What is $\operatorname{Pic}(\mathbb{P}^n_{\mathbb{Z}})$?

Question

I would like to know the Picard group of the projective spaces over the integers $\mathbb{Z}$. I know that the projective space over a field $k$ has $\operatorname{Pic}(\mathbb{P}^n_{\mathbb{k}}) \cong \mathbb{Z}$, but what in the case of the integers (or even arbitrary rings)? Are there any results?

Answer 1

If $A$ is a factorial domain, then $\mathrm{Pic}(A)$ is trivial. It follows that also $\mathrm{Pic}(\mathbb{A}^n_A)$ is trivial. Now $\mathbb{P}^n_A$ is covered by the open subschemes $U_i = D_+(x_i) \cong \mathbb{A}^n_A$ for $0 \leq i \leq n$, each having trivial Picard group. It follows that $\mathrm{Pic}(\mathbb{P}^n_A) \cong \check{H}^1(\{U_0,\dotsc,U_n\},\mathcal{O}^\times)$. This Cech cohomology group consists of elements $\eta_{ij} \in \mathcal{O}(U_i \cap U_j)^\times$ which satisfy the coycle condition, modulo coboundaries (if you don't know cohomology: these $\eta_{ij}$ are just the isomorphisms of the trivial restrictions of a line bundle on $\mathbb{P}^n_A$ to $U_i$ on the overlaps $U_i \cap U_j$). One calculates $\mathcal{O}(U_i \cap U_j)^\times = A^\times \cdot \langle \frac{x_j}{x_i} \rangle$. Writing $\eta_{ij} = \lambda_{ij} (x_j/x_i)^{n_{ij}}$, one finds that $n:=n_{ij}$ doesn't depend on $i,j$ and that $\lambda_{ik} = \lambda_{ij} \lambda_{jk}$ (using the cocycle condition). But then $\lambda_{ij} = \lambda_{0j} \lambda_{0i}^{-1}$ is a coboundary and $\eta_{ij}$ is cobordant to $(x_j/x_i)^n$. It follows that $\mathbb{Z} \cong \mathrm{Pic}(\mathbb{P}^n_A)$.

In my opinion this proof is much more elementary and direct than the usual one with divisors (which only works for fields). We just restrict the given line bundle on $\mathbb{P}^n$ to the $n+1$ affine spaces $\mathbb{A}^n$, where they have to be trivial. Then we look what happens on the intersections, and calculate by hand that they are glued by means of $(x_j/x_i)^n$ for some unique $n \in \mathbb{Z}$.