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I'm trying to prove that finitely generated flat modules over a commutative, local, Noetherian ring are free.

I think I've got really close to a proof, but I'm stuck at the last step that finishes the proof.

So, suppose that $M$ is a finitely generated flat $R$-module, and $(R,m,k)$ is a commutative, local ring which is Noetherian. $M/mM$ is a $k$-vector space and so is $k \otimes M$ and we know that they're isomorphic. Therefore, $\dim(M/mM)=\dim(k \otimes M)$ as $k$-vector spaces. A theorem states that $\dim(M/mM)$ is equal to the number elements of a minimal generating set which is a well-defined finite number. Let's say it's equal to $\dim(M/mM)=n$.

I know every $R$-module is the image of a homomorphism $F \to M$ where $F$ is free (just think of the elements of $M$ as a basis for $F$). Therefore, we get a short exact sequence: $$ 0 \rightarrow L \rightarrow F \rightarrow M \rightarrow 0$$ where $L=\ker(F\rightarrow M)$. Tensoring with $k$ we get:

$$ \cdots \rightarrow {\rm Tor}_1^R(k,M)\rightarrow k\otimes L \rightarrow k \otimes F \rightarrow k\otimes M \rightarrow 0$$ But since $M$ is flat, we get the short exact sequence:

$$ 0\rightarrow k\otimes L \rightarrow k \otimes F \rightarrow k\otimes M \rightarrow 0$$

OK. Now here I think I need to compare the dimensions. If I can prove that $k \otimes L = 0$, then Nakayama's lemma gives $L=0$ and that proves $M \cong F$ and I'm done. But I have no idea how to show that $\dim(k\otimes F) = \dim(k \otimes M)$.

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  • $\begingroup$ In fact, one only needs to show that $F=\oplus_{i=1}^n R$ ($n$ copies of $R$). Any ideas? $\endgroup$
    – user246836
    Jun 4, 2016 at 19:01
  • $\begingroup$ Couldn't you use Nakayama to show that you can choose $F$ to be of this form? $\endgroup$
    – Mohan
    Jun 4, 2016 at 19:34
  • $\begingroup$ @Mohan: Do you mean Nakayama's lemma? $\endgroup$
    – user246836
    Jun 4, 2016 at 19:39
  • $\begingroup$ When you choose a minimal system of generators for $M$ then they form a basis in $k\otimes M$. $\endgroup$
    – user26857
    Jun 4, 2016 at 21:01
  • $\begingroup$ @user26857: Yes because if $\{x_1,\cdots,x_n\}$ generates $M$, $\{ 1\otimes x_1, \cdots, 1\otimes x_n \}$ generates $k \otimes M$. My question is why $F$ is exactly $n$ copies of $R$? (Or is it?) where $n$ is the number of elements of a minimal generating set for $M$. I think Mohan suggested that I should Nakayama's lemma. $\endgroup$
    – user246836
    Jun 4, 2016 at 21:05

2 Answers 2

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A finitely generated module over a noetherian ring is finitely presented, and finitely presented flat implies projective. Now use projective finitely generated implies free over local rings (this is a straightforward application of Nakayama).

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  • $\begingroup$ I don't know what a finitely presented module is. It isn't in the topics covered in my course. Can we do it without that? $\endgroup$
    – user246836
    Jun 4, 2016 at 18:16
  • $\begingroup$ @H.Z. A module $M$ is finitely generated if there is an exact sequence $F_0\to M\to 0$ where $F_0$ is free and finitely generated. A module is finitely presented if there is an exact sequence $F_1\to F_0\to M\to 0$ where both $F_1,F_0$ are free and finitely generated. $\endgroup$
    – Pedro
    Jun 4, 2016 at 22:06
  • $\begingroup$ I see. Thanks. But why do they define finitely-presented modules? Is it any useful? Does it have to do with free resolutions? $\endgroup$
    – user246836
    Jun 4, 2016 at 22:11
  • $\begingroup$ @H.Z. Well, I gave you one example: projective always implies flat, and finitely presented flat always implies projective. Finitely generated modules are in general more manageable than arbitrary modules. $\endgroup$
    – Pedro
    Jun 4, 2016 at 22:52
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In your solution, one can choose $F$ to be minimal, i.e. choose a minimal generating set of $M$ and projected by basis elements of $F$ respectively. Then by NAK, $M\otimes k=F\otimes k$ as $k$-vector space.

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