Lemma 52.19.7. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$ and $d \geq 1$. Assume

$A$ is $I$-adically complete, has a dualizing complex, and $\text{cd}(A, I) \leq d$,

$(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module,

$(\mathcal{F}_ n)$ satisfies the strict $(1, 1 + d)$-inequalities.

Then there exists a unique coherent $\mathcal{O}_ U$-module $\mathcal{F}$ whose completion is $(\mathcal{F}_ n)$ such that for $x \in U$ with $\overline{\{ x\} } \cap Y \subset Z$ we have $\text{depth}(\mathcal{F}_ x) \geq 2$.

**Proof.**
Choose a coherent $\mathcal{O}_ U$-module $\mathcal{F}$ whose completion is $(\mathcal{F}_ n)$. Let $T = \{ x \in U \mid \overline{\{ x\} } \cap Y \subset Z\} $. We will construct $\mathcal{F}$ by applying Local Cohomology, Lemma 51.15.4 with $\mathcal{F}$ and $T$. Then uniqueness will follow from the mapping property of Lemma 52.19.6.

Since $T$ is stable under specialization in $U$ the only thing to check is the following. If $x' \leadsto x$ is an immediate specialization of points of $U$ with $x \in T$ and $x' \not\in T$, then $\text{depth}(\mathcal{F}_{x'}) \geq 1$. Set $W = \overline{\{ x\} }$ and $W' = \overline{\{ x'\} }$. Since $x' \not\in T$ we see that $W' \cap Y$ is not contained in $Z$. If $W' \cap Y$ contains an irreducible component contained in $Z$, then we are done by Lemma 52.19.4. If not, we choose an irreducible component $W_1$ of $W \cap Y$ and an irreducible component $W'_1$ of $W' \cap Y$ with $W_1 \subset W'_1$. Let $z \in W_1$ be the generic point. Let $y \leadsto z$, $y \in W'_1$ be an immediate specialization with $y \not\in Z$; existence of $y$ follows from $W'_1 \not\subset Z$ (see above) and Properties, Lemma 28.6.4. Then we have the following $z \in Z$, $x \leadsto z$, $x' \leadsto y \leadsto z$, $y \in Y \setminus Z$, and $\delta ^ Y_ Z(y) = 1$. By Local Cohomology, Lemma 51.4.10 and the fact that $z$ is a generic point of $W \cap Y$ we have $\dim (\mathcal{O}_{W, z}) \leq d$. Since $x' \leadsto x$ is an immediate specialization we have $\dim (\mathcal{O}_{W', z}) \leq d + 1$. Since $y \not= z$ we conclude $\dim (\mathcal{O}_{W', y}) \leq d$. If $\text{depth}(\mathcal{F}_{x'}) = 0$ then we would get a contradiction with assumption (3); details about passage from $\mathcal{O}_{X, y}$ to its completion omitted. This finishes the proof.
$\square$

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