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?
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$\begingroup$ What if $IM = M$ but $M$ is not free? $\endgroup$– Akhil MathewDec 12, 2009 at 18:00
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$\begingroup$ 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
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$\begingroup$ 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). $\endgroup$– Akhil MathewDec 12, 2009 at 18:14
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$\begingroup$ I don't think that's a very good assumption. $\endgroup$– Ben Webster ♦Dec 12, 2009 at 18:20
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1$\begingroup$ If "$I$-adic" implies complete in your statement, could you add it explicitly? $\endgroup$– Mariano Suárez-ÁlvarezDec 12, 2009 at 21:57
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1 Answer
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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 $M \cong F$.
answered Dec 27, 2009 at 8:55