The Stacks project

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

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

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

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

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. $\square$