[proofplan] We define the rank-determinant map on virtual projective modules and verify that it respects the direct-sum relation defining $K_0(A)$. The Steinitz classification over a Dedekind domain identifies every positive-rank projective module with $A^{r-1}\oplus I$ for an invertible ideal, and the determinant records precisely the class of $I$ in the Picard group. This proves that the map is bijective. Finally, extension of scalars preserves rank and carries determinants to base-changed determinants, which gives the stated naturality formula. [/proofplan]

[step:Define the rank-determinant homomorphism on virtual projective modules]Let $\mathcal P(A)$ denote the commutative monoid of isomorphism classes of finitely generated projective $A$-modules under direct sum. In the proof, write $\operatorname{rank}(P)$ as shorthand for the theorem's notation $\operatorname{rank}_A(P)$ when $P$ is an $A$-module. For a finitely generated projective $A$-module $P$ of constant rank $r$, define $\det(P):=\bigwedge_A^r P$, with the convention $\det(P)=A$ when $r=0$. The standard determinant-line construction for finite locally free modules says that $\bigwedge_A^r P$ is locally free of rank $1$, hence an invertible $A$-module. Define a map \begin{align*} \delta_A:\mathcal P(A)\to \mathbb Z\oplus \operatorname{Pic}(A) \end{align*} by \begin{align*} \delta_A([P])=(\operatorname{rank}(P),[\det(P)]). \end{align*} Here $\operatorname{Pic}(A)$ is written multiplicatively through [tensor product](/page/Tensor%20Product) of invertible $A$-modules, while $\mathbb Z\oplus \operatorname{Pic}(A)$ is the direct product abelian group. If $P$ and $Q$ are finitely generated projective $A$-modules, then \begin{align*} \operatorname{rank}(P\oplus Q)=\operatorname{rank}(P)+\operatorname{rank}(Q) \end{align*} and the determinant direct-sum formula gives an isomorphism of invertible $A$-modules \begin{align*} \det(P\oplus Q)\cong \det(P)\otimes_A \det(Q). \end{align*} Thus \begin{align*} \delta_A([P\oplus Q])=\delta_A([P])+\delta_A([Q]). \end{align*} So $\delta_A$ is a monoid homomorphism. By the universal property of the Grothendieck group of the monoid of finite projective modules, $\delta_A$ extends uniquely to a [group homomorphism](/page/Group%20Homomorphism) \begin{align*} \Phi_A:K_0(A)\to \mathbb Z\oplus \operatorname{Pic}(A) \end{align*} satisfying \begin{align*} \Phi_A([P])=(\operatorname{rank}(P),[\det(P)]) \end{align*} for every finitely generated projective $A$-module $P$. Therefore, for a virtual class $[P]-[Q]$, \begin{align*} \Phi_A([P]-[Q])=(\operatorname{rank}(P)-\operatorname{rank}(Q),[\det(P)\otimes_A \det(Q)^{-1}]). \end{align*}[/step]

[guided]The target has two pieces, and each piece is additive in the sense required by $K_0$. The first piece is rank. Since $A$ is a domain, every finitely generated projective $A$-module has a well-defined constant rank, and direct sums add ranks: \begin{align*} \operatorname{rank}(P\oplus Q)=\operatorname{rank}(P)+\operatorname{rank}(Q). \end{align*} The second piece is the determinant. If $P$ has constant rank $r$, define $\det(P):=\bigwedge_A^r P$, with $\det(P)=A$ in rank $0$. Since a finitely generated projective module is finite locally free and exterior powers commute with localization, $\det(P)$ is locally free of rank $1$, hence is an invertible $A$-module. The determinant is compatible with direct sums through the standard determinant isomorphism \begin{align*} \det(P\oplus Q)\cong \det(P)\otimes_A \det(Q). \end{align*} Passing to isomorphism classes in $\operatorname{Pic}(A)$, this says \begin{align*} [\det(P\oplus Q)]=[\det(P)]+[\det(Q)] \end{align*} when the Picard group is written additively, or equivalently tensor product is the group operation. Therefore the assignment \begin{align*} [P]\mapsto(\operatorname{rank}(P),[\det(P)]) \end{align*} is compatible with direct sum. This is exactly the condition needed for the assignment to extend from the monoid of finite projective modules to its Grothendieck group $K_0(A)$. On a formal difference, the resulting formula is forced: \begin{align*} \Phi_A([P]-[Q])=(\operatorname{rank}(P)-\operatorname{rank}(Q),[\det(P)\otimes_A \det(Q)^{-1}]). \end{align*} The inverse determinant appears because subtraction in $K_0(A)$ corresponds to taking the inverse class in the Picard group.[/guided]

[step:Prove surjectivity using the Steinitz classification]Let $(n,[L])\in \mathbb Z\oplus \operatorname{Pic}(A)$. Choose an invertible $A$-module representative \begin{align*} L \end{align*} of the Picard class $[L]$. First suppose $n\ge 1$. The module \begin{align*} P_n:=A^{n-1}\oplus L \end{align*} is finitely generated projective, because $A^{n-1}$ is finite free and $L$ is invertible. Its rank is $n$, and the determinant direct-sum formula gives \begin{align*} \det(P_n)\cong \det(A^{n-1})\otimes_A \det(L)\cong A\otimes_A L\cong L. \end{align*} Hence \begin{align*} \Phi_A([P_n])=(n,[L]). \end{align*} For $n=0$, the virtual class \begin{align*} [L]-[A]\in K_0(A) \end{align*} satisfies \begin{align*} \Phi_A([L]-[A])=(0,[L]). \end{align*} For $n<0$, the class \begin{align*} ([L]-[A])+n[A]\in K_0(A) \end{align*} satisfies \begin{align*} \Phi_A(([L]-[A])+n[A])=(n,[L]), \end{align*} because $\Phi_A([A])=(1,[A])$ and $[A]$ is the identity element of $\operatorname{Pic}(A)$. Thus $\Phi_A$ is surjective.[/step]

[guided]To realize an arbitrary pair $(n,[L]) \in \mathbb Z \oplus \operatorname{Pic}(A)$, first choose an invertible $A$-module $L$ representing the class $[L]$. If $n \ge 1$, define $P_n := A^{n-1} \oplus L$. This module is finitely generated projective because it is a direct sum of a finite free module and an invertible module. Its rank is $n$, and the determinant formula gives \begin{align*} \det(P_n) \cong \det(A^{n-1}) \otimes_A \det(L) \cong A \otimes_A L \cong L. \end{align*} Therefore $\Phi_A([P_n]) = (n,[L])$. If $n = 0$, then the virtual class $[L] - [A]$ satisfies \begin{align*} \Phi_A([L] - [A]) = (0,[L]). \end{align*} If $n < 0$, the class $([L]-[A]) + n[A]$ has image \begin{align*} \Phi_A(([L]-[A]) + n[A]) = (n,[L]), \end{align*} because $\Phi_A([A]) = (1,[A])$ and $[A]$ is the identity element of $\operatorname{Pic}(A)$. These three cases cover every integer $n$, so every element of $\mathbb Z \oplus \operatorname{Pic}(A)$ lies in the image of $\Phi_A$.[/guided]

[step:Prove injectivity from equality of rank and determinant]Let $x\in K_0(A)$ satisfy $\Phi_A(x)=(0,[A])$. Write \begin{align*} x=[P]-[Q] \end{align*} for finitely generated projective $A$-modules $P$ and $Q$. The equality $\Phi_A(x)=(0,[A])$ means \begin{align*} \operatorname{rank}(P)=\operatorname{rank}(Q) \end{align*} and \begin{align*} [\det(P)]=[\det(Q)] \end{align*} in $\operatorname{Pic}(A)$. Choose any integer $m\ge 1$. Then both modules \begin{align*} P\oplus A^m \end{align*} and \begin{align*} Q\oplus A^m \end{align*} are finitely generated projective $A$-modules of positive rank. Their ranks are equal. Their determinant classes are also equal in $\operatorname{Pic}(A)$, since the determinant direct-sum formula gives \begin{align*} \det(P\oplus A^m)\cong \det(P)\otimes_A \det(A^m)\cong \det(P) \end{align*} and \begin{align*} \det(Q\oplus A^m)\cong \det(Q)\otimes_A \det(A^m)\cong \det(Q), \end{align*} and since $[\det(P)]=[\det(Q)]$ by the kernel assumption. We now use the [Steinitz Classification Theorem](/theorems/8646) in the precise form proved earlier in these notes: over a Dedekind domain, finitely generated projective modules of the same positive constant rank are isomorphic exactly when their determinant classes in the Picard group agree. The hypotheses apply here because $A$ is a Dedekind domain and both $P\oplus A^m$ and $Q\oplus A^m$ are finitely generated projective modules of the same positive rank. Therefore \begin{align*} P\oplus A^m\cong Q\oplus A^m. \end{align*} It follows in $K_0(A)$ that \begin{align*} [P]+m[A]=[Q]+m[A]. \end{align*} Subtracting $m[A]$ gives \begin{align*} [P]=[Q], \end{align*} so $x=0$. Hence $\Phi_A$ is injective.[/step]

[guided]Let $x\in K_0(A)$ satisfy $\Phi_A(x)=(0,[A])$. Choose finitely generated projective $A$-modules $P$ and $Q$ such that \begin{align*} x=[P]-[Q]. \end{align*} The injectivity argument uses the Steinitz classification, which applies to finitely generated projective modules of positive constant rank over a Dedekind domain. Choose an integer $m\ge 1$. Since $P$ and $Q$ are finitely generated projective, so are $P \oplus A^m$ and $Q \oplus A^m$. Because $m \ge 1$, both stabilized modules have positive rank. Moreover, the assumption $\Phi_A(x) = (0,[A])$ gives \begin{align*} \operatorname{rank}(P) = \operatorname{rank}(Q) \end{align*} and \begin{align*} [\det(P)] = [\det(Q)] \end{align*} in $\operatorname{Pic}(A)$. The stabilized modules therefore have the same rank. Their determinants also have the same Picard class, because the determinant direct-sum formula gives \begin{align*} \det(P\oplus A^m)\cong \det(P)\otimes_A \det(A^m)\cong \det(P)\otimes_A A\cong \det(P) \end{align*} and \begin{align*} \det(Q\oplus A^m)\cong \det(Q)\otimes_A \det(A^m)\cong \det(Q)\otimes_A A\cong \det(Q). \end{align*} Thus the Steinitz Classification Theorem applies to $P\oplus A^m$ and $Q\oplus A^m$, so they are isomorphic: \begin{align*} P \oplus A^m \cong Q \oplus A^m. \end{align*} Passing to $K_0(A)$ gives \begin{align*} [P] + m[A] = [Q] + m[A], \end{align*} and cancellation yields $[P] = [Q]$. Therefore $x = 0$.[/guided]

[step:Identify the inverse map explicitly] The preceding two steps show that $\Phi_A$ is a bijective group homomorphism, hence an isomorphism. Its inverse may be written as \begin{align*} \Psi_A:\mathbb Z\oplus \operatorname{Pic}(A)\to K_0(A). \end{align*} For an invertible $A$-module $L$, it is determined by \begin{align*} \Psi_A(n,[L])=[L]_{K_0}+(n-1)[A] \end{align*} when $n\in\mathbb Z$, where $[L]_{K_0}$ denotes the class of the finitely generated projective module $L$ in $K_0(A)$. Indeed, \begin{align*} \Phi_A([L]_{K_0}+(n-1)[A])=(n,[L]). \end{align*} This also confirms that the formula is independent of the chosen representative of the Picard class, since isomorphic invertible modules have the same class in $K_0(A)$. [/step]

[step:Check compatibility with extension of scalars]Let $f:A\to B$ be a unital homomorphism of Dedekind domains satisfying the stated hypothesis. Let $\operatorname{Proj}_{\mathrm{fg}}(A)$ denote the category whose objects are finitely generated projective $A$-modules and whose morphisms are $A$-linear maps, and define $\operatorname{Proj}_{\mathrm{fg}}(B)$ analogously. Let \begin{align*} B\otimes_A -:\operatorname{Proj}_{\mathrm{fg}}(A)\to \operatorname{Proj}_{\mathrm{fg}}(B) \end{align*} denote the extension-of-scalars functor, where $B$ is viewed as an $(B,A)$-bimodule through $f$. The hypothesis on $f$ ensures that this functor has codomain $\operatorname{Proj}_{\mathrm{fg}}(B)$. It is additive because, for finitely generated projective $A$-modules $P$ and $Q$, the canonical tensor-product map gives an isomorphism of $B$-modules \begin{align*} B\otimes_A(P\oplus Q)\cong (B\otimes_A P)\oplus(B\otimes_A Q). \end{align*} By the universal property of the Grothendieck group, this additive functor induces the homomorphism \begin{align*} K_0(f):K_0(A)\to K_0(B),\qquad [P]\mapsto [B\otimes_A P]. \end{align*} Here $\operatorname{rank}_A(P)$ and $\det_A(P)$ mean the constant rank and determinant of a finitely generated projective $A$-module $P$, and $\operatorname{rank}_B(M)$ and $\det_B(M)$ mean the corresponding constructions for a finitely generated projective $B$-module $M$. If $L$ is an invertible $A$-module, then $B\otimes_A L$ is an invertible $B$-module, with inverse $B\otimes_A L^{-1}$, because \begin{align*} (B\otimes_A L)\otimes_B(B\otimes_A L^{-1})\cong B\otimes_A(L\otimes_A L^{-1})\cong B. \end{align*} Thus the expression $[B\otimes_A L]$ defines a class in $\operatorname{Pic}(B)$. For every finitely generated projective $A$-module $P$, extension of scalars preserves rank. Indeed, put $r=\operatorname{rank}_A(P)$. For every prime ideal $\mathfrak q\in\operatorname{Spec}(B)$, let $\mathfrak p=f^{-1}(\mathfrak q)\in\operatorname{Spec}(A)$. Since $P$ has constant rank $r$, there is an isomorphism of $A_{\mathfrak p}$-modules \begin{align*} P_{\mathfrak p}\cong A_{\mathfrak p}^{r}. \end{align*} Localizing the base-changed module at $\mathfrak q$ and using the canonical localization isomorphism 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*} Thus \begin{align*} \operatorname{rank}_B(B\otimes_A P)=\operatorname{rank}_A(P). \end{align*} The determinant is compatible with base change: if $r=\operatorname{rank}_A(P)$, the canonical exterior-power base-change isomorphism gives an isomorphism of invertible $B$-modules \begin{align*} \det_B(B\otimes_A P)=\bigwedge_B^r(B\otimes_A P)\cong B\otimes_A\bigwedge_A^r P=B\otimes_A \det_A(P). \end{align*} Therefore \begin{align*} \Phi_B(K_0(f)([P]))=(\operatorname{rank}_A(P),[B\otimes_A \det_A(P)]). \end{align*} For a virtual class represented under $\Phi_A$ by $(n,[L])$, this formula becomes \begin{align*} (n,[L])\longmapsto (n,[B\otimes_A L]). \end{align*} This is the asserted naturality formula, and the proof is complete.[/step]

[guided]The hypothesis on $f$ is exactly what is needed to make base change land in the category used to define $K_0(B)$. Explicitly, if $P$ is a finitely generated projective $A$-module, then the hypothesis says that $B\otimes_A P$ is a finitely generated projective $B$-module, so extension of scalars defines a functor from $\operatorname{Proj}_{\mathrm{fg}}(A)$ to $\operatorname{Proj}_{\mathrm{fg}}(B)$. The tensor functor is also additive. Indeed, for finitely generated projective $A$-modules $P$ and $Q$, the canonical map induced by the two inclusions and two projections of the direct sum is an isomorphism of $B$-modules \begin{align*} B\otimes_A(P\oplus Q)\cong (B\otimes_A P)\oplus(B\otimes_A Q). \end{align*} Thus \begin{align*} B\otimes_A -:\operatorname{Proj}_{\mathrm{fg}}(A)\to \operatorname{Proj}_{\mathrm{fg}}(B) \end{align*} is an additive functor, so the universal property of the Grothendieck group induces the homomorphism $K_0(f)$. Now fix a finitely generated projective $A$-module $P$ and put $r=\operatorname{rank}_A(P)$. To see why rank is preserved, localize at an arbitrary prime ideal $\mathfrak q\in\operatorname{Spec}(B)$ and set $\mathfrak p=f^{-1}(\mathfrak q)$. Since $P$ has constant rank $r$, there is an isomorphism \begin{align*} P_{\mathfrak p}\cong A_{\mathfrak p}^{r}. \end{align*} After base change and localization, this becomes \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*} Because this holds at every prime ideal of $B$, the base-changed module has constant rank $r$ over $B$. For determinants, the relevant construction is the canonical exterior-power base-change isomorphism. Since $P$ has rank $r$, \begin{align*} \det_B(B\otimes_A P)=\bigwedge_B^r(B\otimes_A P)\cong B\otimes_A\bigwedge_A^rP=B\otimes_A\det_A(P). \end{align*} In particular, if $L$ is an invertible $A$-module, then $B\otimes_A L$ is an invertible $B$-module; its inverse is $B\otimes_A L^{-1}$ because \begin{align*} (B \otimes_A L) \otimes_B (B \otimes_A L^{-1}) \cong B \otimes_A (L \otimes_A L^{-1}) \cong B. \end{align*} Therefore the Picard-class component transforms as $[L] \mapsto [B \otimes_A L]$. Putting the rank and determinant components together gives the formula \begin{align*} (n,[L]) \mapsto (n,[B \otimes_A L]). \end{align*}[/guided]