# Exercise 2.25 in Atiyah & Macdonald

Exercise 25 of Atiyah & Macdonald asks:

Let $$0 \to N' \to N \to N'' \to 0$$ be an exact sequence, with $$N''$$ flat. Then $$N'$$ is flat iff $$N$$ is flat.

One way to prove this (from this post) is to use the exact sequence $$\mathrm{Tor}_2(N'', M) \to \mathrm{Tor}_1(N', M) \to \mathrm{Tor}_1(N, M) \to \mathrm{Tor}_1(N'', M)$$ for an arbitrary module $$M$$. We have $$\mathrm{Tor}_2(N'', M) = \mathrm{Tor}_1(N'', M) = 0$$ because $$N''$$ is flat. If $$N'$$ is flat, then $$\mathrm{Tor}_1(N', M) = 0$$, which implies $$\mathrm{Tor}_1(N, M) = 0$$, so $$N$$ is flat. Similarly for the converse.

But this proof seems to implicitly use the commutativity of $$\mathrm{Tor}$$ (i.e., $$\mathrm{Tor}_i(M, N) \cong \mathrm{Tor}_i(N, M)$$). In particular, we used the fact that $$P$$ is flat iff $$\mathrm{Tor}_1(P, M) = 0$$ for all modules $$M$$. However, I have only seen a proof that $$P$$ is flat iff $$\mathrm{Tor}_1(M, P) = 0$$ for all modules $$M$$. Here, $$\mathrm{Tor}_i(P, M)$$ is computed using the $$- \otimes M$$ functor, while $$\mathrm{Tor}_i(M, P)$$ is computed using the $$- \otimes P$$ functor. In the above sequence, $$M$$ is always in the second slot because its construction uses the $$- \otimes M$$ functor.

I have not seen a proof of the commutativity of $$\mathrm{Tor}$$, and it seems like it is a rather non-trivial result. Is there a way to do the exercise without using commutativity?

Here is an alternative that avoids the $$\rm Tor$$ functor.
Recall that a (left) $$R$$-module $$M$$ is flat if, and only if, the (right) $$R$$-module $$M^* := \text{Hom}_\Bbb Z(M,\Bbb Q/\Bbb Z)$$ is injective. See, for instance, this question.
Therefore, if $$0 \to A \to B \to C \to 0$$ is exact and $$C$$ is flat, then $$0 \to C^* \to B^* \to A^* \to 0$$ is exact ($$\Bbb Q/\Bbb Z$$ is an injective $$\Bbb Z$$-module) and $$C^*$$ is injective, hence splits. Thus, $$B^* \cong C^* \oplus A^*,$$ from which follows that $$A^*$$ is injective if and only if $$B^*$$ is injective (direct products/summands of injectives are injective), equivalently, $$A$$ is flat if and only if $$B$$ is flat.