The Stacks project

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

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. $\square$

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

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

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). $\square$

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

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

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

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. $\square$

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

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

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

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

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

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

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

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

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

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

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

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

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

Proof. Consider the long exact sequences

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

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

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). $\square$

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

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

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

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

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

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

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

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 visualize the argument:

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

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

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. $\square$

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

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. $\square$