Lemma 55.9.5. In Situation 55.9.3 there is an exact sequence

$0 \to \mathbf{Z} \to \mathbf{Z}^{\oplus n} \to \mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (C) \to 0$

where the first map sends $1$ to $(m_1, \ldots , m_ n)$ and the second maps sends the $i$th basis vector to $\mathcal{O}_ X(C_ i)$.

Proof. Observe that $C \subset X$ is an open subscheme. The restriction map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (C)$ is surjective by Divisors, Lemma 31.28.3. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module such that there is an isomorphism $s : \mathcal{O}_ C \to \mathcal{L}|_ C$. Then $s$ is a regular meromorphic section of $\mathcal{L}$ and we see that $\text{div}_\mathcal {L}(s) = \sum a_ i C_ i$ for some $a_ i \in \mathbf{Z}$ (Divisors, Definition 31.27.4). By Divisors, Lemma 31.27.6 (and the fact that $X$ is normal) we conclude that $\mathcal{L} = \mathcal{O}_ X(\sum a_ iC_ i)$. Finally, suppose that $\mathcal{O}_ X(\sum a_ i C_ i) \cong \mathcal{O}_ X$. Then there exists an element $g$ of the function field of $X$ with $\text{div}_ X(g) = \sum a_ i C_ i$. In particular the rational function $g$ has no zeros or poles on the generic fibre $C$ of $X$. Since $C$ is a normal scheme this implies $g \in H^0(C, \mathcal{O}_ C) = K$. Thus $g = \pi ^ a u$ for some $a \in \mathbf{Z}$ and $u \in R^*$. We conclude that $\text{div}_ X(g) = a \sum m_ i C_ i$ and the proof is complete. $\square$

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