Lemma 55.9.5. In Situation 55.9.3 there is an exact sequence
where the first map sends 1 to (m_1, \ldots , m_ n) and the second maps sends the ith basis vector to \mathcal{O}_ X(C_ i).
Lemma 55.9.5. In Situation 55.9.3 there is an exact sequence
where the first map sends 1 to (m_1, \ldots , m_ n) and the second maps sends the ith 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
Comments (0)