Let $A = \mathbb{C}[x,y,z]/(x y - z^k)$. In fact $A$ is the ring of $\mu\_k$ invariants: $A = \mathbb{C}[u,v]^{\mu\_k}$ where $g \in \mu\_k$ acts by $g(u,v) = (g u, g^{-1} v)$.


This allows one to understand vector bundles on the smooth surface $Spec\ A - (0,0,0)$ as $\mu\_k$ equivariant bundles on $Spec\ \mathbb{C}[u,v]$.


However I would like to understand how many of these vector bundles extend to all of $Spec\ A$.


**QUESTION**


It's well known for $k = 2$ that $Pic(Spec\ A)$ is trivial. But what about higher $k$ and higher rank?


One approach is to consider the natural map $\mathbb{C}[x,y]\to A$ which is finite and flat and then (identifying vector bundles with projective modules) work with modules over $\mathbb{C}[x,y]$ equipped with an endomorphism $\phi$ such that $\phi^k = x y$. But I didn't get very far with this. 



vector bundles on $\mathbb{C}[x,y,z]/(x y - z^k)$
The only invertible sheaf on $U := \text{Spec}A \setminus \langle x,y,z\rangle$ that extends to all of $\text{Spec} A$ is the trivial invertible sheaf. As you correctly surmise, the invertible sheaves on $W$ are precisely the $\mu\_k$-linearized invertible sheaves on $V := \text{Spec}\mathbb{C}[u,v] \setminus \langle u,v \rangle$. Of course every invertible sheaf on $V$ is trivial. So these invertible sheaves are just the characters of $\mu\_k$. Since the characteristic is $0$, the character group is $\mathbb{Z}/k\mathbb{Z}$ (I guess even in characteristics dividing $k$, this is true). Thus, denoting by $$q:V\to U,$$ the natural quotient morphism, the invertible sheaves on $U$ are just the $\mu\_k$-eigensheaves of the locally free sheaf $q\_\*\mathcal{O}\_U$. The reflexive extensions of these sheaves to all of $\text{Spec} A$ are just the $\mu\_k$-eigensheaves of $\mathbb{C}[u,v]$ considered as an $A$-module.


For every $r=0,\dots,k$, denote by $B\_r \subset \mathbb{C}[u,v]$ the $A$-submodule of elements that have weight $r$ with respect to the given action. For every $r=1,\dots,k$,
both $u^r$ and $v^{k-r}$ give linearly independent, nonzero elements in $B\_r/\langle u^k,v^k,uv \rangle B\_r.$ Thus, since the "fiber" of $B\_r$ at $\langle u^k, v^k, uv \rangle$ is not rank $1$, the module $B\_r$ is not projective of rank $1$. Thus, the only eigenmodule that is projective is $B\_0$, i.e., $A$.


