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Suppose $R \neq 0$ is a commutative ring with $1$. The following is well known:

(Isomorphism Theorem for Finitely Generated Free Modules) [FGFM] $R^{n}\cong R^{m}$ as $R$-modules if and only if $n=m$.

One proof of this result (if I recall correctly) was somehow based on the idea that we can quotient out by a maximal ideal of $R$ (whose existence is guaranteed by Zorn's Lemma), and reduce the situation to that of vector spaces. However, I am interested in different approach. Recall

(Nakayama's Lemma) Suppose $M$ is a finitely generated $R$-module, and $M=IM$ where $I$ is an ideal contained in Jacobson radical of $R$. Then, $M=0$.

My question is:

Can we prove FGFM using Nakayama's Lemma?

My main motivation in asking this question is two-fold:

1) to see the power and usefulness of Nakayama's Lemma, and

2) to see a nice and short proof of FGFM that does not involve reducing the problem to a result from linear algebra.

I appreciate any input :)

Martin Brandenburg
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Prism
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  • I think the first move is yours: where do you expect to begin with Nakayama's lemma? Presumably you have something in mind to apply it to. (Or did you just say: "Nakayama's lemma: I like it! Let's try to prove random theorem X with it..." – rschwieb May 29 '13 at 13:18
  • @rschwieb: Well, it was probably closer to "Nakayama's lemma: I like it! But let's try to see if we can deduce this particular theorem". I had a hunch that Nakayama would be powerful enough, and I was right :) – Prism May 29 '13 at 14:32
  • Well what I was getting at is that the connection seemed rather whimsical. For example, you could also say "Can we prove this with the axiom of choice?" Indeed the axiom is powerful but that doesn't mean it proves everything. In that case, you would be met with a lot more skepticism. In this case, there turned out to be a connection, but it doesn't have to be that way all the time. – rschwieb May 29 '13 at 16:23
  • And yes, I guess that using the Zorn's lemma argument along with bases of vector spaces would be using the axiom of choice (twice) to prove it. But you get the idea. You could say "prove this with the four color theorem" and your question would be closed. I just wondered why you thought there would be a connection. – rschwieb May 29 '13 at 16:26
  • @rschwieb Good point. To answer the question of why I thought there would be connection, I will go with intuition and sheer luck. Also, I like to see lemmas in action. I would love to ask a question titled "What are your favourite applications of Nakayama's Lemma?", though not sure if this would be well-received by the community. – Prism May 29 '13 at 16:33
  • OK, intuition and sheer luck it is :) If I were you, I would try out your "favorite applications" question. There are certainly worse questions sticking around. The worst that could happen is that it gets closed, but you have nothing to lose. Personally, I wouldn't vote to close unless it is a duplicate. Be sure to check the site for a similar question first! – rschwieb May 29 '13 at 16:38
  • As for 2), there are lots of proofs which do not use linear algebra over fields, but prove it directly. For example, when $n<m$, then $0 = \Lambda^m R^n$ is not isomorphic to $R \cong \Lambda^m R^m$, hence $R^n$ is not isomorphic to $R^m$. – Martin Brandenburg May 29 '13 at 19:52
  • I just wanted to say that you can do it directly without using the special case of fields. – Martin Brandenburg May 30 '13 at 09:40
  • @MartinBrandenburg Dear Martin, thanks for your comment. Unfortunately, I don't have any knowledge of multi-linear algebra. Hoping to learn soon! – Prism May 30 '13 at 21:54
  • See http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf for instance. – Martin Brandenburg May 31 '13 at 00:38

1 Answers1

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Here's one idea. Use the following consequence of Nakayama:

If $M$ is a finitely generated $R$-module and $f\colon M\to M$ is a surjective module homomorphism, then $f$ is an isomorphism.

Proof: View $M$ as an $R[x]$-module, where the action of $x$ on $M$ is given by $f$. By assumption $xM = M$. The proof of Nakayama's lemma then gives that there is some element $P(x)\in R[x]$ such that $(1 - P(x)x)M = 0$. Then for any $m\in M$, $$0 = (1 - P(x)x)m = m - P(f)(f(m))\implies m = P(f)(f(m)).$$ This shows that $P(f)$ is an inverse of $f$.

To use this to prove FGFM, do the following. Suppose $n\geq m$, and that $R^n\cong R^m$. First, fix a basis $e_1,\ldots, e_n$ of $R^n$, and identify $R^m$ with the submodule of $R^n$ generated by $e_1,\ldots, e_m$, so $R^m\subseteq R^n$. Let $\pi\colon R^n\to R^m$ be the projection map. Since by assumption $R^n\cong R^m$, there is some isomorphism $f\colon R^m\to R^n$. But then $\pi\circ f\colon R^m\to R^m$ is surjective, and hence by the result above, is also an isomorphism. However, since $f$ is surjective, the only way that $\pi\circ f$ can be an isomorphism is if $\pi$ is injective. This of course only happens when $n = m$.

froggie
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    Yeah, that's one way to do it :) +1 for overcoming my negativity about the connection between the two. – rschwieb May 29 '13 at 13:34
  • This is so cute! Thank you very much. I wasn't aware of the particular consequence of Nakayama's Lemma you used above. Do you have a reference for it? Perhaps it appeared as an exercise in some textbook? – Prism May 29 '13 at 14:34
  • I am going to wait for some time before accepting this as answer, so that it gathers more upvotes, as it surely deserves. – Prism May 29 '13 at 14:36
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    @Prism The result about a f.g. module over a commutative ring is attributed to Vasconcelos. It is also true that a surjective endomorphism of a Noetherian module is injective. You can read about these two things at this solution. – rschwieb May 29 '13 at 14:50
  • @rschwieb: That's perfect! Georges's answer in that thread lead to original article by Vasconcelos who proves other cool stuff. Thanks a lot. :) – Prism May 29 '13 at 14:56
  • @rschwieb: Didn't know the attribution. Interesting answer you link to! – froggie May 29 '13 at 15:57
  • Sorry for commenting on such an old post! Yet I’m wondering how we can deduce that $\pi$ is injective from the fact that $\pi \circ f$ is an isomorphism and $f$ is surjective? It seems that it is the composition $f \circ \pi$ that we should consider? And in the last sentence, the of course part might be not that trivial (at least for me). It is an exercise in A&M, and the discussion on https://mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-11/47846#47846 may be useful. Thank you for your great answer. That really helps a lot! – Hetong Xu Sep 26 '20 at 06:02