## 51.11 Finiteness of local cohomology, II

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

1. $H^ i_ T(M)$ is a finite $A$-module for $i \leq s$, and

2. 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

51.11.1.1
$$\label{local-cohomology-equation-cutoff} s_{A, I}(M) = \min \{ \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \mid \mathfrak p \in X \setminus Z, \mathfrak q \in Z, \mathfrak p \subset \mathfrak q \}$$

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

$\text{depth}(M_\mathfrak p) = \min \text{depth}(M_{\mathfrak p_ i})$

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

$\dim ((B/\mathfrak p_ i)_{\mathfrak q_{ij}}) = \dim ((A/\mathfrak p)_\mathfrak q)$

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

$\text{depth}_{A_{\mathfrak p'}}(M'_{\mathfrak p'}) \geq \text{depth}_{A_\mathfrak p}(M_\mathfrak p)$

and

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

Proof. We have

$\text{depth}(M'_{\mathfrak p'}) = \text{depth}(M_\mathfrak p) + \text{depth}(A'_{\mathfrak p'}/\mathfrak p A'_{\mathfrak p'}) \geq \text{depth}(M_\mathfrak p)$

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

\begin{align*} \text{depth}(M'_{\mathfrak p'}) + \dim ((A'/\mathfrak p')_{\mathfrak q'}) & = \text{depth}(M_\mathfrak p) + \dim (A'_{\mathfrak p'}/\mathfrak p A'_{\mathfrak p'}) + \dim ((A'/\mathfrak p')_{\mathfrak q'}) \\ & = \text{depth}(M_\mathfrak p) + \dim ((A'/\mathfrak p A')_{\mathfrak q'}) \\ & = \text{depth}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \end{align*}

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

$s_{A, I}(M) \geq s_{A^\wedge , I^\wedge }(M^\wedge )$

If the formal fibres of $A$ are $(S_ n)$, then $\min (n + 1, s_{A, I}(M)) \leq s_{A^\wedge , I^\wedge }(M^\wedge )$.

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

$\xymatrix{ \mathfrak p' \ar[r] & \mathfrak q' \ar[r] & A^\wedge \\ \mathfrak p \ar[r] \ar@{-}[u] & \mathfrak q \ar[r] \ar@{-}[u] & A \ar[u] }$

Let us write

\begin{align*} a & = \dim (A/\mathfrak p) = \dim (A^\wedge /\mathfrak pA^\wedge ),\\ b & = \dim (A/\mathfrak q) = \dim (A^\wedge /\mathfrak qA^\wedge ),\\ a' & = \dim (A^\wedge /\mathfrak p'),\\ b' & = \dim (A^\wedge /\mathfrak q') \end{align*}

Equalities by More on Algebra, Lemma 15.43.1. We also write

\begin{align*} p & = \dim (A^\wedge _{\mathfrak p'}/\mathfrak p A^\wedge _{\mathfrak p'}) = \dim ((A^\wedge /\mathfrak p A^\wedge )_{\mathfrak p'}) \\ q & = \dim (A^\wedge _{\mathfrak q'}/\mathfrak p A^\wedge _{\mathfrak q'}) = \dim ((A^\wedge /\mathfrak q A^\wedge )_{\mathfrak q'}) \end{align*}

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

$\text{depth}(M^\wedge _{\mathfrak p'}) = \text{depth}(M_\mathfrak p) + \text{depth}(A^\wedge _{\mathfrak p'} / \mathfrak p A^\wedge _{\mathfrak p'})$

The quantity we are minimizing for $s_{A, I}(M)$ is

$s(\mathfrak p, \mathfrak q) = \text{depth}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) = \text{depth}(M_\mathfrak p) + a - b$

(last equality as $A$ is catenary). The quantity we are minimizing for $s_{A^\wedge , I^\wedge }(M^\wedge )$ is

$s(\mathfrak p', \mathfrak q') = \text{depth}(M^\wedge _{\mathfrak p'}) + \dim ((A^\wedge /\mathfrak p')_{\mathfrak q'}) = \text{depth}(M^\wedge _{\mathfrak p'}) + a' - b'$

(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

$s(\mathfrak p', \mathfrak q') \geq s(\mathfrak p, \mathfrak q) + q + \min (n, p) - p \geq s_{A, I}(M) + q + \min (n, p) - p$

Thus the only way we can get in trouble is if $p > n$. If this happens then

\begin{align*} s(\mathfrak p', \mathfrak q') & = \text{depth}(M^\wedge _{\mathfrak p'}) + \dim ((A^\wedge /\mathfrak p')_{\mathfrak q'}) \\ & = \text{depth}(M_\mathfrak p) + \text{depth}(A^\wedge _{\mathfrak p'} / \mathfrak p A^\wedge _{\mathfrak p'}) + \dim ((A^\wedge /\mathfrak p')_{\mathfrak q'}) \\ & \geq 0 + n + 1 \end{align*}

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

1. $A$ is universally catenary,

2. 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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