Lemma 55.9.8. In Situation 55.9.3 set $d = \gcd (m_1, \ldots , m_ n)$ and let $D = \sum (m_ i/d)C_ i$ as an effective Cartier divisor. Then $\mathcal{O}_ X(D)$ has order dividing $d$ in $\mathop{\mathrm{Pic}}\nolimits (X)$ and $\mathcal{C}_{D/X}$ an invertible $\mathcal{O}_ D$-module of order dividing $d$ in $\mathop{\mathrm{Pic}}\nolimits (D)$.

Proof. We have

$\mathcal{O}_ X(D)^{\otimes d} = \mathcal{O}_ X(dD) = \mathcal{O}_ X(X_ k) = \mathcal{O}_ X$

by Lemma 55.9.1. We conclude as $\mathcal{C}_{D/X}$ is the pullback of $\mathcal{O}_ X(-D)$. $\square$

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