Lemma 31.28.5. Let $R$ be a UFD. The Picard group of $\mathbf{P}^ n_ R$ is $\mathbf{Z}$. More precisely, there is an isomorphism
In particular, the Picard group of $\mathbf{P}^ n_ k$ of projective space over a field $k$ is $\mathbf{Z}$.
Lemma 31.28.5. Let $R$ be a UFD. The Picard group of $\mathbf{P}^ n_ R$ is $\mathbf{Z}$. More precisely, there is an isomorphism
In particular, the Picard group of $\mathbf{P}^ n_ k$ of projective space over a field $k$ is $\mathbf{Z}$.
Proof. By Lemma 31.28.4 the Picard groups of the opens $D_+(T_ i) \cong \mathbf{A}^ n_ R$ are trivial. Thus if $\mathcal{L}$ is an invertible module on $\mathbf{P}^ n_ R$, then it is given by a $1$-cocycle with values in the sheaf of invertible functors $\mathcal{O}^*$ for the open covering
Observe that for $i \not= j$ we have
Thus such a cocycle $(g_{ij})$ is given by units
with $u_{ij} \in R^*$ and $e_{ij} \in \mathbf{Z}$ satisfying the cocycle condition. The cocycle condition over $D_+(T_ iT_ jT_ k)$ for $\# \{ i, j, k\} = 3$ tell us that
Whence $u_{ij} = u_{i1} u_{j1}^{-1}$ is a boundary. Thus all isomorphism classes of invertible modules are given by taking the cocycle with
for some $e \in \mathbf{Z}$. Since $\mathcal{O}(n)$ has trivializing section $T_ i^ n$ over $D_+(T_ i)$ we see that the corresponding cocycle of $\mathcal{O}(n)$ is $(T_ i/T_ j)^ n$ and the proof is complete. $\square$
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Comments (3)
Comment #8749 by Lukas Heger on
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