The Stacks project

Lemma 10.97.2. Let $I \subset A$ be a finite generated ideal of a ring. Let $(M_ n)$ be an inverse system of $A$-modules with $M_ n = M_{n + 1}/I^ nM_{n + 1}$. Then $M/I^ nM = M_ n$ and $M$ is $I$-adically complete.

Proof. By Lemma 10.97.1 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

Not really sure why the images of generate the . Could this be elaborated? In any case, why not just choose the to be generators of instead? Then the images of seem to generate the due to surjectivity of . 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 10.97.2.

(1) First, if we only require that and is finitely generated, let's show that is -adically complete: We have . Taking limit, we get . Hence is a direct summand of . By lemma 10.95.3, is -adically complete, so is .

(2) If moreover , let's show that : Consider short exact sequence . Notice that is killed by . Therefore, by Nakayama's lemma 10.19.1, is surjective. On the other hand, (1) shows that . We conclude that and thus .

Comment #5231 by on

@#4990: OK, I like part (1). But unfortunately, I do not understand part (2). I am stuck at the assertion that (1) implies . Why? Can somebody explain?

Comment #5232 by Tongmu He on

@Comment #5231: This is because: We have an exact sequence . Taking limit we get (where the surjectivity is a bonus from (2), it's not important here). We notice that (1) implies that . Therefore, .

Comment #5233 by on

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

Comment #5234 by Tongmu He on

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

Comment #5235 by on

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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