# Finitely generated flat modules over a commutative, local, Noetherian ring are free

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

• In fact, one only needs to show that $F=\oplus_{i=1}^n R$ ($n$ copies of $R$). Any ideas? Jun 4, 2016 at 19:01
• Couldn't you use Nakayama to show that you can choose $F$ to be of this form? Jun 4, 2016 at 19:34
• @Mohan: Do you mean Nakayama's lemma? Jun 4, 2016 at 19:39
• When you choose a minimal system of generators for $M$ then they form a basis in $k\otimes M$. Jun 4, 2016 at 21:01
• @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. Jun 4, 2016 at 21:05

• @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.
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.