We continue the discussion of finiteness of local cohomology started in Section 51.7. Using Faltings Annihilator Theorem we easily prove the following fundamental result.
Proposition 51.11.1. Let $A$ be a Noetherian ring which has a dualizing complex. Let $T \subset \mathop{\mathrm{Spec}}(A)$ be a subset stable under specialization. Let $s \geq 0$ an integer. Let $M$ be a finite $A$-module. The following are equivalent
$H^ i_ T(M)$ is a finite $A$-module for $i \leq s$, and
for all $\mathfrak p \not\in T$, $\mathfrak q \in T$ with $\mathfrak p \subset \mathfrak q$ we have
\[ \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > s \]
Proof. Formal consequence of Proposition 51.10.1 and Lemma 51.7.1. $\square$
Besides some lemmas for later use, the rest of this section is concerned with the question to what extend the condition in Proposition 51.11.1 that $A$ has a dualizing complex can be weakened. The answer is roughly that one has to assume the formal fibres of $A$ are $(S_ n)$ for sufficiently large $n$.
Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Z = V(I) \subset X$. Let $M$ be a finite $A$-module. We define
Our conventions on depth are that the depth of $0$ is $\infty $ thus we only need to consider primes $\mathfrak p$ in the support of $M$. It will turn out that $s_{A, I}(M)$ is an important invariant of the situation.
Lemma 51.11.2. Let $A \to B$ be a finite homomorphism of Noetherian rings. Let $I \subset A$ be an ideal and set $J = IB$. Let $M$ be a finite $B$-module. If $A$ is universally catenary, then $s_{B, J}(M) = s_{A, I}(M)$.
Proof. Let $\mathfrak p \subset \mathfrak q \subset A$ be primes with $I \subset \mathfrak q$ and $I \not\subset \mathfrak p$. Since $A \to B$ is finite there are finitely many primes $\mathfrak p_ i$ lying over $\mathfrak p$. By Algebra, Lemma 10.72.11 we have
Let $\mathfrak p_ i \subset \mathfrak q_{ij}$ be primes lying over $\mathfrak q$. By going up for $A \to B$ (Algebra, Lemma 10.36.22) there is at least one $\mathfrak q_{ij}$ for each $i$. Then we see that
by the dimension formula, see Algebra, Lemma 10.113.1. This implies that the minimum of the quantities used to define $s_{B, J}(M)$ for the pairs $(\mathfrak p_ i, \mathfrak q_{ij})$ is equal to the quantity for the pair $(\mathfrak p, \mathfrak q)$. This proves the lemma. $\square$
Lemma 51.11.3. Let $A$ be a Noetherian ring which has a dualizing complex. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Let $A', M'$ be the $I$-adic completions of $A, M$. Let $\mathfrak p' \subset \mathfrak q'$ be prime ideals of $A'$ with $\mathfrak q' \in V(IA')$ lying over $\mathfrak p \subset \mathfrak q$ in $A$. Then and
\[ \text{depth}_{A_{\mathfrak p'}}(M'_{\mathfrak p'}) \geq \text{depth}_{A_\mathfrak p}(M_\mathfrak p) \]
\[ \text{depth}_{A_{\mathfrak p'}}(M'_{\mathfrak p'}) + \dim ((A'/\mathfrak p')_{\mathfrak q'}) = \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \]
Proof. We have
by flatness of $A \to A'$, see Algebra, Lemma 10.163.1. Since the fibres of $A \to A'$ are Cohen-Macaulay (Dualizing Complexes, Lemma 47.23.2 and More on Algebra, Section 15.51) we see that $\text{depth}(A'_{\mathfrak p'}/\mathfrak p A'_{\mathfrak p'}) = \dim (A'_{\mathfrak p'}/\mathfrak p A'_{\mathfrak p'})$. Thus we obtain
Second equality because $A'$ is catenary and third equality by More on Algebra, Lemma 15.43.1 as $(A/\mathfrak p)_\mathfrak q$ and $(A'/\mathfrak p A')_{\mathfrak q'}$ have the same $I$-adic completions. $\square$
Lemma 51.11.4. Let $A$ be a universally catenary Noetherian local ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Then If the formal fibres of $A$ are $(S_ n)$, then $\min (n + 1, s_{A, I}(M)) \leq s_{A^\wedge , I^\wedge }(M^\wedge )$.
\[ s_{A, I}(M) \geq s_{A^\wedge , I^\wedge }(M^\wedge ) \]
Proof. Write $X = \mathop{\mathrm{Spec}}(A)$, $X^\wedge = \mathop{\mathrm{Spec}}(A^\wedge )$, $Z = V(I) \subset X$, and $Z^\wedge = V(I^\wedge )$. Let $\mathfrak p' \subset \mathfrak q' \subset A^\wedge $ be primes with $\mathfrak p' \not\in Z^\wedge $ and $\mathfrak q' \in Z^\wedge $. Let $\mathfrak p \subset \mathfrak q$ be the corresponding primes of $A$. Then $\mathfrak p \not\in Z$ and $\mathfrak q \in Z$. Picture
Let us write
Equalities by More on Algebra, Lemma 15.43.1. We also write
Since $A$ is universally catenary we see that $A^\wedge /\mathfrak pA^\wedge = (A/\mathfrak p)^\wedge $ is equidimensional of dimension $a$ (More on Algebra, Proposition 15.109.5). Hence $a = a' + p$. Similarly $b = b' + q$. By Algebra, Lemma 10.163.1 applied to the flat local ring map $A_\mathfrak p \to A^\wedge _{\mathfrak p'}$ we have
The quantity we are minimizing for $s_{A, I}(M)$ is
(last equality as $A$ is catenary). The quantity we are minimizing for $s_{A^\wedge , I^\wedge }(M^\wedge )$ is
(last equality as $A^\wedge $ is catenary). Now we have enough notation in place to start the proof.
Let $\mathfrak p \subset \mathfrak q \subset A$ be primes with $\mathfrak p \not\in Z$ and $\mathfrak q \in Z$ such that $s_{A, I}(M) = s(\mathfrak p, \mathfrak q)$. Then we can pick $\mathfrak q'$ minimal over $\mathfrak q A^\wedge $ and $\mathfrak p' \subset \mathfrak q'$ minimal over $\mathfrak p A^\wedge $ (using going down for $A \to A^\wedge $). Then we have four primes as above with $p = 0$ and $q = 0$. Moreover, we have $\text{depth}(A^\wedge _{\mathfrak p'} / \mathfrak p A^\wedge _{\mathfrak p'})=0$ also because $p = 0$. This means that $s(\mathfrak p', \mathfrak q') = s(\mathfrak p, \mathfrak q)$. Thus we get the first inequality.
Assume that the formal fibres of $A$ are $(S_ n)$. Then $\text{depth}(A^\wedge _{\mathfrak p'} / \mathfrak p A^\wedge _{\mathfrak p'}) \geq \min (n, p)$. Hence
Thus the only way we can get in trouble is if $p > n$. If this happens then
because $(A^\wedge /\mathfrak p')_{\mathfrak q'}$ has at least two primes. This proves the second inequality. $\square$
The method of proof of the following lemma works more generally, but the stronger results one gets will be subsumed in Theorem 51.11.6 below.
Lemma 51.11.5. Let $A$ be a Gorenstein Noetherian local ring. Let $I \subset A$ be an ideal and set $Z = V(I) \subset \mathop{\mathrm{Spec}}(A)$. Let $M$ be a finite $A$-module. Let $s = s_{A, I}(M)$ as in (51.11.1.1). Then $H^ i_ Z(M)$ is finite for $i < s$, but $H^ s_ Z(M)$ is not finite.
Proof. Since a Gorenstein local ring has a dualizing complex, this is a special case of Proposition 51.11.1. It would be helpful to have a short proof of this special case, which will be used in the proof of a general finiteness theorem below. $\square$
Observe that the hypotheses of the following theorem are satisfied by excellent Noetherian rings (by definition), by Noetherian rings which have a dualizing complex (Dualizing Complexes, Lemma 47.17.4 and Dualizing Complexes, Lemma 47.23.2), and by quotients of regular Noetherian rings.
Theorem 51.11.6. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Set $Z = V(I) \subset \mathop{\mathrm{Spec}}(A)$. Let $M$ be a finite $A$-module. Set $s = s_{A, I}(M)$ as in (51.11.1.1). Assume that
$A$ is universally catenary,
the formal fibres of the local rings of $A$ are Cohen-Macaulay.
Then $H^ i_ Z(M)$ is finite for $0 \leq i < s$ and $H^ s_ Z(M)$ is not finite.
Proof. By Lemma 51.7.2 we may assume that $A$ is a local ring.
If $A$ is a Noetherian complete local ring, then we can write $A$ as the quotient of a regular complete local ring $B$ by Cohen's structure theorem (Algebra, Theorem 10.160.8). Using Lemma 51.11.2 and Dualizing Complexes, Lemma 47.9.2 we reduce to the case of a regular local ring which is a consequence of Lemma 51.11.5 because a regular local ring is Gorenstein (Dualizing Complexes, Lemma 47.21.3).
Let $A$ be a Noetherian local ring. Let $\mathfrak m$ be the maximal ideal. We may assume $I \subset \mathfrak m$, otherwise the lemma is trivial. Let $A^\wedge $ be the completion of $A$, let $Z^\wedge = V(IA^\wedge )$, and let $M^\wedge = M \otimes _ A A^\wedge $ be the completion of $M$ (Algebra, Lemma 10.97.1). Then $H^ i_ Z(M) \otimes _ A A^\wedge = H^ i_{Z^\wedge }(M^\wedge )$ by Dualizing Complexes, Lemma 47.9.3 and flatness of $A \to A^\wedge $ (Algebra, Lemma 10.97.2). Hence it suffices to show that $H^ i_{Z^\wedge }(M^\wedge )$ is finite for $i < s$ and not finite for $i = s$, see Algebra, Lemma 10.83.2. Since we know the result is true for $A^\wedge $ it suffices to show that $s_{A, I}(M) = s_{A^\wedge , I^\wedge }(M^\wedge )$. This follows from Lemma 51.11.4. $\square$
Remark 51.11.7. The astute reader will have realized that we can get away with a slightly weaker condition on the formal fibres of the local rings of $A$. Namely, in the situation of Theorem 51.11.6 assume $A$ is universally catenary but make no assumptions on the formal fibres. Suppose we have an $n$ and we want to prove that $H^ i_ Z(M)$ are finite for $i \leq n$. Then the exact same proof shows that it suffices that $s_{A, I}(M) > n$ and that the formal fibres of local rings of $A$ are $(S_ n)$. On the other hand, if we want to show that $H^ s_ Z(M)$ is not finite where $s = s_{A, I}(M)$, then our arguments prove this if the formal fibres are $(S_{s - 1})$.
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