Lemma 52.8.5. Let $I, J$ be ideals of a Noetherian ring $A$. Let $M$ be a finite $A$-module. Let $s$ and $d$ be integers. With $T$ as in (52.8.0.1) assume

1. $A$ is $I$-adically complete and has a dualizing complex,

2. if $\mathfrak p \in V(I)$ no condition,

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

4. if $\mathfrak p \not\in V(I)$, $\mathfrak p \not\in T$ then

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

for all $\mathfrak q \in V(\mathfrak p) \cap V(J) \cap V(I)$,

5. if $\mathfrak p \not\in V(I)$, $\mathfrak p \not\in T$, $V(\mathfrak p) \cap V(J) \cap V(I) \not= \emptyset$, and $\text{depth}(M_\mathfrak p) < s$, then one of the following holds1:

1. $\dim (\text{Supp}(M_\mathfrak p)) < s + 2$2, or

2. $\delta (\mathfrak p) > d + \delta _{max} - 1$ where $\delta$ is a dimension function and $\delta _{max}$ is the maximum of $\delta$ on $V(J) \cap V(I)$, or

3. $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > d + s + \delta _{max} - \delta _{min} - 2$ for all $\mathfrak q \in V(\mathfrak p) \cap V(J) \cap V(I)$.

Then there exists an ideal $J_0 \subset J$ with $V(J_0) \cap V(I) = V(J) \cap V(I)$ such that for any $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$ the map

$R\Gamma _{J'}(M) \longrightarrow R\Gamma _ J(M)^\wedge$

induces an isomorphism on cohomology in degrees $\leq s$. Here ${}^\wedge$ denotes derived $I$-adic completion.

Proof. For an ideal $\mathfrak a \subset A$ we have $R\Gamma _\mathfrak a = R\Gamma _{V(\mathfrak a)}$, see Dualizing Complexes, Lemma 47.10.1. Next, we observe that

$R\Gamma _ J(M)^\wedge = R\Gamma _ I(R\Gamma _ J(M))^\wedge = R\Gamma _{I + J}(M)^\wedge = R\Gamma _{I + J'}(M)^\wedge = R\Gamma _ I(R\Gamma _{J'}(M))^\wedge = R\Gamma _{J'}(M)^\wedge$

by Dualizing Complexes, Lemmas 47.9.6 and 47.12.1. This explains how we define the arrow in the statement of the lemma.

We claim that the hypotheses of Lemma 52.8.2 are implied by our current hypotheses on $M$. The only thing to verify is hypothesis (3). Thus let $\mathfrak p \not\in V(I)$, $\mathfrak p \in T$. Then $V(\mathfrak p) \cap V(I)$ is nonempty as $I$ is contained in the Jacobson radical of $A$ (Algebra, Lemma 10.96.6). Since $\mathfrak p \in T$ we have $V(\mathfrak p) \cap V(I) = V(\mathfrak p) \cap V(J) \cap V(I)$. Let $\mathfrak q \in V(\mathfrak p) \cap V(I)$ be the generic point of an irreducible component. We have $\text{cd}(A_\mathfrak q, I_\mathfrak q) \leq d$ by Local Cohomology, Lemma 51.4.6. We have $V(\mathfrak pA_\mathfrak q) \cap V(I_\mathfrak q) = \{ \mathfrak qA_\mathfrak q\}$ by our choice of $\mathfrak q$ and we conclude $\dim ((A/\mathfrak p)_\mathfrak q) \leq d$ by Local Cohomology, Lemma 51.4.10.

Observe that the lemma holds for $s < 0$. This is not a trivial case because it is not a priori clear that $H^ i(R\Gamma _ J(M)^\wedge )$ is zero for $i < 0$. However, this vanishing was established in Dualizing Complexes, Lemma 47.12.4. We will prove the lemma by induction for $s \geq 0$.

The lemma for $s = 0$ follows immediately from the conclusion of Lemma 52.8.2 and Dualizing Complexes, Lemma 47.12.5.

Assume $s > 0$ and the lemma has been shown for smaller values of $s$. Let $M' \subset M$ be the maximal submodule whose support is contained in $V(I) \cup T$. Then $M'$ is a finite $A$-module whose support is contained in $V(J') \cup V(I)$ for some ideal $J' \subset J$ with $V(J') \cap V(I) = V(J) \cap V(I)$. We claim that

$R\Gamma _{J'}(M') \to R\Gamma _ J(M')^\wedge$

is an isomorphism for any choice of $J'$. Namely, we can choose a short exact sequence $0 \to M_1 \oplus M_2 \to M' \to N \to 0$ with $M_1$ annihilated by a power of $J'$, with $M_2$ annihilated by a power of $I$, and with $N$ annihilated by a power of $I + J'$. Thus it suffices to show that the claim holds for $M_1$, $M_2$, and $N$. In the case of $M_1$ we see that $R\Gamma _{J'}(M_1) = M_1$ and since $M_1$ is a finite $A$-module and $I$-adically complete we have $M_1^\wedge = M_1$. This proves the claim for $M_1$ by the initial remarks of the proof. In the case of $M_2$ we see that $H^ i_ J(M_2) = H^ i_{I + J}(M) = H^ i_{I + J'}(M) = H^ i_{J'}(M_2)$ are annihilated by a power of $I$ and hence derived complete. Thus the claim in this case also. For $N$ we can use either of the arguments just given. Considering the short exact sequence $0 \to M' \to M \to M/M' \to 0$ we see that it suffices to prove the lemma for $M/M'$. Thus we may assume $\text{Ass}(M) \cap (V(I) \cup T) = \emptyset$.

Let $\mathfrak p \in \text{Ass}(M)$ be such that $V(\mathfrak p) \cap V(J) \cap V(I) = \emptyset$. Since $I$ is contained in the Jacobson radical of $A$ this implies that $V(\mathfrak p) \cap V(J') = \emptyset$ for any $J' \subset J$ with $V(J') \cap V(I) = V(J) \cap V(I)$. Thus setting $N = H^0_\mathfrak p(M)$ we see that $R\Gamma _ J(N) = R\Gamma _{J'}(N) = 0$ for all $J' \subset J$ with $V(J') \cap V(I) = V(J) \cap V(I)$. In particular $R\Gamma _ J(N)^\wedge = 0$. Thus we may replace $M$ by $M/N$ as this changes the structure of $M$ only in primes which do not play a role in conditions (4) or (5). Repeating we may assume that $V(\mathfrak p) \cap V(J) \cap V(I) \not= \emptyset$ for all $\mathfrak p \in \text{Ass}(M)$.

Assume $\text{Ass}(M) \cap (V(I) \cup T) = \emptyset$ and that $V(\mathfrak p) \cap V(J) \cap V(I) \not= \emptyset$ for all $\mathfrak p \in \text{Ass}(M)$. Let $\mathfrak p \in \text{Ass}(M)$. We want to show that we may apply Lemma 52.8.1. It is in the verification of this that we will use the supplemental condition (5). Choose $\mathfrak p' \subset \mathfrak p$ and $\mathfrak q' \subset V(\mathfrak p) \cap V(J) \cap V(I)$.

1. If $M_{\mathfrak p'} = 0$, then $\text{depth}(M_{\mathfrak p'}) = \infty$ and $\text{depth}(M_{\mathfrak p'}) + \dim ((A/\mathfrak p')_{\mathfrak q'}) > d + s$.

2. If $\text{depth}(M_{\mathfrak p'}) < s$, then $\text{depth}(M_{\mathfrak p'}) + \dim ((A/\mathfrak p')_{\mathfrak q'}) > d + s$ by (4).

In the remaining cases we have $M_{\mathfrak p'} \not= 0$ and $\text{depth}(M_{\mathfrak p'}) \geq s$. In particular, we see that $\mathfrak p'$ is in the support of $M$ and we can choose $\mathfrak p'' \subset \mathfrak p'$ with $\mathfrak p'' \in \text{Ass}(M)$.

1. Observe that $\dim ((A/\mathfrak p'')_{\mathfrak p'}) \geq \text{depth}(M_{\mathfrak p'})$ by Algebra, Lemma 10.72.9. If equality holds, then we have

$\text{depth}(M_{\mathfrak p'}) + \dim ((A/\mathfrak p')_{\mathfrak q'}) = \text{depth}(M_{\mathfrak p''}) + \dim ((A/\mathfrak p'')_{\mathfrak q'}) > s + d$

by (4) applied to $\mathfrak p''$ and we are done. This means we are only in trouble if $\dim ((A/\mathfrak p'')_{\mathfrak p'}) > \text{depth}(M_{\mathfrak p'})$. This implies that $\dim (M_\mathfrak p) \geq s + 2$. Thus if (5)(a) holds, then this does not occur.

2. If (5)(b) holds, then we get

$\text{depth}(M_{\mathfrak p'}) + \dim ((A/\mathfrak p')_{\mathfrak q'}) \geq s + \delta (\mathfrak p') - \delta (\mathfrak q') \geq s + 1 + \delta (\mathfrak p) - \delta _{max} > s + d$

as desired.

3. If (5)(c) holds, then we get

\begin{align*} \text{depth}(M_{\mathfrak p'}) + \dim ((A/\mathfrak p')_{\mathfrak q'}) & \geq s + \delta (\mathfrak p') - \delta (\mathfrak q') \\ & \geq s + 1 + \delta (\mathfrak p) - \delta (\mathfrak q') \\ & = s + 1 + \delta (\mathfrak p) - \delta (\mathfrak q) + \delta (\mathfrak q) - \delta (\mathfrak q') \\ & > s + 1 + (s + d + \delta _{max} - \delta _{min} - 2) + \delta (\mathfrak q) - \delta (\mathfrak q') \\ & \geq 2s + d - 1 \geq s + d \end{align*}

as desired. Observe that this argument works because we know that a prime $\mathfrak q \in V(\mathfrak p) \cap V(J) \cap V(I)$ exists.

Now we are ready to do the induction step.

Choose an ideal $J_0$ as in Lemma 52.8.2 and an integer $t > 0$ such that $(J_0I)^ t$ annihilates $H^ s_ J(M)$. The assumptions of Lemma 52.8.1 are satisfied for every $\mathfrak p \in \text{Ass}(M)$ (see previous paragraph). Thus the annihilator $\mathfrak a \subset A$ of $H^ s(R\Gamma _ J(M)^\wedge )$ is not contained in $\mathfrak p$ for $\mathfrak p \in \text{Ass}(M)$. Thus we can find an $f \in \mathfrak a(J_0I)^ t$ not in any associated prime of $M$ which is an annihilator of both $H^ s(R\Gamma _ J(M)^\wedge )$ and $H^ s_ J(M)$. Then $f$ is a nonzerodivisor on $M$ and we can consider the short exact sequence

$0 \to M \xrightarrow {f} M \to M/fM \to 0$

Our choice of $f$ shows that we obtain

$\xymatrix{ H^{s - 1}_{J'}(M) \ar[d] \ar[r] & H^{s - 1}_{J'}(M/fM) \ar[d] \ar[r] & H^ s_{J'}(M) \ar[d] \ar[r] & 0 \\ H^{s - 1}(R\Gamma _ J(M)^\wedge ) \ar[r] & H^{s - 1}(R\Gamma _ J(M/fM)^\wedge ) \ar[r] & H^ s(R\Gamma _ J(M)^\wedge ) \ar[r] & 0 }$

for any $J' \subset J_0$ with $V(J') \cap V(I) = V(J) \cap V(I)$. Thus if we choose $J'$ such that it works for $M$ and $M/fM$ and $s - 1$ (possible by induction hypothesis – see next paragraph), then we conclude that the lemma is true.

To finish the proof we have to show that the module $M/fM$ satisfies the hypotheses (4) and (5) for $s - 1$. Thus we let $\mathfrak p$ be a prime in the support of $M/fM$ with $\text{depth}((M/fM)_\mathfrak p) < s - 1$ and with $V(\mathfrak p) \cap V(J) \cap V(I)$ nonempty. Then $\dim (M_\mathfrak p) = \dim ((M/fM)_\mathfrak p) + 1$ and $\text{depth}(M_\mathfrak p) = \text{depth}((M/fM)_\mathfrak p) + 1$. In particular, we know (4) and (5) hold for $\mathfrak p$ and $M$ with the original value $s$. The desired inequalities then follow by inspection. $\square$

 For example if $M$ satisfies Serre's condition $(S_ s)$ on the complement of $V(I) \cup T$.

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