The Stacks project

31.14 Effective Cartier divisors and invertible sheaves

Since an effective Cartier divisor has an invertible ideal sheaf (Definition 31.13.1) the following definition makes sense.

Definition 31.14.1. Let $S$ be a scheme. Let $D \subset S$ be an effective Cartier divisor with ideal sheaf $\mathcal{I}_ D$.

  1. The invertible sheaf $\mathcal{O}_ S(D)$ associated to $D$ is defined by

    \[ \mathcal{O}_ S(D) = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(\mathcal{I}_ D, \mathcal{O}_ S) = \mathcal{I}_ D^{\otimes -1}. \]
  2. The canonical section, usually denoted $1$ or $1_ D$, is the global section of $\mathcal{O}_ S(D)$ corresponding to the inclusion mapping $\mathcal{I}_ D \to \mathcal{O}_ S$.

  3. We write $\mathcal{O}_ S(-D) = \mathcal{O}_ S(D)^{\otimes -1} = \mathcal{I}_ D$.

  4. Given a second effective Cartier divisor $D' \subset S$ we define $\mathcal{O}_ S(D - D') = \mathcal{O}_ S(D) \otimes _{\mathcal{O}_ S} \mathcal{O}_ S(-D')$.

Some comments. We will see below that the assignment $D \mapsto \mathcal{O}_ S(D)$ turns addition of effective Cartier divisors (Definition 31.13.6) into addition in the Picard group of $S$ (Lemma 31.14.4). However, the expression $D - D'$ in the definition above does not have any geometric meaning. More precisely, we can think of the set of effective Cartier divisors on $S$ as a commutative monoid $\text{EffCart}(S)$ whose zero element is the empty effective Cartier divisor. Then the assignment $(D, D') \mapsto \mathcal{O}_ S(D - D')$ defines a group homomorphism

\[ \text{EffCart}(S)^{gp} \longrightarrow \mathop{\mathrm{Pic}}\nolimits (S) \]

where the left hand side is the group completion of $\text{EffCart}(S)$. In other words, when we write $\mathcal{O}_ S(D - D')$ we may think of $D - D'$ as an element of $\text{EffCart}(S)^{gp}$.

Lemma 31.14.2. Let $S$ be a scheme and let $D \subset S$ be an effective Cartier divisor. Then the conormal sheaf is $\mathcal{C}_{D/S} = \mathcal{I}_ D|D = \mathcal{O}_ S(-D)|_ D$ and the normal sheaf is $\mathcal{N}_{D/S} = \mathcal{O}_ S(D)|_ D$.

Proof. This follows from Morphisms, Lemma 29.31.2. $\square$

Lemma 31.14.3. Let $X$ be a scheme. Let $D, C \subset X$ be effective Cartier divisors with $C \subset D$ and let $D' = D + C$. Then there is a short exact sequence

\[ 0 \to \mathcal{O}_ X(-D)|_ C \to \mathcal{O}_{D'} \to \mathcal{O}_ D \to 0 \]

of $\mathcal{O}_ X$-modules.

Proof. In the statement of the lemma and in the proof we use the equivalence of Morphisms, Lemma 29.4.1 to think of quasi-coherent modules on closed subschemes of $X$ as quasi-coherent modules on $X$. Let $\mathcal{I}$ be the ideal sheaf of $D$ in $D'$. Then there is a short exact sequence

\[ 0 \to \mathcal{I} \to \mathcal{O}_{D'} \to \mathcal{O}_ D \to 0 \]

because $D \to D'$ is a closed immersion. There is a canonical surjection $\mathcal{I} \to \mathcal{I}/\mathcal{I}^2 = \mathcal{C}_{D/D'}$. We have $\mathcal{C}_{D/X} = \mathcal{O}_ X(-D)|_ D$ by Lemma 31.14.2 and there is a canonical surjective map

\[ \mathcal{C}_{D/X} \longrightarrow \mathcal{C}_{D/D'} \]

see Morphisms, Lemmas 29.31.3 and 29.31.4. Thus it suffices to show: (a) $\mathcal{I}^2 = 0$ and (b) $\mathcal{I}$ is an invertible $\mathcal{O}_ C$-module. Both (a) and (b) can be checked locally, hence we may assume $X = \mathop{\mathrm{Spec}}(A)$, $D = \mathop{\mathrm{Spec}}(A/fA)$ and $C = \mathop{\mathrm{Spec}}(A/gA)$ where $f, g \in A$ are nonzerodivisors (Lemma 31.13.2). Since $C \subset D$ we see that $f \in gA$. Then $I = fA/fgA$ has square zero and is invertible as an $A/gA$-module as desired. $\square$

Lemma 31.14.4. Let $S$ be a scheme. Let $D_1$, $D_2$ be effective Cartier divisors on $S$. Let $D = D_1 + D_2$. Then there is a unique isomorphism

\[ \mathcal{O}_ S(D_1) \otimes _{\mathcal{O}_ S} \mathcal{O}_ S(D_2) \longrightarrow \mathcal{O}_ S(D) \]

which maps $1_{D_1} \otimes 1_{D_2}$ to $1_ D$.

Proof. Omitted. $\square$

Lemma 31.14.5. Let $f : S' \to S$ be a morphism of schemes. Let $D$ be a effective Cartier divisors on $S$. If the pullback of $D$ is defined then $f^*\mathcal{O}_ S(D) = \mathcal{O}_{S'}(f^*D)$ and the canonical section $1_ D$ pulls back to the canonical section $1_{f^*D}$.

Proof. Omitted. $\square$

Definition 31.14.6. Let $(X, \mathcal{O}_ X)$ be a locally ringed space. Let $\mathcal{L}$ be an invertible sheaf on $X$. A global section $s \in \Gamma (X, \mathcal{L})$ is called a regular section if the map $\mathcal{O}_ X \to \mathcal{L}$, $f \mapsto fs$ is injective.

Lemma 31.14.7. Let $X$ be a locally ringed space. Let $f \in \Gamma (X, \mathcal{O}_ X)$. The following are equivalent:

  1. $f$ is a regular section, and

  2. for any $x \in X$ the image $f \in \mathcal{O}_{X, x}$ is a nonzerodivisor.

If $X$ is a scheme these are also equivalent to

  1. for any affine open $\mathop{\mathrm{Spec}}(A) = U \subset X$ the image $f \in A$ is a nonzerodivisor,

  2. there exists an affine open covering $X = \bigcup \mathop{\mathrm{Spec}}(A_ i)$ such that the image of $f$ in $A_ i$ is a nonzerodivisor for all $i$.

Proof. Omitted. $\square$

Note that a global section $s$ of an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ may be seen as an $\mathcal{O}_ X$-module map $s : \mathcal{O}_ X \to \mathcal{L}$. Its dual is therefore a map $s : \mathcal{L}^{\otimes -1} \to \mathcal{O}_ X$. (See Modules, Definition 17.24.6 for the definition of the dual invertible sheaf.)

Definition 31.14.8. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible sheaf. Let $s \in \Gamma (X, \mathcal{L})$ be a global section. The zero scheme of $s$ is the closed subscheme $Z(s) \subset X$ defined by the quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$ which is the image of the map $s : \mathcal{L}^{\otimes -1} \to \mathcal{O}_ X$.

Lemma 31.14.9. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible sheaf. Let $s \in \Gamma (X, \mathcal{L})$.

  1. Consider closed immersions $i : Z \to X$ such that $i^*s \in \Gamma (Z, i^*\mathcal{L})$ is zero ordered by inclusion. The zero scheme $Z(s)$ is the maximal element of this ordered set.

  2. For any morphism of schemes $f : Y \to X$ we have $f^*s = 0$ in $\Gamma (Y, f^*\mathcal{L})$ if and only if $f$ factors through $Z(s)$.

  3. The zero scheme $Z(s)$ is a locally principal closed subscheme.

  4. The zero scheme $Z(s)$ is an effective Cartier divisor if and only if $s$ is a regular section of $\mathcal{L}$.

Proof. Omitted. $\square$

slogan

Lemma 31.14.10. Let $X$ be a scheme.

  1. If $D \subset X$ is an effective Cartier divisor, then the canonical section $1_ D$ of $\mathcal{O}_ X(D)$ is regular.

  2. Conversely, if $s$ is a regular section of the invertible sheaf $\mathcal{L}$, then there exists a unique effective Cartier divisor $D = Z(s) \subset X$ and a unique isomorphism $\mathcal{O}_ X(D) \to \mathcal{L}$ which maps $1_ D$ to $s$.

The constructions $D \mapsto (\mathcal{O}_ X(D), 1_ D)$ and $(\mathcal{L}, s) \mapsto Z(s)$ give mutually inverse maps

\[ \left\{ \begin{matrix} \text{effective Cartier divisors on }X \end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} \text{isomorphism classes of pairs }(\mathcal{L}, s) \\ \text{consisting of an invertible } \mathcal{O}_ X\text{-module} \\ \mathcal{L}\text{ and a regular global section }s \end{matrix} \right\} \]

Proof. Omitted. $\square$

Remark 31.14.11. Let $X$ be a scheme, $\mathcal{L}$ an invertible $\mathcal{O}_ X$-module, and $s$ a regular section of $\mathcal{L}$. Then the zero scheme $D = Z(s)$ is an effective Cartier divisor on $X$ and there are short exact sequences

\[ 0 \to \mathcal{O}_ X \to \mathcal{L} \to i_*(\mathcal{L}|_ D) \to 0 \quad \text{and}\quad 0 \to \mathcal{L}^{\otimes -1} \to \mathcal{O}_ X \to i_*\mathcal{O}_ D \to 0. \]

Given an effective Cartier divisor $D \subset X$ using Lemmas 31.14.10 and 31.14.2 we get

\[ 0 \to \mathcal{O}_ X \to \mathcal{O}_ X(D) \to i_*(\mathcal{N}_{D/X}) \to 0 \quad \text{and}\quad 0 \to \mathcal{O}_ X(-D) \to \mathcal{O}_ X \to i_*(\mathcal{O}_ D) \to 0 \]


Comments (2)

Comment #4210 by Che Shen on

0B3P follows immediately from 01R3 (1). Maybe we can use this as proof instead of "omitted"?


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 0C4S. Beware of the difference between the letter 'O' and the digit '0'.