Email correspondence between Janos Kollar, Sandor Kovacs, and Johan de Jong of 23/02/2018.
Lemma 15.100.11. Let $A \to B$ be a flat homomorphism of Noetherian rings. Let $I \subset A$ be an ideal. Let $M, N$ be $A$-modules. Set $B_ n = B/I^ nB$, $M_ n = M/I^ nM$, $N_ n = N/I^ nN$. If $M$ is flat over $A$, then we have for all $i \in \mathbf{Z}$.
Proof.
Choose a resolution
by finite free $B$-modues $P_ i$. Set $P_{i, n} = P_ i/I^ nP_ i$. Since $M$ and $B$ are flat over $A$, the sequence
is exact. We see that on the one hand the complex
computes the modules $\mathop{\mathrm{Ext}}\nolimits ^ i_ B(M, N)$ and on the other hand the complex
computes the modules $\mathop{\mathrm{Ext}}\nolimits ^ i_{B_ n}(M_ n, N_ n)$. Since
we obtain the result from Lemma 15.100.1 part (2).
$\square$
\[ \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ i_ B(M, N)/I^ n \mathop{\mathrm{Ext}}\nolimits ^ i_ B(M, N) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Ext}}\nolimits ^ i_{B_ n}(M_ n, N_ n) \]
\[ \ldots \to P_2 \to P_1 \to P_0 \to M \to 0 \]
\[ \ldots \to P_{2, n} \to P_{1, n} \to P_{0, n} \to M_ n \to 0 \]
\[ \mathop{\mathrm{Hom}}\nolimits _ B(P_0, N) \to \mathop{\mathrm{Hom}}\nolimits _ B(P_1, N) \to \mathop{\mathrm{Hom}}\nolimits _ B(P_2, N) \to \ldots \]
\[ \mathop{\mathrm{Hom}}\nolimits _{B_ n}(P_{0, n}, N_ n) \to \mathop{\mathrm{Hom}}\nolimits _{B_ n}(P_{1, n}, N_ n) \to \mathop{\mathrm{Hom}}\nolimits _{B_ n}(P_{2, n}, N_ n) \to \ldots \]
\[ \mathop{\mathrm{Hom}}\nolimits _{B_ n}(P_{i, n}, N_ n) = \mathop{\mathrm{Hom}}\nolimits _ B(P_ i, N)/I^ n \mathop{\mathrm{Hom}}\nolimits _ B(P_ i, N) \]
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