For every finitely generated ideal $I\subset A$ the map $I\otimes_A M\rightarrow A\otimes_A M$ is injective, then $M$ is flat.

Question

In the book I'm reading, there haven't been given any proof for the following theorem, since it's "obvious". I'm trying to prove it myself, but seem to have a bit of troubles.

Let $A$ ring and $M$ an $A$-module. If for every finitely generated ideal $I\subset A$ the map $I\otimes_A M\rightarrow A\otimes_A M$ is injective, then $M$ is flat.

So I know, what I have to prove is, that given an exact sequence $\ldots\rightarrow N'\rightarrow N \rightarrow N''\rightarrow \ldots $, I want to show that $\ldots\rightarrow N'\otimes_A M\rightarrow N \otimes_A M \rightarrow N''\otimes_A M\rightarrow \ldots $ is exact. I know that you can split long exact sequences into short exact sequences, and that tensor product is right exact. So I guess it would be enough to show that given $N\rightarrow N'$ injective, then $N \otimes_A M\rightarrow N'\otimes_A M$ should be injective?

I'd also say that submodules of $A$ are ideals of $A$, hence we have injective $N \otimes_A M\rightarrow A \otimes_A M\cong M$ and $N'\otimes_A M\rightarrow A\otimes_A M\cong M$. However this is where I'm stuck.

Answer 1

Hint:

First, as any ideal of $A$ is the union of its finitely generated sub-ideals and the direct limit is an exact functor which commutes with tensor products, the same hypothesis is true foe any ideal.

Next, any finitely generated $A$-module is a quotient of a finitely generated free module. So you can try to prove that, if $K$ is a submodule of a finitely generated free module $L$ of rank $n$, then $$K\otimes_A M\longrightarrow L\otimes_A M$$ is injective. This can be done by induction on the rank $n$.

For the case $n=2$, consider a submodule $K$ of $A^2$. Denote $K_1$ the intersection of $K$ with the first factor $A\times\{0\}$ of $A^2$, and $M_K2$ the canonical projection of $M$ onto the second factor $\{0\}\times A$. Then do some diagram hunting in the commutative diagram: \begin{alignat}6 K_1&\otimes_A M&{}\longrightarrow{}& &K&\otimes_A M&{}\longrightarrow {} K_2&\otimes_A M\\ &\downarrow &&&&\downarrow &&\downarrow\\[-0.5ex] L_1&\otimes_A M&{}\longrightarrow{}& &A^2&\otimes_A M&\longrightarrow L_2&\otimes_A M \end{alignat}

Answer 2

Note that the hypothesis is that for every finitely generated ideal $I$ of $A$, $\mathrm{Tor}_1^A(A/I,M) = 0$. Indeed, from the exact sequence $0\to I\to A\to A/I\to 0$ we get the exact sequence

$$0\to \mathrm{Tor}_1^A(A/I,M)\to I\otimes_A M\to A\otimes_A M \to A/I \otimes_A M\to0$$

since $A$ is flat. Thus $\mathrm{Tor}_1^A(A/I,M)$ is the kernel of your map, and hence zero iff it is injective. Now I claim that

1. Since every ideal is the filtered colimit of its f.g. ideals, tensor products commute with these, and filtered colimits are exact, your hypothesis holds for every ideal.
2. Every module is the filtered colimit of its cyclic submodules, which are all isomorphic to $A/I$ for some ideal $I$.
3. Because $\otimes_A M$ commutes with filtered colimits, so does $\mathrm{Tor}_1^A(?,M)$, which means that for any module $M'$, $$\mathrm{Tor}_1^A(M',M) = \mathrm{colim}_{I} \mathrm{Tor}_A^1(A/I,M) = 0$$
4. Finally, since $\mathrm{Tor}_1^A(?,M)$ vanishes identically, this means that $M$ is flat.

Certainly you would have to prove $1$ (which is not hard), $2$ follows from this, while $3$ follows by the LES of a SES, in a similar way as the first part of this answer.