Lemma 10.98.2 (09B8)—The Stacks project

Lemma 10.98.2. Let $I \subset A$ be a finitely generated ideal of a ring. Let $(M_ n)$ be an inverse system of $A$ -modules with $M_ n = M_{n + 1}/I^ nM_{n + 1}$ . Set $M = \mathop{\mathrm{lim}}\nolimits M_ n$ . Then $M/I^ nM = M_ n$ and $M$ is $I$ -adically complete.

Proof. By Lemma 10.98.1 we see that $M$ is $I$ -adically complete. Since the transition maps are surjective, the maps $M \to M_ n$ are surjective. Consider the inverse system of short exact sequences

$0 \to N_ n \to M \to M_ n \to 0$

defining $N_ n$ . Since $M_ n = M_{n + 1}/I^ nM_{n + 1}$ the map $N_{n + 1} + I^ nM \to N_ n$ is surjective. Hence $N_{n + 1}/(N_{n + 1} \cap I^{n + 1}M) \to N_ n/(N_ n \cap I^ nM)$ is surjective. Taking the inverse limit of the short exact sequences

$0 \to N_ n/(N_ n \cap I^ nM) \to M/I^ nM \to M_ n \to 0$

we obtain an exact sequence

$0 \to \mathop{\mathrm{lim}}\nolimits N_ n/(N_ n \cap I^ nM) \to M^\wedge \to M$

Since $M$ is $I$ -adically complete we conclude that $\mathop{\mathrm{lim}}\nolimits N_ n/(N_ n \cap I^ nM) = 0$ and hence by the surjectivity of the transition maps we get $N_ n/(N_ n \cap I^ nM) = 0$ for all $n$ . Thus $M_ n = M/I^ nM$ as desired. $\square$

Comments (9)

Comment #4730 by Hammerhead on December 01, 2019 at 18:51

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 Johan on December 01, 2019 at 19:03

OK, we should add a reference to Nakayama's lemma 10.20.1 or we should do what you said.

Comment #4817 by Johan on December 10, 2019 at 12:11

Thanks and fixed here.

Comment #4990 by Tongmu He on March 17, 2020 at 08:44

I would like to give an another proof of lemma 10.98.2.

(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 10.96.3, $M^\wedge$ 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 10.20.1, $N_{n+1}/(N_{n+1}\cap I^{n+1}M)\to N_{n}/(N_{n}\cap I^{n}M)$ 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 $M/I^nM=M_n$ .

Comment #5231 by Johan on June 11, 2020 at 11:23

@#4990: OK, I like part (1). But unfortunately, I do not understand part (2). I am stuck at the assertion that (1) implies $\lim N_n/(N_n \cap I^nM) = 0$ . Why? Can somebody explain?

Comment #5232 by Tongmu He on June 11, 2020 at 15:48

@Comment #5231: This is because: We have an exact sequence $0\to N_{n}/(N_{n}\cap I^{n}M)\to M/I^nM\to M_n\to 0$ . Taking limit we get $0\to \lim N_{n}/(N_{n}\cap I^{n}M)\to M^\wedge \to \lim M_n\to 0$ (where the surjectivity is a bonus from (2), it's not important here). We notice that (1) implies that $M^\wedge = M=\lim M_n$ . Therefore, $\lim N_{n}/(N_{n}\cap I^{n}M)=0$ .

Comment #5233 by Johan on June 11, 2020 at 16:37

OK, I got it. I will replace the now proof by yours soon! Thanks very much.

Comment #5234 by Tongmu He on June 11, 2020 at 16:45

It's my pleasure to contribute a little bit to this wonderful textbook from which I have benefited a lot :)

Comment #5235 by Johan on June 13, 2020 at 15:37

Thanks! I have added your proof in this commit. It will be online after I go through all the remaining comments.

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