It is known that
Let $R$ be a commutative ring with unit and $S \subset R$ a multiplicative sistem. If $M$ and $N$ are $R$-modules there is a isomorphism of $S^{-1}R$-modules:
$$S^{-1}\mathrm{Tor}_i^R(M,N) \simeq\mathrm{Tor}_i^{S^{-1}R}(S^{-1}M,S^{-1}N).$$
It can be proven using flat base change.
I was wondering if it is true that there is also an isomorphism of $R$-modules
$$S^{-1}\mathrm{Tor}_i^R(M,N) \simeq\mathrm{Tor}_i^R(S^{-1}M,S^{-1}N).$$
I think i found a proof of this fact but i'm not completely sure that is right.
Take a flat resolution of $R$-modules for M
$\rightarrow F_2 \rightarrow F_1 \rightarrow F_0 \rightarrow M\rightarrow 0 $
Its localization $\rightarrow S^{-1}F_2 \rightarrow S^{-1}F_1 \rightarrow S^{-1}F_0 \rightarrow S^{-1}M\rightarrow 0 $ is a flat resolution of $R$-modules for $S^{-1}M$.
Indeed $S^{-1}F_i \cong S^{-1}R \otimes_R F_i$ is flat because tensor product of flat $R$-modules. The sequence is exact because the localization is an exact functor from $R-mod$ to $R-mod$.
Then $Tor_i^R(S^{-1}M,S^{-1}N)$ is the $i$-th homology module of the sequence $\rightarrow S^{-1}F_2 \otimes_R S^{-1}N \rightarrow S^{-1}F_1 \otimes_R S^{-1}N \rightarrow S^{-1}F_0 \otimes_R S^{-1}N \rightarrow 0 $.
But $S^{-1}F_i \otimes_R S^{-1}N \cong S^{-1}R \otimes_R F_i \otimes_R S^{-1}R \otimes_R N \cong S^{-1}R \otimes_R F_i \otimes_R N \cong S^{-1}(F_i \otimes_R N)$
Therefore the precedent sequence is isomorphic to $\rightarrow S^{-1}(F_2 \otimes_R N) \rightarrow S^{-1}(F_1 \otimes_R N) \rightarrow S^{-1}(F_0 \otimes_R N) \rightarrow 0 $
And by the exactness of localization the the $i$-th homology module of this sequence is $S^{-1}Tor_i^R(M,N)$
This isomorphism sound strange to me but I cannot find any mistake in the proof above. Can you check this proof or show me a counterexample?