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?


1 Answer 1


To me the proof seems correct. Also I find it not really surprising since $Tor_\ast^R(S^{-1}M,S^{-1}N)$ is naturally a $S^{-1}R$-module which makes $$Tor_\ast^R(S^{-1}M,S^{-1}N) \cong Tor_\ast^{S^{-1}R}(S^{-1}M,S^{-1}N)$$ reasonable. This isomorphism also follows from flat base change (Rotmann, Theorem 10.73):

Let $R \to T$ be a hom. of rings such that $T$ is flat as $R$-module and let $A$ be a right $T$-module and $B$ a left $R$-module. Then $$Tor^R_\ast(A,B) \cong Tor^T_\ast(A,T\otimes_R B).$$

By taking $T = S^{-1}R, A=S^{-1}M, B=S^{-1}N$ the isomorphism above follows (as in OP) from $S^{-1}R\otimes_R S^{-1}M=(S^{-1}R\otimes_R S^{-1}R)\otimes_R M\cong S^{-1}R\otimes_R M= S^{-1}M$ (as $S^{-1}R$-modules).

Remark: $Tor^R_\ast(-,-)$ is a module over the center $Z(R)$ of $R$. Going into the proof one can show that Rotmann's isomorphism is an isomorphism of $Z(R)$-modules.


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