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. $\square$


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"?

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