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
Comments (0)