# Non-finite version of Nakayama's lemma?

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.

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The question should be revised to say $S$ is a set of representatives of $\overline{S}$, not that it is the preimage of $\overline{S}$ (as the latter doesn't recover usual Nakayama, and is both useless and trivial to prove). Should also clarify the meaning of "nilpotent ideal" (confusing away from the noetherian case) since might mean just that each element of $\mathfrak{m}$ is nilpotent (perhaps that's called a nil-ideal?). In this more general sense of "nilpotent ideal" the answer is negative: take $A = k[[x]][x^{1/2}, x^{1/3}, \dots]/(x)$, $M = \mathfrak{m}$, $\overline{S} = 0$, $S = 0$. – BCnrd Jul 27 '10 at 14:48
Dear BCnrd, a) is it true that $A=k[x,x^{1/2}, x^{1/3},...]/(x)$ ? b) Is it true that in $B=k[[x]][x^{1/2}]$ the infinite formal sum $x^{1/2}+x^{3/2}+x^{5/2}+...$ doesn't exist and so cannot replace the product $[1+x+x^2+...]. x^{1/2}$ ? Let me emphasize that I'm not in the least saying that you are not correct: I am just trying to check if I understend these two rings by asking myself elementary questions. – Georges Elencwajg Jul 27 '10 at 18:01
BCnrd rightly caught me being less than careful with the preimage business. Thanks. – ashpool Jul 27 '10 at 19:56
Atiyah defines the set of nilpotent elements a nilradical. I was using the word nilpotent as an adjective meaning some power of the object being described is zero. Sorry I should have clarified that in the beginning. – ashpool Jul 27 '10 at 20:01
Dear kwan: I took the liberty of promoting your confirmation of the "typos" to corrections in the formulation of the question, and I also included a more specific reference within Matsumura's CRT book. Dear Georges: yes for (a), and no for (b) (think of the analogy with $\mathbf{Q}_ p(p^{1/2})$, or more rigorously the ring in (b) is $k[[x]][T]/(T^2 - x)$ and so...). – BCnrd Jul 27 '10 at 22:04

Dear Kwan,

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.S. I've found this to be quite a useful fact!

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I didn't know this was true. Thanks! – Keenan Kidwell Jul 27 '10 at 14:27
This is, needless to say, perfectly correct. There is a caveat though: to say that $\mathfrak m$ is nilpotent means indeed that some power of this ideal is zero. This is stronger (for non-noetherian rings) than just asserting that $\mathfrak m$ consists of nilpotent elements: the terminology for this last condition is "nil ideal". – Georges Elencwajg Jul 27 '10 at 14:43
Dear Georges, Thanks for making this important point clear! – Emerton Jul 27 '10 at 14:52
Dear Emerton: I find it very cold to address someone as friendly as you by surname. May I use the opportunity to ask for permission to call you Matt? ( I do this automatically if the first name is part of the username) – Georges Elencwajg Jul 27 '10 at 15:23
Dear Georges, Of course! Regards, Matt – Emerton Jul 27 '10 at 15:38

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.

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