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.**
Observe that the local rings of $X = \mathbf{P}^ n_ R$ are UFDs because $X$ is covered by affine pieces isomorphic to $\mathbf{A}^ n_ R$ and $R[x_1, \ldots , x_ n]$ is a UFD (Algebra, Lemma 10.120.10). Hence $X$ is an integral Noetherian scheme all of whose local rings are UFDs and we see that $\mathop{\mathrm{Pic}}\nolimits (X) = \text{Cl}(X)$ by Lemma 31.27.7.

The displayed map is a group homomorphism by Constructions, Lemma 27.10.3. The map is injective because $H^0$ of $\mathcal{O}_ X$ and $\mathcal{O}_ X(m)$ are non-isomorphic $R$-modules if $m > 0$, see Cohomology of Schemes, Lemma 30.8.1. Let $\mathcal{L}$ be an invertible module on $X$. Consider the open $U = D_+(T_0) \cong \mathbf{A}^ n_ R$. The complement $H = X \setminus U$ is a prime divisor because it is isomorphic to $\text{Proj}(R[T_1, \ldots , T_ n])$ which is integral by the discussion in the previous paragraph. In fact $H$ is the zero scheme of the regular global section $T_0$ of $\mathcal{O}_ X(1)$ hence $\mathcal{O}_ X(1)$ maps to the class of $H$ in $\text{Cl}(X)$. By Lemma 31.28.4 we see that $\mathcal{L}|_ U \cong \mathcal{O}_ U$. Let $s \in \mathcal{L}(U)$ be a trivializing section. Then we can think of $s$ as a regular meromorphic section of $\mathcal{L}$ and we see that necessarily $\text{div}_\mathcal {L}(s) = m[H]$ for some $m \in \mathbf{Z}$ as $H$ is the only prime divisor of $X$ not meeting $U$. In other words, we see that $\mathcal{L}$ and $\mathcal{O}_ X(m)$ map to the same element of $\text{Cl}(X)$ and hence $\mathcal{L} \cong \mathcal{O}_ X(m)$ as desired. $\square$

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