Similarly to the case of modules over rings (More on Algebra, Section 15.117) we have the following definition.
Definition 17.25.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. An invertible $\mathcal{O}_ X$-module is a sheaf of $\mathcal{O}_ X$-modules $\mathcal{L}$ such that the functor is an equivalence of categories. We say that $\mathcal{L}$ is trivial if it is isomorphic as an $\mathcal{O}_ X$-module to $\mathcal{O}_ X$.
\[ \textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ X),\quad \mathcal{F} \longmapsto \mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{F} \]
Lemma 17.25.4 below explains the relationship with locally free modules of rank $1$.
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
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. $\square$
Lemma 17.25.3. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. The pullback $f^*\mathcal{L}$ of an invertible $\mathcal{O}_ Y$-module is invertible.
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. $\square$
Lemma 17.25.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Any locally free $\mathcal{O}_ X$-module of rank $1$ is invertible. If all stalks $\mathcal{O}_{X, x}$ are local rings, then the converse holds as well (but in general this is not the case).
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. $\square$
Lemma 17.25.5. Let $(X, \mathcal{O}_ X)$ be a ringed space.
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 $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{L}, \mathcal{O}_ X)$ and the evaluation map $\mathcal{L} \otimes _{\mathcal{O}_ X} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{L}, \mathcal{O}_ X) \to \mathcal{O}_ X$ is an isomorphism.
Proof. Part (1) is clear from the definition and part (2) follows from Lemma 17.25.2 and its proof. $\square$
Definition 17.25.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Given an invertible sheaf $\mathcal{L}$ on $X$ and $n \in \mathbf{Z}$ we define the $n$th tensor power $\mathcal{L}^{\otimes n}$ of $\mathcal{L}$ as the image of $\mathcal{O}_ X$ under applying the equivalence $\mathcal{F} \mapsto \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}$ exactly $n$ times.
This makes sense also for negative $n$ as we've defined an invertible $\mathcal{O}_ X$-module as one for which tensoring is an equivalence. More explicitly, we have
see Lemma 17.25.5. With this definition we have canonical isomorphisms $\mathcal{L}^{\otimes n} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes m} \to \mathcal{L}^{\otimes n + m}$, and these isomorphisms satisfy a commutativity and an associativity constraint (formulation omitted).
Let $(X, \mathcal{O}_ X)$ be a ringed space. We can define a $\mathbf{Z}$-graded ring structure on $\bigoplus \Gamma (X, \mathcal{L}^{\otimes n})$ by mapping $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ and $t \in \Gamma (X, \mathcal{L}^{\otimes m})$ to the section corresponding to $s \otimes t$ in $\Gamma (X, \mathcal{L}^{\otimes n + m})$. We omit the verification that this defines a commutative and associative ring with $1$. However, by our conventions in Algebra, Section 10.56 a graded ring has no nonzero elements in negative degrees. This leads to the following definition.
Definition 17.25.7. Let $(X, \mathcal{O}_ X)$ be a ringed space. Given an invertible sheaf $\mathcal{L}$ on $X$ we define the associated graded ring to be Given a sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$ we set which we think of as a graded $\Gamma _*(X, \mathcal{L})$-module.
\[ \Gamma _*(X, \mathcal{L}) = \bigoplus \nolimits _{n \geq 0} \Gamma (X, \mathcal{L}^{\otimes n}) \]
\[ \Gamma _*(X, \mathcal{L}, \mathcal{F}) = \bigoplus \nolimits _{n \in \mathbf{Z}} \Gamma (X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n}) \]
We often write simply $\Gamma _*(\mathcal{L})$ and $\Gamma _*(\mathcal{F})$ (although this is ambiguous if $\mathcal{F}$ is invertible). The multiplication of $\Gamma _*(\mathcal{L})$ on $\Gamma _*(\mathcal{F})$ is defined using the isomorphisms above. If $\gamma : \mathcal{F} \to \mathcal{G}$ is a $\mathcal{O}_ X$-module map, then we get an $\Gamma _*(\mathcal{L})$-module homomorphism $\gamma : \Gamma _*(\mathcal{F}) \to \Gamma _*(\mathcal{G})$. If $\alpha : \mathcal{L} \to \mathcal{N}$ is an $\mathcal{O}_ X$-module map between invertible $\mathcal{O}_ X$-modules, then we obtain a graded ring homomorphism $\Gamma _*(\mathcal{L}) \to \Gamma _*(\mathcal{N})$. If $f : (Y, \mathcal{O}_ Y) \to (X, \mathcal{O}_ X)$ is a morphism of ringed spaces and if $\mathcal{L}$ is invertible on $X$, then we get an invertible sheaf $f^*\mathcal{L}$ on $Y$ (Lemma 17.25.3) and an induced homomorphism of graded rings
Furthermore, there are some compatibilities between the constructions above whose statements we omit.
Lemma 17.25.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. There exists a set of invertible modules $\{ \mathcal{L}_ i\} _{i \in I}$ such that each invertible module on $X$ is isomorphic to exactly one of the $\mathcal{L}_ i$.
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. $\square$
This lemma says roughly speaking that the collection of isomorphism classes of invertible sheaves forms a set. Lemma 17.25.5 says that tensor product defines the structure of an abelian group on this set.
Definition 17.25.9. Let $(X, \mathcal{O}_ X)$ be a ringed space. The Picard group $\mathop{\mathrm{Pic}}\nolimits (X)$ of $X$ is the abelian group whose elements are isomorphism classes of invertible $\mathcal{O}_ X$-modules, with addition corresponding to tensor product.
Lemma 17.25.10. Let $X$ be a ringed space. Assume that each stalk $\mathcal{O}_{X, x}$ is a local ring with maximal ideal $\mathfrak m_ x$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. For any section $s \in \Gamma (X, \mathcal{L})$ the set is open in $X$. The map $s : \mathcal{O}_{X_ s} \to \mathcal{L}|_{X_ s}$ is an isomorphism, and there exists a section $s'$ of $\mathcal{L}^{\otimes -1}$ over $X_ s$ such that $s' (s|_{X_ s}) = 1$.
\[ X_ s = \{ x \in X \mid \text{image }s \not\in \mathfrak m_ x\mathcal{L}_ x\} \]
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. $\square$
It is also true that, given a morphism of locally ringed spaces $f : Y \to X$ (see Schemes, Definition 26.2.1) that the inverse image $f^{-1}(X_ s)$ is equal to $Y_{f^*s}$, where $f^*s \in \Gamma (Y, f^*\mathcal{L})$ is the pullback of $s$.
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