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 $\dim(k\otimes F) = \dim(k \otimes M)$.
Best Answer
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).