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Let $R$ be a commutative ring, let $A$ be an $R$-algebra, let $S\subset A$ be a multiplicative subset, and let $M$ be an $R$-module. Is it true that

$\mathrm{Tor}_p^R(S^{-1}A,M) \cong S^{-1}\mathrm{Tor}_p^R(A,M)$ ?

The first problem is that I don't even see how the right hand side makes sense. For that, I would need $\mathrm{Tor}_p^R(A,M)$ to be an $A$-module, but how is it so? For $p=0$, this is $A\otimes_R M$ and it is, indeed, an $A$-module, but if I take a projective resolution of $R$-modules $P_\bullet \to A$, the $P_i$ need not be $A$-modules, so this line of thought doesn't go through...

user46225
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1 Answers1

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This is true, to see that this is indeed an isomorphism and why the right $\operatorname{Tor}_p^R(A,M)$ is indeed an $A$-module, you can take a projective resolution $P_\bullet\rightarrow M$ of $M$. Then $\operatorname{Tor}_p^R(S^{-1}A,M)=H_p(S^{-1}A\otimes_R P_\bullet)$. But we have the following isomorphisms : $$ S^{-1}A\otimes_R P_\bullet\simeq S^{-1}A\otimes_A (A\otimes_R P_\bullet) $$ and $$ H_p(S^{-1}A\otimes_A (A\otimes_R P_\bullet))\simeq S^{-1}A\otimes_A H_p(A\otimes_R P_\bullet)=S^{-1}H_p(A\otimes_R P_\bullet)$$ (this is because the functor $S^{-1}A\otimes_A\cdot$ is exact and commutes with homology).

Finally $S^{-1}H_p(A\otimes_R P_\bullet)=S^{-1}\operatorname{Tor}_p^R(A,M)$.

Roland
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