Lemma 52.10.5. Let $I \subset \mathfrak a$ be ideals of a Noetherian ring $A$. Let $M$ be a finite $A$-module. Let $s$ and $d$ be integers. If we assume

$A$ has a dualizing complex,

$\text{cd}(A, I) \leq d$,

if $\mathfrak p \not\in V(I)$ and $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$ then $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) > s$ or $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > d + s$.

Then $A, I, V(\mathfrak a), M, s, d$ are as in Situation 52.10.1.

**Proof.**
We have to show that assumptions (1), (3), (4), and (6) of Situation 52.10.1 hold. It is clear that (a) $\Rightarrow $ (1), (b) $\Rightarrow $ (3), and (c) $\Rightarrow $ (4). To finish the proof in the next paragraph we show (6) holds.

Let $\mathfrak q \in V(\mathfrak a)$. Denote $A', I', \mathfrak m', M'$ the $I$-adic completions of $A_\mathfrak q, I_\mathfrak q, \mathfrak qA_\mathfrak q, M_\mathfrak q$. Let $\mathfrak p' \subset A'$ be a nonmaximal prime with $V(\mathfrak p') \cap V(I') = \{ \mathfrak m'\} $. Observe that this implies $\dim (A'/\mathfrak p') \leq d$ by Local Cohomology, Lemma 51.4.10. Denote $\mathfrak p \subset A$ the image of $\mathfrak p'$. We have $\text{depth}(M'_{\mathfrak p'}) \geq \text{depth}(M_\mathfrak p)$ and $\text{depth}(M'_{\mathfrak p'}) + \dim (A'/\mathfrak p') = \text{depth}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q)$ by Local Cohomology, Lemma 51.11.3. By assumption (c) either we have $\text{depth}(M'_{\mathfrak p'}) \geq \text{depth}(M_\mathfrak p) > s$ and we're done or we have $\text{depth}(M'_{\mathfrak p'}) + \dim (A'/\mathfrak p') > s + d$ which implies $\text{depth}(M'_{\mathfrak p'}) > s$ because of the already shown inequality $\dim (A'/\mathfrak p') \leq d$. In both cases we obtain what we want.
$\square$

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