The Stacks project

Proof. If (1) holds, then $X$ is normal by Algebra, Lemma 10.120.11. Hence the map (31.27.5.1) is injective by Lemma 31.27.6. Moreover, every prime divisor $D \subset X$ is an effective Cartier divisor by Lemma 31.15.7. In this case the canonical section $1_ D$ of $\mathcal{O}_ X(D)$ (Definition 31.14.1) vanishes exactly along $D$ and we see that the class of $D$ is the image of $\mathcal{O}_ X(D)$ under the map (31.27.5.1). Thus the map is surjective as well.

Assume (2) holds. Pick a prime divisor $D \subset X$. Since (31.27.5.1) is surjective there exists an invertible sheaf $\mathcal{L}$, a regular meromorphic section $s$, and $f \in R(X)^*$ such that $\text{div}_\mathcal {L}(s) + \text{div}(f) = [D]$. In other words, $\text{div}_\mathcal {L}(fs) = [D]$. Let $x \in X$ and let $A = \mathcal{O}_{X, x}$. Thus $A$ is a Noetherian local normal domain with fraction field $K = R(X)$. Every height $1$ prime of $A$ corresponds to a prime divisor on $X$ and every invertible $\mathcal{O}_ X$-module restricts to the trivial invertible module on $\mathop{\mathrm{Spec}}(A)$. It follows that for every height $1$ prime $\mathfrak p \subset A$ there exists an element $f \in K$ such that $\text{ord}_{A_\mathfrak p}(f) = 1$ and $\text{ord}_{A_{\mathfrak p'}}(f) = 0$ for every other height one prime $\mathfrak p'$. Then $f \in A$ by Algebra, Lemma 10.157.6. Arguing in the same fashion we see that every element $g \in \mathfrak p$ is of the form $g = af$ for some $a \in A$. Thus we see that every height one prime ideal of $A$ is principal and $A$ is a UFD by Algebra, Lemma 10.120.6. $\square$