**Proof.**
Proof of (E). We have to prove assumptions (1), (3), (4), (6) of Situation 52.10.1 hold for $A, I, T, M$. Shrinking $T$ to $T'$ weakens assumption (6) and strengthens assumption (4). However, if we have $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ with $\mathfrak p \not\in V(I)$, $\mathfrak r \in V(I) \setminus T'$, $\mathfrak q \in T'$ as in assumption (4) for $A, I, T', M$, then either we can pick $\mathfrak r \in V(I) \setminus T$ and condition (4) for $A, I, T, M$ kicks in or we cannot find such an $\mathfrak r$ in which case we get $\text{depth}(M_\mathfrak p) > s$ by Lemma 52.10.2. This proves (4) holds for $A, I, T', M$ as desired.

Proof of (F). This is straightforward and we omit the details.

Proof of (G). We have to prove assumptions (1), (3), (4), (6) of Situation 52.10.1 hold for the $I$-adic completions $A', I', T', M'$. Please keep in mind that $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ induces an isomorphism $V(I') \to V(I)$.

Assumption (1): The ring $A'$ has a dualizing complex, see Dualizing Complexes, Lemma 47.22.4.

Assumption (3): Since $I' = IA'$ this follows from Local Cohomology, Lemma 51.4.5.

Assumption (4): If we have primes $\mathfrak p' \subset \mathfrak r' \subset \mathfrak q'$ in $A'$ with $\mathfrak p' \not\in V(I')$, $\mathfrak r' \in V(I') \setminus T'$, $\mathfrak q' \in T'$ then their images $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ in the spectrum of $A$ satisfy $\mathfrak p \not\in V(I)$, $\mathfrak r \in V(I) \setminus T$, $\mathfrak q \in T$. Then we have

\[ \text{depth}_{A_\mathfrak p}(M_\mathfrak p) \geq s \quad \text{or}\quad \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > d + s \]

by assumption (4) for $A, I, T, M$. We have $\text{depth}(M'_{\mathfrak p'}) \geq \text{depth}(M_\mathfrak p)$ and $\text{depth}(M'_{\mathfrak p'}) + \dim ((A'/\mathfrak p')_{\mathfrak q'}) = \text{depth}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q)$ by Local Cohomology, Lemma 51.11.3. Thus assumption (4) holds for $A', I', T', M'$.

Assumption (6): Let $\mathfrak q' \in T'$ lying over the prime $\mathfrak q \in T$. Then $A'_{\mathfrak q'}$ and $A_\mathfrak q$ have isomorphic $I$-adic completions and similarly for $M_\mathfrak q$ and $M'_{\mathfrak q'}$. Thus assumption (6) for $A', I', T', M'$ is equivalent to assumption (6) for $A, I, T, M$.

Proof of (A). We have to check conditions (1), (2), (3), (4), and (6) of Lemmas 52.8.2 and 52.8.4 for $(A, I, \mathfrak a, M)$. Warning: the set $T$ in the statement of these lemmas is not the same as the set $T$ above.

Condition (1): This holds because we have assumed $A$ has a dualizing complex in Situation 52.10.1.

Condition (2): This is empty.

Condition (3): Let $\mathfrak p \subset A$ with $V(\mathfrak p) \cap V(I) \subset V(\mathfrak a)$. Since $I$ is contained in the Jacobson radical of $A$ we see that $V(\mathfrak p) \cap V(I) \not= \emptyset $. Let $\mathfrak q \in V(\mathfrak p) \cap V(I)$ be a generic point. Since $\text{cd}(A_\mathfrak q, I_\mathfrak q) \leq d$ (Local Cohomology, Lemma 51.4.6) and since $V(\mathfrak p A_\mathfrak q) \cap V(I_\mathfrak q) = \{ \mathfrak q A_\mathfrak q\} $ we get $\dim ((A/\mathfrak p)_\mathfrak q) \leq d$ by Local Cohomology, Lemma 51.4.10 which proves (3).

Condition (4): Suppose $\mathfrak p \not\in V(I)$ and $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$. It suffices to show

\[ \text{depth}_{A_\mathfrak p}(M_\mathfrak p) \geq s \quad \text{or}\quad \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > d + s \]

If there exists a prime $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ with $\mathfrak r \in V(I) \setminus T$, then this follows immediately from assumption (4) in Situation 52.10.1. If not, then $\text{depth}(M_\mathfrak p) > s$ by Lemma 52.10.2.

Condition (6): Let $\mathfrak p \not\in V(I)$ with $V(\mathfrak p) \cap V(I) \subset V(\mathfrak a)$. Since $I$ is contained in the Jacobson radical of $A$ we see that $V(\mathfrak p) \cap V(I) \not= \emptyset $. Choose $\mathfrak q \in V(\mathfrak p) \cap V(I) \subset V(\mathfrak a)$. It is clear there does not exist a prime $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ with $\mathfrak r \in V(I) \setminus T$. By Lemma 52.10.2 we have $\text{depth}(M_\mathfrak p) > s$ which proves (6).

Proof of (B). We have to check conditions (1), (2), (3), (4) of Lemma 52.8.5. Warning: the set $T$ in the statement of this lemma is not the same as the set $T$ above.

Condition (1): This holds because $A$ is complete and has a dualizing complex.

Condition (2): This is empty.

Condition (3): This is the same as assumption (3) in Situation 52.10.1.

Condition (4): This is the same as assumption (4) in Lemma 52.8.2 which we proved in (A).

Proof of (C). This is true because the assumptions in Lemmas 52.9.2 and 52.9.4 are the same as the assumptions in Lemmas 52.8.2 and 52.8.4 in the local case and we proved these hold in (A).

Proof of (D). This is true because the assumptions in Lemma 52.9.5 are the same as the assumptions (1), (2), (3), (4) in Lemma 52.8.5 and we proved these hold in (B).
$\square$

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