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

\[ \mathbf{Z} \longrightarrow \mathop{\mathrm{Pic}}\nolimits (\mathbf{P}^ n_ R),\quad m \longmapsto \mathcal{O}_{\mathbf{P}^ n_ R}(m) \]

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

\[ \mathbf{P}^ n_ R = \bigcup \nolimits _{i = 0, \ldots , n} D_+(T_ i) \]

Observe that for $i \not= j$ we have

\[ \mathcal{O}^*(D_+(T_ i) \cap D_+(T_ j)) = \mathcal{O}^*(D_+(T_ iT_ j)) = \left(R[T_0, \ldots , T_ n]_{(T_ iT_ j)}\right)^* = R^* \times (T_ i/T_ j)^\mathbf {Z} \]

Thus such a cocycle $(g_{ij})$ is given by units

\[ g_{ij} = u_{ij} (T_ i/T_ j)^{e_{ij}} \]

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

\[ u_{ik} = u_{ij} u_{jk} \quad \text{and}\quad e_{ik} = e_{ij} = e_{jk} \]

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

\[ g_{ij} = (T_ i/T_ j)^ e \]

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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (3)

Comment #8749 by Lukas Heger on

Comment #8764 by Lukas Heger on

Comment #9306 by Stacks project on