1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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 $3)\Rightarrow 1)$? $\endgroup$
    – Peter John
    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. $\endgroup$
    – Peter John
    May 16, 2021 at 15:00

2 Answers 2

3
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ 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 $pure$? $\endgroup$
    – Peter John
    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 $\endgroup$
    – Peter John
    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? $\endgroup$
    – Peter John
    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
1
$\begingroup$

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 $K=0$.

$\endgroup$
1
  • $\begingroup$ Thank you for your effort. I really appreciated :) $\endgroup$
    – Peter John
    May 16, 2021 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.