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?

  • $\begingroup$ 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) ? $\endgroup$ May 16, 2021 at 9:54
  • $\begingroup$ @MaximeRamzi Yes. I know $2)\Rightarrow 3)$. How can I prove $3)\Rightarrow 1)$? $\endgroup$ May 16, 2021 at 10:20
  • $\begingroup$ One could add a fifth condition: $5) M$ is projective. $\endgroup$ May 16, 2021 at 14:51
  • $\begingroup$ @GeoffreyTrang Actually the Exercise only states up to 4) and projective module is not covered in that book. $\endgroup$ May 16, 2021 at 15:00

2 Answers 2



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?

  • $\begingroup$ 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 $pure$? $\endgroup$ May 16, 2021 at 12:31
  • $\begingroup$ $L$ 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. $\endgroup$
    – Bernard
    May 16, 2021 at 12:42
  • $\begingroup$ 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 $\endgroup$ May 16, 2021 at 13:07
  • $\begingroup$ 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? $\endgroup$ May 16, 2021 at 13:32
  • $\begingroup$ 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. $\endgroup$
    – Bernard
    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 $K=0$.

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

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