## 51.6 Filtrations on local cohomology

Some tricks related to the spectral sequence of Lemma 51.5.8.

Lemma 51.6.1. Let $A$ be a Noetherian ring. Let $T \subset \mathop{\mathrm{Spec}}(A)$ be a subset stable under specialization. Let $T' \subset T$ be the set of nonminimal primes in $T$. Then $T'$ is a subset of $\mathop{\mathrm{Spec}}(A)$ stable under specialization and for every $A$-module $M$ there is an exact sequence

\[ 0 \to \mathop{\mathrm{colim}}\nolimits _{Z, f} H^1_ f(H^{i - 1}_ Z(M)) \to H^ i_{T'}(M) \to H^ i_ T(M) \to \bigoplus \nolimits _{\mathfrak p \in T \setminus T'} H^ i_{\mathfrak p A_\mathfrak p}(M_\mathfrak p) \]

where the colimit is over closed subsets $Z \subset T$ and $f \in A$ with $V(f) \cap Z \subset T'$.

**Proof.**
For every $Z$ and $f$ the spectral sequence of Dualizing Complexes, Lemma 47.9.6 degenerates to give short exact sequences

\[ 0 \to H^1_ f(H^{i - 1}_ Z(M)) \to H^ i_{Z \cap V(f)}(M) \to H^0_ f(H^ i_ Z(M)) \to 0 \]

We will use this without further mention below.

Let $\xi \in H^ i_ T(M)$ map to zero in the direct sum. Then we first write $\xi $ as the image of some $\xi ' \in H^ i_ Z(M)$ for some closed subset $Z \subset T$, see Lemma 51.5.3. Then $\xi '$ maps to zero in $H^ i_{\mathfrak p A_\mathfrak p}(M_\mathfrak p)$ for every $\mathfrak p \in Z$, $\mathfrak p \not\in T'$. Since there are finitely many of these primes, we may choose $f \in A$ not contained in any of these such that $f$ annihilates $\xi '$. Then $\xi '$ is the image of some $\xi '' \in H^ i_{Z'}(M)$ where $Z' = Z \cap V(f)$. By our choice of $f$ we have $Z' \subset T'$ and we get exactness at the penultimate spot.

Let $\xi \in H^ i_{T'}(M)$ map to zero in $H^ i_ T(M)$. Choose closed subsets $Z' \subset Z$ with $Z' \subset T'$ and $Z \subset T$ such that $\xi $ comes from $\xi ' \in H^ i_{Z'}(M)$ and maps to zero in $H^ i_ Z(M)$. Then we can find $f \in A$ with $V(f) \cap Z = Z'$ and we conclude.
$\square$

Lemma 51.6.2. Let $A$ be a Noetherian ring of finite dimension. Let $T \subset \mathop{\mathrm{Spec}}(A)$ be a subset stable under specialization. Let $\{ M_ n\} _{n \geq 0}$ be an inverse system of $A$-modules. Let $i \geq 0$ be an integer. Assume that for every $m$ there exists an integer $m'(m) \geq m$ such that for all $\mathfrak p \in T$ the induced map

\[ H^ i_{\mathfrak p A_\mathfrak p}(M_{k, \mathfrak p}) \longrightarrow H^ i_{\mathfrak p A_\mathfrak p}(M_{m, \mathfrak p}) \]

is zero for $k \geq m'(m)$. Let $m'' : \mathbf{N} \to \mathbf{N}$ be the $2^{\dim (T)}$-fold self-composition of $m'$. Then the map $H^ i_ T(M_ k) \to H^ i_ T(M_ m)$ is zero for all $k \geq m''(m)$.

**Proof.**
We first make a general remark: suppose we have an exact sequence

\[ (A_ n) \to (B_ n) \to (C_ n) \]

of inverse systems of abelian groups. Suppose that for every $m$ there exists an integer $m'(m) \geq m$ such that

\[ A_ k \to A_ m \quad \text{and}\quad C_ k \to C_ m \]

are zero for $k \geq m'(m)$. Then for $k \geq m'(m'(m))$ the map $B_ k \to B_ m$ is zero.

We will prove the lemma by induction on $\dim (T)$ which is finite because $\dim (A)$ is finite. Let $T' \subset T$ be the set of nonminimal primes in $T$. Then $T'$ is a subset of $\mathop{\mathrm{Spec}}(A)$ stable under specialization and the hypotheses of the lemma apply to $T'$. Since $\dim (T') < \dim (T)$ we know the lemma holds for $T'$. For every $A$-module $M$ there is an exact sequence

\[ H^ i_{T'}(M) \to H^ i_ T(M) \to \bigoplus \nolimits _{\mathfrak p \in T \setminus T'} H^ i_{\mathfrak p A_\mathfrak p}(M_\mathfrak p) \]

by Lemma 51.6.1. Thus we conclude by the initial remark of the proof.
$\square$

Lemma 51.6.3. Let $A$ be a Noetherian ring. Let $T \subset \mathop{\mathrm{Spec}}(A)$ be a subset stable under specialization. Let $\{ M_ n\} _{n \geq 0}$ be an inverse system of $A$-modules. Let $i \geq 0$ be an integer. Assume the dimension of $A$ is finite and that for every $m$ there exists an integer $m'(m) \geq m$ such that for all $\mathfrak p \in T$ we have

$H^{i - 1}_{\mathfrak p A_\mathfrak p}(M_{k, \mathfrak p}) \to H^{i - 1}_{\mathfrak p A_\mathfrak p}(M_{m, \mathfrak p})$ is zero for $k \geq m'(m)$, and

$ H^ i_{\mathfrak p A_\mathfrak p}(M_{k, \mathfrak p}) \to H^ i_{\mathfrak p A_\mathfrak p}(M_{m, \mathfrak p})$ has image $G(\mathfrak p, m)$ independent of $k \geq m'(m)$ and moreover $G(\mathfrak p, m)$ maps injectively into $H^ i_{\mathfrak p A_\mathfrak p}(M_{0, \mathfrak p})$.

Then there exists an integer $m_0$ such that for every $m \geq m_0$ there exists an integer $m''(m) \geq m$ such that for $k \geq m''(m)$ the image of $H^ i_ T(M_ k) \to H^ i_ T(M_ m)$ maps injectively into $H^ i_ T(M_{m_0})$.

**Proof.**
We first make a general remark: suppose we have an exact sequence

\[ (A_ n) \to (B_ n) \to (C_ n) \to (D_ n) \]

of inverse systems of abelian groups. Suppose that there exists an integer $m_0$ such that for every $m \geq m_0$ there exists an integer $m'(m) \geq m$ such that the maps

\[ \mathop{\mathrm{Im}}(B_ k \to B_ m) \longrightarrow B_{m_0} \quad \text{and}\quad \mathop{\mathrm{Im}}(D_ k \to D_ m) \longrightarrow D_{m_0} \]

are injective for $k \geq m'(m)$ and $A_ k \to A_ m$ is zero for $k \geq m'(m)$. Then for $m \geq m'(m_0)$ and $k \geq m'(m'(m))$ the map

\[ \mathop{\mathrm{Im}}(C_ k \to C_ m) \to C_{m'(m_0)} \]

is injective. Namely, let $c_0 \in C_ m$ be the image of $c_3 \in C_ k$ and say $c_0$ maps to zero in $C_{m'(m_0)}$. Picture

\[ C_ k \to C_{m'(m'(m))} \to C_{m'(m)} \to C_ m \to C_{m'(m_0)},\quad c_3 \mapsto c_2 \mapsto c_1 \mapsto c_0 \mapsto 0 \]

We have to show $c_0 = 0$. The image $d_3$ of $c_3$ maps to zero in $C_{m_0}$ and hence we see that the image $d_1 \in D_{m'(m)}$ is zero. Thus we can choose $b_1 \in B_{m'(m)}$ mapping to the image $c_1$. Since $c_3$ maps to zero in $C_{m'(m_0)}$ we find an element $a_{-1} \in A_{m'(m_0)}$ which maps to the image $b_{-1} \in B_{m'(m_0)}$ of $b_1$. Since $a_{-1}$ maps to zero in $A_{m_0}$ we conclude that $b_1$ maps to zero in $B_{m_0}$. Thus the image $b_0 \in B_ m$ is zero which of course implies $c_0 = 0$ as desired.

We will prove the lemma by induction on $\dim (T)$ which is finite because $\dim (A)$ is finite. Let $T' \subset T$ be the set of nonminimal primes in $T$. Then $T'$ is a subset of $\mathop{\mathrm{Spec}}(A)$ stable under specialization and the hypotheses of the lemma apply to $T'$. Since $\dim (T') < \dim (T)$ we know the lemma holds for $T'$. For every $A$-module $M$ there is an exact sequence

\[ 0 \to \mathop{\mathrm{colim}}\nolimits _{Z, f} H^1_ f(H^{i - 1}_ Z(M)) \to H^ i_{T'}(M) \to H^ i_ T(M) \to \bigoplus \nolimits _{\mathfrak p \in T \setminus T'} H^ i_{\mathfrak p A_\mathfrak p}(M_\mathfrak p) \]

by Lemma 51.6.1. Thus we conclude by the initial remark of the proof and the fact that we've seen the system of groups

\[ \left\{ \mathop{\mathrm{colim}}\nolimits _{Z, f} H^1_ f(H^{i - 1}_ Z(M_ n))\right\} _{n \geq 0} \]

is pro-zero in Lemma 51.6.2; this uses that the function $m''(m)$ in that lemma for $H^{i - 1}_ Z(M)$ is independent of $Z$.
$\square$

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