Proof.
Consider the long exact sequences
\[ 0 \to H^0_ T(I^ nM) \to H^0_ T(M) \to H^0_ T(M/I^ nM) \to H^1_ T(I^ nM) \to H^1_ T(M) \to \ldots \]
Parts (1) and (3) follows easily from this and Lemma 52.10.6.
Let $m_0$ and $m'(-)$ be as in Lemma 52.10.6. For $m \geq m_0$ consider the long exact sequence
\[ H^ s_ T(I^ mM) \to H^ s_ T(I^{m_0}M) \to H^ s_ T(I^{m_0}M/I^ mM) \to H^{s + 1}_ T(I^ mM) \to H^1_ T(I^{m_0}M) \]
Then for $k \geq m'(m)$ the image of $H^{s + 1}_ T(I^ kM) \to H^{s + 1}_ T(I^ mM)$ maps injectively to $H^{s + 1}_ T(I^{m_0}M)$. Hence the image of $H^ s_ T(I^{m_0}M/I^ kM) \to H^ s_ T(I^{m_0}M/I^ mM)$ maps to zero in $H^{s + 1}_ T(I^ mM)$ for all $k \geq m'(m)$. We conclude that (2) and (4) hold.
Consider the short exact sequences $0 \to I^{m_0}M \to M \to M/I^{m_0} M \to 0$ and $0 \to I^{m_0}M/I^ nM \to M/I^ nM \to M/I^{m_0} M \to 0$. We obtain a diagram
\[ \xymatrix{ H^{s - 1}_ T(M/I^{m_0}M) \ar[r] & \mathop{\mathrm{lim}}\nolimits H^ s_ T(I^{m_0}M/I^ nM) \ar[r] & \mathop{\mathrm{lim}}\nolimits H^ s_ T(M/I^ nM) \ar[r] & H^ s_ T(M/I^{m_0}M) \\ H^{s - 1}_ T(M/I^{m_0}M) \ar[r] \ar@{=}[u] & H^ s_ T(I^{m_0}M) \ar[r] \ar[u]_{\cong } & H^ s_ T(M) \ar[r] \ar[u] & H^ s_ T(M/I^{m_0}M) \ar@{=}[u] } \]
whose lower row is exact. The top row is also exact (at the middle two spots) by Homology, Lemma 12.31.4. Part (5) follows.
Write $B_ n = H^ s_ T(M/I^ nM)$. Let $A_ n \subset B_ n$ be the image of $H^ s_ T(I^{m_0}M/I^ nM) \to H^ s_ T(M/I^ nM)$. Then $(A_ n)$ satisfies the Mittag-Leffler condition by (2) and Homology, Lemma 12.31.3. Also $C_ n = B_ n/A_ n$ is killed by $I^{m_0}$. Thus $R^1\mathop{\mathrm{lim}}\nolimits B_ n \cong R^1\mathop{\mathrm{lim}}\nolimits C_ n$ is killed by $I^{m_0}$ and we get (6).
$\square$
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