# $M$ is flat iff $Tor_1(M,N)=0$?

Given a $$R$$-module $$M$$, it's flat iff $$Tor_1(N,M)=0$$ for all $$R$$-module $$N$$, which can be deduced from a free resolution of $$N$$, tensoring with $$M$$ and applying the definition of flatness.

But there is equivalent statement that $$M$$ is flat iff $$Tor_1(M,N)=0$$ for all $$R$$-module $$N$$, which interchanges the position of $$M$$ and $$N$$. Now, it's tensoring the free resolution of $$M$$ with $$N$$! I don't know how to prove this from definition or use other ways. Hope someone could help. Thanks!

• $M\otimes N\cong N\otimes M$. – Mohan Sep 28 '20 at 14:14

## 1 Answer

Use the balancing of $$\mathrm{Tor}$$: for all modules $$M,N$$ and all $$i\in \mathbb{Z}$$ you have $$\mathrm{Tor}_i(M,N)\cong \mathrm{Tor}_i(N,M)$$. For $$i=0$$ this is clear, as per Mohan's comment. For $$i>0$$ you can find a proof in Weibel. The idea is to take flat resolutions of both $$M$$ and $$N$$ and tensor them together to get a double complex, which you can then analyse.