[proofplan] We construct the rank-determinant map first on the commutative monoid of isomorphism classes of finitely generated projective $A$-modules, then extend it to $K_0(A)$ by the universal property of group completion. The key input is the Steinitz classification: every positive-rank finitely generated projective module over a Dedekind domain is determined by its rank and determinant. This proves injectivity and surjectivity, while the final step checks that extension of scalars preserves both rank and determinant in the required way. [/proofplan]

[step:Define the rank-determinant homomorphism on the projective monoid]Let $V(A)$ denote the commutative monoid of isomorphism classes of finitely generated projective $A$-modules under direct sum. Its identity element is $[0]$, and its operation is \begin{align*} [P]+[Q]=[P\oplus Q]. \end{align*} Define the map \begin{align*} \rho_A:V(A)&\longrightarrow \mathbb Z\oplus \operatorname{Pic}(A) \end{align*} by sending the isomorphism class of a finitely generated projective $A$-module $P$ to \begin{align*} \rho_A([P])=\bigl(\operatorname{rank}(P),[\det(P)]\bigr). \end{align*} This is well-defined on isomorphism classes because isomorphic projective modules have the same constant rank and isomorphic determinant module. Here $\operatorname{Pic}(A)$ is written additively, so addition is induced by [tensor product](/page/Tensor%20Product) of invertible $A$-modules: \begin{align*} [L]+[M]=[L\otimes_A M]. \end{align*} The determinant of a direct sum satisfies the canonical isomorphism \begin{align*} \det(P\oplus Q)\cong \det(P)\otimes_A \det(Q). \end{align*} Ranks also add under direct sums: \begin{align*} \operatorname{rank}(P\oplus Q)=\operatorname{rank}(P)+\operatorname{rank}(Q). \end{align*} Therefore \begin{align*} \rho_A([P\oplus Q])=\rho_A([P])+\rho_A([Q]). \end{align*} Also $\rho_A([0])=(0,[A])$, which is the identity element of $\mathbb Z\oplus \operatorname{Pic}(A)$. Thus $\rho_A:V(A)\to \mathbb Z\oplus \operatorname{Pic}(A)$ is a monoid homomorphism. By the universal property of group completion [citetheorem:8633], and by the definition of $K_0(A)$ as the group completion of $V(A)$, there is a unique [group homomorphism](/page/Group%20Homomorphism) \begin{align*} \Phi_A:K_0(A)\longrightarrow \mathbb Z\oplus \operatorname{Pic}(A) \end{align*} such that \begin{align*} \Phi_A([P])=\bigl(\operatorname{rank}(P),[\det(P)]\bigr) \end{align*} for every finitely generated projective $A$-module $P$.[/step]

[guided]We first build the desired map before trying to prove that it is an isomorphism. Let $V(A)$ be the commutative monoid whose elements are isomorphism classes $[P]$ of finitely generated projective $A$-modules, with operation \begin{align*} [P]+[Q]=[P\oplus Q]. \end{align*} The zero module gives the identity element $[0]$. Define the map \begin{align*} \rho_A:V(A)&\longrightarrow \mathbb Z\oplus \operatorname{Pic}(A) \end{align*} by \begin{align*} \rho_A([P])=\bigl(\operatorname{rank}(P),[\det(P)]\bigr) \end{align*} for each finitely generated projective $A$-module $P$. This is well-defined on isomorphism classes because an isomorphism $P\cong P'$ preserves the local ranks and induces an isomorphism $\det(P)\cong\det(P')$. The target is the group $\mathbb Z\oplus \operatorname{Pic}(A)$, where $\operatorname{Pic}(A)$ consists of isomorphism classes of invertible $A$-modules and its operation is tensor product: \begin{align*} [L]+[M]=[L\otimes_A M]. \end{align*} We now check that $\rho_A$ respects the direct-sum operation. The rank part is additive: \begin{align*} \operatorname{rank}(P\oplus Q)=\operatorname{rank}(P)+\operatorname{rank}(Q). \end{align*} The determinant part is additive in the Picard group because determinant lines satisfy the canonical direct-sum formula \begin{align*} \det(P\oplus Q)\cong \det(P)\otimes_A \det(Q). \end{align*} Hence \begin{align*} \rho_A([P\oplus Q])=\rho_A([P])+\rho_A([Q]). \end{align*} The zero module has rank $0$ and determinant $A$, so \begin{align*} \rho_A([0])=(0,[A]), \end{align*} which is the identity element of $\mathbb Z\oplus \operatorname{Pic}(A)$. Thus $\rho_A$ is a monoid homomorphism. Since $K_0(A)$ is the group completion of $V(A)$, the universal property of group completion [citetheorem:8633] gives a unique group homomorphism \begin{align*} \Phi_A:K_0(A)\longrightarrow \mathbb Z\oplus \operatorname{Pic}(A) \end{align*} extending $\rho_A$. This means precisely that \begin{align*} \Phi_A([P])=\bigl(\operatorname{rank}(P),[\det(P)]\bigr) \end{align*} for every finitely generated projective $A$-module $P$.[/guided]

[step:Use invertible modules to prove surjectivity]Let $(n,[L])\in \mathbb Z\oplus \operatorname{Pic}(A)$, where $L$ is an invertible $A$-module. An invertible $A$-module is finite locally free of rank $1$; hence $L$ is a finitely generated projective $A$-module with $\operatorname{rank}(L)=1$ and $\det(L)\cong L$. Thus $[L]$ and $[A]$ define elements of $K_0(A)$. Define \begin{align*} x=n[A]+[L]-[A]. \end{align*} Then additivity of $\Phi_A$ gives \begin{align*} \Phi_A(x)=n(1,[A])+(1,[L])-(1,[A])=(n,[L]). \end{align*} This computation is valid for every integer $n$, including negative $n$, because $K_0(A)$ is an abelian group. Therefore every element of $\mathbb Z\oplus \operatorname{Pic}(A)$ lies in the image of $\Phi_A$, so $\Phi_A$ is surjective.[/step]

[guided]We want to hit an arbitrary pair $(n,[L])$ in $\mathbb Z\oplus \operatorname{Pic}(A)$. The first coordinate is controlled by free modules, since \begin{align*} \Phi_A([A])=(1,[A]). \end{align*} The second coordinate is controlled by the invertible module $L$. By definition of the Picard group, $L$ is an invertible $A$-module, and invertible modules are finite locally free of rank $1$. Consequently $L$ is a finitely generated projective $A$-module, its rank is $1$, and its determinant is isomorphic to $L$: \begin{align*} \Phi_A([L])=(1,[L]). \end{align*} Subtracting the free rank-one class removes the unwanted rank contribution while preserving the Picard component: \begin{align*} \Phi_A([L]-[A])=(1,[L])-(1,[A])=(0,[L]). \end{align*} Now define \begin{align*} x=n[A]+[L]-[A] \end{align*} in the abelian group $K_0(A)$. This definition makes sense for all $n\in\mathbb Z$, because integer multiples are defined in an abelian group. Applying the group homomorphism $\Phi_A$ gives \begin{align*} \Phi_A(x)=n(1,[A])+(1,[L])-(1,[A])=(n,[L]). \end{align*} Thus every pair $(n,[L])$ has a preimage under $\Phi_A$, proving surjectivity.[/guided]

[step:Use rank and determinant to prove injectivity]Let $x\in K_0(A)$ satisfy $\Phi_A(x)=0$. Since $K_0(A)$ is the group completion of $V(A)$, there exist finitely generated projective $A$-modules $P$ and $Q$ such that \begin{align*} x=[P]-[Q]. \end{align*} The equality $\Phi_A(x)=0$ says \begin{align*} \operatorname{rank}(P)=\operatorname{rank}(Q) \end{align*} and \begin{align*} [\det(P)]=[\det(Q)] \end{align*} in $\operatorname{Pic}(A)$. Let $r=\operatorname{rank}(P)=\operatorname{rank}(Q)$. If $r=0$, then $P=0$ and $Q=0$ because a finitely generated projective module over a domain with rank $0$ is zero. Hence $x=0$. Assume now that $r>0$. By the Steinitz classification [citetheorem:8646], a finitely generated projective $A$-module of positive rank $r$ is isomorphic to $A^{r-1}\oplus I$ for an invertible ideal $I$ of $A$, and its isomorphism class is determined by $r$ and the Picard class of $I$. For such a decomposition, the determinant is isomorphic to $I$, because $\det(A^{r-1})\cong A$ and determinants are multiplicative under direct sums. Hence the equality $[\det(P)]=[\det(Q)]$ gives equality of the Steinitz Picard classes, so \begin{align*} P\cong Q. \end{align*} Therefore \begin{align*} x=[P]-[Q]=0. \end{align*} Thus $\ker \Phi_A=0$, so $\Phi_A$ is injective.[/step]

[guided]We prove injectivity by showing that a formal difference whose rank and determinant vanish must already be zero in $K_0(A)$. Let $x\in K_0(A)$ and assume \begin{align*} \Phi_A(x)=0. \end{align*} Because $K_0(A)$ is the group completion of the monoid $V(A)$, the element $x$ can be represented as a formal difference \begin{align*} x=[P]-[Q] \end{align*} for finitely generated projective $A$-modules $P$ and $Q$. Applying $\Phi_A$ gives \begin{align*} 0=\Phi_A([P]-[Q])=\bigl(\operatorname{rank}(P)-\operatorname{rank}(Q),[\det(P)]-[\det(Q)]\bigr). \end{align*} Hence the ranks agree: \begin{align*} \operatorname{rank}(P)=\operatorname{rank}(Q). \end{align*} The determinant classes also agree: \begin{align*} [\det(P)]=[\det(Q)] \end{align*} in $\operatorname{Pic}(A)$. Let $r=\operatorname{rank}(P)=\operatorname{rank}(Q)$. If $r=0$, then $P$ and $Q$ vanish. Indeed, a finitely generated projective module over a domain is torsion-free, and rank zero means that its localization at the fraction field is zero; a torsion-free finitely generated module with zero localization must be zero. Therefore $P=Q=0$, so $x=0$ in this case. Now suppose $r>0$. The Steinitz classification [citetheorem:8646] applies because $A$ is a Dedekind domain and $P,Q$ are finitely generated projective $A$-modules of positive rank. It says that such a module is classified by its rank together with the Picard class of its rank-one determinant factor. Since $P$ and $Q$ have the same rank and the same determinant class, Steinitz gives an isomorphism \begin{align*} P\cong Q. \end{align*} Therefore their classes in $K_0(A)$ are equal, and \begin{align*} x=[P]-[Q]=0. \end{align*} So every element in the kernel of $\Phi_A$ is zero, which proves that $\Phi_A$ is injective.[/guided]

[step:Conclude the classification isomorphism] The homomorphism \begin{align*} \Phi_A:K_0(A)\longrightarrow \mathbb Z\oplus \operatorname{Pic}(A) \end{align*} is both surjective and injective. Hence it is an isomorphism. By construction, it sends the class of every finitely generated projective $A$-module $P$ to \begin{align*} \bigl(\operatorname{rank}(P),[\det(P)]\bigr). \end{align*} This proves the stated classification of $K_0(A)$. [/step]

[step:Check compatibility with extension of scalars]Let $f:A\to B$ be a unital ring homomorphism between Dedekind domains, and regard $B$ as an $(B,A)$-bimodule through the right action $b\cdot a=bf(a)$. For every finitely generated projective $A$-module $P$, choose a finitely generated projective $A$-module $Q$ and an integer $m\ge 0$ with $P\oplus Q\cong A^m$. Tensoring this isomorphism over $A$ with $B$ gives an isomorphism of $B$-modules \begin{align*} (B\otimes_A P)\oplus (B\otimes_A Q)\cong B^m. \end{align*} Thus $B\otimes_A P$ is a direct summand of a finite free $B$-module, so it is a finitely generated projective $B$-module. Since extension of scalars preserves direct sums, it induces a group homomorphism \begin{align*} K_0(f):K_0(A)\longrightarrow K_0(B) \end{align*} with $K_0(f)([P])=[B\otimes_A P]$. Let $P$ be a finitely generated projective $A$-module of rank $r$. For each prime ideal $\mathfrak q\in\operatorname{Spec}(B)$, let $\mathfrak p=f^{-1}(\mathfrak q)\in\operatorname{Spec}(A)$. Since $P$ is locally free of rank $r$, there is an isomorphism $P_{\mathfrak p}\cong A_{\mathfrak p}^r$. Localizing the base-changed module at $\mathfrak q$ gives \begin{align*} (B\otimes_A P)_{\mathfrak q}\cong B_{\mathfrak q}\otimes_{A_{\mathfrak p}}P_{\mathfrak p}\cong B_{\mathfrak q}^r. \end{align*} Therefore $B\otimes_A P$ is locally free of constant rank $r$ over $B$, and \begin{align*} \operatorname{rank}(B\otimes_A P)=\operatorname{rank}(P). \end{align*} Because $P$ and $B\otimes_A P$ are finite locally free of constant rank $r$, the determinant construction is compatible with base change through the canonical isomorphism \begin{align*} \det(B\otimes_A P)\cong B\otimes_A \det(P). \end{align*} Therefore \begin{align*} \Phi_B(K_0(f)([P]))=\bigl(\operatorname{rank}(P),[\det(P)\otimes_A B]\bigr). \end{align*} In particular, for an invertible $A$-module $L$, \begin{align*} K_0(f)([L]-[A])=[B\otimes_A L]-[B], \end{align*} and under the rank-determinant isomorphisms this element maps to \begin{align*} (0,[L\otimes_A B]). \end{align*} Also $K_0(f)([A])=[B]$, which maps to $(1,[B])$. Since every element of $\mathbb Z\oplus \operatorname{Pic}(A)$ can be written as \begin{align*} (n,[L])=n(1,[A])+(0,[L]), \end{align*} the induced map on the classified form is \begin{align*} (n,[L])\longmapsto (n,[L\otimes_A B]). \end{align*} This proves the naturality statement and completes the proof.[/step]

[guided]We must first prove that extension of scalars really lands in finitely generated projective $B$-modules; this is not an extra hypothesis. Let $P$ be a finitely generated projective $A$-module. By finite projectivity, there are a finitely generated projective $A$-module $Q$ and an integer $m\ge 0$ such that \begin{align*} P\oplus Q\cong A^m. \end{align*} We regard $B$ as an $(B,A)$-bimodule using $f$, so $B\otimes_A P$ is a left $B$-module. Tensoring the displayed decomposition with $B$ over $A$ and using distributivity of tensor product over direct sums gives \begin{align*} (B\otimes_A P)\oplus (B\otimes_A Q)\cong B\otimes_A A^m\cong B^m. \end{align*} Thus $B\otimes_A P$ is a direct summand of a finite free $B$-module. This proves that $B\otimes_A P$ is finitely generated projective over $B$, and therefore extension of scalars defines the homomorphism \begin{align*} K_0(f):K_0(A)\longrightarrow K_0(B) \end{align*} by $K_0(f)([P])=[B\otimes_A P]$. Now we check what this homomorphism does under the rank-determinant classification. Let $P$ have constant rank $r$ over $A$. For a prime ideal $\mathfrak q\in\operatorname{Spec}(B)$, define $\mathfrak p=f^{-1}(\mathfrak q)\in\operatorname{Spec}(A)$. Since $P$ is locally free of rank $r$, there is an isomorphism of $A_{\mathfrak p}$-modules \begin{align*} P_{\mathfrak p}\cong A_{\mathfrak p}^r. \end{align*} Localizing after base change gives \begin{align*} (B\otimes_A P)_{\mathfrak q}\cong B_{\mathfrak q}\otimes_{A_{\mathfrak p}}P_{\mathfrak p}\cong B_{\mathfrak q}\otimes_{A_{\mathfrak p}}A_{\mathfrak p}^r\cong B_{\mathfrak q}^r. \end{align*} So $B\otimes_A P$ is locally free of constant rank $r$, and hence \begin{align*} \operatorname{rank}(B\otimes_A P)=\operatorname{rank}(P). \end{align*} The determinant can now be compared because $P$ is finite locally free of constant rank over $A$, and the previous localization computation shows that $B\otimes_A P$ is finite locally free of constant rank over $B$. The determinant base-change isomorphism gives \begin{align*} \det(B\otimes_A P)\cong B\otimes_A \det(P). \end{align*} Consequently \begin{align*} \Phi_B(K_0(f)([P]))=\bigl(\operatorname{rank}(P),[\det(P)\otimes_A B]\bigr). \end{align*} Apply this to the two generators used in the classification form. First $K_0(f)([A])=[B]$, and this maps to $(1,[B])$. Second, if $L$ is an invertible $A$-module, then \begin{align*} K_0(f)([L]-[A])=[B\otimes_A L]-[B], \end{align*} and the rank-determinant coordinates of this class are \begin{align*} (0,[L\otimes_A B]). \end{align*} Since every element of $\mathbb Z\oplus\operatorname{Pic}(A)$ has the form \begin{align*} (n,[L])=n(1,[A])+(0,[L]), \end{align*} the classified map is \begin{align*} (n,[L])\longmapsto (n,[L\otimes_A B]). \end{align*} This proves the compatibility with extension of scalars.[/guided]