# Is $M$ is a flat module over a Noetherian local ring, then it's free

In Atiyah-Macdonald introduction to commutative algebra Exercise 7.15, it states that

Let $$A$$ be a Noetherian local ring, $$\mathfrak{m}$$ its maximal ideal and $$k$$ its residue field, and let $$M$$ be a finitely generated $$A$$-module. Then TFAE:
$$1)$$ $$M$$ is free
$$2)$$ $$M$$ is flat
$$3)$$ the mapping of $$\mathfrak{m}\otimes M$$ into $$A\otimes M$$ is injective
$$4)$$ $$\operatorname{Tor}^A_1(k,M)=0$$.

I think the proof of this is $$1)\Rightarrow 2)\Rightarrow 3)\Rightarrow 4) \Rightarrow 1)$$. But I need is $$2)\Rightarrow 1)$$. I already proved $$1)\Rightarrow 2)$$. How can I prove $$2)\Rightarrow 1)$$ directly?

• Let me start by mentioning that 2) => 3) and 2) => 4) are immediate (by definition of flatness, and of Tor in the second case). Therefore if you have a proof of 3) => 1) or of 4) => 1), that would constitute what most would call a direct proof of 2) => 1). Do you explicitly want something avoiding 4) ? May 16, 2021 at 9:54
• @MaximeRamzi Yes. I know $2)\Rightarrow 3)$. How can I prove $3)\Rightarrow 1)$? May 16, 2021 at 10:20
• One could add a fifth condition: $5) M$ is projective. May 16, 2021 at 14:51
• @GeoffreyTrang Actually the Exercise only states up to 4) and projective module is not covered in that book. May 16, 2021 at 15:00

Hint:

You can directly prove that 2) implies 1): consider a free $$A$$-module $$L$$ such that $$L/\mathfrak mL\simeq M/\mathfrak m M$$. We have an exact sequence $$0\longrightarrow K\longrightarrow L\longrightarrow M\longrightarrow 0$$ which is pure (i.e. universally exact) since $$M$$ is flat. In particular, the sequence $$0\longrightarrow K/\mathfrak mK\longrightarrow L/\mathfrak mL\xrightarrow{\enspace\simeq\enspace} M/\mathfrak m M\longrightarrow 0$$ is exact, so $$K=\mathfrak m K$$. Can you conclude now?

• Thank you for your answer. But I need more detail about the hint. First, how do you know such free $A$-module $L$ exists? Also, what is $K$ in that sequence? Why such sequence is exact? What do you mean $pure$? May 16, 2021 at 12:31
• $L$ exists because you only have to lift a basis of the $k$-vector space $M/\mathfrak m M$. $K$ is the kernel of the (surjective) map $L\longrightarrow M$, so the sequence is exact by construction. To be more precise, a short exact sequence if it remains exact after tensorisation by any module. It is known to be so if in a short exact sequence, the last module is flat. May 16, 2021 at 12:42
• Ok I understood. The second exact sequence is obtained by tensoring $A/\mathfrak{m}$ right? And by Nakayama, $K =0$ so $L\simeq M$. So $M$ is free module May 16, 2021 at 13:07
• So far I prove that: Choose a basis $x_1+mM,...,x_n+mM$ of $M/mM$. Now let $L$ be a free $A$-module generated by $x_1,...,x_n$. Define $\phi:L\to M/mM$ by $x_i\mapsto x_i+mM$. Then I need to show the kernel $\{\sum_{i=1}^na_ix_i| \sum_{i=1}^na_ix_i \in mM\}=mL$. $\supset$ is clear but how can I prove the reverse inclusion? May 16, 2021 at 13:32
• It is simpler to check that you have an exact sequence $0 \longrightarrow K/\mathfrak mK\longrightarrow L/\mathfrak mL \longrightarrow M/\mathfrak mM$. As the last map if an isomorphism by construction, there will result that $K/\mathfrak mK=0$, and you can apply Nakayama. May 16, 2021 at 14:28

Given the comments, here's how one can prove 3) => 1) without mentioning Tor. The point is that we need to prove a special case of so-called "Tor-balancing", namely that if $$\operatorname{Tor}^A_1(M,k) = 0$$, then tensoring with $$k$$ preserves any exact sequence of the form $$0\to C\to D\to M\to 0$$.

This is not immediate from the definition of flatness of $$M$$, but follows exactly from this "Tor-balancing" thing, that I won't make precise here. So let me prove that lemma:

Lemma : suppose $$M$$ is flat. Then any short exact sequence of the above form is preserved by tensoring with $$k$$

(or in fact any $$A$$-module)

Proof: Write down the following commutative diagram, which can be seen as "tensoring our short exact sequence with $$0\to m\to A\to k\to 0$$:

$$\require{AMScd}\begin{CD}& & & &&& 0 \\ & & & && @VVV\\&&C\otimes_A m @>>> D\otimes_A m @>>> M\otimes_A m @>>> 0 \\ &@VVV @VVV @VVV @VVV\\ 0@>>>C @>>> D @>>> M @>>> 0 \\ && @VVV @VVV @VVV @VVV \\ &&C\otimes_A k @>>> D\otimes_A k @>>> M\otimes_A k@>>> 0 \\ && @VVV @VVV @VVV @VVV \\ && 0 @>>> 0 @>>> 0 @>>> 0\end{CD}$$

Each column and each line is exact, the exact column with a $$0$$ on top is due to $$M$$'s flatness.

Now let $$x\in C\otimes_A k$$ be mapped to $$0$$ in $$D\otimes_A k$$. Lift it to $$y\in C$$ by surjectivity of $$C\to C\otimes_A k$$. Then $$y$$ is mapped to $$0$$ if you go to $$D$$ and then $$D\otimes_A k$$, so that the image of $$y$$ in $$D$$ must lift to some $$t\in D\otimes_A m$$ by exactness of that column.

This $$t$$ goes to $$0$$ if you go to $$M\otimes_A m$$ and then to $$M$$ (because its image is then the image of $$y$$ along $$C\to D\to M$$), and thus by injectivity of $$M\otimes_A m\to M$$, $$t$$ must map to $$0$$ in $$M\otimes_A m$$. Thus $$t$$ must lift to some $$u\in C\otimes_A m$$.

But now by injectivity of $$C\to D$$, $$u$$ must be a lift of $$y$$, so that $$y$$ maps to $$0$$ in $$C\otimes_A k$$, so $$x=0$$, so $$C\otimes_A k \to D\otimes_A k$$ is injective; which is what we needed.

From there on you can follow Bernard's answer, namely : find a basis of $$M\otimes_A k$$, and lift it to a family in $$M$$, and so a surjective morphism $$A^n\to M$$, with kernel $$K$$. This surjective morphism is further an isomorphism upon tensoring with $$k$$, so that you have an exact sequence $$0\to K\to A^n\to M\to 0$$ of the form above, which must be preserved by tensoring with $$k$$.

Therefore, since $$k^n\to M\otimes_A k$$ is an isomorphism, $$K\otimes_A k = 0$$. Because $$A$$ is noetherian, $$K$$ is finitely generated, and so this implies (Nakayama's lemma) that $$K=0$$.

• Thank you for your effort. I really appreciated :) May 16, 2021 at 14:38