Lemma 52.20.6. 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 $\mathfrak a = \mathfrak m$,

$A$ has a dualizing complex,

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

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, and

one of the following is true

$A_ f$ is $(S_2)$ and every irreducible component of $X$ not contained in $Y$ has dimension $\geq 4$, or

if $\mathfrak p \not\in V(f)$ and $V(\mathfrak p) \cap V(f) \not= \{ \mathfrak m\} $, then $\text{depth}(A_\mathfrak p) + \dim (A/\mathfrak p) > 3$.

Then $(\mathcal{F}_ n)$ satisfies the $(2, 3)$-inequalities.

**Proof.**
We will use the criterion of Lemma 52.20.5. Let $y \in U \cap Y$ with $\dim (\overline{\{ y\} } = 1$ and let $\mathfrak p$ be a prime $\mathfrak p \subset \mathcal{O}_{X, y}^\wedge $ with $\mathfrak p \not\in V(I\mathcal{O}_{X, y}^\wedge )$. Condition (4) shows that $\text{depth}((\mathcal{F}_ y^\wedge )_\mathfrak p) = \text{depth}((\mathcal{O}_{X, y}^\wedge )_\mathfrak p)$. Thus we have to prove

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

Let $\mathfrak p_0 \subset A$ be the image of $\mathfrak p$. Let $\mathfrak q \subset A$ be the prime corresponding to $y$. By Local Cohomology, Lemma 51.11.3 we have

\begin{align*} \text{depth}((\mathcal{O}_{X, y}^\wedge )_\mathfrak p) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) & = \text{depth}(A_{\mathfrak p_0}) + \dim ((A/\mathfrak p_0)_\mathfrak q) \\ & = \text{depth}(A_{\mathfrak p_0}) + \dim (A/\mathfrak p_0) - 1 \end{align*}

If (5)(a) holds, then we get that this is

\[ \geq \min (2, \dim (A_{\mathfrak p_0})) + \dim (A/\mathfrak p_0) - 1 \]

Note that in any case $\dim (A/\mathfrak p_0) \geq 2$. Hence if we get $2$ for the minimum, then we are done. If not we get

\[ \dim (A_{\mathfrak p_0}) + \dim (A/\mathfrak p_0) - 1 \geq 4 - 1 \]

because every component of $\mathop{\mathrm{Spec}}(A)$ passing through $\mathfrak p_0$ has dimension $\geq 4$. If (5)(b) holds, then we win immediately.
$\square$

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