
[Theorem 15, Matlis]. The slick proof given here is from an email of Bjorn Poonen dated Nov 5, 2016.

Lemma 10.95.3. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$. Let $M$ be an $R$-module. Then

1. the completion $M^\wedge$ is $I$-adically complete, and

2. $I^ nM^\wedge = \mathop{\mathrm{Ker}}(M^\wedge \to M/I^ nM) = (I^ nM)^\wedge$ for all $n \geq 1$.

In particular $R^\wedge$ is $I$-adically complete, $I^ nR^\wedge = (I^ n)^\wedge$, and $R^\wedge /I^ nR^\wedge = R/I^ n$.

Proof. Since $I$ is finitely generated, $I^ n$ is finitely generated, say by $f_1, \ldots , f_ r$. Applying Lemma 10.95.1 part (2) to the surjection $(f_1, \ldots , f_ r) : M^{\oplus r} \to I^ n M$ yields a surjection

$(M^\wedge )^{\oplus r} \xrightarrow {(f_1, \ldots , f_ r)} (I^ n M)^\wedge = \mathop{\mathrm{lim}}\nolimits _{m \geq n} I^ n M/I^ m M = \mathop{\mathrm{Ker}}(M^\wedge \to M/I^ n M).$

On the other hand, the image of $(f_1, \ldots , f_ r) : (M^\wedge )^{\oplus r} \to M^\wedge$ is $I^ n M^\wedge$. Thus $M^\wedge / I^ n M^\wedge \simeq M/I^ n M$. Taking inverse limits yields $(M^\wedge )^\wedge \simeq M^\wedge$; that is, $M^\wedge$ is $I$-adically complete. $\square$

Comment #3282 by Nicolas on

The proof doesn't seem to use that $I$ is finitely generated.

Comment #3283 by Dario on

Rather subtle: The terms 'finite direct sum' and 'finite product' coincide. As limits commute we have $(M^r)^\wedge = (M^\wedge)^r$.

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