Lemma 52.20.5. In Situation 52.16.1 let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(U, I\mathcal{O}_ U)$. Assume

$A$ is local with maximal ideal $\mathfrak a = \mathfrak m$,

$\text{cd}(A, I) = 1$.

Then $(\mathcal{F}_ n)$ satisfies the $(2, 3)$-inequalities if and only if for all $y \in U \cap Y$ with $\dim (\{ y\} ) = 1$ and every prime $\mathfrak p \subset \mathcal{O}_{X, y}^\wedge $, $\mathfrak p \not\in V(I\mathcal{O}_{X, y}^\wedge )$ we have

\[ \text{depth}((\mathcal{F}_ y^\wedge )_\mathfrak p) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) > 2 \]

**Proof.**
We will use Lemma 52.19.3 without further mention. In particular, we see the condition is necessary. Conversely, suppose the condition is true. Note that $\delta ^ Y_ Z(y) = \dim (\overline{\{ y\} })$ by Lemma 52.18.1. Let us write $\delta $ for this function. Let $y \in U \cap Y$. If $\delta (y) > 2$, then the inequality of Lemma 52.19.3 holds. Finally, suppose $\delta (y) = 2$. We have to show that

\[ \text{depth}((\mathcal{F}_ y^\wedge )_\mathfrak p) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) > 1 \]

Choose a specialization $y \leadsto y'$ with $\delta (y') = 1$. Then there is a ring map $\mathcal{O}_{X, y'}^\wedge \to \mathcal{O}_{X, y}^\wedge $ which identifies the target with the completion of the localization of $\mathcal{O}_{X, y'}^\wedge $ at a prime $\mathfrak q$ with $\dim (\mathcal{O}_{X, y'}^\wedge /\mathfrak q) = 1$. Moreover, we then obtain

\[ \mathcal{F}_ y^\wedge = \mathcal{F}_{y'}^\wedge \otimes _{\mathcal{O}_{X, y'}^\wedge } \mathcal{O}_{X, y}^\wedge \]

Let $\mathfrak p' \subset \mathcal{O}_{X, y'}^\wedge $ be the image of $\mathfrak p$. By Local Cohomology, Lemma 51.11.3 we have

\begin{align*} \text{depth}((\mathcal{F}_ y^\wedge )_\mathfrak p) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) & = \text{depth}((\mathcal{F}_{y'}^\wedge )_{\mathfrak p'}) + \dim ((\mathcal{O}_{X, y}^\wedge /\mathfrak p)_{\mathfrak p'}) \\ & = \text{depth}((\mathcal{F}_{y'}^\wedge )_{\mathfrak p'}) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p') - 1 \end{align*}

the last equality because the specialization is immediate. Thus the lemma is prove by the assumed inequality for $y', \mathfrak p'$.
$\square$

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