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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!

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  • $\begingroup$ $M\otimes N\cong N\otimes M$. $\endgroup$ – Mohan Sep 28 at 14:14
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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.

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