[proofplan] We prove the two implications separately. If $P$ is stably free, then after adding a finite free summand it becomes free, and multiplicativity of the determinant under direct sums forces $\det(P)$ to be isomorphic to $A$. Conversely, if $\det(P)\cong A$, the Steinitz classification over a Dedekind domain writes $P$ as $A^{r-1}\oplus I$, where the invertible ideal $I$ represents the determinant class; the fact that the determinant is isomorphic to $A$ then gives $I\cong A$, so $P$ is actually free. Since every free module is stably free, this proves the criterion. [/proofplan]

[step:Take determinants of a stable free decomposition]Assume first that $P$ is stably free. By definition, there exist integers $m,n\geq 0$ and an $A$-module isomorphism $P\oplus A^m \cong A^n$. Since $P$ has positive constant rank, write $r=\operatorname{rank}(P)\geq 1$. Comparing ranks in the isomorphism gives $n=r+m$. For a finitely generated projective module $M$ of constant rank $s$, let $\det(M):=\bigwedge_A^s M$ denote its determinant line. We use the standard multiplicativity of determinant lines under direct sums: $\det(M\oplus N)\cong \det(M)\otimes_A \det(N)$ for finite projective modules $M$ and $N$ of constant rank. Applying this to $M=P$ and $N=A^m$, and using $\det(A^m)\cong A$ and $\det(A^n)\cong A$, we obtain $\det(P)\otimes_A A \cong \det(P\oplus A^m)\cong \det(A^n)\cong A$. Since $\det(P)\otimes_A A\cong \det(P)$, it follows that $\det(P)\cong A$.[/step]

[guided]Assume that $P$ is stably free. The phrase means precisely that there are integers $m,n\geq 0$ such that adding the free module $A^m$ to $P$ produces a free module: $P\oplus A^m \cong A^n$. Let $r=\operatorname{rank}(P)$. The theorem assumes that $P$ has positive constant rank, so $r\geq 1$. Taking ranks on both sides of the isomorphism gives $n=r+m$. Now we use the determinant line. For a finitely generated projective $A$-module $M$ of constant rank $s$, its determinant is the rank-one projective module $\det(M):=\bigwedge_A^s M$. The key formal property is multiplicativity under direct sums: if $M$ and $N$ are finite projective modules of constant ranks, then $\det(M\oplus N)\cong \det(M)\otimes_A \det(N)$. Apply this property to the direct sum $P\oplus A^m$. Since a free module has determinant isomorphic to $A$, we have $\det(A^m)\cong A$ and also $\det(A^n)\cong A$. Therefore the stable free isomorphism gives $\det(P)\otimes_A A \cong \det(P\oplus A^m)\cong \det(A^n)\cong A$. Finally, tensoring any $A$-module with $A$ over $A$ does not change it, so $\det(P)\cong \det(P)\otimes_A A\cong A$. Thus stable freeness forces the determinant line of $P$ to be isomorphic to $A$.[/guided]

[step:Use Steinitz classification when the determinant line is isomorphic to $A$]Conversely, assume that $\det(P)\cong A$. Let $r=\operatorname{rank}(P)$, so $r\geq 1$. By the Steinitz Classification [citetheorem:8646], which applies because $A$ is a Dedekind domain and $P$ is a finitely generated projective $A$-module of positive constant rank, there exists an invertible ideal $I$ of $A$ such that $P\cong A^{r-1}\oplus I$, where $A^{r-1}$ is interpreted as the zero module when $r=1$. Taking determinants of this decomposition gives $\det(P)\cong \det(A^{r-1})\otimes_A \det(I)$. Since $\det(A^{r-1})\cong A$ and since $I$ has rank $1$, so $\det(I)\cong I$, this becomes $\det(P)\cong I$. The hypothesis $\det(P)\cong A$ therefore implies $I\cong A$. Substituting this into the Steinitz decomposition yields $P\cong A^{r-1}\oplus A\cong A^r$. Thus $P$ is free, hence stably free.[/step]

[guided]We now use the Steinitz classification. The theorem applies exactly in this situation because $A$ is a Dedekind domain and $P$ is a finitely generated projective $A$-module of positive constant rank. Therefore there exists an invertible ideal $I$ of $A$ such that $P\cong A^{r-1}\oplus I$. The case $r=1$ is included in this formula, since then $A^{r-1}=A^0=0$. To connect this decomposition to the determinant, we take determinant lines. For a free module $A^{r-1}$ we have $\det(A^{r-1})\cong A$. Because $I$ is an invertible ideal, it is a rank-one projective module, so its determinant line is naturally isomorphic to $I$ itself: $\det(I)\cong I$. Multiplicativity of determinant lines under direct sums then gives $\det(P)\cong \det(A^{r-1})\otimes_A \det(I)\cong A\otimes_A I\cong I$. The hypothesis of the theorem says that $\det(P)\cong A$, so the preceding identification forces $I\cong A$. Substituting this back into the Steinitz decomposition yields $P\cong A^{r-1}\oplus A\cong A^r$. Thus $P$ is free, and every free module is stably free, so the converse implication is proved.[/guided]

[step:Conclude the equivalence] The first step proved that every stably free finitely generated projective $A$-module of positive constant rank has determinant line isomorphic to $A$. The second step proved that a determinant line isomorphic to $A$ implies that $P$ is free, and therefore stably free. Hence $P$ is stably free if and only if $\det(P)\cong A$. [/step]