If \(M\) is classically \(I\)-complete, we have
\begin{equation*}
\Hom_A(A_f, M) = \Hom_A(A_f, \lim_n M/I^n M) = 0.
\end{equation*}
To show that \(\Ext^1_A(A_f, M) = 0\text{,}\) consider an extension
\begin{equation}
0 \to M \to E \to A_f \to 0.\tag{6.3}
\end{equation}
For each \(n \geq 0\text{,}\) pick \(e_n \in E\) mapping to \(f^{-n} \in A_f\) and set \(\delta_n = fe_{n+1} - e_n \in M\text{.}\) Since \(M\) is complete, we may define the elements
\begin{equation*}
e_n' = e_n + \delta_n + f \delta_{n+1} + f^2 \delta_{n+2} + \cdots
\end{equation*}
which satisfy \(f e_{n+1}' = e_n'\text{;}\) we thus obtain a map \(A_f \to E\) splitting the sequence by mapping \(f^{-n}\) to \(e_n'\text{.}\)
In the converse direction, by an elementary argument (
Exercise 6.7.1) we can reduce to the case where
\(I = (f)\text{.}\) That, we must show that if
(6.1) holds, then for any
\(x_0, x_1, \ldots \in M\) there exists
\(x \in M\) such that
\(x \equiv x_0 + fx_1 + \cdots + f^{n-1} x_{n-1} \pmod{f^nM}\) for each
\(n\text{.}\) To this end, form an extension as in
(6.3) by taking
\begin{equation*}
E = M \oplus \bigoplus_n Ae_n/(x_n - fe_{n+1} + e_n)
\end{equation*}
with \(e_n\) mapping to \(f^{-n}\) in \(A_f\text{;}\) note that by the snake lemma, \(M/f^n M = E/f^n E\) for all \(n\text{.}\) Since the extension splits by hypothesis, there is an element \(x + e_0 \in E\) which generates a copy of \(A_f\) in \(E\text{.}\) We then have
\begin{equation*}
x + e_0 = x - x_0 + fe_1 = x - x_0 + x_1 + f^2 e_2 = \cdots;
\end{equation*}
this yields the desired result. (Compare
[117], tag 091R.)