Lemma 33.38.3. Let $X$ be a scheme. Let $U \subset X$ be an open. Assume

1. $U$ is a retrocompact open of $X$,

2. $X \setminus U$ is discrete, and

3. for $x \in X \setminus U$ the local ring $\mathcal{O}_{X, x}$ is Noetherian of dimension $\leq 1$.

Then (1) there exists an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ and a section $s$ such that $U = X_ s$ and (2) the map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (U)$ is surjective.

Proof. Let $X \setminus U = \{ x_ i; i \in I\}$. Choose affine opens $U_ i \subset X$ with $x_ i \in U_ i$ and $x_ j \not\in U_ i$ for $j \not= i$. This is possible by condition (2). Say $U_ i = \mathop{\mathrm{Spec}}(A_ i)$. Let $\mathfrak m_ i \subset A_ i$ be the maximal ideal corresponding to $x_ i$. By our assumption on the local rings there are only a finite number of prime ideals $\mathfrak q \subset \mathfrak m_ i$, $\mathfrak q \not= \mathfrak m_ i$ (see Algebra, Lemma 10.31.6). Thus by prime avoidance (Algebra, Lemma 10.15.2) we can find $f_ i \in \mathfrak m_ i$ not contained in any of those primes. Then $V(f_ i) = \{ \mathfrak m_ i\} \amalg Z_ i$ for some closed subset $Z_ i \subset U_ i$ because $Z_ i$ is a retrocompact open subset of $V(f_ i)$ closed under specialization, see Algebra, Lemma 10.41.7. After shrinking $U_ i$ we may assume $V(f_ i) = \{ x_ i\}$. Then

$\mathcal{U} : X = U \cup \bigcup U_ i$

is an open covering of $X$. Consider the $2$-cocycle with values in $\mathcal{O}_ X^*$ given by $f_ i$ on $U \cap U_ i$ and by $f_ i/f_ j$ on $U_ i \cap U_ j$. This defines a line bundle $\mathcal{L}$ such that the section $s$ defined by $1$ on $U$ and $f_ i$ on $U_ i$ is as in the statement of the lemma.

Let $\mathcal{N}$ be an invertible $\mathcal{O}_ U$-module. Let $N_ i$ be the invertible $(A_ i)_{f_ i}$ module such that $\mathcal{N}|_{U \cap U_ i}$ is equal to $\tilde N_ i$. Observe that $(A_{\mathfrak m_ i})_{f_ i}$ is an Artinian ring (as a dimension zero Noetherian ring, see Algebra, Lemma 10.60.5). Thus it is a product of local rings (Algebra, Lemma 10.53.6) and hence has trivial Picard group. Thus, after shrinking $U_ i$ (i.e., after replacing $A_ i$ by $(A_ i)_ g$ for some $g \in A_ i$, $g \not\in \mathfrak m_ i$) we can assume that $N_ i = (A_ i)_{f_ i}$, i.e., that $\mathcal{N}|_{U \cap U_ i}$ is trivial. In this case it is clear how to extend $\mathcal{N}$ to an invertible sheaf over $X$ (by extending it by a trivial invertible module over each $U_ i$). $\square$

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