The Stacks project

Proof. Let $\mathcal{F}$ be a rank $1$ coherent reflexive $\mathcal{O}_ X$-module. Choose an open $U \subset X$ such that every irreducible component of $X \setminus U$ has codimension $\geq 2$ in $X$ and such that $\mathcal{F}|_ U$ is invertible, see Lemma 31.12.13. Observe that $\text{Cl}(U) = \text{Cl}(X)$ as the Weil divisor class group of $X$ only depends on its field of rational functions and the points of codimension $1$ and their local rings. Thus we can define the Weil divisor class of $\mathcal{F}$ to be the Weil divisor class of $\mathcal{F}|_ U$ in $\text{Cl}(U)$. We omit the verification that this is independent of the choice of $U$.

Denote $\text{Cl}'(X)$ the set of isomorphism classes of rank $1$ coherent reflexive $\mathcal{O}_ X$-modules. The construction above gives a group homorphism

because for any pair $\mathcal{F}, \mathcal{G}$ of elements of $\text{Cl}'(X)$ we can choose a $U$ which works for both and the assignment (31.27.5.1) sending an invertible module to its Weil divisor class is a homorphism. If $\mathcal{F}$ is in the kernel of this map, then we find that $\mathcal{F}|_ U$ is trivial (Lemma 31.27.6) and hence $\mathcal{F}$ is trivial too by Lemma 31.12.12 and Serre's criterion Properties, Lemma 28.12.5. To finish the proof it suffices to check the map is surjective.

Let $D = \sum n_ Z Z$ be a Weil divisor on $X$. We claim that there is an open $U \subset X$ such that every irreducible component of $X \setminus U$ has codimension $\geq 2$ in $X$ and such that $Z|_ U$ is an effective Cartier divisor for $n_ Z \not= 0$. To prove the claim we may assume $X$ is affine. Then we may assume $D = n_1 Z_1 + \ldots + n_ r Z_ r$ is a finite sum with $Z_1, \ldots , Z_ r$ pairwise distinct. After throwing out $Z_ i \cap Z_ j$ for $i \not= j$ we may assume $Z_1, \ldots , Z_ r$ are pairwise disjoint. This reduces us to the case of a single prime divisor $Z$ on $X$. As $X$ is $(R_1)$ by Properties, Lemma 28.12.5 the local ring $\mathcal{O}_{X, \xi }$ at the generic point $\xi$ of $Z$ is a discrete valuation ring. Let $f \in \mathcal{O}_{X, \xi }$ be a uniformizer. Let $V \subset X$ be an open neighbourhood of $\xi$ such that $f$ is the image of an element $f \in \mathcal{O}_ X(V)$. After shrinking $V$ we may assume that $Z \cap V = V(f)$ scheme theoretically, since this is true in the local ring at $\xi$. In this case taking

gives the desired open, thereby proving the claim.

In order to show that the divisor class of $D$ is in the image, we may write $D = \sum _{n_ Z < 0} n_ Z Z - \sum _{n_ Z > 0} (-n_ Z) Z$. By additivity of the map constructed above, we may and do assume $n_ Z \leq 0$ for all prime divisors $Z$ (this step may be avoided if the reader so desires). Let $U \subset X$ be as in the claim above. If $U$ is quasi-compact, then we write $D|_ U = -n_1 Z_1 - \ldots - n_ r Z_ r$ for pairwise distinct prime divisors $Z_ i$ and $n_ i > 0$ and we consider the invertible $\mathcal{O}_ U$-module

where $\mathcal{I}_ i$ is the ideal sheaf of $Z_ i$. This is invertible by our choice of $U$ and Lemma 31.13.7. Also $\text{div}_\mathcal {L}(1) = D|_ U$. Since $\mathcal{L} = \mathcal{F}|_ U$ for some rank $1$ coherent reflexive $\mathcal{O}_ X$-module $\mathcal{F}$ by Lemma 31.12.12 we find that $D$ is in the image of our map.

If $U$ is not quasi-compact, then we define $\mathcal{L} \subset \mathcal{O}_ U$ locally by the displayed formula above. The reader shows that the construction glues and finishes the proof exactly as before. Details omitted. $\square$