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. $\square$

Proof. By Lemma 17.25.2 there exists an $\mathcal{O}_ Y$-module $\mathcal{N}$ such that $\mathcal{L} \otimes _{\mathcal{O}_ Y} \mathcal{N} \cong \mathcal{O}_ Y$. Pulling back we get $f^*\mathcal{L} \otimes _{\mathcal{O}_ X} f^*\mathcal{N} \cong \mathcal{O}_ X$ by Lemma 17.16.4. Thus $f^*\mathcal{L}$ is invertible by Lemma 17.25.2. $\square$

Proof. The parenthetical statement follows by considering a one point space $X$ with sheaf of rings $\mathcal{O}_ X$ given by a ring $R$. Then invertible $\mathcal{O}_ X$-modules correspond to invertible $R$-modules, hence as soon as $\mathop{\mathrm{Pic}}\nolimits (R)$ is not the trivial group, then we get an example.

Assume $\mathcal{L}$ is locally free of rank $1$ and consider the evaluation map

Looking over an open covering trivialization $\mathcal{L}$, we see that this map is an isomorphism. Hence $\mathcal{L}$ is invertible by Lemma 17.25.2.

Assume all stalks $\mathcal{O}_{X, x}$ are local rings and $\mathcal{L}$ invertible. In the proof of Lemma 17.25.2 we have seen that $\mathcal{L}_ x$ is an invertible $\mathcal{O}_{X, x}$-module for all $x \in X$. Since $\mathcal{O}_{X, x}$ is local, we see that $\mathcal{L}_ x \cong \mathcal{O}_{X, x}$ (More on Algebra, Section 15.117). Since $\mathcal{L}$ is of finite presentation by Lemma 17.25.2 we conclude that $\mathcal{L}$ is locally free of rank $1$ by Lemma 17.11.6. $\square$

Proof. Part (1) is clear from the definition and part (2) follows from Lemma 17.25.2 and its proof. $\square$

Proof. Recall that any invertible $\mathcal{O}_ X$-module is locally a direct summand of a finite free $\mathcal{O}_ X$-module, see Lemma 17.25.2. For each open covering $\mathcal{U} : X = \bigcup _{j \in J} U_ j$ and map $r : J \to \mathbf{N}$ consider the sheaves of $\mathcal{O}_ X$-modules $\mathcal{F}$ such that $\mathcal{F}_ j = \mathcal{F}|_{U_ j}$ is a direct summand of $\mathcal{O}_{U_ j}^{\oplus r(j)}$. The collection of isomorphism classes of $\mathcal{F}_ j$ is a set, because $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_ U^{\oplus r}, \mathcal{O}_ U^{\oplus r})$ is a set. The sheaf $\mathcal{F}$ is gotten by glueing $\mathcal{F}_ j$, see Sheaves, Section 6.33. Note that the collection of all glueing data forms a set. The collection of all coverings $\mathcal{U} : X = \bigcup _{j \in J} U_ i$ where $J \to \mathcal{P}(X)$, $j \mapsto U_ j$ is injective forms a set as well. For each covering there is a set of maps $r : J \to \mathbf{N}$. Hence the collection of all $\mathcal{F}$ forms a set. $\square$

Proof. Suppose $x \in X_ s$. We have an isomorphism

by Lemma 17.25.5. Both $\mathcal{L}_ x$ and $(\mathcal{L}^{\otimes -1})_ x$ are free $\mathcal{O}_{X, x}$-modules of rank $1$. We conclude from Algebra, Nakayama's Lemma 10.20.1 that $s_ x$ is a basis for $\mathcal{L}_ x$. Hence there exists a basis element $t_ x \in (\mathcal{L}^{\otimes -1})_ x$ such that $s_ x \otimes t_ x$ maps to $1$. Choose an open neighbourhood $U$ of $x$ such that $t_ x$ comes from a section $t$ of $\mathcal{L}^{\otimes -1}$ over $U$ and such that $s \otimes t$ maps to $1 \in \mathcal{O}_ X(U)$. Clearly, for every $x' \in U$ we see that $s$ generates the module $\mathcal{L}_{x'}$. Hence $U \subset X_ s$. This proves that $X_ s$ is open. Moreover, the section $t$ constructed over $U$ above is unique, and hence these glue to give the section $s'$ of the lemma. $\square$