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Let $X$ be a noetherian local normal scheme, we may even assume that $X$ is complete if necessary.

Consider $X\times (\mathbb{A}^2-\{0\})$, is it true that the Picard group of this scheme vanishes?

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  • $\begingroup$ I doubt very much that this is true. The Zariski tangent space of the Picard functor, $H^1(\mathbb{A}^2_k\setminus\{0\},\mathcal{O})$ is huge, and the obstruction group, $H^2(\mathbb{A}^2_k\setminus\{0\},\mathcal{O})$, is zero. So over the formal scheme $\text{Spf}\ k[[t]]$, there are many nontrivial elements. Using homogeneity and the quotient projective line, I suspect that we can algebraize some of these formal invertible sheaves. $\endgroup$ Oct 14, 2017 at 11:16
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    $\begingroup$ I think this is true. Let $X = \mathrm{Spec}(R)$, then I want to say that $\mathrm{Spec}\ R[x, y^{\pm 1}]$, $\mathrm{Spec}\ R[x^{\pm 1}, y]$ and $\mathrm{Spec}\ R[x^{\pm 1}, y^{\pm 1}]$ all have trivial $\mathrm{Pic}$, so we can compute using the Cech cover $\mathrm{Spec}(R)[x, y^{\pm 1}] \cup \mathrm{Spec}(R)[x^{\pm 1}, y]$. As long as $R$ is a domain, the units of $R[x^{\pm 1}, y^{\pm 1}]$ are just $R^{\times} x^i y^j$, so they are generated by the units of $R[x, y^{\pm 1}]$ and $R[x^{\pm 1}, y]$. The only thing I'm missing is the justification for the Pic vanishing in the first sentence. $\endgroup$ Oct 14, 2017 at 16:39
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    $\begingroup$ For example, the above argument works whenever $R$ is a UFD. $\endgroup$ Oct 14, 2017 at 16:40
  • $\begingroup$ I now agree with David Speyer. We can trivialize the invertible sheaf on the "$(1,1)$"-cross-section. Then the problem reduces to extending this trivialization to all of $\mathbb{A}^2_R\setminus \text{Zero}_R$. If we restrict over DVRs inside the fraction field of $R$ that dominate $R$, the trivialization extends uniquely. So I think we can use the fact that a normal integral domain equals the intersection of such DVRs to extend the trivialization over $\mathbb{A}^2_R\setminus\text{Zero}_R$. $\endgroup$ Oct 14, 2017 at 23:14

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My doubt in the first comment was incorrect. There are, indeed, many formal invertible sheaves without trivializations over $\text{Spf} \ k[[t]].$ Thus, for instance, over $R=\text{Spec}\ k[t]/\langle t^n\rangle$ there are invertible sheaves on $\mathbb{A}^2_R\setminus\{0\}$ that are not isomorphic to the structure sheaf. Nonetheless, for every locally Noetherian scheme $S$ that is normal, every invertible sheaf is the pullback of a unique invertible sheaf from $S$.

A bit more generally, let $T$ be a locally Noetherian scheme, let $\mathbb{A}_{T}$ be a finite type, separated $T$-scheme, let $\mathbb{A}^{*}_{T}\subset \mathbb{A}_{T}$ be an open subset with closed complement $C,$ and let $\sigma:T\to \mathbb{A}_{T}$ be a section.

Hypothesis 0. The $T$-scheme $\mathbb{A}_T$ is smooth, and for every morphism from a field, $t:\text{Spec}\ \kappa \to T,$ the base change scheme $\mathbb{A}_\kappa$ is integral, $C_\kappa$ has codimension $\geq 2$, every invertible sheaf on $\mathbb{A}_\kappa$ is isomorphic to the structure sheaf, and the following pullback homomorphism is an isomorphism, $$\kappa^\times \to \mathcal{O}_{\mathbb{A}_\kappa}(\mathbb{A}_\kappa)^\times.$$

For every $T$-scheme $S$, denote by $\mathbb{A}_S$ the fiber product $\mathbb{A}_T\times_T S$ with its base change section $\sigma_S:S\to \mathbb{A}_S.$ Denote by $\pi_S:\mathbb{A}_S \to S$ the projection morphism.

Definition 1. For every invertible sheaf $\mathcal{L}$ on $\mathbb{A}^{*}_S$, a normalized trivialization is an isomorphism of $\mathcal{L}$ with $\pi_S^*\sigma_S^*\mathcal{L}$ whose pullback by $\sigma_S$ equals the tautological trivialization. The normalized invertible sheaf $\mathcal{N}$ is the invertible sheaf $$\mathcal{L}\otimes_{\mathcal{O}}\pi_S^*\sigma_S^*\mathcal{L}^\vee.$$ By construction, there is a canonical trivializing global section $n:\mathcal{O}_S\to \sigma_S^*\mathcal{N}.$

Lemma 2. There exists a normalized trivialization of $\mathcal{L}$ if and only there exists a normalized trivialization of $\mathcal{N}.$

Proof This is just a matter of unwinding definitions. QED

Lemma 3. If $S$ is Noetherian, then there exists a coherent sheaf $\overline{\mathcal{N}}$ on $\mathbb{A}_S$ whose restriction to $\mathbb{A}^{*}_S$ is isomorphic to $\mathcal{N}.$ If $S$ is also locally integral, then there exists such a coherent sheaf that is isomorphic to $\textit{Hom}_{\mathcal{O}}(\mathcal{M},\mathcal{O}_{\mathbb{A}^2_S})$ for a coherent, reflexive sheaf $\mathcal{M}.$

Proof. By an Exercise II.5.15 of Hartshorne's Algebraic geometry, there exists a coherent sheaf $\mathcal{N}^{\text{pre}}$ on $\mathbb{A}_S$ whose restriction to $\mathbb{A}^{*}_S$ equals $\mathcal{N}.$ Then both of the following sheaves are also coherent, $$ \mathcal{M} := (\mathcal{N}^{\text{pre}})^\vee=\textit{Hom}_{\mathcal{O}}(\mathcal{N}^\text{pre},\mathcal{O}_{\mathbb{A}_S}),\ \ \ \overline{\mathcal{N}}=\textit{Hom}_{\mathcal{O}}(\mathcal{M},\mathcal{O}_{\mathbb{A}_S}). $$ QED

Lemma 4. Let $X$ be a scheme that is locally Noetherian and normal. Let $\phi:\mathcal{E}\to \mathcal{F}$ be an injective homomorphism of reflexive, coherent $\mathcal{O}_X$-modules whose cokernel has support of codimension $\geq 2$. Then $\phi$ is an isomorphism.

Proof. Denote by $U$ the open complement of the support $B$ of $\text{Coker}(\phi)$. The restriction $\phi|_U$ is an isomorphism of $\mathcal{O}_U$-modules. Denote by $\psi_U:\mathcal{F}|_U\to \mathcal{E}|_U$ the inverse of $\phi|_U$. Since $X$ is normal, by Serre's criterion, $X$ is $S2$. By hypothesis, the closed subset $B$ has codimension $\geq 2$ everywhere. Thus, by Exercise III.3.5, p. 282, of Hartshorne's Algebraic geometry, there is a unique homomorphism of coherent $\mathcal{O}_X$-modules, $$\psi:\mathcal{F}\to \mathcal{E},$$ whose restriction to $U$ equals $\psi_U.$ By construction, both $\psi\circ \phi$ and $\phi\circ \psi$ restrict on $U$ to equal the respective identity homomorphisms. Thus, by the uniqueness from Exercise III.3.5, these compositions equal the respective identity homomorphisms on all of $X$. QED

Proposition 5. For every Noetherian, normal $T$-scheme $S,$ every invertible sheaf on $\mathbb{A}^{*}_S$ has a unique normalized trivialization.

Proof. Algebraic geometry</I>. -> Assume that $S=\text{Spec}\ R$ is affine and connected: the uniqueness guarantees the glueing hypothesis relative to an open affine cover of $S.$ By hypothesis, $R$ is an integral domain. Denote by $K$ the fraction field of $R.$ Since Hom for finitely presented modules commutes with localization, the base change $\overline{\mathcal{N}}_K$ is the reflexive extension of $\mathcal{N}_K.$ By hypothesis, there is a unique normalized trivialization of $\mathcal{N}_K.$ Clearing denominators, there exists a unique normalized trivialization $s_r$ after base change to $\text{Spec}\ R[r^{-1}]$ for some nonzero $r\in R.$

By Krull's Hauptidealsatz, for every minimal prime $\mathfrak{p}$ over the principal ideal $\langle r \rangle$, the height of $\mathfrak{p}$ equals one. Since $R$ is normal, it is $R1$. Thus, the local ring $R_{\mathfrak{p}}$ is a DVR, and the scheme $$\mathbb{A}_{\mathfrak{p}}:=\mathbb{A}_{T}\times_T \text{Spec}\ R_{\mathfrak{p}}$$ is a smooth scheme over the regular ring $R_{\mathfrak{p}}$. So $\mathbb{A}_{\mathfrak{p}}$ is a regular scheme. Hence it is locally factorial by the Auslander-Buchsbaum-Nagata theorem. Thus, the base change of the reflexive, rank one sheaf $\overline{\mathcal{N}}$ is an invertible sheaf, cf. Theorem II.6.11 of Hartshorne's Algebraic geometry. In particular, the unique normalized trivialization over the residue field $\mathbb{A}_{\kappa(\mathfrak{p})}$ lifts to a section over $\mathbb{A}_{\mathfrak{p}}$.

Since this section is surjective after base change to $\kappa(\mathfrak{p}),$ the section is surjective on $\mathbb{A}_{\kappa(\mathfrak{p})}$ by Nakayama's Lemma. By Lemma 4, the section is an isomorphism, i.e., it is a trivialization. After adjusting the section by the multiplicative inverse of the pullback by $\sigma,$ the section is a normalized trivialization. Since the normalized trivialization is unique, this agrees with $s_r$.

Therefore $s_r$ extends over every minimal prime over $\langle r \rangle$. By the usual limit arguments, the maximal open subset on which $s_r$ extends to an $\mathcal{O}_{\mathbb{A}_R}$-module homomorphism that is surjective is an open subscheme of $\mathbb{A}_R$ that contains $\mathbb{A}_{R[r^{-1}]}$ as well as the generic point of every fiber $\mathbb{A}_{\mathfrak{p}}.$ Thus, the complement in $\mathbb{A}_{R[r^{-1}]}$ of this open is a closed subset $B$ of codimension $\geq 2.$ By Lemma 4 once more, $s_r$ extends to a normalized trivialization over all of $\mathbb{A}_R.$ Since the normalized trivialization over $K$ is unique, this normalized trivialization over $R$ is unique. QED

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  • $\begingroup$ There was a typo and one of the proofs was not displaying (it got caught between some open HTML tags). It should display correctly now. $\endgroup$ Oct 16, 2017 at 1:09
  • $\begingroup$ I reorganized the proof to make it more clear. $\endgroup$ Oct 16, 2017 at 2:20

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