Lemma 10.96.3 (05GG)—The Stacks project

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

Lemma 10.96.3. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$ . Let $M$ be an $R$ -module. Then

the completion $M^\wedge$ is $I$ -adically complete, and
$I^ nM^\wedge = \mathop{\mathrm{Ker}}(M^\wedge \to M/I^ nM) = (I^ nM)^\wedge$ 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 10.96.1 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. $\square$

Comments (6)

Comment #3282 by Nicolas on May 03, 2018 at 00:17

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

Comment #3283 by Dario on May 04, 2018 at 19:50

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

Comment #4550 by 羽山籍真 on September 19, 2019 at 03:15

The comment left by Nicolas is reasonable. In fact, this Lemma is true in more general case, provided that the $I$ -adically completions exist. See, for instance, [Fujiwara. K and Kato. F, The foundation of Rigid Geometry I, Def. 7.2.6 and Prop. 7.2.7]

Comment #4551 by Johan on September 19, 2019 at 06:31

Dear 羽山籍真, I think this lemma corresponds to 7.2.16 in their text and that one isn't more general.

Comment #4552 by 羽山籍真 on September 19, 2019 at 12:21

Dear John, yes of course, so this Prop. has the mild condition (finitely generated ideal) to make sure that I-adic completion exists. (But I still don't know if finitely generated ideal is necessary)

Comment #4553 by Johan on September 19, 2019 at 14:40

A non-complete completion can be found in Section 110.7.

Comments (6)

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