Lemma 52.22.5. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Assume

$A$ is a local ring which has a dualizing complex,

all irreducible components of $X$ have the same dimension,

the scheme $X \setminus Y$ is Cohen-Macaulay,

$I$ is generated by $d$ elements,

$\dim (X) - \dim (Z) > d + 2$, and

for $y \in U \cap Y$ the module $\mathcal{F}_ y^\wedge $ is finite locally free outside $V(I\mathcal{O}_{X, y}^\wedge )$, for example if $\mathcal{F}_ n$ is a finite locally free $\mathcal{O}_ U/I^ n\mathcal{O}_ U$-module.

Then $(\mathcal{F}_ n)$ extends to $X$. In particular if $A$ is $I$-adically complete, then $(\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_ U$-module.

**Proof.**
We will show that the hypotheses (1), (2), (3)(b) of Proposition 52.22.4 are satisfied. This is clear for (1) and (2).

Let $y \in U \cap Y$ and let $\mathfrak p$ be a prime $\mathfrak p \subset \mathcal{O}_{X, y}^\wedge $ with $\mathfrak p \not\in V(I\mathcal{O}_{X, y}^\wedge )$. The last condition shows that $\text{depth}((\mathcal{F}_ y^\wedge )_\mathfrak p) = \text{depth}((\mathcal{O}_{X, y}^\wedge )_\mathfrak p)$. Since $X \setminus Y$ is Cohen-Macaulay we see that $(\mathcal{O}_{X, y}^\wedge )_\mathfrak p$ is Cohen-Macaulay. Thus we see that

\begin{align*} & \text{depth}((\mathcal{F}^\wedge _ y)_\mathfrak p) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) + \delta ^ Y_ Z(y) \\ & = \dim ((\mathcal{O}_{X, y}^\wedge )_\mathfrak p) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) + \delta ^ Y_ Z(y) \\ & = \dim (\mathcal{O}_{X, y}^\wedge ) + \delta ^ Y_ Z(y) \end{align*}

The final equality because $\mathcal{O}_{X, y}$ is equidimensional by the second condition. Let $\delta (y) = \dim (\overline{\{ y\} })$. This is a dimension function as $A$ is a catenary local ring. By Lemma 52.18.1 we have $\delta ^ Y_ Z(y) \geq \delta (y) - \dim (Z)$. Since $X$ is equidimensional we get

\[ \dim (\mathcal{O}_{X, y}^\wedge ) + \delta ^ Y_ Z(y) \geq \dim (\mathcal{O}_{X, y}^\wedge ) + \delta (y) - \dim (Z) = \dim (X) - \dim (Z) \]

Thus we get the desired inequality and we win.
$\square$

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