I've been working through Atiyah & Macdonald's text, and came across the following problem from an old commutative algebra qualifying exam.
Let $M,N$ be $R$-modules over a commutative ring $R$. Suppose that every $R$-submodule of every free $R$-module is free. Prove that $\text{Tor}_2^R(M,N)$ vanishes for all $R$-modules $M$ and $N$.
Here are my thoughts so far:
Take a free resolution $\cdots \rightarrow F_3 \xrightarrow{f_3} F_2 \xrightarrow{f_2} F_1 \rightarrow F_0 \rightarrow M \rightarrow 0$ of $M$, where each $F_i$ is a free $R$-module. Removing $M$ and applying the functor $-\otimes_R N$, we obtain the chain complex $\cdots \rightarrow F_3 \otimes_R N \xrightarrow{d_3 = f_3 \otimes 1} F_2 \otimes_R N \xrightarrow{d_2 = f_2 \otimes 1} F_1 \otimes_R N \xrightarrow{d_1} F_0 \otimes_R N \xrightarrow{d_0} 0$. Since $\text{Tor}_2^R(M,N) = \ker(d_2)/\text{Im}(d_3)$, we need to show that either $\ker(d_2)$ is trivial or $\ker(d_2) = \text{Im}(d_3)$.
I'm not quite sure how to proceed -- I could take the route of showing that $\ker(d_2)$ is trivial (i.e., $d_2$ is injective), or I could take the route of showing that $\ker(d_2) = \text{Im}(d_3)$.
I know that the assumption that every $R$-submodule of every free $R$-module is free tells us that both $\ker(d_2)$ and $\text{Im}(d_3)$ are free $R$-modules. We also know that $\text{Im}(d_3) \subseteq \ker(d_2)$ by the chain complex above, so to show that $\ker(d_2) = \text{Im}(d_3)$, it's only left to show the reverse inclusion.
One difficulty I found is that I don't think that $f_2$ being injective would imply that $d_2$ is injective (as would be the case if we knew, for instance, that $N$ was a flat $R$-module).
Any help would be greatly appreciated! Thanks!