin{align*} u_{\mathfrak m}v_{\mathfrak m}c\mapsto u_{\mathfrak m}\otimes v_{\mathfrak m}c. \end{align*} The modules $I$ and $J$ are finitely generated because they are invertible $A$-modules, and $IJ$ is a nonzero fractional ideal, hence finitely generated after clearing denominators inside the Noetherian ring $A$. Since an $A$-linear map between finitely generated $A$-modules is an isomorphism if and only if its localization at every maximal ideal is an isomorphism, $\widetilde{\mu}_{I,J}$ is an isomorphism. Therefore \begin{align*} \Phi(IJ)=[IJ]=[I\otimes_A J]=[I]+[J] \end{align*} in $\operatorname{Pic}(A)$, where the group law in $\operatorname{Pic}(A)$ is induced by tensor product. Thus $\Phi$ is a [group homomorphism](/page/Group%20Homomorphism). [guided] For a maximal ideal $\mathfrak m \subset A$ and an $A$-module $M$, write $M_{\mathfrak m}:=A_{\mathfrak m}\otimes_A M$ for localization at $\mathfrak m$; in particular, $I_{\mathfrak m}$ and $J_{\mathfrak m}$ are the localizations of $I$ and $J$. The source group multiplies fractional ideals, while the Picard group multiplies classes by tensor product. So the essential compatibility is the statement that the natural map $I\otimes_A J\to IJ$ is not just a surjection but an isomorphism. Let $I,J\in \mathcal I(A)$. Multiplication in the field $K$ gives an $A$-bilinear map \begin{align*} \mu_{I,J}:I\times J\to IJ,\qquad (x,y)\mapsto xy. \end{align*} Because $\mu_{I,J}$ is $A$-bilinear, the [universal property of the tensor product](/theorems/3971) gives a unique $A$-linear map \begin{align*} \widetilde{\mu}_{I,J}:I\otimes_A J\to IJ,\qquad x\otimes y\mapsto xy. \end{align*} Why should this be an isomorphism? The point is that over a Dedekind domain, fractional ideals become principal after localizing at a maximal ideal. To see this, first clear denominators: choose $d,e\in A\setminus\{0\}$ with $dI\subset A$ and $eJ\subset A$. Then $dI$ and $eJ$ are ordinary nonzero ideals of $A$. After localizing at a maximal ideal $\mathfrak m\subset A$, they become nonzero ideals of the discrete valuation ring $A_{\mathfrak m}$, so they are principal. Multiplication by $d^{-1}$ and $e^{-1}$ inside $K$ converts these local generators for $(dI)_{\mathfrak m}$ and $(eJ)_{\mathfrak m}$ into local generators for $I_{\mathfrak m}$ and $J_{\mathfrak m}$. Thus, for the fixed maximal ideal $\mathfrak m\subset A$, there exist elements $u_{\mathfrak m},v_{\mathfrak m}\in K^\times$ such that \begin{align*} I_{\mathfrak m}=u_{\mathfrak m}A_{\mathfrak m}\quad\text{and}\quad J_{\mathfrak m}=v_{\mathfrak m}A_{\mathfrak m}. \end{align*} Localization commutes with products of fractional ideals, so $(IJ)_{\mathfrak m}=I_{\mathfrak m}J_{\mathfrak m}=u_{\mathfrak m}v_{\mathfrak m}A_{\mathfrak m}$. After localization at $\mathfrak m$, the map $\widetilde{\mu}_{I,J}$ becomes multiplication of two free rank-one $A_{\mathfrak m}$-modules: \begin{align*} u_{\mathfrak m}A_{\mathfrak m}\otimes_{A_{\mathfrak m}}v_{\mathfrak m}A_{\mathfrak m}\to u_{\mathfrak m}v_{\mathfrak m}A_{\mathfrak m},\qquad u_{\mathfrak m}a\otimes v_{\mathfrak m}b\mapsto u_{\mathfrak m}v_{\mathfrak m}ab. \end{align*} This map is an isomorphism. Its inverse sends $u_{\mathfrak m}v_{\mathfrak m}c$ to $u_{\mathfrak m}\otimes v_{\mathfrak m}c$, and this is well-defined because both sides are free rank-one modules over $A_{\mathfrak m}$ with the displayed generators. Now $I$, $J$, and $IJ$ are finitely generated $A$-modules, since nonzero fractional ideals over a Noetherian domain are finitely generated after clearing denominators. An $A$-linear map between finitely generated modules is an isomorphism exactly when its localization at every maximal ideal is an isomorphism. Therefore $\widetilde{\mu}_{I,J}$ is an isomorphism globally. Consequently the class of $IJ$ in $\operatorname{Pic}(A)$ is the tensor-product class of $I$ and $J$: \begin{align*} [IJ]=[I\otimes_A J]=[I]+[J]. \end{align*} This proves that $\Phi:\mathcal I(A)\to\operatorname{Pic}(A)$ is a group homomorphism. [/guided] [/step] [step:Show that principal fractional ideals map to the identity] Let $a\in K^\times$. Define an $A$-linear map \begin{align*} m_a:A\to aA,\qquad x\mapsto ax. \end{align*} The map $m_a$ is injective because $A$ is a domain and $a\ne0$, and it is surjective by the definition of $aA$. Hence $aA\cong A$ as $A$-modules. Therefore \begin{align*} \Phi(aA)=[aA]=[A], \end{align*} which is the identity element of $\operatorname{Pic}(A)$. Thus $\mathcal P(A)\subseteq \ker\Phi$, so $\Phi$ induces a group homomorphism \begin{align*} \overline{\Phi}:\operatorname{Cl}(A)\to\operatorname{Pic}(A),\qquad [I]\mapsto [I]. \end{align*} [/step] [step:Identify isomorphic fractional ideals up to principal multiplication] Let $I,J\in\mathcal I(A)$, and suppose $f:I\to J$ is an $A$-module isomorphism. Extend scalars from $A$ to $K$ and define the $K$-linear map \begin{align*} f_K:K\otimes_A I\to K\otimes_A J,\qquad c\otimes x\mapsto c\otimes f(x). \end{align*} Since $I$ and $J$ are nonzero fractional ideals, multiplication gives canonical $K$-linear isomorphisms $K\otimes_A I\cong K$ and $K\otimes_A J\cong K$. Under these identifications, $f_K$ is a $K$-linear automorphism of the one-dimensional $K$-[vector space](/page/Vector%20Space) $K$. Hence there exists a unique element $b\in K^\times$ such that \begin{align*} f_K(z)=bz \end{align*} for every $z\in K$. For each $x\in I$, the element $f(x)\in J\subset K$ equals $f_K(x)=bx$. Therefore \begin{align*} J=f(I)=bI. \end{align*} Thus $I$ and $J$ represent the same class in $\operatorname{Cl}(A)$. In particular, if $\overline{\Phi}([I])=\overline{\Phi}([J])$, then $[I]=[J]$ in $\operatorname{Cl}(A)$, so $\overline{\Phi}$ is injective. [/step] [step:Represent every invertible module by a fractional ideal] Let $L$ be an invertible $A$-module. By definition, there exists an invertible $A$-module $M$ such that \begin{align*} L\otimes_A M\cong A. \end{align*} In particular, $L$ is finitely generated projective of constant rank $1$. Hence $K\otimes_A L$ is a one-dimensional $K$-vector space. Choose a $K$-linear isomorphism \begin{align*} \theta:K\otimes_A L\to K. \end{align*} Define the $A$-linear map \begin{align*} \lambda:L\to K,\qquad \ell\mapsto \theta(1\otimes \ell). \end{align*} Since $L$ is projective over the domain $A$, it is torsion-free: if $a\in A\setminus\{0\}$ and $a\ell=0$ for some $\ell\in L$, then the same equality holds after embedding $L$ as a direct summand of a free $A$-module, and multiplication by $a$ is injective on a free module over the domain $A$. A torsion-free $A$-module injects into its localization at $A\setminus\{0\}$, so the natural map $L\to K\otimes_A L$ is injective. Hence $\lambda$ is injective. Let \begin{align*} I_L=\lambda(L)\subset K. \end{align*} Then $I_L$ is a nonzero finitely generated $A$-submodule of $K$. Choose generators $y_1,\dots,y_n\in I_L$. Since each $y_i\in K$, there exists a nonzero element $d_i\in A$ such that $d_i y_i\in A$. For \begin{align*} d=d_1d_2\cdots d_n\in A\setminus\{0\}, \end{align*} we have $dI_L\subset A$. Thus $I_L$ is a nonzero fractional ideal of $A$. The map $\lambda:L\to I_L$ is an $A$-module isomorphism by construction, so $[L]=[I_L]$ in $\operatorname{Pic}(A)$. Therefore every element of $\operatorname{Pic}(A)$ lies in the image of $\overline{\Phi}$, and $\overline{\Phi}$ is surjective. [/step] [step:Conclude that the induced map is an isomorphism] We have constructed a group homomorphism \begin{align*} \overline{\Phi}:\operatorname{Cl}(A)\to\operatorname{Pic}(A),\qquad [I]\mapsto [I]. \end{align*} The preceding steps show that $\overline{\Phi}$ is well-defined, injective, and surjective. Hence $\overline{\Phi}$ is a group isomorphism. This proves \begin{align*} \operatorname{Cl}(A)\cong\operatorname{Pic}(A). \end{align*} [/step]