Invertible sheaf

Remarks

The structure sheaf $\mathcal{O}_X$ is invertible, with $U = X$ for every point. On $X = \mathbb{A}^2_k = \operatorname{Spec} k[x, y]$ over a field $k$, the ideal sheaf $\mathcal{I}$ of the origin, given by the ideal $(x, y)$, is not invertible. Its stalk at the origin is the maximal ideal of $k[x, y]_{(x, y)}$, which needs two generators, so it is not free of rank $1$. In other words, an invertible sheaf is a locally free sheaf of rank $1$. For invertible sheaves $\mathcal{L}$ and $\mathcal{M}$, the tensor product $\mathcal{L} \otimes_{\mathcal{O}_X} \mathcal{M}$ is invertible. The dual $\mathcal{L}^\vee = \mathcal{H}om(\mathcal{L}, \mathcal{O}_X)$ satisfies $\mathcal{L} \otimes \mathcal{L}^\vee \cong \mathcal{O}_X$, which explains the name.