The Stacks project

Lemma 31.27.6. Let $X$ be a locally Noetherian integral scheme. If $X$ is normal, then the map ( $\mathop{\mathrm{Pic}}\nolimits (X) \to \text{Cl}(X)$ is injective.

Proof. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module whose associated Weil divisor class is trivial. Let $s$ be a regular meromorphic section of $\mathcal{L}$. The assumption means that $\text{div}_\mathcal {L}(s) = \text{div}(f)$ for some $f \in R(X)^*$. Then we see that $t = f^{-1}s$ is a regular meromorphic section of $\mathcal{L}$ with $\text{div}_\mathcal {L}(t) = 0$, see Lemma 31.27.3. We will show that $t$ defines a trivialization of $\mathcal{L}$ which finishes the proof of the lemma. In order to prove this we may work locally on $X$. Hence we may assume that $X = \mathop{\mathrm{Spec}}(A)$ is affine and that $\mathcal{L}$ is trivial. Then $A$ is a Noetherian normal domain and $t$ is an element of its fraction field such that $\text{ord}_{A_\mathfrak p}(t) = 0$ for all height $1$ primes $\mathfrak p$ of $A$. Our goal is to show that $t$ is a unit of $A$. Since $A_\mathfrak p$ is a discrete valuation ring for height one primes of $A$ (Algebra, Lemma 10.157.4), the condition signifies that $t \in A_\mathfrak p^*$ for all primes $\mathfrak p$ of height $1$. This implies $t \in A$ and $t^{-1} \in A$ by Algebra, Lemma 10.157.6 and the proof is complete. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

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.

All contributions are licensed under the GNU Free Documentation License.

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 0BE8. Beware of the difference between the letter 'O' and the digit '0'.