The Stacks project

72.7 The Weil divisor class associated to an invertible module

In this section we go through exactly the same progression as in Section 72.6 to define a canonical map $\mathop{\mathrm{Pic}}\nolimits (X) \to \text{Cl}(X)$ on a locally Noetherian integral algebraic space.

Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. By Divisors on Spaces, Lemma 71.10.11 there exists a regular meromorphic section $s \in \Gamma (X, \mathcal{K}_ X(\mathcal{L}))$. In fact, by Divisors on Spaces, Lemma 71.10.8 this is the same thing as a nonzero element in $\mathcal{L}_\eta $ where $\eta \in |X|$ is the generic point. The same lemma tells us that if $\mathcal{L} = \mathcal{O}_ X$, then $s$ is the same thing as a nonzero rational function on $X$ (so what we will do below matches the construction in Section 72.6).

Let $Z \subset X$ be a prime divisor and let $\xi \in |Z|$ be the generic point. We are going to define the order of vanishing of $s$ along $Z$. Consider the canonical morphism

\[ c_\xi : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h) \longrightarrow X \]

whose source is the spectrum of the henselian local ring of $X$ as $\xi $ (Decent Spaces, Definition 68.11.7). The pullback $\mathcal{L}_\xi = c_\xi ^*\mathcal{L}$ is an invertible module and hence trivial; choose a generator $s_\xi $ of $\mathcal{L}_\xi $. Since $c_\xi $ is flat, pullbacks of meromorphic functions and (regular) sections are defined for $c_\xi $, see Divisors on Spaces, Definition 71.10.6 and Lemmas 71.10.7 and 71.10.10. Thus we get

\[ c_\xi ^*(s) = f s_\xi \]

for some nonzerodivisor $f \in Q(\mathcal{O}_{X, \xi }^ h)$. Here we are using Divisors, Lemma 31.24.2 to identify the space of meromorphic sections of $\mathcal{L}_\xi \cong \mathcal{O}_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h)}$ in terms of the total ring of fractions of $\mathcal{O}_{X, \xi }^ h$. Let us agree to denote this element

\[ s/s_\xi = f \in Q(\mathcal{O}_{X, \xi }^ h) \]

Observe that $f = s/s_\xi $ is replaced by $uf$ where $u \in \mathcal{O}_{X, \xi }^ h$ is a unit if we change our choice of $s_\xi $.

Definition 72.7.1. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic algebraic space over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{K}_ X(\mathcal{L}))$ be a regular meromorphic section of $\mathcal{L}$. For every prime divisor $Z \subset X$ with generic point $\xi \in |Z|$ we define the order of vanishing of $s$ along $Z$ as the integer

\[ \text{ord}_{Z, \mathcal{L}}(s) = \text{length}_{\mathcal{O}_{X, \xi }^ h} (\mathcal{O}_{X, \xi }^ h/a \mathcal{O}_{X, \xi }^ h) - \text{length}_{\mathcal{O}_{X, \xi }^ h} (\mathcal{O}_{X, \xi }^ h/b \mathcal{O}_{X, \xi }^ h) \]

where $a, b \in \mathcal{O}_{X, \xi }^ h$ are nonzerodivisors such that the element $s/s_\xi $ of $Q(\mathcal{O}_{X, \xi }^ h)$ constructed above is equal to $a/b$. This is well defined by the above and Algebra, Lemma 10.121.1.

As explained above, a regular meromorphic section $s$ of $\mathcal{O}_ X$ can be written $s = f \cdot 1$ where $f$ is a nonzero rational function on $X$ and we have $\text{ord}_ Z(f) = \text{ord}_{Z, \mathcal{O}_ X}(s)$. As in the case of principal divisors we have the following lemma.

Lemma 72.7.2. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s \in \mathcal{K}_ X(\mathcal{L})$ be a regular (i.e., nonzero) meromorphic section of $\mathcal{L}$. Then the sets

\[ \{ Z \subset X \mid Z \text{ a prime divisor with generic point }\xi \text{ and }s\text{ not in }\mathcal{L}_\xi \} \]

and

\[ \{ Z \subset X \mid Z \text{ is a prime divisor and } \text{ord}_{Z, \mathcal{L}}(s) \not= 0\} \]

are locally finite in $X$.

Proof. There exists a nonempty open subspace $U \subset X$ such that $s$ corresponds to a section of $\Gamma (U, \mathcal{L})$ which generates $\mathcal{L}$ over $U$. Hence the prime divisors which can occur in the sets of the lemma all correspond to irreducible components of $|X| \setminus |U|$. Hence Lemma 72.6.1. gives the desired result. $\square$

Lemma 72.7.3. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$ Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s, s' \in \mathcal{K}_ X(\mathcal{L})$ be nonzero meromorphic sections of $\mathcal{L}$. Then $f = s/s'$ is an element of $R(X)^*$ and we have

\[ \sum \text{ord}_{Z, \mathcal{L}}(s)[Z] = \sum \text{ord}_{Z, \mathcal{L}}(s')[Z] + \text{div}(f) \]

as Weil divisors.

Proof. This is clear from the definitions. Note that Lemma 72.7.2 guarantees that the sums are indeed Weil divisors. $\square$

Definition 72.7.4. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module.

  1. For any nonzero meromorphic section $s$ of $\mathcal{L}$ we define the Weil divisor associated to $s$ as

    \[ \text{div}_\mathcal {L}(s) = \sum \text{ord}_{Z, \mathcal{L}}(s) [Z] \in \text{Div}(X) \]

    where the sum is over prime divisors. This is well defined by Lemma 72.7.2.

  2. We define Weil divisor class associated to $\mathcal{L}$ as the image of $\text{div}_\mathcal {L}(s)$ in $\text{Cl}(X)$ where $s$ is any nonzero meromorphic section of $\mathcal{L}$ over $X$. This is well defined by Lemma 72.7.3.

As expected this construction is additive in the invertible module.

Lemma 72.7.5. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. Let $\mathcal{L}$, $\mathcal{N}$ be invertible $\mathcal{O}_ X$-modules. Let $s$, resp. $t$ be a nonzero meromorphic section of $\mathcal{L}$, resp. $\mathcal{N}$. Then $st$ is a nonzero meromorphic section of $\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{N}$ and

\[ \text{div}_{\mathcal{L} \otimes \mathcal{N}}(st) = \text{div}_\mathcal {L}(s) + \text{div}_\mathcal {N}(t) \]

in $\text{Div}(X)$. In particular, the Weil divisor class of $\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{N}$ is the sum of the Weil divisor classes of $\mathcal{L}$ and $\mathcal{N}$.

Proof. Let $s$, resp. $t$ be a nonzero meromorphic section of $\mathcal{L}$, resp. $\mathcal{N}$. Then $st$ is a nonzero meromorphic section of $\mathcal{L} \otimes \mathcal{N}$. Let $Z \subset X$ be a prime divisor. Let $\xi \in |Z|$ be its generic point. Choose generators $s_\xi \in \mathcal{L}_\xi $, and $t_\xi \in \mathcal{N}_\xi $ with notation as described earlier in this section. Then $s_\xi \otimes t_\xi $ is a generator for $(\mathcal{L} \otimes \mathcal{N})_\xi $. So $st/(s_\xi t_\xi ) = (s/s_\xi )(t/t_\xi )$ in $Q(\mathcal{O}_{X, \xi }^ h)$. Applying the additivity of Algebra, Lemma 10.121.1 we conclude that

\[ \text{div}_{\mathcal{L} \otimes \mathcal{N}, Z}(st) = \text{div}_{\mathcal{L}, Z}(s) + \text{div}_{\mathcal{N}, Z}(t) \]

Some details omitted. $\square$

Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. By the constructions and lemmas above we obtain a homomorphism of abelian groups

72.7.5.1
\begin{equation} \label{spaces-over-fields-equation-c1} \mathop{\mathrm{Pic}}\nolimits (X) \longrightarrow \text{Cl}(X) \end{equation}

which assigns to an invertible module its Weil divisor class.

Lemma 72.7.6. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. If $X$ is normal, then the map (72.7.5.1) $\mathop{\mathrm{Pic}}\nolimits (X) \to \text{Cl}(X)$ is injective.

Proof. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module whose associated Weil divisor class is trivial. Let $s$ be a regular meromorphic section of $\mathcal{L}$. The assumption means that $\text{div}_\mathcal {L}(s) = \text{div}(f)$ for some $f \in R(X)^*$. Then we see that $t = f^{-1}s$ is a regular meromorphic section of $\mathcal{L}$ with $\text{div}_\mathcal {L}(t) = 0$, see Lemma 72.7.3. We claim that $t$ defines a trivialization of $\mathcal{L}$. The claim finishes the proof of the lemma. Our proof of the claim is a bit awkward as we don't yet have a lot of theory at our dispposal; we suggest the reader skip the proof.

We may check our claim étale locally. Let $U \in X_{\acute{e}tale}$ be affine such that $\mathcal{L}|_ U$ is trivial. Say $s_ U \in \Gamma (U, \mathcal{L}|_ U)$ is a trivialization. By Properties, Lemma 28.7.5 we may also assume $U$ is integral. Write $U = \mathop{\mathrm{Spec}}(A)$ as the spectrum of a normal Noetherian domain $A$ with fraction field $K$. We may write $t|_ U = f s_ U$ for some element $f$ of $K$, see Divisors on Spaces, Lemma 71.10.4 for example. Let $\mathfrak p \subset A$ be a height one prime corresponding to a codimension $1$ point $u \in U$ which maps to a codimension $1$ point $\xi \in |X|$. Choose a trivialization $s_\xi $ of $\mathcal{L}_\xi $ as in the beginning of this section. Choose a geometric point $\overline{u}$ of $U$ lying over $u$. Then

\[ (\mathcal{O}_{X, \xi }^ h)^{sh} = \mathcal{O}_{X, \overline{u}} = \mathcal{O}_{U, u}^{sh} = (A_\mathfrak p)^{sh} \]

see Decent Spaces, Lemmas 68.11.9 and Properties of Spaces, Lemma 66.22.1. The normality of $X$ shows that all of these are discrete valuation rings. The trivializations $s_ U$ and $s_\xi $ differ by a unit as sections of $\mathcal{L}$ pulled back to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \overline{u}})$. Write $t = f_\xi s_\xi $ with $f_\xi \in Q(\mathcal{O}_{X, \xi }^ h)$. We conclude that $f_\xi $ and $f$ differ by a unit in $Q(\mathcal{O}_{X, \overline{u}})$. If $Z \subset X$ denotes the prime divisor corresponding to $\xi $ (Lemma 72.4.7), then $0 = \text{ord}_{Z, \mathcal{L}}(t) = \text{ord}_{\mathcal{O}_{X, \xi }^ h}(f_\xi )$ and since $\mathcal{O}_{X, \xi }^ h$ is a discrete valuation ring we see that $f_\xi $ is a unit. Thus $f$ is a unit in $\mathcal{O}_{X, \overline{u}}$ and hence in particular $f \in A_\mathfrak p$. This implies $f \in A$ by Algebra, Lemma 10.157.6. We conclude that $t \in \Gamma (X, \mathcal{L})$. Repeating the argument with $t^{-1}$ viewed as a meromorphic section of $\mathcal{L}^{\otimes -1}$ finishes the proof. $\square$


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