**Proof.**
Let $\mathcal{L}$ be an invertible $\mathcal{O}_ U$-module. Observe that $\mathcal{L}$ maps to $0$ in $\mathop{\mathrm{Pic}}\nolimits (U_0)$ if and only if we can extend $\mathcal{L}$ to an invertible coherent triple $(\mathcal{L}, \mathcal{L}_0, \lambda )$ as in Section 52.25. By Proposition 52.25.4 the function

\[ n \longmapsto \chi ((\mathcal{L}, \mathcal{L}_0, \lambda )^{\otimes n}) \]

is a polynomial $P$. By Lemma 52.25.5 we have $P(n) \geq 0$ for all $n \in \mathbf{Z}$ with equality if and only if $\mathcal{L}^{\otimes n}$ is trivial. In particular $P(0) = 0$ and $P$ is either identically zero and we win or $P$ has even degree $\geq 2$.

Set $M = \Gamma (U, \mathcal{L})$ and $M_0 = \Gamma (X_0, \mathcal{L}_0) = \Gamma (U_0, \mathcal{L}_0)$. Then $M$ is a finite $A$-module of depth $\geq 2$ and $M_0 \cong A/fA$, see proof of Lemma 52.25.5. Note that $H^2_\mathfrak m(M)$ is finite $A$-module by Local Cohomology, Lemma 51.7.4 and the fact that $H^ i_\mathfrak m(A) = 0$ for $i = 0, 1, 2$ since $\text{depth}(A) \geq 3$. Consider the short exact sequence

\[ 0 \to M/fM \to M_0 \to Q \to 0 \]

Lemma 52.25.5 tells us $Q$ has finite length equal to $\chi (\mathcal{L}, \mathcal{L}_0, \lambda )$. We obtain $Q = H^1_\mathfrak m(M/fM)$ and $H^ i_\mathfrak m(M/fM) = H^ i_\mathfrak m(M_0) \cong H^ i_\mathfrak m(A/fA)$ for $i > 1$ from the long exact sequence of local cohomology associated to the displayed short exact sequence. Consider the long exact sequence of local cohomology associated to the sequence $0 \to M \to M \to M/fM \to 0$. It starts with

\[ 0 \to Q \to H^2_\mathfrak m(M) \to H^2_\mathfrak m(M) \to H^2_\mathfrak m(A/fA) \]

Using additivity of lengths we see that $\chi (\mathcal{L}, \mathcal{L}_0, \lambda )$ is equal to the length of the image of $H^2_\mathfrak m(M) \to H^2_\mathfrak m(A/fA)$.

Let prove the lemma in a special case to elucidate the rest of the proof. Namely, assume for a moment that $H^2_\mathfrak m(A/fA)$ is a finite length module. Then we would have $P(1) \leq \text{length}_ A H^2_\mathfrak m(A/fA)$. The exact same argument applied to $\mathcal{L}^{\otimes n}$ shows that $P(n) \leq \text{length}_ A H^2_\mathfrak m(A/fA)$ for all $n$. Thus $P$ cannot have positive degree and we win. In the rest of the proof we will modify this argument to give a linear upper bound for $P(n)$ which suffices.

Let us study the map $H^2_\mathfrak m(M) \to H^2_\mathfrak m(M_0) \cong H^2_\mathfrak m(A/fA)$. Choose a normalized dualizing complex $\omega _ A^\bullet $ for $A$. By local duality (Dualizing Complexes, Lemma 47.18.4) this map is Matlis dual to the map

\[ \text{Ext}^{-2}_ A(M, \omega _ A^\bullet ) \longleftarrow \text{Ext}^{-2}_ A(M_0, \omega _ A^\bullet ) \]

whose image therefore has the same (finite) length. The support (if nonempty) of the finite $A$-module $\text{Ext}^{-2}_ A(M_0, \omega _ A^\bullet )$ consists of $\mathfrak m$ and a finite number of primes $\mathfrak p_1, \ldots , \mathfrak p_ r$ containing $f$ with $\dim (A/\mathfrak p_ i) = 1$. Namely, by Local Cohomology, Lemma 51.9.4 the support is contained in the set of primes $\mathfrak p \subset A$ with $\text{depth}_{A_\mathfrak p}(M_{0, \mathfrak p}) + \dim (A/\mathfrak p) \leq 2$. Thus it suffices to show there is no prime $\mathfrak p$ containing $f$ with $\dim (A/\mathfrak p) = 2$ and $\text{depth}_{A_\mathfrak p}(M_{0, \mathfrak p}) = 0$. However, because $M_{0, \mathfrak p} \cong (A/fA)_\mathfrak p$ this would give $\text{depth}(A_\mathfrak p) = 1$ which contradicts assumption (4). Choose a section $t \in \Gamma (U, \mathcal{L}^{\otimes -1})$ which does not vanish in the points $\mathfrak p_1, \ldots , \mathfrak p_ r$, see Properties, Lemma 28.29.7. Multiplication by $t$ on global sections determines a map $t : M \to A$ which defines an isomorphism $M_{\mathfrak p_ i} \to A_{\mathfrak p_ i}$ for $i = 1, \ldots , r$. Denote $t_0 = t|_{U_0}$ the corresponding section of $\Gamma (U_0, \mathcal{L}_0^{\otimes -1})$ which similarly determines a map $t_0 : M_0 \to A/fA$ compatible with $t$. We conclude that there is a commutative diagram

\[ \xymatrix{ \text{Ext}^{-2}_ A(M, \omega _ A^\bullet ) & \text{Ext}^{-2}_ A(M_0, \omega _ A^\bullet ) \ar[l] \\ \text{Ext}^{-2}_ A(A, \omega _ A^\bullet ) \ar[u]^ t & \text{Ext}^{-2}_ A(A/fA, \omega _ A^\bullet ) \ar[l] \ar[u]_{t_0} } \]

It follows that the length of the image of the top horizontal map is at most the length of $\text{Ext}^{-2}_ A(A/fA, \omega _ A^\bullet )$ plus the length of the cokernel of $t_0$.

However, if we replace $\mathcal{L}$ by $\mathcal{L}^ n$ for $n > 1$, then we can use

\[ t^ n : M_ n = \Gamma (U, \mathcal{L}^{\otimes n}) \longrightarrow \Gamma (U, \mathcal{O}_ U) = A \]

instead of $t$. This replaces $t_0$ by its $n$th power. Thus the length of the image of the map $\text{Ext}^{-2}_ A(M_ n, \omega _ A^\bullet ) \leftarrow \text{Ext}^{-2}_ A(M_{n, 0}, \omega _ A^\bullet )$ is at most the length of $\text{Ext}^{-2}_ A(A/fA, \omega _ A^\bullet )$ plus the length of the cokernel of

\[ t_0^ n : \text{Ext}^{-2}_ A(A/fA, \omega _ A^\bullet ) \longrightarrow \text{Ext}^{-2}_ A(M_{n, 0}, \omega _ A^\bullet ) \]

Via the isomorphism $M_0 \cong A/fA$ the map $t_0$ becomes $g : A/fA \to A/fA$ for some $g \in A/fA$ and via the corresponding isomorphisms $M_{n, 0} \cong A/fA$ the map $t_0^ n$ becomes $g^ n : A/fA \to A/fA$. Thus the length of the cokernel above is the length of the quotient of $\text{Ext}^{-2}_ A(A/fA, \omega _ A^\bullet )$ by $g^ n$. Since $\text{Ext}^{-2}_ A(A/fA, \omega _ A^\bullet )$ is a finite $A$-module with support $T$ of dimension $1$ and since $V(g) \cap T$ consists of the closed point by our choice of $t$ this length grows linearly in $n$ by Algebra, Lemma 10.62.6.

To finish the proof we prove the final assertion. Assume $f \in \mathfrak m \subset A$ satisfies (1), (2), (3), $A$ is $(S_2)$, and $\dim (A) \geq 4$. Condition (1) implies $A$ is catenary, see Dualizing Complexes, Lemma 47.17.4. Then $\mathop{\mathrm{Spec}}(A)$ is equidimensional by Local Cohomology, Lemma 51.3.2. Thus $\dim (A_\mathfrak p) + \dim (A/\mathfrak p) \geq 4$ for every prime $\mathfrak p$ of $A$. Then $\text{depth}(A_\mathfrak p) \geq \min (2, \dim (A_\mathfrak p)) \geq \min (2, 4 - \dim (A/\mathfrak p))$ and hence (4) holds.
$\square$

## Comments (2)

Comment #3968 by Kestutis Cesnavicius on

Comment #4103 by Johan on