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

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

2. $H^ i_{\mathfrak p A_\mathfrak p}(M_{k, \mathfrak p}) \to H^ i_{\mathfrak p A_\mathfrak p}(M_{m, \mathfrak p})$ 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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