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This question already has an answer here:

I know that $Tor^{\mathbb z}_1(\mathbb Z, N) = 0$ for any $\mathbb Z-$module, because free modules are flat. Then because $Tor_1$ is local, we have $Tor_1^{\mathbb Q}(\mathbb Q, S^{-1}N) = 0$, which is not useful at all. How should I use localization to prove this statement?

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marked as duplicate by Yanior Weg, Jyrki Lahtonen abstract-algebra May 11 at 16:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I am seeking for a solution which does not use the theorem about torsion because we never did that @AlvaroMartinez $\endgroup$ – z.z May 11 at 1:10
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You can think think of this as a result of the two following more general statements, which themselves should not be too difficult to prove: throughout the following discussion let $S$ be a multiplicative subset of a ring $A$.

$1$) Localization is exact; that is, if $0\to M'\to M\to M''\to 0$ is an exact sequence of $A$-modules, then the induced sequence $0\to S^{-1}M'\to S^{-1}M\to S^{-1}M''\to 0$ of $S^{-1}A$-modules is exact.

Just to be clear--what do we mean by "induced" sequence? If $\varphi:M\to N$ is a homomorphism of $A$-modules, then there is a natural map $S^{-1}\varphi:S^{-1}M\to S^{-1}N$ defined by $(S^{-1}\varphi)(m/s):=f(m)/s$.

$2)$ For any $M$ there is an isomorphism $S^{-1}A\otimes_A M\cong S^{-1}M$ of $S^{-1}A$-modules, which is "natural in $M$"; this means that if we have $\varphi:M\to N$, then there is a commutative diagram $$\require{AMScd} \begin{CD} S^{-1}A\otimes_A M @>{\text{id}\otimes\varphi}>> S^{-1}A\otimes_A N\\ @V{\cong}VV @VV{\cong}V\\ S^{-1}M @>>{S^{-1}\varphi}> S^{-1}N \end{CD}.$$

Combining these facts, you see that for any localization $S^{-1}A$, if we have an injective homomorphism $\varphi:M\to N$, then we will get a commutative diagram as above by fact $(2)$; the morphism $S^{-1}\varphi$ will be injective by $(1)$, and then commutativity will imply that $\text{id}\otimes\varphi$ is injective as well, proving $S^{-1}A$ is flat. Of course for your example we are just taking $A=\Bbb Z$ and $S=\Bbb Z\smallsetminus\{0\}$.

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Any localization of a commutative ring $R$ is a flat $R$-module. For this special case, $R= \mathbb Z$ and the multiplicatively closed set is $S=R\setminus \{0\}$.

This is likely proved before proving that Tor "commutes" with localization so trying looking earlier on in whatever book/reference you are using.

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