[proofplan]
We use the standard structure theory of finitely generated torsion-free modules over a Dedekind domain: such a module decomposes as a direct sum of invertible fractional ideals. The [Steinitz exchange lemma](/theorems/373) compresses a direct sum of rank-one invertible modules into a free summand of rank one less together with the product ideal. The top exterior power identifies the product ideal intrinsically, and comparing top exterior powers gives the uniqueness criterion in the Picard group.
[/proofplan]
[step:Decompose $P$ into invertible fractional ideals]
Let $K=\operatorname{Frac}(A)$ denote the fraction field of $A$. Since $P$ is finitely generated projective, it is finitely generated and torsion-free as an $A$-module. We use the standard structure theorem for finitely generated torsion-free modules over a Dedekind domain: there exist invertible fractional ideals $I_1,\dots,I_r \subset K$ and an $A$-module isomorphism
\begin{align*}
P \cong I_1 \oplus \cdots \oplus I_r.
\end{align*}
The number of summands is $r$ because localizing at any nonzero prime ideal $\mathfrak p \trianglelefteq A$ gives
\begin{align*}
P_{\mathfrak p} \cong A_{\mathfrak p}^r,
\end{align*}
and each nonzero invertible fractional ideal becomes a free rank-one $A_{\mathfrak p}$-module.
[guided]
Let $K=\operatorname{Frac}(A)$ be the fraction field of $A$. The reason fractional ideals enter the proof is that rank-one projective modules over a Dedekind domain are naturally realized as invertible fractional ideals inside $K$. Since $P$ is projective, it is torsion-free: if $a \in A$ is nonzero and $ap=0$ for some $p \in P$, then after embedding $P$ as a direct summand of a free module, the same equality holds in a torsion-free free module, forcing $p=0$.
We now use the standard structure theorem for finitely generated torsion-free modules over a Dedekind domain. It states that every finitely generated torsion-free $A$-module is isomorphic to a finite direct sum of invertible fractional ideals. Applying this theorem to $P$, there exist invertible fractional ideals $I_1,\dots,I_m \subset K$ and an isomorphism of $A$-modules
\begin{align*}
P \cong I_1 \oplus \cdots \oplus I_m.
\end{align*}
It remains to identify $m$ with the rank $r$ in the theorem statement. For every nonzero prime ideal $\mathfrak p \trianglelefteq A$, localization gives
\begin{align*}
P_{\mathfrak p} \cong (I_1)_{\mathfrak p} \oplus \cdots \oplus (I_m)_{\mathfrak p}.
\end{align*}
Because each $I_k$ is invertible, each localization $(I_k)_{\mathfrak p}$ is a free rank-one $A_{\mathfrak p}$-module. Hence the right-hand side is free of rank $m$ over $A_{\mathfrak p}$. Since $P$ has constant rank $r$, the left-hand side is free of rank $r$ over $A_{\mathfrak p}$. Therefore $m=r$, and we have
\begin{align*}
P \cong I_1 \oplus \cdots \oplus I_r.
\end{align*}
[/guided]
[/step]
[step:Compress the direct sum to one ideal summand]
We use the Steinitz exchange lemma for invertible fractional ideals over a Dedekind domain: for invertible fractional ideals $L$ and $M$ of $A$, there is an $A$-module isomorphism
\begin{align*}
L \oplus M \cong A \oplus LM.
\end{align*}
Applying this lemma successively to $I_1,\dots,I_r$ gives
\begin{align*}
I_1 \oplus \cdots \oplus I_r \cong A^{r-1} \oplus I_1\cdots I_r.
\end{align*}
Combining this with the decomposition of $P$ yields
\begin{align*}
P \cong A^{r-1} \oplus I_1\cdots I_r.
\end{align*}
The product $I_1\cdots I_r$ is an invertible fractional ideal.
[/step]
[step:Replace the fractional ideal by an integral invertible ideal]
Choose a nonzero element $a \in A$ such that
\begin{align*}
aI_1\cdots I_r \subset A.
\end{align*}
Define the integral ideal $I \trianglelefteq A$ by
\begin{align*}
I:=aI_1\cdots I_r.
\end{align*}
Since $I_1\cdots I_r$ is an invertible fractional ideal, $I$ is an invertible integral ideal. Multiplication by $a$ defines an $A$-module isomorphism
\begin{align*}
I_1\cdots I_r \to I.
\end{align*}
Therefore
\begin{align*}
P \cong A^{r-1}\oplus I.
\end{align*}
This proves the existence assertion.
[/step]
[step:Compute the determinant of the normal form]
For a finitely generated projective $A$-module $M$ of constant rank $r$, define its determinant line by
\begin{align*}
\det(M):=\bigwedge_A^r M.
\end{align*}
If $L$ is an invertible ideal, then the direct-sum formula for exterior powers gives
\begin{align*}
\det(A^{r-1}\oplus L) \cong \bigwedge_A^{r-1} A^{r-1}\otimes_A L.
\end{align*}
Since $\bigwedge_A^{r-1}A^{r-1}\cong A$, this becomes
\begin{align*}
\det(A^{r-1}\oplus L)\cong L.
\end{align*}
Thus the determinant of the normal form $A^{r-1}\oplus L$ is exactly the invertible rank-one module represented by $L$.
[/step]
[step:Use determinants to prove uniqueness]
Suppose first that
\begin{align*}
A^{r-1}\oplus I \cong A^{r-1}\oplus J.
\end{align*}
Taking determinant lines and using the computation above gives an isomorphism
\begin{align*}
I \cong J
\end{align*}
of invertible $A$-modules. Hence $I$ and $J$ represent the same class in $\operatorname{Pic}(A)$.
Conversely, suppose $I$ and $J$ represent the same class in $\operatorname{Pic}(A)$. By the definition of the Picard group, there is an $A$-module isomorphism
\begin{align*}
I \cong J.
\end{align*}
Taking the direct sum with the identity map on $A^{r-1}$ gives an $A$-module isomorphism
\begin{align*}
A^{r-1}\oplus I \cong A^{r-1}\oplus J.
\end{align*}
This proves the stated equivalence and completes the proof.
[/step]
