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I was trying to solve this problem:

Let $R$ be a commutative Noetherian ring. Prove that a finitely generated $R$-module $M$ is flat if and only if $Tor_1(R/m,M)=0 $ for any maximal ideal $m$ of $R$.

Actually i found a similar statement in Eisenbud, Proposition 6.1. It does not require $R$ to be Noetherian and $M$ to be finitely generated, but instead it requires $m$ to be any finitely generated ideal(not necessarily maximal).

I tried to mimic the proof in Eisenbud, but soon figured out that if i use this proof, then I need one of the following statements:

  1. For any finitely generated ideal(actually $R$ is Noetherian, this just means every ideal) I, the obvious morphism $I\otimes_R M \rightarrow R\otimes_R M=M$ is injective, or

  2. For any finitely generated $R$-module $N'$ and its submodule $N$, we can find a chain of $R$-modules $N=N_0\subset N_1 \subset \cdots \subset N_t=N'$ such that for each $i$, $N_i/N_{i-1}\simeq R/{m_i}$ for a maximal ideal $m_i$ of $R$ or $N_i/N_{i-1}\simeq R$.

If at least one of these statements is true, then it seems like I don't need $R$ to be Noetherian and $M$ to be finitely generated for any other part of the proof. So these assumptions must be related to one of these statements, but I can't immediately see which one of these is true and why. I'm expecting that I can do something with Chinese remainder theorem to prove 2, but I'm not sure. Maybe both statements are not true but a statement in somewhere between statement can be true:

  1. For some collection of ideals, the morphism $I\otimes_R \rightarrow M$ is injective for any ideal $I$ in that collection, and for any finitely generated $R$-module $N'$ and its submodule $N$, we can find a chain of $R$-modules $N=N_0\subset N_1\subset \cdots \subset N_t=N'$ such that for each $i$, $N_i/N_{i-1}\simeq R/I_i$ for an ideal $I$ in that collection.

So can you explain which of these is true and why? If none of these are true, can you suggest how should I approach to the original problem at the top of this question?

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When you see a statement like this, where you characterize some property of a module in terms of a statement for each maximal (or prime) ideal, your first thought should always be to localize and thus reduce to the case of a local ring.

In this case, to show $M$ is flat, it suffices to show the localization $M_m$ is flat over $R_m$ for each maximal ideal $m$ (proof sketch: if $0\to A\to B$ is exact but $0\to M\otimes A\to M\otimes B$ is not exact, let $K$ be the kernel of $M\otimes A\to M\otimes B$ and localize everything at some maximal ideal $m$ such that $K_m\neq 0$). So we may assume $R$ is local with maximal ideal $m$, and we wish to show that if $\mathrm{Tor}_1(M,R/m)=0$, then $M$ is flat.

In fact, we can show something even stronger: $M$ is free (to be clear, this conclusion is only true assuming $R$ is local; under the original assumptions, we only conclude that $M$ is locally free, or projective). To prove something like this from a statement about Tor, typically you want to find some particular short exact sequence involving $M$ and then consider the long exact sequence you get on Tor. In particular, since we want to prove $M$ is free, let's take a presentation of $M$: a short exact sequence $0\to K \to F\to M\to 0$ where $F$ is a free module. Since $M$ is finitely generated, we can take $F$ to be finitely generated. Since $R$ is Noetherian, $K$ will also be finitely generated.

Now, since we're given information about Tor with $R/m$, we're going to want to tensor with $R/m$, which means we'll probably be using Nakayama's lemma. We'd like to find that $F\to M$ is actually an isomorphism, so as a first step we need $F\otimes R/m\to M\otimes R/m$ to be an isomorphism. This might not actually be true (maybe $F$ has a bunch of extra unnecessary generators), but it is possible to choose a presentation so that it is true.

At this point, I'll encourage you to try to finish the proof on your own: prove it is possible to find a presentation $0\to K\to F\to M\to 0$ such that $F\otimes R/m\to M\otimes R/m$ is an isomorphism, and then prove that for such a presentation, $K=0$ so $F\to M$ is an isomorphism and therefore $M$ is free. The details are hidden below.

First, note that $M\otimes R/m$ is a vector space over the field $R/m$, so we can pick some basis. Lifting those basis elements to elements of $M$, we get a set of elements which generate $M$ by Nakayama's lemma. So, let this be the generating set of our presentation, so the map $F\to M$ maps the free generators of $F$ to these generators of $M$. The map $F\otimes R/m\to M\otimes R/m$ then sends the free generators of $F\otimes R/m$ to the basis of $M\otimes R/m$ we chose originally, and hence is an isomorphism.

Now given such a presentation $0\to K\to F\to M\to 0$, let us tensor with $R/m$. We get a long exact sequence $$\mathrm{Tor}_1(M,R/m)\to K\otimes R/m\to F\otimes R/m\to M\otimes R/m\to 0.$$ But $\mathrm{Tor}_1(M,R/m)=0$, and $F\otimes R/m\to M\otimes R/m$ is injective, so this exact sequence implies $K\otimes R/m=0$. Since $K$ is finitely generated, by Nakayama's lemma this implies $K=0$.

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  • $\begingroup$ Thanks! So it seems like localizing the ring at the start makes me possible to avoid some annoying problems such as I wrote above. I'm not used to commutative algebraic tools like localization, so I tried to solve the problem without it, but maybe I better study commutative algebra as soon as possible. $\endgroup$ – Absol Dec 11 '17 at 18:41

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