Non-finite version of Nakayama's lemma? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:26:41Z http://mathoverflow.net/feeds/question/33513 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33513/non-finite-version-of-nakayamas-lemma Non-finite version of Nakayama's lemma? ashpool 2010-07-27T14:07:03Z 2010-07-27T22:01:45Z <p>Let $A$ be a local ring with nilpotent maximal ideal $\mathfrak{m}$ (i.e., some power of $\mathfrak{m}$ vanishes), and $M$ an $A$-module (not necessarily finitely generated). Let $\bar{S}\subset M/\mathfrak{m}M$ be a set of generators and $S$ a set of representatives of $\bar{S}$ in $M$. Then is it true that $S$ is a set of generators of $M$? This is a common form of Nakayama's lemma with the assumption of finite generation of $M$ replacing the nilpotence of $\mathfrak{m}$. A passage in Matsumura's book "Commutative Ring Theory" (see Theorem 7.10) seems to imply this result, and I can't figure out why.</p> http://mathoverflow.net/questions/33513/non-finite-version-of-nakayamas-lemma/33517#33517 Answer by Emerton for Non-finite version of Nakayama's lemma? Emerton 2010-07-27T14:25:19Z 2010-07-27T14:25:19Z <p>Dear Kwan,</p> <p>Let $N$ be the submodule of $M$ generated by $S$. Then by assumption $M = N +\mathfrak m M.$ Iterating this, we find that $$M = N + \mathfrak m (N + \mathfrak m M) = N + \mathfrak m^2 M = \cdots = N + \mathfrak m^n M$$ for any $n > 0.$ If we take $n$ large enough then $\mathfrak m^n = 0$ (by hypothesis). Thus $M = N,$ as desired. </p> <p>P.S. I've found this to be quite a useful fact!</p> http://mathoverflow.net/questions/33513/non-finite-version-of-nakayamas-lemma/33519#33519 Answer by Wilberd van der Kallen for Non-finite version of Nakayama's lemma? Wilberd van der Kallen 2010-07-27T14:32:03Z 2010-07-27T14:32:03Z <p>Let $N$ be the $A$-module generated by $S$. Now $M$ is contained in $N+\mathfrak{m}M$, which is contained in $N+\mathfrak{m}(N+\mathfrak{m}M)$, hence in $N+\mathfrak{m}^2M$. Repeat.</p>