Let $\mathcal{O}\_k$ be the ring of integers in an algebraic number field $k$. Let $M$ be a rank $1$ projective module over $\mathcal{O}\_k$ (in other words, $M$ is a projective module such that $k \otimes M \cong k$). Let $X\_M$ be the set of all direct summands of $(\mathcal{O}\_k)^2$ which are isomorphic to $M$. Using the classification of finitely generated projective modules over Dedekind domains, we see that $X\_M$ is nonempty. The group $\text{GL}\_2(\mathcal{O}\_k)$ acts on $X\_M$.


**Question** : Does $\text{GL}\_2(\mathcal{O}\_k)$ act transitively on $X\_M$?


I should remark that this is true if $M \cong \mathcal{O}\_k$. Indeed, let $L \in X\_{\mathcal{O}\_k}$. Then there exists some $\vec{v} \in (\mathcal{O}\_k)^2$ such that $M = \mathcal{O}\_k \cdot \vec{v}$, and writing $\vec{v} = (v\_1,v\_2)$ we have that $\mathcal{O}\_k \cdot v\_1 + \mathcal{P}\_k \cdot v\_2 = \mathcal{O}\_k$. We can thus find $w\_1,w\_2 \in \mathcal{O}\_k$ such that $w\_1 v\_1 + w\_2 v\_2 = 1$, and hence the matrix $\left(\begin{smallmatrix} v\_1 & -w\_2 \\ v\_2 & w\_1 \end{smallmatrix}\right)$ lies in $\text{SL}\_2(\mathcal{O}\_k)$ and takes $(1,0)$ to $\vec{v}$.



Action of GL(2,O_k) on 1d subspaces of (O_k)^2
I think this is clear; the complementary summand to M will be its inverse M' in the Picard group, so if you have have a second such summand N, you just pick isomorphisms $M\cong N$ and $M'\cong N'$, and this will induce an automorphism of $\mathcal{O}\_k^2$.


