Lemma 52.27.3. Let $(A, \mathfrak m)$ be a Noetherian local ring and $f \in \mathfrak m$. Assume

1. the conditions of Lemma 52.27.1 hold, and

2. for every maximal ideal $\mathfrak p \subset A_ f$ the punctured spectrum of $(A_ f)_\mathfrak p$ has trivial Picard group.

Let $U$, resp. $U_0$ be the punctured spectrum of $A$, resp. $A/fA$. Then

$\mathop{\mathrm{Pic}}\nolimits (U) \longrightarrow \mathop{\mathrm{Pic}}\nolimits (U_0)$

is surjective.

Proof. Let $\mathcal{L}_0 \in \mathop{\mathrm{Pic}}\nolimits (U_0)$. By Lemma 52.27.1 there exists an open $U_0 \subset U' \subset U$ and $\mathcal{L}' \in \mathop{\mathrm{Pic}}\nolimits (U')$ whose restriction to $U_0$ is $\mathcal{L}_0$. Since $U' \supset U_0$ we see that $U \setminus U'$ consists of points corresponding to prime ideals $\mathfrak p_1, \ldots , \mathfrak p_ n$ as in (2). By assumption we can find invertible modules $\mathcal{L}'_ i$ on $\mathop{\mathrm{Spec}}(A_{\mathfrak p_ i})$ agreeing with $\mathcal{L}'$ over the punctured spectrum $U' \times _ U \mathop{\mathrm{Spec}}(A_{\mathfrak p_ i})$ since trivial invertible modules always extend. By Limits, Lemma 32.19.2 applied $n$ times we see that $\mathcal{L}'$ extends to an invertible module on $U$. $\square$

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