The Stacks project

Proof. Any polynomial ring over $R$ is a UFD by Algebra, Lemma 10.120.10. Thus the question is to show that $\mathop{\mathrm{Pic}}\nolimits (U)$ is trivial if $U \subset \mathop{\mathrm{Spec}}(R)$ is a (nonempty) open.

In the Noetherian case, we may use Lemma 31.28.3 to reduce to showing $\mathop{\mathrm{Pic}}\nolimits (\mathop{\mathrm{Spec}}(R))$ is trivial. This in turn is proved in More on Algebra, Lemma 15.117.3. We urge the reader to skip the rest of the proof.

Let $U \subset \mathop{\mathrm{Spec}}(R)$ be a nonempty open. Choose a nonzero $f \in R$ with $D(f) \subset U$. Since $f$ is a product of prime elements of $R$ we conclude that there are finitely many prime elements $a_1, \ldots , a_ n \in R$ such that $(a_ i) \not\in U$ (or equivalentaly $U \cap V(a_ i) = \emptyset$). After replacing $R$ by $R_{a_1 \ldots a_ n}$ we may assume that $U$ contains all principal prime ideals. (Note that a localization of a UFD is a UFD, for example by Algebra, Lemma 10.120.7.)

Assume $U$ contains all principal prime ideals. Write $U = \mathop{\mathrm{Spec}}(R) \setminus V(I)$ for some ideal $I \subset R$. Pick $f \in I$ nonzero and write $f = a_1 \ldots a_ n$ as a product of prime elements. Since $(a_ i) \not\in V(I)$ we may choose $g \in I$ with $g \not\in (a_ i)$ for all $i$ (prime avoidance). Picture

By More on Algebra, Lemma 15.117.3 the Picard groups of $D(f)$ and $D(g)$ are trivial and by an elementary argument about UFDs the map $R_ f^* \times R_ g^* \to R_{fg}^*$, $(u, v) \mapsto uv^{-1}$ is surjective. We conclude that $\mathop{\mathrm{Pic}}\nolimits (U')$ is trivial.

Thus it suffices to show that an invertible module $\mathcal{L}$ on $U$ whose pullback by $j : U' \to U'$ is trivial, is trivial. To do this it suffices to show that the map $\mathcal{L} \to j_*j^*\mathcal{L}$ is an isomorphism. For this in turn, let $u \in U \setminus U'$. Then $f, g \in \mathfrak m_ u \subset \mathcal{O}_{U, u} = A$ are elements of a local UFD which do not have a prime factor in common. The reader easily shows that this means that

is exact by looking at prime factors. Since $A = \mathcal{L}_ u$ because $\mathcal{L}$ is invertible and since $(j_*j^*\mathcal{L})_ u$ is the kernel of $A_ f \times A_ g \to A_{fg}$ we conclude. $\square$