## 17.24 Invertible modules

Similarly to the case of modules over rings (More on Algebra, Section 15.117) we have the following definition.

Definition 17.24.1. 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

$\textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ X),\quad \mathcal{F} \longmapsto \mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{F}$

is an equivalence of categories. We say that $\mathcal{L}$ is trivial if it is isomorphic as an $\mathcal{O}_ X$-module to $\mathcal{O}_ X$.

Lemma 17.24.4 below explains the relationship with locally free modules of rank $1$.

Lemma 17.24.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{L}$ be an $\mathcal{O}_ X$-module. Equivalent are

1. $\mathcal{L}$ is invertible, and

2. 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. $\square$

Lemma 17.24.3. 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 17.24.2 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 17.24.2. $\square$

Lemma 17.24.4. 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

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

Looking over an open covering trivialization $\mathcal{L}$, we see that this map is an isomorphism. Hence $\mathcal{L}$ is invertible by Lemma 17.24.2.

Assume all stalks $\mathcal{O}_{X, x}$ are local rings and $\mathcal{L}$ invertible. In the proof of Lemma 17.24.2 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 17.24.2 we conclude that $\mathcal{L}$ is locally free of rank $1$ by Lemma 17.11.6. $\square$

Lemma 17.24.5. Let $(X, \mathcal{O}_ X)$ be a ringed space.

1. If $\mathcal{L}$, $\mathcal{N}$ are invertible $\mathcal{O}_ X$-modules, then so is $\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{N}$.

2. 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 17.24.2 and its proof. $\square$

Definition 17.24.6. 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

$\mathcal{L}^{\otimes n} = \left\{ \begin{matrix} \mathcal{O}_ X & \text{if} & n = 0 \\ \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{L}, \mathcal{O}_ X) & \text{if} & n = -1 \\ \mathcal{L} \otimes _{\mathcal{O}_ X} \ldots \otimes _{\mathcal{O}_ X} \mathcal{L} & \text{if} & n > 0 \\ \mathcal{L}^{\otimes -1} \otimes _{\mathcal{O}_ X} \ldots \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes -1} & \text{if} & n < -1 \end{matrix} \right.$

see Lemma 17.24.5. 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 17.24.7. 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

$\Gamma _*(X, \mathcal{L}) = \bigoplus \nolimits _{n \geq 0} \Gamma (X, \mathcal{L}^{\otimes n})$

Given a sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$ we set

$\Gamma _*(X, \mathcal{L}, \mathcal{F}) = \bigoplus \nolimits _{n \in \mathbf{Z}} \Gamma (X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes n})$

which we think of as a graded $\Gamma _*(X, \mathcal{L})$-module.

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 17.24.3) and an induced homomorphism of graded rings

$f^* : \Gamma _*(X, \mathcal{L}) \longrightarrow \Gamma _*(Y, f^*\mathcal{L})$

Furthermore, there are some compatibilities between the constructions above whose statements we omit.

Lemma 17.24.8. 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 17.24.2. 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 17.24.5 says that tensor product defines the structure of an abelian group on this set.

Definition 17.24.9. 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 17.24.10. 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

$X_ s = \{ x \in X \mid \text{image }s \not\in \mathfrak m_ x\mathcal{L}_ x\}$

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 $s' (s|_{X_ s}) = 1$.

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

$\mathcal{L}_ x \otimes _{\mathcal{O}_{X, x}} (\mathcal{L}^{\otimes -1})_ x \longrightarrow \mathcal{O}_{X, x}$

by Lemma 17.24.5. 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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