Lemma 15.91.3 (091R)—The Stacks project

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

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

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

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 (5)

Comment #7376 by Sriram on May 23, 2022 at 06:39

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

Comment #7578 by Stacks Project on August 12, 2022 at 14:52

THanks, fixed here.

Comment #9966 by Elías Guisado on February 04, 2025 at 10:43

In the proof of (2), here is the argument for the short exactness of $0\to M\to E\to A_f\to 0,$ where $E = M \oplus \bigoplus Ae_n\Big/\langle x_n - fe_{n + 1} + e_n\rangle.$ On the one hand, $E\to A_f$ is clearly onto, the composite $M\to E\to A_f$ vanishes and it is not difficult to see that $M\to E$ is injective. It is left to verify that an element from $E$ that vanishes in $A_f$ comes from an element in $M$ . We proceed with the following strategy. Note that we have a commutative diagram as this. \xymatrix{ 0 \ar@{->}[r] & \bigoplus_{n\geq 0} A \ar@{->}[d] \ar@{->}[r] & \bigoplus_{n\geq 0} A \ar@{->}[d] \ar@{->}[r] & A_f \ar@{->}[d] \ar@{->}[r] & 0 \ 0 \ar@{->}[r] & M \ar@{->}[r] & E \ar@{->}[r] & A_f \ar@{->}[r] & 0 }

Where the top sequence is described in the proof of Lemma 15.91.1 (and is short exact), the map $\bigoplus_{n\geq 0}A\to M$ sends $e_n$ to $-x_n$ , the map $\bigoplus_{n\geq 0}A\to E$ sends $e_n$ to the coset of $e_n$ in $E$ and the map $A_f\to A_f$ is the identity. Since $E$ is generated by the cosets of the $e_n$ 's, the map $\bigoplus_{n\geq 0}A\to E$ is onto. Thus one may apply the following result, whose proof is an easy diagram chase.

Lemma. Suppose we have a commutative diagram like this. \xymatrix{ A \ar@{->}[r] \ar@{->}[d] & B \ar@{->}[r] \ar@{->>}[d] & C \ar@{^{(}->}[d] \ A' \ar@{->}[r] & B' \ar@{->}[r] & C' } such that the top row is exact and the bottom row is a complex. Then the bottom row is exact.

Comment #9969 by Elías Guisado on February 05, 2025 at 06:51

For anyone interested in the proof of (2), here are the details behind the assertion “hence we obtain an element $x+e_0$ that generates a copy of $A_f$ in $E$ .”

We have a section $A_f\to E$ of $E\to A_f$ . Let $z\in E$ be the image of $1\in A_f$ along this section. We are going to find $x\in M$ such that $z=x+e_0$ . Write $z=y + \sum_{n=0}^Na_ne_n$ , $y\in M$ , $a_n\in A$ . Then $1=\sum_{n=0}^N\dfrac{a_n}{f^n}$ in $A_f$ . Thus there is $k\geq 0$ such that $\begin{equation} \label{eq:rel_a_i}\tag{1} 0=f^k(f^Na_0+f^{N-1}a_1+\cdots+ a_N-f^N)\quad\text{in }A. \end{equation}$ Inside $E$ , we have $\begin{align*} f^N(a_0e_0+a_1e_1+\cdots+a_Ne_N) &=\phantom{+}f^Na_0e_0\\ &\phantom{=}+f^{N-1}a_1(x_0+e_0)\\ &\phantom{=}+f^{N-2}a_2(fx_1+x_0+e_0)\\ &\phantom{=}+\cdots\\ &\phantom{=}+a_N(f^{N-1}x_{N-1}f^{N-2}x_{N-2}+\cdots+e_0)\\ &=\phantom{+}(f^Na_0+f^{N-1}a_1+\cdots+ a_N)e_0\\ &\phantom{=}+(f^{N-1}a_1+f^{N-2}a_2+\cdots+ a_N)x_0\\ &\phantom{=}+(f^{N-2}a_2+f^{N-3}a_3+\cdots +a_N)fx_1\\ &\phantom{=}+\cdots\\ &\phantom{=}+a_Nf^{N-1}x_{N-1}. \end{align*}$ Therefore, using \eqref{eq:rel_a_i}, we get $f^{k+N}z=f^{k+N}(x'+e_0)$ , where $x'\in M$ . Thus the elements $z$ and $x'+e_0$ of $E$ have same image in $A_f$ . Therefore $x'+e_0-z=w\in M$ . Calling $x=x'+w$ , we get $z=x+e_0$ , as desired.

Comment #9977 by Elías Guisado on February 10, 2025 at 04:08

Ignore #9966. The easy way to see short exactness of $0\to M\to E\to A_f\to 0$ is because it is the pushout (in the sense defined in 12.6) of the s.e.s. $0\to\bigoplus_{n\geq 0}A\to\bigoplus_{n\geq 0}A\to A_f\to 0$ (given in the proof of 15.91.1) along the map $\bigoplus_{n\geq 0}A\to M$ that sends $e_n$ to $-x_n$ .

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