The Stacks project

81.17 The divisor associated to an invertible sheaf

This section is the analogue of Chow Homology, Section 42.24. The following definition is the analogue of Spaces over Fields, Definition 71.7.4 in our current setup.

Definition 81.17.1. In Situation 81.2.1 let $X/B$ be good. Assume $X$ is integral and $n = \dim _\delta (X)$. 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 $s$ is the $(n - 1)$-cycle

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

    defined in Spaces over Fields, Definition 71.7.4. This makes sense because Weil divisors have $\delta $-dimension $n - 1$ by Lemma 81.12.1.

  2. We define Weil divisor associated to $\mathcal{L}$ as

    \[ c_1(\mathcal{L}) \cap [X] = \text{class of }\text{div}_\mathcal {L}(s) \in \mathop{\mathrm{CH}}\nolimits _{n - 1}(X) \]

    where $s$ is any nonzero meromorphic section of $\mathcal{L}$ over $X$. This is well defined by Spaces over Fields, Lemma 71.7.3.

The zero scheme of a nonzero section is an effective Cartier divisor whose Weil divisor class computes the Weil divisor associated to the invertible module.

Lemma 81.17.2. In Situation 81.2.1 let $X/B$ be good. Assume $X$ is integral and $n = \dim _\delta (X)$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{L})$ be a nonzero global section. Then

\[ \text{div}_\mathcal {L}(s) = [Z(s)]_{n - 1} \]

in $Z_{n - 1}(X)$ and

\[ c_1(\mathcal{L}) \cap [X] = [Z(s)]_{n - 1} \]

in $\mathop{\mathrm{CH}}\nolimits _{n - 1}(X)$.

Proof. Let $Z \subset X$ be an integral closed subspace of $\delta $-dimension $n - 1$. Let $\xi \in |Z|$ be its generic point. To prove the first equality we compare the coefficients of $Z$ on both sides. Choose an elementary ├ętale neighbourhood $(U, u) \to (X, \xi )$, see Decent Spaces, Section 67.11 and recall that $\mathcal{O}_{X, \xi }^ h = \mathcal{O}_{U, u}^ h$ in this case. After replacing $U$ by an open neighbourhood of $u$ we may assume there is a trivializing section $s_ U$ of $\mathcal{L}|_ U$. Write $s|_ U = f s_ U$ for some $f \in \Gamma (U, \mathcal{O}_ U)$. Then $Z \times _ X U$ is equal to $V(f)$ as a closed subscheme of $U$, see Divisors on Spaces, Definition 70.7.6. As in Spaces over Fields, Section 71.7 denote $\mathcal{L}_\xi $ the pullback of $\mathcal{L}$ under the canonical morphism $c_\xi : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h) \to X$. Denote $s_\xi $ the pullback of $s_ U$; it is a trivialization of $\mathcal{L}_\xi $. Then we see that $c_\xi ^*(s) = fs_\xi $. The coefficient of $Z$ in $[Z(s)]_{n - 1}$ is by definition

\[ \text{length}_{\mathcal{O}_{U, u}}(\mathcal{O}_{U, u}/f\mathcal{O}_{U, u}) \]

Since $\mathcal{O}_{U, u} \to \mathcal{O}_{X, \xi }^ h$ is flat and identifies residue fields this is equal to

\[ \text{length}_{\mathcal{O}_{X, \xi }^ h} (\mathcal{O}_{X, \xi }^ h/f\mathcal{O}_{X, \xi }^ h) \]

by Algebra, Lemma 10.52.13. This final quantity is equal to $\text{ord}_{Z, \mathcal{L}}(s)$ by Spaces over Fields, Definition 71.7.1, i.e., to the coefficient of $Z$ in $\text{div}_\mathcal {L}(s)$ as desired. $\square$

Lemma 81.17.3. In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. The morphism

\[ q : T = \underline{\mathop{\mathrm{Spec}}}\left( \bigoplus \nolimits _{n \in \mathbf{Z}} \mathcal{L}^{\otimes n}\right) \longrightarrow X \]

has the following properties:

  1. $q$ is surjective, smooth, affine, of relative dimension $1$,

  2. there is an isomorphism $\alpha : q^*\mathcal{L} \cong \mathcal{O}_ T$,

  3. formation of $(q : T \to X, \alpha )$ commutes with base change,

  4. $q^* : Z_ k(X) \to Z_{k + 1}(T)$ is injective,

  5. if $Z \subset X$ is an integral closed subspace, then $q^{-1}(Z) \subset T$ is an integral closed subspace,

  6. if $Z \subset X$ is a closed subspace of $X$ of $\delta $-dimension $\leq k$, then $q^{-1}(Z)$ is a closed subspace of $T$ of $\delta $-dimension $\leq k + 1$ and $q^*[Z]_ k = [q^{-1}(Z)]_{k + 1}$,

  7. if $\xi ' \in |T|$ is the generic point of the fibre of $|T| \to |X|$ over $\xi $, then the ring map $\mathcal{O}_{X, \xi }^ h \to \mathcal{O}_{T, \xi '}^ h$ is flat, we have $\mathfrak m_{\xi '}^ h = \mathfrak m_\xi ^ h \mathcal{O}_{T, \xi '}^ h$, and the residue field extension is purely transcendental of transcendence degree $1$, and

  8. add more here as needed.

Proof. Let $U \to X$ be an ├ętale morphism such that $\mathcal{L}|_ U$ is trivial. Then $T \times _ X U \to U$ is isomorphic to the projection morphism $\mathbf{G}_ m \times U \to U$, where $\mathbf{G}_ m$ is the multipliciative group scheme, see Groupoids, Example 39.5.1. Thus (1) is clear.

To see (2) observe that $q_*q^*\mathcal{L} = \bigoplus _{n \in \mathbf{Z}} \mathcal{L}^{\otimes n + 1}$. Thus there is an obvious isomorphism $q_*q^*\mathcal{L} \to q_*\mathcal{O}_ T$ of $q_*\mathcal{O}_ T$-modules. By Morphisms of Spaces, Lemma 66.20.10 this determines an isomorphism $q^*\mathcal{L} \to \mathcal{O}_ T$.

Part (3) holds because forming the relative spectrum commutes with arbitrary base change and the same thing is clearly true for the isomorphism $\alpha $.

Part (4) follows immediately from (1) and the definitions.

Part (5) follows from the fact that if $Z$ is an integral algebraic space, then $\mathbf{G}_ m \times Z$ is an integral algebraic space.

Part (6) follows from the fact that lengths are preserved: if $(A, \mathfrak m)$ is a local ring and $B = A[x]_{\mathfrak m A[x]}$ and if $M$ is an $A$-module, then $\text{length}_ A(M) = \text{length}_ B(M \otimes _ A B)$. This implies that if $\mathcal{F}$ is a coherent $\mathcal{O}_ X$-module and $\xi \in |X|$ with $\xi ' \in |T|$ the generic point of the fibre over $\xi $, then the length of $\mathcal{F}$ at $\xi $ is the same as the length of $q^*\mathcal{F}$ at $\xi '$. Tracing through the definitions this gives (6) and more.

The map in part (7) comes from Decent Spaces, Remark 67.11.11. However, in our case we have

\[ \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h) \times _ X T = \mathbf{G}_ m \times \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h) = \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h[t, t^{-1}]) \]

and $\xi '$ corresponds to the generic point of the special fibre of this over $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi }^ h)$. Thus $\mathcal{O}_{T, \xi '}^ h$ is the henselization of the localization of $\mathcal{O}_{X, \xi }^ h[t, t^{-1}]$ at the corresponding prime. Part (7) follows from this and some commutative algebra; details omitted. $\square$

Lemma 81.17.4. In Situation 81.2.1 let $X/B$ be good. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume $X$ is integral. Let $s$ be a nonzero meromorphic section of $\mathcal{L}$. Let $q : T \to X$ be the morphism of Lemma 81.17.3. Then

\[ q^*\text{div}_\mathcal {L}(s) = \text{div}_ T(q^*(s)) \]

where we view the pullback $q^*(s)$ as a nonzero meromorphic function on $T$ using the isomorphism $q^*\mathcal{L} \to \mathcal{O}_ T$

Proof. Observe that $\text{div}_ T(q^*(s)) = \text{div}_{\mathcal{O}_ T}(q^*(s))$ by the compatibility between the constructions given in Spaces over Fields, Sections 71.6 and 71.7. We will show the agreement with $\text{div}_{\mathcal{O}_ T}(q^*(s))$ in this proof. We will use all the properties of $q : T \to X$ stated in Lemma 81.17.3 without further mention. Let $Z \subset T$ be a prime divisor. Then either $Z \to X$ is dominant or $Z = q^{-1}(Z')$ for some prime divisor $Z' \subset X$. If $Z \to X$ is dominant, then the coefficient of $Z$ in either side of the equality of the lemma is zero. Thus we may assume $Z = q^{-1}(Z')$ where $Z' \subset X$ is a prime divisor. Let $\xi ' \in |Z'|$ and $\xi \in |Z|$ be the generic points. Then we obtain a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(\mathcal{O}_{T, \xi }^ h) \ar[r]_-{c_\xi } \ar[d]_ h & T \ar[d]^ q \\ \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \xi '}^ h) \ar[r]^-{c_{\xi '}} & X } \]

see Decent Spaces, Remark 67.11.11. Choose a trivialization $s_{\xi '}$ of $\mathcal{L}_{\xi '} = c_{\xi '}^*\mathcal{L}$. Then we can use the pullback $s_\xi $ of $s_{\xi '}$ via $h$ as our trivialization of $\mathcal{L}_\xi = c_\xi ^* q^*\mathcal{L}$. Write $s/s_{\xi '} = a/b$ for $a, b \in \mathcal{O}_{X, \xi '}$ nonzerodivisors. By definition the coefficient of $Z'$ in $\text{div}_\mathcal {L}(s)$ is

\[ \text{length}_{\mathcal{O}_{X, \xi '}^ h}( \mathcal{O}_{X, \xi '}^ h/a \mathcal{O}_{X, \xi '}^ h) - \text{length}_{\mathcal{O}_{X, \xi '}^ h}( \mathcal{O}_{X, \xi '}^ h/b \mathcal{O}_{X, \xi '}^ h) \]

Since $Z = q^{-1}(Z')$, this is also the coefficient of $Z$ in $q^*\text{div}_\mathcal {L}(s)$. Since $\mathcal{O}_{X, \xi '}^ h \to \mathcal{O}_{T, \xi }^ h$ is flat the elements $a, b$ map to nonzerodivisors in $\mathcal{O}_{T, \xi }^ h$. Thus $q^*(s)/s_\xi = a/b$ in the total quotient ring of $\mathcal{O}_{T, \xi }^ h$. By definition the coefficient of $Z$ in $\text{div}_ T(q^*(s))$ is

\[ \text{length}_{\mathcal{O}_{T, \xi }^ h}( \mathcal{O}_{T, \xi }^ h/a \mathcal{O}_{T, \xi }^ h) - \text{length}_{\mathcal{O}_{T, \xi }^ h}( \mathcal{O}_{T, \xi }^ h/b \mathcal{O}_{T, \xi }^ h) \]

The proof is finished because these lengths are the same as before by Algebra, Lemma 10.52.13 and the fact that $\mathfrak m_\xi ^ h = \mathfrak m_{\xi '}^ h\mathcal{O}_{T, \xi }^ h$ shown in Lemma 81.17.3. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EQD. Beware of the difference between the letter 'O' and the digit '0'.