## 52.10 Algebraization of local cohomology, III

In this section we bootstrap the material in Sections 52.8 and 52.11 to give a stronger result the following situation.

Situation 52.10.1. Here $A$ is a Noetherian ring. We have an ideal $I \subset A$, a finite $A$-module $M$, and a subset $T \subset V(I)$ stable under specialization. We have integers $s$ and $d$. We assume

1. $A$ has a dualizing complex,

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

3. given primes $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ with $\mathfrak p \not\in V(I)$, $\mathfrak r \in V(I) \setminus T$, $\mathfrak q \in T$ we have

$\text{depth}_{A_\mathfrak p}(M_\mathfrak p) \geq s \quad \text{or}\quad \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > d + s$
4. given $\mathfrak q \in T$ denoting $A', \mathfrak m', I', M'$ are the usual $I$-adic completions of $A_\mathfrak q, \mathfrak qA_\mathfrak q, I_\mathfrak q, M_\mathfrak q$ we have

$\text{depth}(M'_{\mathfrak p'}) > s$

for all $\mathfrak p' \in \mathop{\mathrm{Spec}}(A') \setminus V(I')$ with $V(\mathfrak p') \cap V(I') = \{ \mathfrak m'\}$.

The following lemma explains why in Situation 52.10.1 it suffices to look at triples $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ of primes in (4) even though the actual assumption only involves $\mathfrak p$ and $\mathfrak q$.

Lemma 52.10.2. In Situation 52.10.1 let $\mathfrak p \subset \mathfrak q$ be primes of $A$ with $\mathfrak p \not\in V(I)$ and $\mathfrak q \in T$. If there does not exist an $\mathfrak r \in V(I) \setminus T$ with $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ then $\text{depth}(M_\mathfrak p) > s$.

Proof. Choose $\mathfrak q' \in T$ with $\mathfrak p \subset \mathfrak q' \subset \mathfrak q$ such that there is no prime in $T$ strictly in between $\mathfrak p$ and $\mathfrak q'$. To prove the lemma we may and do replace $\mathfrak q$ by $\mathfrak q'$. Next, let $\mathfrak p' \subset A_\mathfrak q$ be the prime corresponding to $\mathfrak p$. After doing this we obtain that $V(\mathfrak p') \cap V(IA_\mathfrak q) = \{ \mathfrak q A_\mathfrak q\}$ because of the nonexistence of a prime $\mathfrak r$ as in the lemma. Let $A', I', \mathfrak m', M'$ be the $I$-adic completions of $A_\mathfrak q, I_\mathfrak q, \mathfrak qA_\mathfrak q, M_\mathfrak q$. Since $A_\mathfrak q \to A'$ is faithfully flat (Algebra, Lemma 10.97.3) we can choose $\mathfrak p'' \subset A'$ lying over $\mathfrak p'$ with $\dim (A'_{\mathfrak p''}/\mathfrak p' A'_{\mathfrak p''}) = 0$. Then we see that

$\text{depth}(M'_{\mathfrak p''}) = \text{depth}((M_\mathfrak q \otimes _{A_\mathfrak q} A')_{\mathfrak p''}) = \text{depth}(M_\mathfrak p \otimes _{A_\mathfrak p} A'_{\mathfrak p''}) = \text{depth}(M_\mathfrak p)$

by flatness of $A \to A'$ and our choice of $\mathfrak p''$, see Algebra, Lemma 10.163.1. Since $\mathfrak p''$ lies over $\mathfrak p'$ we have $V(\mathfrak p'') \cap V(I') = \{ \mathfrak m'\}$. Thus condition (6) in Situation 52.10.1 implies $\text{depth}(M'_{\mathfrak p''}) > s$ which finishes the proof. $\square$

The following tedious lemma explains the relationships between various collections of conditions one might impose.

Lemma 52.10.3. In Situation 52.10.1 we have

1. if $T' \subset T$ is a smaller specialization stable subset, then $A, I, T', M$ satisfies the assumptions of Situation 52.10.1,

2. if $S \subset A$ is a multiplicative subset, then $S^{-1}A, S^{-1}I, T', S^{-1}M$ satisfies the assumptions of Situation 52.10.1 where $T' \subset V(S^{-1}I)$ is the inverse image of $T$,

3. the quadruple $A', I', T', M'$ satisfies the assumptions of Situation 52.10.1 where $A', I', M'$ are the usual $I$-adic completions of $A, I, M$ and $T' \subset V(I')$ is the inverse image of $T$.

Let $I \subset \mathfrak a \subset A$ be an ideal such that $V(\mathfrak a) \subset T$. Then

1. if $I$ is contained in the Jacobson radical of $A$, then all hypotheses of Lemmas 52.8.2 and 52.8.4 are satisfied for $A, I, \mathfrak a, M$,

2. if $A$ is complete with respect to $I$, then all hypotheses except for possibly (5) of Lemma 52.8.5 are satisfied for $A, I, \mathfrak a, M$,

3. if $A$ is local with maximal ideal $\mathfrak m = \mathfrak a$, then all hypotheses of Lemmas 52.9.2 and 52.9.4 hold for $A, \mathfrak m, I, M$,

4. if $A$ is local with maximal ideal $\mathfrak m = \mathfrak a$ and $I$-adically complete, then all hypotheses of Lemma 52.9.5 hold for $A, \mathfrak m, I, M$,

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). $\square$

Lemma 52.10.4. In Situation 52.10.1 assume $A$ is local with maximal ideal $\mathfrak m$ and $T = \{ \mathfrak m\}$. Then $H^ i_\mathfrak m(M) \to \mathop{\mathrm{lim}}\nolimits H^ i_\mathfrak m(M/I^ nM)$ is an isomorphism for $i \leq s$ and these modules are annihilated by a power of $I$.

Proof. Let $A', I', \mathfrak m', M'$ be the usual $I$-adic completions of $A, I, \mathfrak m, M$. Recall that we have $H^ i_\mathfrak m(M) \otimes _ A A' = H^ i_{\mathfrak m'}(M')$ by flatness of $A \to A'$ and Dualizing Complexes, Lemma 47.9.3. Since $H^ i_\mathfrak m(M)$ is $\mathfrak m$-power torsion we have $H^ i_\mathfrak m(M) = H^ i_\mathfrak m(M) \otimes _ A A'$, see More on Algebra, Lemma 15.89.3. We conclude that $H^ i_\mathfrak m(M) = H^ i_{\mathfrak m'}(M')$. The exact same arguments will show that $H^ i_\mathfrak m(M/I^ nM) = H^ i_{\mathfrak m'}(M'/(I')^ nM')$ for all $n$ and $i$.

Lemmas 52.9.5, 52.9.2, and 52.9.4 apply to $A', \mathfrak m', I', M'$ by Lemma 52.10.3 parts (C) and (D). Thus we get an isomorphism

$H^ i_{\mathfrak m'}(M') \longrightarrow H^ i(R\Gamma _{\mathfrak m'}(M')^\wedge )$

for $i \leq s$ where ${}^\wedge$ is derived $I'$-adic completion and these modules are annihilated by a power of $I'$. By Lemma 52.5.4 we obtain isomorphisms

$H^ i_{\mathfrak m'}(M') \longrightarrow \mathop{\mathrm{lim}}\nolimits H^ i_{\mathfrak m'}(M'/(I')^ nM'))$

for $i \leq s$. Combined with the already established comparison with local cohomology over $A$ we conclude the lemma is true. $\square$

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

1. $A$ has a dualizing complex,

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

3. 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. $\square$

Lemma 52.10.6. In Situation 52.10.1 the inverse systems $\{ H^ i_ T(I^ nM)\} _{n \geq 0}$ are pro-zero for $i \leq s$. Moreover, there exists an integer $m_0$ such that for all $m \geq m_0$ there exists an integer $m'(m) \geq m$ such that for $k \geq m'(m)$ the image of $H^{s + 1}_ T(I^ kM) \to H^{s + 1}_ T(I^ mM)$ maps injectively to $H^{s + 1}_ T(I^{m_0}M)$.

Proof. Fix $m$. Let $\mathfrak q \in T$. By Lemmas 52.10.3 and 52.10.4 we see that

$H^ i_\mathfrak q(M_\mathfrak q) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^ i_\mathfrak q(M_\mathfrak q/I^ nM_\mathfrak q)$

is an isomorphism for $i \leq s$. The inverse systems $\{ H^ i_\mathfrak q(I^ nM_\mathfrak q)\} _{n \geq 0}$ and $\{ H^ i_\mathfrak q(M/I^ nM)\} _{n \geq 0}$ satisfy the Mittag-Leffler condition for all $i$, see Lemma 52.5.2. Thus looking at the inverse system of long exact sequences

$0 \to H^0_\mathfrak q(I^ nM_\mathfrak q) \to H^0_\mathfrak q(M_\mathfrak q) \to H^0_\mathfrak q(M_\mathfrak q/I^ nM_\mathfrak q) \to H^1_\mathfrak q(I^ nM_\mathfrak q) \to H^1_\mathfrak q(M_\mathfrak q) \to \ldots$

we conclude (some details omitted) that there exists an integer $m'(m, \mathfrak q) \geq m$ such that for all $k \geq m'(m, \mathfrak q)$ the map $H^ i_\mathfrak q(I^ kM_\mathfrak q) \to H^ i_\mathfrak q(I^ mM_\mathfrak q)$ is zero for $i \leq s$ and the image of $H^{s + 1}_\mathfrak q(I^ kM_\mathfrak q) \to H^{s + 1}_\mathfrak q(I^ mM_\mathfrak q)$ is independent of $k \geq m'(m, \mathfrak q)$ and maps injectively into $H^{s + 1}_\mathfrak q(M_\mathfrak q)$.

Suppose we can show that $m'(m, \mathfrak q)$ can be chosen independently of $\mathfrak q \in T$. Then the lemma follows immediately from Local Cohomology, Lemmas 51.6.2 and 51.6.3.

Let $\omega _ A^\bullet$ be a dualizing complex. Let $\delta : \mathop{\mathrm{Spec}}(A) \to \mathbf{Z}$ be the corresponding dimension function. Recall that $\delta$ attains only a finite number of values, see Dualizing Complexes, Lemma 47.17.4. Claim: for each $d \in \mathbf{Z}$ the integer $m'(m, \mathfrak q)$ can be chosen independently of $\mathfrak q \in T$ with $\delta (\mathfrak q) = d$. Clearly the claim implies the lemma by what we said above.

Pick $\mathfrak q \in T$ with $\delta (\mathfrak q) = d$. Consider the ext modules

$E(n, j) = \text{Ext}^ j_ A(I^ nM, \omega _ A^\bullet )$

A key feature we will use is that these are finite $A$-modules. Recall that $(\omega _ A^\bullet )_\mathfrak q[-d]$ is a normalized dualizing complex for $A_\mathfrak q$ by definition of the dimension function associated to a dualizing complex, see Dualizing Complexes, Section 47.17. The local duality theorem (Dualizing Complexes, Lemma 47.18.4) tells us that the $\mathfrak qA_\mathfrak q$-adic completion of $E(n, -d - i)_\mathfrak q$ is Matlis dual to $H^ i_\mathfrak q(I^ nM_\mathfrak q)$. Thus the choice of $m'(m, \mathfrak q)$ for $i \leq s$ in the first paragraph tells us that for $k \geq m'(m, \mathfrak q)$ and $j \geq -d - s$ the map

$E(m, j)_\mathfrak q \to E(k, j)_\mathfrak q$

is zero. Since these modules are finite and nonzero only for a finite number of possible $j$ (small detail omitted), we can find an open neighbourhood $W \subset \mathop{\mathrm{Spec}}(A)$ of $\mathfrak q$ such that

$E(m, j)_{\mathfrak q'} \to E(m'(m, \mathfrak q), j)_{\mathfrak q'}$

is zero for $j \geq -d - s$ for all $\mathfrak q' \in W$. Then of course the maps $E(m, j)_{\mathfrak q'} \to E(k, j)_{\mathfrak q'}$ for $k \geq m'(m, \mathfrak q)$ are zero as well.

For $i = s + 1$ corresponding to $j = - d - s - 1$ we obtain from local duality and the results of the first paragraph that

$K_{k, \mathfrak q} = \mathop{\mathrm{Ker}}(E(m, -d - s - 1)_\mathfrak q \to E(k, -d - s - 1)_\mathfrak q)$

is independent of $k \geq m'(m, \mathfrak q)$ and that

$E(0, -d - s - 1)_\mathfrak q \to E(m, -d - s - 1)_\mathfrak q/K_{m'(m, \mathfrak q), \mathfrak q}$

is surjective. For $k \geq m'(m, \mathfrak q)$ set

$K_ k = \mathop{\mathrm{Ker}}(E(m, -d - s - 1) \to E(k, -d - s - 1))$

Since $K_ k$ is an increasing sequence of submodules of the finite module $E(m, -d - s - 1)$ we see that, at the cost of increasing $m'(m, \mathfrak q)$ a little bit, we may assume $K_{m'(m, \mathfrak q)} = K_ k$ for $k \geq m'(m, \mathfrak q)$. After shrinking $W$ further if necessary, we may also assume that

$E(0, -d - s - 1)_{\mathfrak q'} \to E(m, -d - s - 1)_{\mathfrak q'}/K_{m'(m, \mathfrak q), \mathfrak q'}$

is surjective for all $\mathfrak q' \in W$ (as before use that these modules are finite and that the map is surjective after localization at $\mathfrak q$).

Any subset, in particular $T_ d = \{ \mathfrak q \in T \text{ with }\delta (\mathfrak q) = d\}$, of the Noetherian topological space $\mathop{\mathrm{Spec}}(A)$ with the endowed topology is Noetherian and hence quasi-compact. Above we have seen that for every $\mathfrak q \in T_ d$ there is an open neighbourhood $W$ where $m'(m, \mathfrak q)$ works for all $\mathfrak q' \in T_ d \cap W$. We conclude that we can find an integer $m'(m, d)$ such that for all $\mathfrak q \in T_ d$ we have

$E(m, j)_\mathfrak q \to E(m'(m, d), j)_\mathfrak q$

is zero for $j \geq -d - s$ and with $K_{m'(m, d)} = \mathop{\mathrm{Ker}}(E(m, -d - s - 1) \to E(m'(m, d), -d - s - 1))$ we have

$K_{m'(m, d), \mathfrak q} = \mathop{\mathrm{Ker}}(E(m, -d - s - 1)_{\mathfrak q} \to E(k, -d - s - 1)_{\mathfrak q})$

for all $k \geq m'(m, d)$ and the map

$E(0, -d - s - 1)_\mathfrak q \to E(m, -d - s - 1)_\mathfrak q/K_{m'(m, d), \mathfrak q}$

is surjective. Using the local duality theorem again (in the opposite direction) we conclude that the claim is correct. This finishes the proof. $\square$

Lemma 52.10.7. In Situation 52.10.1 there exists an integer $m_0 \geq 0$ such that

1. $\{ H^ i_ T(M/I^ nM)\} _{n \geq 0}$ satisfies the Mittag-Leffler condition for $i < s$.

2. $\{ H^ i_ T(I^{m_0}M/I^ nM)\} _{n \geq m_0}$ satisfies the Mittag-Leffler condition for $i \leq s$,

3. $H^ i_ T(M) \to \mathop{\mathrm{lim}}\nolimits H^ i_ T(M/I^ nM)$ is an isomorphism for $i < s$,

4. $H^ s_ T(I^{m_0}M) \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(I^{m_0}M/I^ nM)$ is an isomorphism for $i \leq s$,

5. $H^ s_ T(M) \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM)$ is injective with cokernel killed by $I^{m_0}$, and

6. $R^1\mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM)$ is killed by $I^{m_0}$.

Proof. Consider the long exact sequences

$0 \to H^0_ T(I^ nM) \to H^0_ T(M) \to H^0_ T(M/I^ nM) \to H^1_ T(I^ nM) \to H^1_ T(M) \to \ldots$

Parts (1) and (3) follows easily from this and Lemma 52.10.6.

Let $m_0$ and $m'(-)$ be as in Lemma 52.10.6. For $m \geq m_0$ consider the long exact sequence

$H^ s_ T(I^ mM) \to H^ s_ T(I^{m_0}M) \to H^ s_ T(I^{m_0}M/I^ mM) \to H^{s + 1}_ T(I^ mM) \to H^1_ T(I^{m_0}M)$

Then for $k \geq m'(m)$ the image of $H^{s + 1}_ T(I^ kM) \to H^{s + 1}_ T(I^ mM)$ maps injectively to $H^{s + 1}_ T(I^{m_0}M)$. Hence the image of $H^ s_ T(I^{m_0}M/I^ kM) \to H^ s_ T(I^{m_0}M/I^ mM)$ maps to zero in $H^{s + 1}_ T(I^ mM)$ for all $k \geq m'(m)$. We conclude that (2) and (4) hold.

Consider the short exact sequences $0 \to I^{m_0}M \to M \to M/I^{m_0} M \to 0$ and $0 \to I^{m_0}M/I^ nM \to M/I^ nM \to M/I^{m_0} M \to 0$. We obtain a diagram

$\xymatrix{ H^{s - 1}_ T(M/I^{m_0}M) \ar[r] & \mathop{\mathrm{lim}}\nolimits H^ s_ T(I^{m_0}M/I^ nM) \ar[r] & \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM) \ar[r] & H^ s_ T(M/I^{m_0}M) \\ H^{s - 1}_ T(M/I^{m_0}M) \ar[r] \ar@{=}[u] & H^ s_ T(I^{m_0}M) \ar[r] \ar[u]_{\cong } & H^ s_ T(M) \ar[r] \ar[u] & H^ s_ T(M/I^{m_0}M) \ar@{=}[u] }$

whose lower row is exact. The top row is also exact (at the middle two spots) by Homology, Lemma 12.31.4. Part (5) follows.

Write $B_ n = H^ s_ T(M/I^ nM)$. Let $A_ n \subset B_ n$ be the image of $H^ s_ T(I^{m_0}M/I^ nM) \to H^ s_ T(M/I^ nM)$. Then $(A_ n)$ satisfies the Mittag-Leffler condition by (2) and Homology, Lemma 12.31.3. Also $C_ n = B_ n/A_ n$ is killed by $I^{m_0}$. Thus $R^1\mathop{\mathrm{lim}}\nolimits B_ n \cong R^1\mathop{\mathrm{lim}}\nolimits C_ n$ is killed by $I^{m_0}$ and we get (6). $\square$

Theorem 52.10.8. In Situation 52.10.1 the inverse system $\{ H^ i_ T(M/I^ nM)\} _{n \geq 0}$ satisfies the Mittag-Leffler condition for $i \leq s$, the map

$H^ i_ T(M) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^ i_ T(M/I^ nM)$

is an isomorphism for $i \leq s$, and $H^ i_ T(M)$ is annihilated by a power of $I$ for $i \leq s$.

Proof. To prove the final assertion of the theorem we apply Local Cohomology, Proposition 51.10.1 with $T \subset V(I) \subset \mathop{\mathrm{Spec}}(A)$. Namely, suppose that $\mathfrak p \not\in V(I)$, $\mathfrak q \in T$ with $\mathfrak p \subset \mathfrak q$. Then either there exists a prime $\mathfrak p \subset \mathfrak r \subset \mathfrak q$ with $\mathfrak r \in V(I) \setminus T$ and we get

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

by (4) in Situation 52.10.1 or there does not exist an $\mathfrak r$ and we get $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) > s$ by Lemma 52.10.2. In all three cases we see that $\text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > s$. Thus Local Cohomology, Proposition 51.10.1 (2) holds and we find that a power of $I$ annihilates $H^ i_ T(M)$ for $i \leq s$.

We already know the other two assertions of the theorem hold for $i < s$ by Lemma 52.10.7 and for the module $I^{m_0}M$ for $i = s$ and $m_0$ large enough. To finish of the proof we will show that in fact these assertions for $i = s$ holds for $M$.

Let $M' = H^0_ I(M)$ and $M'' = M/M'$ so that we have a short exact sequence

$0 \to M' \to M \to M'' \to 0$

and $M''$ has $H^0_ I(M') = 0$ by Dualizing Complexes, Lemma 47.11.6. By Artin-Rees (Algebra, Lemma 10.51.2) we get short exact sequences

$0 \to M' \to M/I^ n M \to M''/I^ n M'' \to 0$

for $n$ large enough. Consider the long exact sequences

$H^ s_ T(M') \to H^ s_ T(M/I^ nM) \to H^ s_ T(M''/I^ nM'') \to H^{s + 1}_ T(M')$

Now it is a simple matter to see that if we have Mittag-Leffler for the inverse system $\{ H^ s_ T(M''/I^ nM'')\} _{n \geq 0}$ then we have Mittag-Leffler for the inverse system $\{ H^ s_ T(M/I^ nM)\} _{n \geq 0}$. (Note that the ML condition for an inverse system of groups $G_ n$ only depends on the values of the inverse system for sufficiently large $n$.) Moreover the sequence

$H^ s_ T(M') \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM) \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(M''/I^ nM'') \to H^{s + 1}_ T(M')$

is exact because we have ML in the required spots, see Homology, Lemma 12.31.4. Hence, if $H^ s_ T(M'') \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(M''/I^ nM'')$ is an isomorphism, then $H^ s_ T(M) \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM)$ is an isomorphism too by the five lemma (Homology, Lemma 12.5.20). This reduces us to the case discussed in the next paragraph.

Assume that $H^0_ I(M) = 0$. Choose generators $f_1, \ldots , f_ r$ of $I^{m_0}$ where $m_0$ is the integer found for $M$ in Lemma 52.10.7. Then we consider the exact sequence

$0 \to M \xrightarrow {f_1, \ldots , f_ r} (I^{m_0}M)^{\oplus r} \to Q \to 0$

defining $Q$. Some observations: the first map is injective exactly because $H^0_ I(M) = 0$. The cokernel $Q$ of this injection is a finite $A$-module such that for every $1 \leq j \leq r$ we have $Q_{f_ j} \cong (M_{f_ j})^{\oplus r - 1}$. In particular, for a prime $\mathfrak p \subset A$ with $\mathfrak p \not\in V(I)$ we have $Q_\mathfrak p \cong (M_\mathfrak p)^{\oplus r - 1}$. Similarly, given $\mathfrak q \in T$ and $\mathfrak p' \subset A' = (A_\mathfrak q)^\wedge$ not contained in $V(IA')$, we have $Q'_{\mathfrak p'} \cong (M'_{\mathfrak p'})^{\oplus r - 1}$ where $Q' = (Q_\mathfrak q)^\wedge$ and $M' = (M_\mathfrak q)^\wedge$. Thus the conditions in Situation 52.10.1 hold for $A, I, T, Q$. (Observe that $Q$ may have nonvanishing $H^0_ I(Q)$ but this won't matter.)

For any $n \geq 0$ we set $F^ nM = M \cap I^ n(I^{m_0}M)^{\oplus r}$ so that we get short exact sequences

$0 \to F^ nM \to I^ n(I^{m_0}M)^{\oplus r} \to I^ nQ \to 0$

By Artin-Rees (Algebra, Lemma 10.51.2) there exists a $c \geq 0$ such that $I^ n M \subset F^ nM \subset I^{n - c}M$ for all $n \geq c$. Let $m_0$ be the integer and let $m'(m)$ be the function defined for $m \geq m_0$ found in Lemma 52.10.6 applied to $M$. Note that the integer $m_0$ is the same as our integer $m_0$ chosen above (you don't need to check this: you can just take the maximum of the two integers if you like). Finally, by Lemma 52.10.6 applied to $Q$ for every integer $m$ there exists an integer $m''(m) \geq m$ such that $H^ s_ T(I^ kQ) \to H^ s_ T(I^ mQ)$ is zero for all $k \geq m''(m)$.

Fix $m \geq m_0$. Choose $k \geq m'(m''(m + c))$. Choose $\xi \in H^{s + 1}_ T(I^ kM)$ which maps to zero in $H^{s + 1}_ T(M)$. We want to show that $\xi$ maps to zero in $H^{s + 1}_ T(I^ mM)$. Namely, this will show that $\{ H^ s_ T(M/I^ nM)\} _{n \geq 0}$ is Mittag-Leffler exactly as in the proof of Lemma 52.10.7. Picture to help vizualize the argument:

$\xymatrix{ & H^{s + 1}_ T(I^ kM) \ar[r] \ar[d] & H^{s + 1}_ T(I^ k(I^{m_0}M)^{\oplus r}) \ar[d] & \\ H^ s_ T(I^{m''(m + c)}Q) \ar[r]_-\delta \ar[d] & H^{s + 1}_ T(F^{m''(m + c)}M) \ar[r] \ar[d] & H^{s + 1}_ T(I^{m''(m + c)}(I^{m_0}M)^{\oplus r}) \\ H^ s_ T(I^{m + c}Q) \ar[r] & H^{s + 1}_ T(F^{m + c}M) \ar[d] & \\ & H^{s + 1}_ T(I^ mM) }$

The image of $\xi$ in $H^{s + 1}_ T(I^ k(I^{m_0}M)^{\oplus r})$ maps to zero in $H^{s + 1}_ T((I^{m_0}M)^{\oplus r})$ and hence maps to zero in $H^{s + 1}_ T(I^{m''(m + c)}(I^{m_0}M)^{\oplus r})$ by choice of $m'(-)$. Thus the image $\xi ' \in H^{s + 1}_ T(F^{m''(m + c)}M)$ maps to zero in $H^{s + 1}_ T(I^{m''(m + c)}(I^{m_0}M)^{\oplus r})$ and hence $\xi ' = \delta (\eta )$ for some $\eta \in H^ s_ T(I^{m''(m + c)}Q)$. By our choice of $m''(-)$ we find that $\eta$ maps to zero in $H^ s_ T(I^{m + c}Q)$. This in turn means that $\xi '$ maps to zero in $H^{s + 1}_ T(F^{m + c}M)$. Since $F^{m + c}M \subset I^ mM$ we conclude.

Finally, we prove the statement on limits. Consider the short exact sequences

$0 \to M/F^ nM \to (I^{m_0}M)^{\oplus r}/I^ n (I^{m_0}M)^{\oplus r} \to Q/I^ nQ \to 0$

We have $\mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM) = \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/F^ nM)$ as these inverse systems are pro-isomorphic. We obtain a commutative diagram

$\xymatrix{ H^{s - 1}_ T(Q) \ar[r] \ar[d] & \mathop{\mathrm{lim}}\nolimits H^{s - 1}_ T(Q/I^ nQ) \ar[d] \\ H^ s_ T(M) \ar[r] \ar[d] & \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM) \ar[d] \\ H^ s_ T((I^{m_0}M)^{\oplus r}) \ar[r] \ar[d] & \mathop{\mathrm{lim}}\nolimits H^ s_ T((I^{m_0}M)^{\oplus r}/I^ n(I^{m_0}M)^{\oplus r}) \ar[d] \\ H^ s_ T(Q) \ar[r] & \mathop{\mathrm{lim}}\nolimits H^ s_ T(Q/I^ nQ) }$

The right column is exact because we have ML in the required spots, see Homology, Lemma 12.31.4. The lowest horizontal arrow is injective (!) by part (5) of Lemma 52.10.7. The horizontal arrow above it is bijective by part (4) of Lemma 52.10.7. The arrows in cohomological degrees $\leq s - 1$ are isomorphisms. Thus we conclude $H^ s_ T(M) \to \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM)$ is an isomorphism by the five lemma (Homology, Lemma 12.5.20). This finishes the proof of the theorem. $\square$

Lemma 52.10.9. Let $I \subset \mathfrak a \subset A$ be ideals of a Noetherian ring $A$ and let $M$ be a finite $A$-module. Let $s$ and $d$ be integers. Suppose that

1. $A, I, V(\mathfrak a), M$ satisfy the assumptions of Situation 52.10.1 for $s$ and $d$, and

2. $A, I, \mathfrak a, M$ satisfy the conditions of Lemma 52.8.5 for $s + 1$ and $d$ with $J = \mathfrak a$.

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

$H^{s + 1}_ J(M) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^{s + 1}_\mathfrak a(M/I^ nM)$

is an isomorphism.

Proof. Namely, we have the existence of $J_0$ and the isomorphism $H^{s + 1}_ J(M) = H^{s + 1}(R\Gamma _\mathfrak a(M)^\wedge )$ by Lemma 52.8.5, we have a short exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^ s_\mathfrak a(M/I^ nM) \to H^{s + 1}(R\Gamma _\mathfrak a(M)^\wedge ) \to \mathop{\mathrm{lim}}\nolimits H^{s + 1}_\mathfrak a(M/I^ nM) \to 0$

by Dualizing Complexes, Lemma 47.12.4, and the module $R^1\mathop{\mathrm{lim}}\nolimits H^ s_\mathfrak a(M/I^ nM)$ is zero because $\{ H^ s_\mathfrak a(M/I^ nM)\} _{n \geq 0}$ has Mittag-Leffler by Theorem 52.10.8. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).