Lemma 31.28.4. Let $R$ be a UFD. The Picard groups of the following are trivial.
$\mathop{\mathrm{Spec}}(R)$ and any open subscheme of it.
$\mathbf{A}^ n_ R = \mathop{\mathrm{Spec}}(R[x_1, \ldots , x_ n])$ and any open subscheme of it.
In particular, the Picard group of any open subscheme of affine $n$-space $\mathbf{A}^ n_ k$ over a field $k$ is trivial.
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. $\square$
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)