[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*}
[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]
[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.
[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]
[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.
[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]
[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.
[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]
[/step]
