Next we turn to surjectivity in local Lefschetz for the Picard group. First to extend an invertible module on $U_0$ to an open neighbourhood we have the following simple criterion.
Lemma 52.27.1. Let $(A, \mathfrak m)$ be a Noetherian local ring and $f \in \mathfrak m$. Assume
$A$ is $f$-adically complete,
$f$ is a nonzerodivisor,
$H^1_\mathfrak m(A/fA)$ and $H^2_\mathfrak m(A/fA)$ are finite $A$-modules, and
$H^3_\mathfrak m(A/fA) = 0$1.
Let $U$, resp. $U_0$ be the punctured spectrum of $A$, resp. $A/fA$. Then
is surjective.
Proof. Let $U_0 \subset U_ n \subset U$ be the $n$th infinitesimal neighbourhood of $U_0$. Observe that the ideal sheaf of $U_ n$ in $U_{n + 1}$ is isomorphic to $\mathcal{O}_{U_0}$ as $U_0 \subset U$ is the principal closed subscheme cut out by the nonzerodivisor $f$. Hence we have an exact sequence of abelian groups
see More on Morphisms, Lemma 37.4.1. Thus every invertible $\mathcal{O}_{U_0}$-module is the restriction of an invertible coherent formal module, i.e., an invertible object of $\textit{Coh}(U, f\mathcal{O}_ U)$. We conclude by applying Lemma 52.24.2. $\square$
Remark 52.27.2. Let $(A, \mathfrak m)$ be a Noetherian local ring and $f \in \mathfrak m$. The conclusion of Lemma 52.27.1 holds if we assume
$A$ has a dualizing complex,
$A$ is $f$-adically complete,
$f$ is a nonzerodivisor,
one of the following is true
$A_ f$ is $(S_2)$ and for $\mathfrak p \subset A$, $f \not\in \mathfrak p$ minimal we have $\dim (A/\mathfrak p) \geq 4$, or
if $\mathfrak p \not\in V(f)$ and $V(\mathfrak p) \cap V(f) \not= \{ \mathfrak m\} $, then $\text{depth}(A_\mathfrak p) + \dim (A/\mathfrak p) > 3$.
$H^3_{\mathfrak m}(A/fA) = 0$.
The proof is exactly the same as the proof of Lemma 52.27.1 using Lemma 52.24.1 instead of Lemma 52.24.2. Two points need to be made here: (a) it seems hard to find examples where one knows $H^3_{\mathfrak m}(A/fA) = 0$ without assuming $\text{depth}(A/fA) \geq 4$, and (b) the proof of Lemma 52.24.1 is a good deal harder than the proof of Lemma 52.24.2.
Lemma 52.27.3. Let $(A, \mathfrak m)$ be a Noetherian local ring and $f \in \mathfrak m$. Assume
the conditions of Lemma 52.27.1 hold, and
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
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 32.20.2 applied $n$ times we see that $\mathcal{L}'$ extends to an invertible module on $U$. $\square$
Lemma 52.27.4. Let $(A, \mathfrak m)$ be a Noetherian local ring of depth $\geq 2$. Let $A^\wedge $ be its completion. Let $U$, resp. $U^\wedge $ be the punctured spectrum of $A$, resp. $A^\wedge $. Then $\mathop{\mathrm{Pic}}\nolimits (U) \to \mathop{\mathrm{Pic}}\nolimits (U^\wedge )$ is injective.
Proof. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ U$-module with pullback $\mathcal{L}^\wedge $ on $U^\wedge $. We have $H^0(U, \mathcal{O}_ U) = A$ by our assumption on depth and Dualizing Complexes, Lemma 47.11.1 and Local Cohomology, Lemma 51.8.2. Thus $\mathcal{L}$ is trivial if and only if $M = H^0(U, \mathcal{L})$ is isomorphic to $A$ as an $A$-module. (Details omitted.) Since $A \to A^\wedge $ is flat we have $M \otimes _ A A^\wedge = \Gamma (U^\wedge , \mathcal{L}^\wedge )$ by flat base change, see Cohomology of Schemes, Lemma 30.5.2. Finally, it is easy to see that $M \cong A$ if and only if $M \otimes _ A A^\wedge \cong A^\wedge $. $\square$
Lemma 52.27.5. Let $(A, \mathfrak m)$ be a regular local ring. Then the Picard group of the punctured spectrum of $A$ is trivial.
Proof. Combine Divisors, Lemma 31.28.3 with More on Algebra, Lemma 15.121.2. $\square$
Now we can bootstrap the earlier results to prove that Picard groups are trivial for punctured spectra of complete intersections of dimension $\geq 4$. Recall that a Noetherian local ring is called a complete intersection if its completion is the quotient of a regular local ring by the ideal generated by a regular sequence. See the discussion in Divided Power Algebra, Section 23.8.
Proposition 52.27.6 (Grothendieck). Let $(A, \mathfrak m)$ be a Noetherian local ring. If $A$ is a complete intersection of dimension $\geq 4$, then the Picard group of the punctured spectrum of $A$ is trivial.
Proof. By Lemma 52.27.4 we may assume that $A$ is a complete local ring. By assumption we can write $A = B/(f_1, \ldots , f_ r)$ where $B$ is a complete regular local ring and $f_1, \ldots , f_ r$ is a regular sequence. We will finish the proof by induction on $r$. The base case is $r = 0$ which follows from Lemma 52.27.5.
Assume that $A = B/(f_1, \ldots , f_ r)$ and that the proposition holds for $r - 1$. Set $A' = B/(f_1, \ldots , f_{r - 1})$ and apply Lemma 52.27.3 to $f_ r \in A'$. This is permissible:
condition (1) of Lemma 52.27.1 holds because our local rings are complete,
condition (2) of Lemma 52.27.1 holds holds as $f_1, \ldots , f_ r$ is a regular sequence,
condition (3) and (4) of Lemma 52.27.1 hold as $A = A'/f_ r A'$ is Cohen-Macaulay of dimension $\dim (A) \geq 4$,
condition (2) of Lemma 52.27.3 holds by induction hypothesis as $\dim ((A'_{f_ r})_\mathfrak p) \geq 4$ for a maximal prime $\mathfrak p$ of $A'_{f_ r}$ and as $(A'_{f_ r})_\mathfrak p = B_\mathfrak q/(f_1, \ldots , f_{r - 1})$ for some prime ideal $\mathfrak q \subset B$ and $B_\mathfrak q$ is regular.
This finishes the proof. $\square$
Example 52.27.7. The dimension bound in Proposition 52.27.6 is sharp. For example the Picard group of the punctured spectrum of $A = k[[x, y, z, w]]/(xy - zw)$ is nontrivial. Namely, the ideal $I = (x, z)$ cuts out an effective Cartier divisor $D$ on the punctured spectrum $U$ of $A$ as it is easy to see that $I_ x, I_ y, I_ z, I_ w$ are invertible ideals in $A_ x, A_ y, A_ z, A_ w$. But on the other hand, $A/I$ has depth $\geq 1$ (in fact $2$), hence $I$ has depth $\geq 2$ (in fact $3$), hence $I = \Gamma (U, \mathcal{O}_ U(-D))$. Thus if $\mathcal{O}_ U(-D)$ were trivial, then we'd have $I \cong \Gamma (U, \mathcal{O}_ U) = A$ which isn't true as $I$ isn't generated by $1$ element.
Example 52.27.8. Proposition 52.27.6 cannot be extended to quotients where $B$ is regular and $\dim (B) - r \geq 4$. In other words, the condition that $f_1, \ldots , f_ r$ be a regular sequence is (in general) needed for vanishing of the Picard group of the punctured spectrum of $A$. Namely, let $k$ be a field and set Observe that $A = A_0[w]/(w^2)$ with $A_0 = k[[a, b, x, y, z, u, v]]/(a^3, b^3, xa^2 + yab + zb^2)$. We will show below that $A_0$ has depth $2$. Denote $U$ the punctured spectrum of $A$ and $U_0$ the punctured spectrum of $A_0$. Observe there is a short exact sequence $0 \to A_0 \to A \to A_0 \to 0$ where the first arrow is given by multiplication by $w$. By More on Morphisms, Lemma 37.4.1 we find that there is an exact sequence Since the depth of $A_0$ and hence $A$ is $2$ we see that $H^0(U_0, \mathcal{O}_{U_0}) = A_0$ and $H^0(U, \mathcal{O}_ U) = A$ and that $H^1(U_0, \mathcal{O}_{U_0})$ is nonzero, see Dualizing Complexes, Lemma 47.11.1 and Local Cohomology, Lemma 51.2.2. Thus the last arrow displayed above is nonzero and we conclude that $\mathop{\mathrm{Pic}}\nolimits (U)$ is nonzero. To show that $A_0$ has depth $2$ it suffices to show that $A_1 = k[[a, b, x, y, z]]/(a^3, b^3, xa^2 + yab + zb^2)$ has depth $0$. This is true because $a^2b^2$ maps to a nonzero element of $A_1$ which is annihilated by each of the variables $a, b, x, y, z$. For example $ya^2b^2 = (yab)(ab) = - (xa^2 + zb^2)(ab) = -xa^3b - yab^3 = 0$ in $A_1$. The other cases are similar.
\[ A = B/(f_1, \ldots , f_ r) \]
\[ A = k[[a, b, x, y, z, u, v, w]]/(a^3, b^3, xa^2 + yab + zb^2, w^2) \]
\[ H^0(U, \mathcal{O}_ U^*) \to H^0(U_0, \mathcal{O}_{U_0}^*) \to H^1(U_0, \mathcal{O}_{U_0}) \to \mathop{\mathrm{Pic}}\nolimits (U) \]
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