Lemma 10.97.1. Let $A$ be a ring. Let $I \subset A$ be an ideal. Let $(M_ n)$ be an inverse system of $A$-modules. Set $M = \mathop{\mathrm{lim}}\nolimits M_ n$. If $M_ n = M_{n + 1}/I^ nM_{n + 1}$ and $I$ is finitely generated then $M/I^ nM = M_ n$ and $M$ is $I$-adically complete.

Proof. As $M_{n + 1} \to M_ n$ is surjective, the map $M \to M_1$ is surjective. Pick $x_ t \in M$, $t \in T$ mapping to generators of $M_1$. This gives a map $\bigoplus _{t \in T} A \to M$. Note that the images of $x_ t$ in $M_ n$ generate $M_ n$ for all $n$ too by Nakayama's lemma (Lemma 10.19.1). Consider the exact sequences

$0 \to K_ n \to \bigoplus \nolimits _{t \in T} A/I^ n \to M_ n \to 0$

We claim the map $K_{n + 1} \to K_ n$ is surjective. Namely, if $y \in K_ n$ choose a lift $y' \in \bigoplus _{t \in T} A/I^{n + 1}$. Then $y'$ maps to an element of $I^ n M_{n + 1}$ by our assumption $M_ n = M_{n + 1}/I^ nM_{n + 1}$. Hence we can modify our choice of $y'$ by an element of $\bigoplus _{t \in T} I^ n/I^{n + 1}$ so that $y'$ maps to zero in $M_{n + 1}$. Then $y' \in K_{n +1}$ maps to $y$. Hence $(K_ n)$ is a sequence of modules with surjective transition maps and we obtain an exact sequence

$0 \to \mathop{\mathrm{lim}}\nolimits K_ n \to \left(\bigoplus \nolimits _{t \in T} A\right)^\wedge \to M \to 0$

by Lemma 10.86.1. Fix an integer $m$. As $I$ is finitely generated, the completion with respect to $I$ is complete and $(\bigoplus _{t \in T} A)^\wedge / I^ m(\bigoplus _{t \in T} A)^\wedge = \bigoplus _{t \in T} A/I^ m$ (Lemma 10.95.3). We obtain a short exact sequence

$(\mathop{\mathrm{lim}}\nolimits K_ n)/I^ m(\mathop{\mathrm{lim}}\nolimits K_ n) \to \bigoplus \nolimits _{t \in T} A/I^ m \to M/I^ mM \to 0$

Since $\mathop{\mathrm{lim}}\nolimits K_ n \to K_ m$ is surjective we conclude that $M/I^ mM = M_ m$. It follows in particular that $M$ is $I$-adically complete. $\square$

Not really sure why the images of $x_t$ generate the $M_n$. Could this be elaborated? In any case, why not just choose the $x_t$ to be generators of $M$ instead? Then the images of $x_t$ seem to generate the $M_n$ due to surjectivity of $M \to M_n$. Did I miss something?

Comment #4731 by on

OK, we should add a reference to Nakayama's lemma 10.19.1 or we should do what you said.

Comment #4990 by Tongmu He on

I would like to give an another proof of lemma 10.97.1.

(1) First, if we only require that $I^n M_n=0$ and $I$ is finitely generated, let's show that $M$ is $I$-adically complete: We have $M\to M/I^nM\to M_n$. Taking limit, we get $M\to M^\wedge\to M$. Hence $M$ is a direct summand of $M^\wedge$. By lemma 10.95.3, $M^\wedge$ is $I$-adically complete, so is $M$.

(2) If moreover $M_ n = M_{n + 1}/I^ nM_{n + 1}$, let's show that $M/I^nM=M_n$: Consider short exact sequence $0\to N_n\to M\to M_n\to 0$. Notice that $\operatorname{Coker}(N_{n+1}\to N_n)=\operatorname{Ker}(M_{n+1}\to M_n)=I^nM_{n+1}$ is killed by $I$. Therefore, by Nakayama's lemma 10.19.1, $N_{n+1}/(N_{n+1}\cap I^{n+1}M)\to N_{n}/(N_{n}\cap I^{n}M)$ is surjective. On the other hand, (1) shows that $\lim N_{n}/(N_{n}\cap I^{n}M)=0$. We conclude that $N_{n}/(N_{n}\cap I^{n}M)=0$ and thus $M/I^nM=M_n$.

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