**Proof.**
Proof of (1). Assume $M$ is $I$-adically complete. By Lemma 15.91.1 it suffices to prove $\mathop{\mathrm{Ext}}\nolimits ^1_ A(A_ f, M) = 0$ and $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, M) = 0$. Since $M = \mathop{\mathrm{lim}}\nolimits M/I^ nM$ and since $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, M/I^ nM) = 0$ it follows that $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, M) = 0$. Suppose we have an extension

\[ 0 \to M \to E \to A_ f \to 0 \]

For $n \geq 0$ pick $e_ n \in E$ mapping to $1/f^ n$. Set $\delta _ n = fe_{n + 1} - e_ n \in M$ for $n \geq 0$. Replace $e_ n$ by

\[ e'_ n = e_ n + \delta _ n + f\delta _{n + 1} + f^2 \delta _{n + 2} + \ldots \]

The infinite sum exists as $M$ is complete with respect to $I$ and $f \in I$. A simple calculation shows that $fe'_{n + 1} = e'_ n$. Thus we get a splitting of the extension by mapping $1/f^ n$ to $e'_ n$.

Proof of (2). Assume that $I = (f_1, \ldots , f_ r)$ and that $T(M, f_ i) = 0$ for $i = 1, \ldots , r$. By Algebra, Lemma 10.96.7 we may assume $I = (f)$ and $T(M, f) = 0$. Let $x_ n \in M$ for $n \geq 0$. Consider the extension

\[ 0 \to M \to E \to A_ f \to 0 \]

given by

\[ E = M \oplus \bigoplus Ae_ n\Big/\langle x_ n - fe_{n + 1} + e_ n\rangle \]

mapping $e_ n$ to $1/f^ n$ in $A_ f$ (see above). By assumption and Lemma 15.91.1 this extension is split, hence we obtain an element $x + e_0$ which generates a copy of $A_ f$ in $E$. Then

\[ x + e_0 = x - x_0 + fe_1 = x - x_0 - f x_1 + f^2 e_2 = \ldots \]

Since $M/f^ nM = E/f^ nE$ by the snake lemma, we see that $x = x_0 + fx_1 + \ldots + f^{n - 1}x_{n - 1}$ modulo $f^ nM$. In other words, the map $M \to \mathop{\mathrm{lim}}\nolimits M/f^ nM$ is surjective as desired.
$\square$

## Comments (2)

Comment #7376 by Sriram on

Comment #7578 by Stacks Project on

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