Lemma 15.91.3. Let $A$ be a ring. Let $I \subset A$ be an ideal. Let $M$ be an $A$-module.

1. If $M$ is $I$-adically complete, then $T(M, f) = 0$ for all $f \in I$.

2. Conversely, if $T(M, f) = 0$ for all $f \in I$ and $I$ is finitely generated, then $M \to \mathop{\mathrm{lim}}\nolimits M/I^ nM$ is surjective.

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

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

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. $\square$

Comment #7376 by Sriram on

Hello!

In the proof of (2), showing the surjection of the canonical map to completion, the sequence of equations must have an "f" in the coefficient of x_1. That is, " ... = x-x_0+f e_1 = x-x_0 -f x_1 + f^2 e_2 = ..."

Thanks

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