According to Hartshorne, an invertible sheaf $\mathcal{L}$ on a ringed space $(X, \mathcal{O}_X)$ is a locally free sheaf of rank 1. So, this means that there exists a cover $\{U_i\}_{i\in I}$ of X such that $\mathcal{L}_{{U}_i}$ is $\mathcal{O}_X(U_i)$. My question is that why is $\mathcal{L}$ not a globally free sheaf of rank 1? In other words, by the gluing property of sheaves $\mathcal{L}_{V}$ should be $\mathcal{O}_X(V)$ for an open set $V$ in $X$. Am I right?
2026-02-23 14:08:12.1771855692
What is the difference between invertible sheaf and sheaf of ring on a ringed space?
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