18.32 Invertible modules
Here is the definition.
Definition 18.32.1. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site.
A finite locally free $\mathcal{O}$-module $\mathcal{F}$ is said to have rank $r$ if for every object $U$ of $\mathcal{C}$ there exists a covering $\{ U_ i \to U\} $ of $U$ such that $\mathcal{F}|_{U_ i}$ is isomorphic to $\mathcal{O}_{U_ i}^{\oplus r}$ as an $\mathcal{O}_{U_ i}$-module.
An $\mathcal{O}$-module $\mathcal{L}$ is invertible if the functor
\[ \textit{Mod}(\mathcal{O}) \longrightarrow \textit{Mod}(\mathcal{O}),\quad \mathcal{F} \longmapsto \mathcal{F} \otimes _\mathcal {O} \mathcal{L} \]
is an equivalence.
The sheaf $\mathcal{O}^*$ is the subsheaf of $\mathcal{O}$ defined by the rule
\[ U \longmapsto \mathcal{O}^*(U) = \{ f \in \mathcal{O}(U) \mid \exists g \in \mathcal{O}(U)\text{ such that }fg = 1\} \]
It is a sheaf of abelian groups with multiplication as the group law.
Lemma 18.40.7 below explains the relationship with locally free modules of rank $1$.
Lemma 18.32.2. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{L}$ be an $\mathcal{O}$-module. The following are equivalent:
$\mathcal{L}$ is invertible, and
there exists an $\mathcal{O}$-module $\mathcal{N}$ such that $\mathcal{L} \otimes _\mathcal {O} \mathcal{N} \cong \mathcal{O}$.
In this case we have
$\mathcal{L}$ is a flat $\mathcal{O}$-module of finite presentation,
for every object $U$ of $\mathcal{C}$ there exists a covering $\{ U_ i \to U\} $ such that $\mathcal{L}|_{U_ i}$ is a direct summand of a finite free module, and
the module $\mathcal{N}$ in (2) is isomorphic to $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O})$.
Proof.
Assume (1). Then the functor $- \otimes _\mathcal {O} \mathcal{L}$ is essentially surjective, hence there exists an $\mathcal{O}$-module $\mathcal{N}$ as in (2). If (2) holds, then the functor $- \otimes _\mathcal {O} \mathcal{N}$ is a quasi-inverse to the functor $- \otimes _\mathcal {O} \mathcal{L}$ and we see that (1) holds.
Assume (1) and (2) hold. Since $- \otimes _\mathcal {O} \mathcal{L}$ is an equivalence, it is exact, and hence $\mathcal{L}$ is flat. Denote $\psi : \mathcal{L} \otimes _\mathcal {O} \mathcal{N} \to \mathcal{O}$ the given isomorphism. Let $U$ be an object of $\mathcal{C}$. We will show that the restriction $\mathcal{L}$ to the members of a covering of $U$ is a direct summand of a free module, which will certainly imply that $\mathcal{L}$ is of finite presentation. By construction of $\otimes $ we may assume (after replacing $U$ by the members of a covering) that there exists an integer $n \geq 1$ and sections $x_ i \in \mathcal{L}(U)$, $y_ i \in \mathcal{N}(U)$ such that $\psi (\sum x_ i \otimes y_ i) = 1$. 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 $x$ to $\sum x_ i \otimes x \otimes y_ i$ and the second arrow sends $x \otimes x' \otimes y$ to $\psi (x' \otimes y)x$. We conclude that $x \mapsto \sum \psi (x \otimes y_ i)x_ i$ 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 $x \mapsto (\psi (x \otimes y_1), \ldots , \psi (x \otimes y_ n))$ and the second arrow by $(a_1, \ldots , a_ n) \mapsto \sum a_ i x_ i$. 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
\[ \mathcal{L} \otimes _\mathcal {O} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O}_ X) \longrightarrow \mathcal{O}_ X \]
To finish the proof of the lemma we will show this is an isomorphism. By Lemma 18.27.6 we have
\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{O}, \mathcal{O}) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {O} (\mathcal{N} \otimes _\mathcal {O} \mathcal{L}, \mathcal{O}) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _\mathcal {O} (\mathcal{N}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O})) \]
The image of $1$ gives a morphism $\mathcal{N} \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O})$. Tensoring with $\mathcal{L}$ we obtain
\[ \mathcal{O} = \mathcal{L} \otimes _\mathcal {O} \mathcal{N} \longrightarrow \mathcal{L} \otimes _\mathcal {O} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O}) \]
This map is the inverse to the evaluation map; computation omitted.
$\square$
Lemma 18.32.3. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ be a morphism of ringed topoi. The pullback $f^*\mathcal{L}$ of an invertible $\mathcal{O}_\mathcal {D}$-module is invertible.
Proof.
By Lemma 18.32.2 there exists an $\mathcal{O}_\mathcal {D}$-module $\mathcal{N}$ such that $\mathcal{L} \otimes _{\mathcal{O}_\mathcal {D}} \mathcal{N} \cong \mathcal{O}_\mathcal {D}$. Pulling back we get $f^*\mathcal{L} \otimes _{\mathcal{O}_\mathcal {C}} f^*\mathcal{N} \cong \mathcal{O}_\mathcal {C}$ by Lemma 18.26.2. Thus $f^*\mathcal{L}$ is invertible by Lemma 18.32.2.
$\square$
Lemma 18.32.4. Let $(\mathcal{C}, \mathcal{O})$ be a ringed space.
If $\mathcal{L}$, $\mathcal{N}$ are invertible $\mathcal{O}$-modules, then so is $\mathcal{L} \otimes _\mathcal {O} \mathcal{N}$.
If $\mathcal{L}$ is an invertible $\mathcal{O}$-module, then so is $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O})$ and the evaluation map $\mathcal{L} \otimes _\mathcal {O} \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O}) \to \mathcal{O}$ is an isomorphism.
Proof.
Part (1) is clear from the definition and part (2) follows from Lemma 18.32.2 and its proof.
$\square$
Lemma 18.32.5. Let $(\mathcal{C}, \mathcal{O})$ be a ringed space. There exists a set of invertible modules $\{ \mathcal{L}_ i\} _{i \in I}$ such that each invertible module on $(\mathcal{C}, \mathcal{O})$ is isomorphic to exactly one of the $\mathcal{L}_ i$.
Proof.
Omitted, but see Sheaves of Modules, Lemma 17.25.8.
$\square$
Lemma 18.32.5 says that the collection of isomorphism classes of invertible sheaves forms a set. Lemma 18.32.4 says that tensor product defines the structure of an abelian group on this set with inverse of $\mathcal{L}$ given by $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O})$.
In fact, given an invertible $\mathcal{O}$-module $\mathcal{L}$ and $n \in \mathbf{Z}$ we define the $n$th tensor power $\mathcal{L}^{\otimes n}$ of $\mathcal{L}$ as the image of $\mathcal{O}$ under applying the equivalence $\mathcal{F} \mapsto \mathcal{F} \otimes _\mathcal {O} \mathcal{L}$ exactly $n$ times. This makes sense also for negative $n$ as we've defined an invertible $\mathcal{O}$-module as one for which tensoring is an equivalence. More explicitly, we have
\[ \mathcal{L}^{\otimes n} = \left\{ \begin{matrix} \mathcal{O}
& \text{if}
& n = 0
\\ \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{L}, \mathcal{O})
& \text{if}
& n = -1
\\ \mathcal{L} \otimes _\mathcal {O} \ldots \otimes _\mathcal {O} \mathcal{L}
& \text{if}
& n > 0
\\ \mathcal{L}^{\otimes -1} \otimes _\mathcal {O} \ldots \otimes _\mathcal {O} \mathcal{L}^{\otimes -1}
& \text{if}
& n < -1
\end{matrix} \right. \]
see Lemma 18.32.4. With this definition we have canonical isomorphisms $\mathcal{L}^{\otimes n} \otimes _\mathcal {O} \mathcal{L}^{\otimes m} \to \mathcal{L}^{\otimes n + m}$, and these isomorphisms satisfy a commutativity and an associativity constraint (formulation omitted).
Definition 18.32.6. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. The Picard group $\mathop{\mathrm{Pic}}\nolimits (\mathcal{O})$ of the ringed site is the abelian group whose elements are isomorphism classes of invertible $\mathcal{O}$-modules, with addition corresponding to tensor product.
Comments (4)
Comment #1188 by Mohamed Hashi on
Comment #1203 by Johan on
Comment #4565 by Martin Olsson on
Comment #4756 by Johan on