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The sheaf of a Cartier divisor is invertible

Statement

Let $X$ be a scheme and let $D$ be a Cartier divisor on $X$ , represented by meromorphic units $f_{i} \in K_{X} (U_{i})^{\times}$ with unit ratios on the overlaps (Cartier divisor), and let $O_{X} (D)$ be the $O_{X}$ -submodule sheaf of $K_{X}$ of Invertible sheaf of cartier divisor. Then $O_{X} (D)$ is an invertible $O_{X}$ -module (Invertible sheaves), and on each $U_{i}$ it is freely generated by $f_{i}^{- 1}$ , so that the map $O_{U_{i}} \to O_{X} (D) ∣_{U_{i}}$ , $a \mapsto a f_{i}^{- 1}$ , is an isomorphism. If $X$ is integral (Integral schemes), then $K_{X}$ is the constant sheaf with value the function field $K (X)$ (Sheaf total quotient rings), and $O_{X} (D)$ is a fractional $O_{X}$ -subsheaf of $K (X)$ : an $O_{X}$ -submodule of the constant sheaf $K (X)$ which is locally of the form $g O_{X}$ with $g \in K (X)^{\times}$ .

Facts & Assumptions

Given: A scheme $X$ , a Cartier divisor $D$ on $X$ with a local-equation datum ${(U_{i}, f_{i})}_{i \in I}$ .

[F1]

$O_{X} (D)$ is the subsheaf of $K_{X}$ whose sections over an open $V$ are the $g \in K_{X} (V)$ with $f_{i} g ∣_{V \cap U_{i}} \in O_{X} (V \cap U_{i})$ for all $i$ ; equivalently $O_{X} (D) ∣_{U_{i}} = f_{i}^{- 1} O_{U_{i}}$ , and this is well defined, independent of the datum, and closed under addition and regular scalar multiplication (Invertible sheaf of cartier divisor).

[F2]

An $O_{X}$ -module is invertible if and only if every point has an open neighbourhood on which it admits a generator, i.e. a section $s$ such that $O_{U} \to L ∣_{U}$ , $a \mapsto a s$ , is an isomorphism (Invertible sheaves).

[F3]

On an integral scheme $X$ the sheaf $K_{X}$ is the constant sheaf with value the function field $K (X)$ , and the structure map $O_{X} \to K_{X}$ is injective (Sheaf total quotient rings, Integral schemes).

Proof

1.1F1

Local generator. For each $i$ the section $f_{i}^{- 1} \in K_{X} (U_{i})$ lies in $O_{X} (D) (U_{i})$ : multiplying it by $f_{i}$ gives $1 \in O_{X} (U_{i})$ . The map $φ_{i} : O_{U_{i}} \to O_{X} (D) ∣_{U_{i}}$ , $a \mapsto a f_{i}^{- 1}$ , is an isomorphism, with inverse induced by $g \mapsto f_{i} g$ : for $g \in O_{X} (D) (V \cap U_{i})$ the product $f_{i} g$ lies in $O_{X} (V \cap U_{i})$ by the defining condition of [F1], and the two maps are mutually inverse because $f_{i}$ is a unit of $K_{X}$ .

1.2F1F3

Integral case. If $X$ is integral, then $K_{X}$ is the constant sheaf with value $K (X)$ by [F3], so $O_{X} (D)$ is a subsheaf of that constant sheaf; it is an $O_{X}$ -submodule by [F1]. On $U_{i}$ the description $O_{X} (D) ∣_{U_{i}} = f_{i}^{- 1} O_{U_{i}}$ of [F1] exhibits it as $g O_{X}$ with $g = f_{i}^{- 1} \in K (X)^{\times}$ , the meromorphic unit $f_{i}$ being an element of the field $K (X)$ . Hence $O_{X} (D)$ is a fractional $O_{X}$ -subsheaf of $K (X)$ .

2.1F2step 1.1

Invertibility. Since the $U_{i}$ cover $X$ , step 1.1 exhibits on every point an open neighbourhood on which $O_{X} (D)$ is freely generated by $f_{i}^{- 1}$ , so $O_{X} (D)$ is invertible.

3.1step 2.1step 1.2∎

Conclusion. $O_{X} (D)$ is invertible and locally freely generated by the sections $f_{i}^{- 1}$ , and on an integral $X$ it is a fractional subsheaf of $K (X)$ .

No choice principle is used: the local generators are the given equations of the divisor, and no trivialisation or atlas is selected. For the zero divisor one may take $f_{i} = 1$ on the whole of $X$ , so $O_{X} (0) = O_{X}$ is generated by $1$ . On the empty scheme the formula gives the unique module sheaf, which satisfies the local rank-one condition vacuously.

Depends on

Used by

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Sources

The Stacks Project, Divisors, §31.15 Definition 15.1 and §31.24 Definition 24.1 (standard reference, not scraped)
Ravi Vakil, The Rising Sea, Ch. 15 Exercise 15.2.D (standard reference, not scraped)