Lemma 17.25.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{L}$ be an $\mathcal{O}_ X$-module. Equivalent are

$\mathcal{L}$ is invertible, and

there exists an $\mathcal{O}_ X$-module $\mathcal{N}$ such that $\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{N} \cong \mathcal{O}_ X$.

In this case $\mathcal{L}$ is locally a direct summand of a finite free $\mathcal{O}_ X$-module and the module $\mathcal{N}$ in (2) is isomorphic to $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{L}, \mathcal{O}_ X)$.

**Proof.**
Assume (1). Then the functor $- \otimes _{\mathcal{O}_ X} \mathcal{L}$ is essentially surjective, hence there exists an $\mathcal{O}_ X$-module $\mathcal{N}$ as in (2). If (2) holds, then the functor $- \otimes _{\mathcal{O}_ X} \mathcal{N}$ is a quasi-inverse to the functor $- \otimes _{\mathcal{O}_ X} \mathcal{L}$ and we see that (1) holds.

Assume (1) and (2) hold. Denote $\psi : \mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{N} \to \mathcal{O}_ X$ the given isomorphism. Let $x \in X$. Choose an open neighbourhood $U$ an integer $n \geq 1$ and sections $s_ i \in \mathcal{L}(U)$, $t_ i \in \mathcal{N}(U)$ such that $\psi (\sum s_ i \otimes t_ i) = 1$. Consider the isomorphisms

\[ \mathcal{L}|_ U \to \mathcal{L}|_ U \otimes _{\mathcal{O}_ U} \mathcal{L}|_ U \otimes _{\mathcal{O}_ U} \mathcal{N}|_ U \to \mathcal{L}|_ U \]

where the first arrow sends $s$ to $\sum s_ i \otimes s \otimes t_ i$ and the second arrow sends $s \otimes s' \otimes t$ to $\psi (s' \otimes t)s$. We conclude that $s \mapsto \sum \psi (s \otimes t_ i)s_ i$ is an automorphism of $\mathcal{L}|_ U$. This automorphism factors as

\[ \mathcal{L}|_ U \to \mathcal{O}_ U^{\oplus n} \to \mathcal{L}|_ U \]

where the first arrow is given by $s \mapsto (\psi (s \otimes t_1), \ldots , \psi (s \otimes t_ n))$ and the second arrow by $(a_1, \ldots , a_ n) \mapsto \sum a_ i s_ i$. In this way we conclude that $\mathcal{L}|_ U$ is a direct summand of a finite free $\mathcal{O}_ U$-module.

Assume (1) and (2) hold. Consider the evaluation map

\[ \mathcal{L} \otimes _{\mathcal{O}_ X} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{L}, \mathcal{O}_ X) \longrightarrow \mathcal{O}_ X \]

To finish the proof of the lemma we will show this is an isomorphism by checking it induces isomorphisms on stalks. Let $x \in X$. Since we know (by the previous paragraph) that $\mathcal{L}$ is a finitely presented $\mathcal{O}_ X$-module we can use Lemma 17.22.4 to see that it suffices to show that

\[ \mathcal{L}_ x \otimes _{\mathcal{O}_{X, x}} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X, x}}(\mathcal{L}_ x, \mathcal{O}_{X, x}) \longrightarrow \mathcal{O}_{X, x} \]

is an isomorphism. Since $\mathcal{L}_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{N}_ x = (\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{N})_ x = \mathcal{O}_{X, x}$ (Lemma 17.16.1) the desired result follows from More on Algebra, Lemma 15.117.2.
$\square$

## Comments (2)

Comment #1705 by Yogesh More on

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