This tag has label modules-lemma-constructions-invertible and it points to
The corresponding content:
Lemma 16.21.2. Let $(X, \mathcal{O}_X)$ be a ringed space. Assume that all stalks $\mathcal{O}_{X, x}$ are local rings.
- If $\mathcal{L}$, $\mathcal{N}$ are invertible $\mathcal{O}_X$-modules, then so is $\mathcal{L} \otimes_{\mathcal{O}_X} \mathcal{N}$.
- If $\mathcal{L}$ is an invertible $\mathcal{O}_X$-module, then so is $\mathcal{L}^{\otimes -1} = \mathop{\mathcal{H}\!{\it om}}\nolimits_{\mathcal{O}_X}(\mathcal{L}, \mathcal{O}_X)$.
- If $\mathcal{L}$ is an invertible $\mathcal{O}_X$-module, then the evaluation map $\mathcal{L} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes -1} \to \mathcal{O}_X$ is an isomorphism.
Proof. Omitted. $\square$
\begin{lemma}
\label{lemma-constructions-invertible}
Let $(X, \mathcal{O}_X)$ be a ringed space.
Assume that all stalks $\mathcal{O}_{X, x}$ are local rings.
\begin{enumerate}
\item If $\mathcal{L}$, $\mathcal{N}$ are invertible
$\mathcal{O}_X$-modules, then so is
$\mathcal{L} \otimes_{\mathcal{O}_X} \mathcal{N}$.
\item If $\mathcal{L}$ is an invertible
$\mathcal{O}_X$-module, then so is
$\mathcal{L}^{\otimes -1}
= \SheafHom_{\mathcal{O}_X}(\mathcal{L}, \mathcal{O}_X)$.
\item If $\mathcal{L}$ is an invertible
$\mathcal{O}_X$-module, then the evaluation map
$\mathcal{L} \otimes_{\mathcal{O}_X} \mathcal{L}^{\otimes -1}
\to \mathcal{O}_X$ is an isomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
Omitted.
\end{proof}
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