6
$\begingroup$

A is an I-adic complete Noetherian ring. M is a finitely generated A module. For any n>0, $M/I^nM$ is a finitely generated locally free A/I^n-module. Is M necessarily a locally free A-module?

$\endgroup$
5
  • $\begingroup$ What if $IM = M$ but $M$ is not free? $\endgroup$ Dec 12, 2009 at 18:00
  • $\begingroup$ Then Nakayama's lemma shows that $M=0$, I believe (I guess it depends what $I$-adic Noetherian ring means). $\endgroup$
    – Ben Webster
    Dec 12, 2009 at 18:11
  • $\begingroup$ Yeah that wasn't clear to me; if $A$ is local and $I$ the maximal ideal, though, the answer is yes. (I had assumed in the comment he meant $I$ was just some ideal). $\endgroup$ Dec 12, 2009 at 18:14
  • $\begingroup$ I don't think that's a very good assumption. $\endgroup$
    – Ben Webster
    Dec 12, 2009 at 18:20
  • 1
    $\begingroup$ If "$I$-adic" implies complete in your statement, could you add it explicitly? $\endgroup$ Dec 12, 2009 at 21:57

1 Answer 1

4
$\begingroup$

The answer is yes.

1) $A$ is $I$-adic complete implies that $I \subset rad(A)$, the intersection of all maximal primes. Indeed, pick any $a \in I$. Look at $1-a + a^2 -a^3 ... \in A $ (this is where we use completeness). The inverse of this is $1+a$, so $1+a$ is an unit. Since this is true for any $a\in I$, we have $I\subset rad(A)$ (see Section 1 Matsumura).

2) It suffices to prove that $M$ is free at any maximal ideal $m$ of $A$. Since $I$ is inside $m$, we may as well replace $A$ by $A_m$ and assume $A$ is local. By assumption then $M/I \cong (A/I)^l$, Nakayama Lemma shows that $l$ is the mimimum number of generators of $M$.

Look at the beginning of a minimal resolution of M: $ N \to F=A^l \to M $.The last map become isomorphism when tensoring with $A/I^n$, so the first map has to become $0$. This means that $N \subset I^nF$ for all $n>0$. This forces $N=0$ (use Artin-Rees lemma), thus $M \cong F$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.