Lemma 70.7.6. Let $S$ be a scheme. Let $X$ be a locally Noetherian integral algebraic space over $S$. If $X$ is normal, then the map (70.7.5.1) $\mathop{\mathrm{Pic}}\nolimits (X) \to \text{Cl}(X)$ 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 70.7.3. We claim that $t$ defines a trivialization of $\mathcal{L}$. The claim finishes the proof of the lemma. Our proof of the claim is a bit awkward as we don't yet have a lot of theory at our dispposal; we suggest the reader skip the proof.

We may check our claim étale locally. Let $U \in X_{\acute{e}tale}$ be affine such that $\mathcal{L}|_ U$ is trivial. Say $s_ U \in \Gamma (U, \mathcal{L}|_ U)$ is a trivialization. By Properties, Lemma 28.7.5 we may also assume $U$ is integral. Write $U = \mathop{\mathrm{Spec}}(A)$ as the spectrum of a normal Noetherian domain $A$ with fraction field $K$. We may write $t|_ U = f s_ U$ for some element $f$ of $K$, see Divisors on Spaces, Lemma 69.10.4 for example. Let $\mathfrak p \subset A$ be a height one prime corresponding to a codimension $1$ point $u \in U$ which maps to a codimension $1$ point $\xi \in |X|$. Choose a trivialization $s_\xi$ of $\mathcal{L}_\xi$ as in the beginning of this section. Choose a geometric point $\overline{u}$ of $U$ lying over $u$. Then

$(\mathcal{O}_{X, \xi }^ h)^{sh} = \mathcal{O}_{X, \overline{u}} = \mathcal{O}_{U, u}^{sh} = (A_\mathfrak p)^{sh}$

see Decent Spaces, Lemmas 66.11.9 and Properties of Spaces, Lemma 64.22.1. The normality of $X$ shows that all of these are discrete valuation rings. The trivializations $s_ U$ and $s_\xi$ differ by a unit as sections of $\mathcal{L}$ pulled back to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, \overline{u}})$. Write $t = f_\xi s_\xi$ with $f_\xi \in Q(\mathcal{O}_{X, \xi }^ h)$. We conclude that $f_\xi$ and $f$ differ by a unit in $Q(\mathcal{O}_{X, \overline{u}})$. If $Z \subset X$ denotes the prime divisor corresponding to $\xi$ (Lemma 70.4.7), then $0 = \text{ord}_{Z, \mathcal{L}}(t) = \text{ord}_{\mathcal{O}_{X, \xi }^ h}(f_\xi )$ and since $\mathcal{O}_{X, \xi }^ h$ is a discrete valuation ring we see that $f_\xi$ is a unit. Thus $f$ is a unit in $\mathcal{O}_{X, \overline{u}}$ and hence in particular $f \in A_\mathfrak p$. This implies $f \in A$ by Algebra, Lemma 10.155.6. We conclude that $t \in \Gamma (X, \mathcal{L})$. Repeating the argument with $t^{-1}$ viewed as a meromorphic section of $\mathcal{L}^{\otimes -1}$ finishes the proof. $\square$

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