Lemma 31.27.7. Let $X$ be a locally Noetherian integral scheme. Consider the map (31.27.5.1) $\mathop{\mathrm{Pic}}\nolimits (X) \to \text{Cl}(X)$. The following are equivalent

1. the local rings of $X$ are UFDs, and

2. $X$ is normal and $\mathop{\mathrm{Pic}}\nolimits (X) \to \text{Cl}(X)$ is surjective.

In this case $\mathop{\mathrm{Pic}}\nolimits (X) \to \text{Cl}(X)$ is an isomorphism.

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. $\square$

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.

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 0BE9. Beware of the difference between the letter 'O' and the digit '0'.