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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