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This is connected with Divisor -- line bundle correspondence in algebraic geometry

The long exact cohomology sequence for a regular scheme $X$ associated to $0 \to \mathcal{O}_X^\times \to \mathcal{M}_X^\times \to \mathcal{M}_X^\times/\mathcal{O}_X^\times \to 0$ induces the isomorphism $\mathrm{CaCl}(X) = H^0(X,\mathcal{M}_X^\times/\mathcal{O}_X^\times)/H^0(X,\mathcal{M}_X^\times) \to \mathrm{Pic}(X)$. Here, $H^0(X,\mathcal{M}_X^\times/\mathcal{O}_X^\times)$ is the group of Cartier divisors.

But it is also true that $H^0(X,\mathcal{M}_X^\times/\mathcal{O}_X^\times) = \{(\mathcal{L} \in \mathrm{Pic}(X), s \text{ invertible meromorphic section of $\mathcal{L}$}\}/\cong$ (the mappings are given by $(\mathcal{L},s) \mapsto Z(s)$ and $D \mapsto (\mathcal{O}(D),1)$.). Is there a cohomological way to deduce this? (A[n invertible] meromorphic section $s$ of a line bundle $\mathcal{L}$ is a section $s \in \Gamma(U,\mathcal{L})$ for $U \subseteq X$ dense open [such that $s$ is a regular function without zeros].)

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  • $\begingroup$ What is the meaning of the equivalence relation in the right-hand side ? $\endgroup$ – Cantlog Sep 22 '13 at 15:40
  • $\begingroup$ $(\mathcal{L},s) \cong (\mathcal{L}',s')$ iff there is an iso $\phi: \mathcal{L} \to \mathcal{L}'$ with $s' = \phi \circ s$. $\endgroup$ – user5262 Sep 22 '13 at 16:21
  • $\begingroup$ In $\textrm{CaCl}(X)$ do not we mod out by the image of $H^0(X,\mathcal M_X^\times)$? $\endgroup$ – Brenin Sep 22 '13 at 16:30
  • $\begingroup$ Yes, thanks. This was a typo. $\endgroup$ – user5262 Sep 22 '13 at 16:30
  • $\begingroup$ Are you looking for a proof of that identification, or you just want to know if there is a "cohomological" proof? $\endgroup$ – Brenin Sep 22 '13 at 16:41

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