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

abstract-algebracommutative-algebraflatness

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?

Best Answer

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?

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